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  • 1. 1 AbstractIn these notes I present an overview of electrodynamics and quantum mechanicswhich (together with statistical mechanics) are the foundation of much of today’stechnology: electronics, chemistry, communication, optics, etc.
  • 2. CONTENTS1 Introduction: the Unity of Science 42 Quantum Mechanics 5 2.1 The puzzles of matter and radiation 6 2.1.1 Planck’s Black-body radiation 8 2.1.2 The photo-electric effect 12 2.1.3 Bohr’s atom 14 2.1.4 Adsorption, stimulated emission and the laser 16 2.2 Quantum Mechanical formalism 19 2.3 Simple QM systems 21 2.3.1 The chiral amonia molecule 21 2.3.2 The amonia molecule in a constant electric field 24 2.3.3 The amonia maser and atomic clocks 26 2.3.4 The energy spectrum of aromatic molecules 28 2.3.5 Conduction bands in solids 29 2.4 Momentum and space operators 31 2.4.1 Heisenberg uncertainty principle 34 2.5 Schroedinger’s equation 35 2.5.1 Diffraction of free particles 35 2.5.2 Quantum interference observed with C60 37 2.5.3 QM tunneling and the Scanning Tunneling Micro- scope 38 2.6 The correspondance principle 41 2.6.1 Gauge invariance and the Aharonov-Bohm effect 42 2.7 Dirac’s equation: antiparticles and spin 44 2.7.1 Angular momentum and spin 49 2.8 The Hydrogen atom and electronic orbitals 52 2.8.1 Spin-orbit coupling 54 2.8.2 Many electron systems 55 2.8.3 The periodic table 56 2.9 The chemical bond 59 ¨ 2.9.1 Huckel’s molecular orbital theory 63 2.9.2 Molecular vibrational spectrum 65 2.9.3 Molecular rotational spectrum 67 2.10 Time independent perturbation theory 68 2.10.1 The polarizability of atoms in an electric field 71 2.10.2 Atom in a constant magnetic field: the Zeeman effect 73 2.10.3 Degenerate eigenstates 76 2.11 Time dependent perturbation theory 77 3
  • 4. 2 QUANTUM MECHANICSQuantum mechanics (QM) is a theory of matter and its interactions with force fields(here we will only care about electromagnetic fields). While classical mechanics andelectromagnetism are intuitive (one has a direct experience of gravitation, light, elec-tricity, magnetism, etc.) quantum mechanics is not. The description of matter that arisesfrom the QM formalism is totally at odds with our daily experience: particles can passthrough walls, can be at two different places and in different states at the same time,can behave as waves and interfere with each other. Worse, QM is a non-deterministicdescription of reality: it only predicts the probability of observing events. This aspectdeeply disturbed Einstein who could not accept that QM was the correct final descrip-tion of reality (as he famously quipped: ”God does not play dice”). He and many otherscame up with alternative descriptions of QM introducing hidden variables (unknow-able to the observer) to account for its non-deterministic aspects. But in 1964 JohnBell showed that if hidden variables existed some measurements would satisfy certaininequalities. The experiments performed by Alain Aspect and his collaborators in the1970’s showed that the Bell inequalities were violated as predicted by QM, but not bythe hidden variable theories thereby falsifying them. Yet, for all its technical prowess Aspect’s experiment was only addressing a philo-sophical issue concerning the foundations and interpretation of QM. The theory itselfhad been amply vindicated earlier by its enormous predictive power: QM explains thestability of atoms, their spectra, the origin of the star’s energy and of the elements andtheir properties, the nature of the chemical bond, the origin of magnetism, conductivity,superconductivity and superfluidity, the behaviour of semi-conductors and lasers, etc.,etc.. All of today’s micro-electronic industry is derived from applications of QM (tran-sistors, diodes, integrated circuits, etc.), the development of the chemical industry is aresult of the QM understanding of the chemical bond and the nuclear industry would ofcourse been impossible without an understanding of the nucleus and the nuclear forcesthat QM provided. So, for all its weirdness Quantum Mechanics is the most successfull explanation ofthe World ever proposed by Mankind. It beats Platonicism, the Uppanishads, Kabbalah,Scholasticism, etc., yet is non-intuitive and cannot be understood except by followingits mathematical formalism to its logical conclusions. ”The great book of Nature iswritten in the language of mathematics”, Galileo’s quip is truer for QM more than forany other scientific theory. More recently one of the founder’s of QM, Eugene Wignerwrote in an article entitled ”the unreasonable effectiveness of mathematics in the naturalsciences”, that ”the miracle of the appropriateness of the language of mathematics forthe formulation of the laws of physics is a wonderful gift which we neither understandnor deserve”. 5
  • 5. 6 QUANTUM MECHANICS It is with this mind set that I would like you to approach the study of QM. Like anapprentice sorcerer learning the tricks of his master without fully understanding them,yet always at awe confronting their power. As we have done with electromagnetism,we will approach QM by following as far as we can the historical narrative. We willsee why the radiation of a Black Body was such a puzzle that it prompted Max Plank tointroduce the idea that energy was quantized; why the stability of atoms and their spectraprompted Bohr, Sommereld and others to suggest that the energy levels of atoms werealso quantized; how the idea that particles could also have wavelike behaviour was firstsuggested by de Broglie and brought to fruition by Schroedinger, Heisenberg and Dirac.And how from then on, QM revolutionized the understanding of matter, the chemicalbond, magnetism, conduction, etc.2.1 The puzzles of matter and radiationAt the end of the 19th century, scientists disposed of a very succesfull theoretical frame-work that could explain many of the problems known at that time and which was tech-nologically revolutionary. Newtonian mechanics was amazingly successful in predict-ing the motion of celestial bodies. Its most striking success was the prediction by LeVerrier in 1846 of the existence of the planet Neptune. Analysing some anomalies inthe motion of Uranus, he predicted Neptune’s precise location in the sky, a predictionwhich was immediately confirmed by German astronomers. In 1861, Maxwell unifiedelectricity, magnetism and optics opening the area of electrical appliances and wirelesscommunication: Edison invented the light bulb in 1879 and founded ”General Electric”in 1892 while Marconi established the ”Marconi Wireless Telegraph Company” in 1897.Finally, thermodynamics was sustaining the advance of the industrial revolution as ther-mal engines were driving industry and railways. In spite of terrible social inequalities(as described by C.Dickens, E.Zola and others) this was a time of peace, prosperity andoptimism, illustrated by the nascent Impressionist movement. Yet, many fondamental scientific questions remained unsolved and paradoxical. Thechemical properties of the various elements were not understood. The periodicity ofthese properties as a function of the mass of the elements as determined in 1869 byMendeleev in his famous Periodic Table of the Elements was a mystery. Nonethelesson the basis of his ad-hoc classification Mendeleev predicted the existence of two newelements, Gallium and Germanium, which were duly discovered in 1875 and 1886 andare essential in today’s semiconductor industry! The existence of atoms (indivisibleparticles of matter characteristic of each element) postulated by Dalton to explain theproperties of molecules was not generally accepted. Because of the successful applica-tions of continuum mechanics (in the design of bridges, buildings (e.g. the Eiffel tower),etc.) and fluid dynamics (in explaining the tides, water waves, etc.), matter was gener-ally believed to be some sort of continuum akin to a gel not a swarm of particles. It wasEinstein who in 1905 finally managed to convince the scientific world of the existenceof atoms and molecules by showing that the erratic motion exhibited by dust particleson the surface of water (first observed by the botanist R.Brown in 1827) was due to theshocks of the water molecules. The continuum pre-conception also sustained the inter-pretation of electromagnetic waves. Since all known waves at the times were observed
  • 6. THE PUZZLES OF MATTER AND RADIATION 7F. 2.1. The emission spectrum in the visible range for a few elements. Notice the fine spectral lines and the different spectral characteristics for the different elements. This was one of the puzzles that QM solved.to propagate in a continuum medium (such as water, air, etc.) at a velocity v = κ/ρ(where κ is the compressibility and ρ the density of the medium), the electromagneticwaves predicted by Maxwell and discovered by Hertz were assumed to propagate insome continuum medium: the ether which properties determined their velocity. How-ever all attempts to detect the ”wind” of ether resulting from the motion of the Earth inthat medium proved negative. This prompted Einstein to formulate his theory of rela-tivity which postulated the constancy of the speed of light and got rid of any notion ofether, see Appendix. Then there were questions related to the emission and absorption spectra of elementsthat exhibited discrete lines rather than an undifferentiated continuum of absorption oremission, as was the case for sound and water waves and as we have seen for scatteredand refracted light. Not only did the elements exhibit specific adsorption lines but thosediffered from element to element. These observations did not fit with the then prevailingconception of matter as a continuum. Finally there was the problem of the radiation from a Black Body, a material (such asto a good approximation graphite) which adsorbs radiation uniformly at all frequenciesand which can therefore also emit radiation uniformly at all frequencies. Notice how-ever that many bodies (e.g. the elements just mentioned) are not black-bodies as theyadsorb/emit only at certain frequencies. At a given temperature, the radiation inside ablack body cavity is at thermal equilibrium with the walls of the cavity that absorb andre-emit it. When computing the electromagnetic radiation energy emitted by a black-body at a given temperature, one found its energy to diverge because the number ofmodes at high frequencies diverged. This was not only absurd but also in contradiction
  • 7. 8 QUANTUM MECHANICSwith the experiments which studied the energy distribution inside the cavity by measur-ing the energy leaking out of the cavity (for example through a small hole) as a functionof frequency. To see how this comes about, consider a square cavity (an oven) of size a at tempera-ture T . As we have seen, a body at a given temperature emits electromagnetic radiation,see Fig.??. Imagine that the walls of the cavity are made of small oscillators emitting ra-diation at frequency ω (like the oscillators we considered when studying the frequencydependence of the refraction index). Stationary waves of the form sin k · r cos ωt willbe present in the cavity if its walls are reflecting (though for the energy to equilibratebetween oscillators the walls cannot be 100% reflective). To satisfy the boundary con-ditions on the walls we shall require that: k x = πl/a, ky = πm/a, kz = πn/a (n, m, l ≥ 0).Hence we have: k2 = (l2 + m2 + n2 )(π/a)2 ≡ (πρ/a)2 . The number of modes dNlnm withsuch wavelength is: 4πρ2 ka adk k dk dNlnm = 2 dρ= π( )2 = 8πa3 ( )2 8 π π 2π 2πThe factor 2 results from the two possible polarizations of the fields, while the factor(4πρ2 /8)dρ counts the number of modes in a shell in the positive octant (n, m, l ≥ 0).According to the equipartition theorem of statistical mechanics (see below) the averageenergy of each oscillatory mode is: < E >= kB T . Using the relations: k = ω/c ≡ 2πν/c(ν like f is the frequency), the energy density of the emitted radiation du = dNlnm <E > /a3 becomes: 8πkB T 2 du = ν dν (2.1) c3hence the total energy, the integral of the energy density over all frequencies, diverges asν3 . This divergence became known as the Jeans’ (or ultra-violet) catastrophy. While thedata agreed with that formula at low frequencies, it differed at high frequencies (smallwavelengths).2.1.1 Planck’s Black-body radiationRather than questioning the equipartition theorem which was verified in other contextsor the possibility of atoms to emit light of arbitrarily high frequencies, Planck suggestedin 1900 that light was emitted by the cavity walls in very small discrete quantities,quantas of energy: e = hν, where h, the Planck constant is: h = 6.626 10−27 erg sec = 4.135 10−15 eV sec, so that light of energy En is made up of n quantas: En = nhν. In that case the averageenergy emitted at frequency ν is the sum over all possible energies En , weighted by theirBoltzmann probability (see below the Chapter on Statistical Mechanics): e−En /kB T P(En ) = −En /kB T neSo that the average energy is:
  • 8. THE PUZZLES OF MATTER AND RADIATION 9 hν < E >= En P(En ) = n ehν/kB T − 1When hν kB T , one recovers the previous result: < E >= kB T , however at largeemission frequencies the average energy decays as < E >∼ hν exp(−hν/kB T ). Plancktherefore suggested to modify the previous result, Eq.2.1 to yield, see Fig.2.2: 8πhν3 1 ρ(ν) ≡ du/dν = (2.2) c3 ehν/kB T − 1Where ρ(ν) is known as the spectral density of radiation. Identifying the smallest quantaof energy with a light particle (a photon of energy hν), Eq.2.2 states that the density ofphotons in a Black-body is: dN p ρ(ν) 8πν2 1 = = 3 hν/k T (2.3) dν hν c e B −1Notice that the total energy density in the cavity is now finite: ∞ 8π(kB T )4 ∞ x3 8π5 k4 4 Utot = ρ(ν)dν = dx = B T (2.4) 0 h3 c3 0 ex − 1 15h3 c3 ∞Where we used the equality: 0 dx x3 /(e x − 1) = π4 /15. The total radiated power perunit area trough a small hole in the cavity becomes, 1 cUtot cUtot Irad = k · ndΩ = ˆ ˆ cos θd(cos θ) = σS B T 4 (2.5) 4π 2 0which is known as Stefan’s law and where the Stefan-Boltzmann constant: 2π5 k4 σS B = B = 5.67 10−5 erg sec−1 cm−2 ◦ K−4 = 5.67 10−8 W m−2 ◦ K−4 15h3 c2Therefore by measuring the total intensity of the radiation leaking out from a cavity(for example an oven) one can measure the temperature of that cavity. One can testthe validity of Planck’s law (actually how close to a black-body the cavity really is)by measuring the dependence of the intensity on the radiation wavelength. From thewavelength λmax at which the intensity is maximal an other estimate of the temperaturecan be deduced: kB T hc/5λmax . For example at 300K (which corresponds to a thermalenergy kB T 25 meV), the maximum of emission is at λmax ∼ 10µm. The thermal cam-eras that visualize humans and warm animals (see Fig.??) must therefore be sensitive tofar-infrared light. The temperature of the Sun and the Earth The Sun is to a very good approximation a black body, see Fig.2.2. The radiationsemitted by the fusion reactions occuring at its core (at temperature of 13 106 K) are
  • 9. 10 QUANTUM MECHANICSF. 2.2. The emission spectra of the sun and the universe. The sun emission spectra is pretty well fit by the spectrum of a lack body at 5770K, however notice the existence of some specific adsorption bands in the visible and UV spectrum. The universe on the other hand presents a spectrum that is perfectly matched by a black-body at 2.726K.at thermal equilibrium with the reacting nuclear particles and diffuse out to the Sun’ssurface which is much cooler. By fitting the spectrum of the sunlight to Planck’s for-mula one can determine the Sun’s surface temperature: T S = 5770K. The total powergenerated at the Sun’s core and emitted at its surface is: PS = 4πR2 σS B T S = 3.85 1026 W S 4where RS = 6.96 108 m is the Sun’s radius. Since the radius of the Sun’s core is es-timated to be ∼ RS /5 the volume of the core is: Vcore = 1.13 1025 m3 and the averagepower per unit volume generated in the Sun’s core is: PS /Vcore = 34 W/m3 . This is lessthan the power generated by our body to keep warm!! There are a few ways to verify that. Let us assume an average daily calory intake of3000kcal 1.2 107 J, which comes to a power consumption of 150W. Approximatinga man as a cylinder of height L = 2m and radius r = 0.2m, the power consumption perunit volume is 600 W/m3 of which about half goes to metabolic activity. Alternativelyone can use Stefan’s law to estimate the losses between a body at 37◦ C (T b = 310◦ K)and an environment at 27◦ C (T e = 300◦ K) (this is a crude estimate since other effectssuch as perspiration regulate our temperature): ∆I = σS B (T b − T e ) 64W/m2 which 4 4yields a power per unit volume 640W/m3 . From these consistent estimates we de-duce that our power consumption per unit volume is much larger than the Sun’s!! Whatmakes the Sun so bright and hot is its huge mass, not its rather inefficient thermonuclearreactions. Let us now estimate the temperature of the Earth T E resulting from its adsorption of
  • 10. THE PUZZLES OF MATTER AND RADIATION 11the Sun’s radiation and its own radiation at T E . The sunlight impinging on the Earth ata distance from the Sun RS E = 1.496 1011 m has an intensity: IE = PS /4πR2 E = (RS /RS E )2 σS B T S = 1.37 kW/m2 S 4Of that radiation a fraction (known as the Earth’s albedo) α ∼30% is reflected, mostlyby the clouds, snow and ice-caps. The Sun radiation power arriving at the surface of theEarth is thus about 1 kW/m2 . It is an important number to remember when designingsolar energy plants: its sets the maximal power per unit area available from the Sun.Notice that by measuring the radiation arriving on Earth and the angle sustained bythe Sun: θS = (RS /RS E ) one can also get an estimate of the Sun’s temperature: T S =(IE /σS B θS )1/4 . The energy absorbed by the Earth heats it and is reradiated (to a good 2approximation) like a black body at temperature T E . We can compute the temperatureof the Earth by a simple energy balance. At steady-state the energy radiated is equal tothe energy absorbed: σS B T E 4πR2 = (1 − α)IE πR2 4 E EFrom which we get T E = ((1 − α)IE /4σS B )1/4 = 255K = -18C. The Earth is actuallyslightly warmer because of the green house effect that reflects part of the emitted energyback to Earth. The Universe as a perfect black-body While it is difficult to design a perfect black-body, since as we shall see belowbound electrons adsorb at their resonance frequency (as is the case for the Sun’s spec-trum for example), the Universe as a whole turned out to be the best known exampleof a black-body, see Fig.2.2. The Universe is bathed in a uniform radiation field ofvery low frequency whose spectral distribution is perfectly matched by a black-bodyat 2.726K. This phenomena was predicted by George Gamow in 1948 and observedserendipitously by Arno Penzias and Robert Wilson in 1964 when measuring the noiseof a microwave antenna they had built. It was higher than they had expected as theywere actually detecting the 3K radiation of the Universe. This background radiation isthe most striking evidence for the existence of the Big-Bang. According to this scenario,the Universe began as a big explosion of matter and radiation. At the beginning lightand matter interacted continuously and were in thermal equilibrium (as they are in theSun’s core). But then as the Universe expanded it cooled. When it reached a temper-ature of ∼ 3000K Hydrogen atoms started to form that could not absorb non-resonantlight: radiation decoupled from matter. At that point the radiation spectrum was that of ablack body at the temperature of decoupling. It is the relics of that original radiation thatwe are observing today as an isotropic cosmic background radiation. Let us see why itexhibits a black-body spectrum at a temperature of 2.7K. Since once hydrogen atoms formed radiation largely stopped to interact with matter,the number of photons at frequency ν (see Eq.2.3): a3 dN p remained constant. But as theuniverse continued to expand to a size a > a so did the radiation wavelength (recallthat in a box k = 2π/λ is a multiple of π/a), i.e. the frequency of the radiation decreased
  • 11. 12 QUANTUM MECHANICSby the expansion factor αe = a /a. So that the energy density du of the backgroundphotons at frequency ν obeys now: 8πν2 1 (a )3 du = hν a3 dN p = hν a3 3 ehν/kB T − 1 dν c 8πν 2 1 = (a )3 α−3 hν α3 3 hν /k (T/α ) e e dν (2.6) c e B e − 1which is the energy density of a black-body at a temperature T , smaller than the tem-perature at decoupling T by the expansion factor αe : T = (T/αe ): 8πhν 3 1 du = 3 hν /kB T − 1 dν c eBecause the Universe expanded by a factor αe ∼ 1100 since the decoupling time, oneobtains a current temperature for the background radiation of T = 2.72◦ K. The preciseagreement on the value of that temperature is not very important as is the observationthat the cosmic background radiation is the best Black-Body ever observed. It is alsohighly isotropic in the rest frame of the Universe. As our galaxy the Milky Way movesat about 600 km/sec with respect to the Cosmic background, the Doppler effect red-shifts the radiation in one direction and blue-shifts it in the opposite one. This effect canbe subtracted from the measured distribution of radiation intensities. One also needs tosubtract the contribution from the stars in the galaxy (which fortunately emit at muchhigher frequencies, in the visible mostly). The measured variations in the temperature ofthe Universe at different angular positions are then smaller than 10−5 K, yet these smallfluctuations served as the nucleation points for the galaxies and can account for theirobserved distribution, see Fig.2.3. As E.Wigner wrote it is a ”miracle ... that we neitherundersand nor deserve” that a theory devised to explain (approximatively) the radia-tion of hot bodies has turned out to provide such an amazingly precise and powerfulldescription of the Universe!2.1.2 The photo-electric effectBesides the emission spectrum of atoms and the black body radiation, an other ex-periment stood in apparent contradiction with Maxwell’s electromagnetic theory: thephoto-electric effect which observed that electron were emitted from a conducting ma-terial with an energy that depended on the color (the frequency) of the radiation not onits intensity. This was at odds with Maxwell’s electromagnetic theory that asserted thatthe energy of radiation was related to its intensity (see Eq.??) not its frequency! Ein-stein knew the solution for Black-body radiation for which Planck had to assume thatthe radiation emitters in the walls’ cavity could only emit light in small quantas. In 1905Einstein went further and assumed that all light actually comes in small bunches, pho-tons, which energy is proportional to their frequency: E = hν. When such a photon isabsorbed by an electron its energy is used to tear the electron from the binding potentialΦ of the material and move it at velocity v: 1 2 hν = mv + Φ (2.7) 2
  • 12. THE PUZZLES OF MATTER AND RADIATION 13F. 2.3. The temperature of the cosmic microwave background measured across the sky by the COsmic Background Explorer (COBE) satellite. The top image is the raw data which is red/blue shifted due to the movement of our galaxy through the universe at ∼ 600km/s. Correcting for this Doppler shift yields the middle image which is still ”polluted” at the equator by the light emitted from the stars in our galaxy. Substracting that measurable emission yields the bottom image where the temperature fluctuations of the microwave background across the Universe are as small as 10µ◦ K. These small fluctuations nonetheless served as nucleation points for the formation of the stars and galaxies shortly after the decoupling time.Hence an electron can only be observed if light of high enough frequency is used toremove it from the material. The kinetic energy of the electron increases then linearlywith the illumination frequency. The current emitted is however proportional to thenumber of adsorbed photons, i.e. to the light intensity. At the time Einstein proposalwas revolutionary since it assumed that energy came in discrete packets that could not beinfinitely divided and it appeared to contradict Maxwell’s equations. It took 16 years andconfirming experiments to establish the validity of his model, for which he got the Nobelprize in 1921 (and not for his more profound and revolutionary theories of relativityand gravitation). Incidently from Einstein’s relation between energy and momentum:E 2 = (mc2 )2 + (pc)2 (see Appendix) one deduces that if the energy of the photon (ofmass m = 0) is quantized so must its momentum be: p = E/c = h/λ ≡ k ( ≡ h/2π).
  • 13. 14 QUANTUM MECHANICS Einstein’s understanding of the photo-electric effect has had enormous technologi-cal impact. All digital cameras are based upon it. These CCD (Charge Coupled Device)cameras consist of an array of small capacitors (a few micron in size) each defininga pixel (= picture element). When light (with frequency in the infrared or higher) im-pinges on a given pixel it kicks off an electron from one side of the capacitor to the otherand charges it with an amount which is proportional to the light intensity. The chargesin a given row of capacitors are then read out by transfering them from one capacitor tothe next along the line like in a ”bucket brigade”. Thus is the image read and stored. Bycovering the array of pixels with a mask-array that filters different colors (Red, Green orBlue) the device can be transformed into a color camera where adjacent pixels respondto different colors. Similarly all of today’s solar cells are based on the photo-electric effect using lightto generate a current by transfering electrons in a semiconductor material from the so-called valence band (and leaving a positively charge ”hole” behind) into the conductionband (on which more below).2.1.3 Bohr’s atomFollowing on the footsteps of Planck and Einstein who proposed that energy and mo-mentum were quantized: E = n ω and p = n k, Niels Bohr suggested in 1913 thatthe angular momentum of electrons in an atom was similarly quantized: L ≡ mvr = nthereby explaining their paradoxical stability (see above). Indeed given the balance ofelectrostatic and centrifugal forces: mv2 /r = e2 /r2 and the assumed quantization of an-gular momentum one deduces that in the hydrogen atom the orbits of the electron arequantized with a radius: r = n2 2 /me2 ≡ n2 r0 (r0 ≡ 2 /me2 = 0.53Å is known as theBohr radius of Hydrogen), velocity v = e2 /n and energy: 1 2 e2 me4 e2 1 13.6eV En = mv − =− 2 2 =− =− (2.8) 2 r 2 n 2r0 n2 n2Thus the energy to ionise a hydrogen atom, i.e. kick off its electron from its ground stateat n = 1 is 13.6 eV. Because of energy quantization an electron orbiting the nucleus willnot radiate continuously, but emit (or absorb) radiation in quantas of energy: 1 1 hνnm = Em − En = 13.6eV( 2 − 2) (2.9) n mHence the emission or adsorption spectra of atoms consists of discrete lines correspond-ing to electronic transitions between states with different quantum numbers. For the hy-drogen atom only the lowest energy transitions to n = 2 (the so-called Balmer series) liein the visible range with wavelengths in the red: 656.3nm (m = 3); in the blue 486.1nm(m = 4) and in the UV range: 434.1nm (m = 5) and 410.2nm (m = 6), see Table 2.1 andFig.2.1. The explanation of the stability of atoms and the emission lines of hydrogenwas a major success of the Bohr model for which he was awarded the Nobel prize in1922.
  • 14. THE PUZZLES OF MATTER AND RADIATION 15 m= 2 3 4 5 6 7 8 Lyman series (n=1) 121.6 102.6 97.2 94.9 93.7 93.0 92.6 Balmer series (n=2) - 656.3 486.1 434.1 410.2 397.0 388.9 Pashen series (n=3) - - 1870 1280 1090 1000 954Table 2.1 The major emission lines in the hydrogen atom. The wavelength (in nm) istabulated for various values of the initial (m) and final state (n) Bohr’s approach to the hydrogen atom was generalized in 1915 by Arnold Som-merfeld who proposed that for any bound particle (atom, harmonic oscillator, etc.) aquantity known in classical mechanics as the action was quantized: p · dq = nh (2.10)where p, q are the momentum and coordinate of the particle. The Wilson-Sommerfeldquantization rule, Eq.2.10, reduces to Bohr’s quantization of the angular momentum inthe case of the hydrogen atom since the integral pdq = 2πmvr. But the same rule alsoexplains why the oscillators of frequency ω assumed by Planck to exist in the walls ofa back-body would emit radiation in quantas of energy ω. For a harmonic oscillator √with frequency ω = k/m, the energy is: p2 kq2 p2 mω2 q2 Eosc = + = + 2m 2 2m 2from which one derives: qm qm pdq = 2 2mEosc − (mωq)2 = 2mω q2 − q2 dq = πmωq2 m m −qm −qmwith q2 ≡ 2Eosc /mω2 . Hence Eq.2.10 implies: Eosc = n ω, which is the assumption mmade by Planck. The Wilson-Sommerfeld quantization rule can also be used to find the energy levelof a quantum rotator rotating at frequency ω about one of its major axes with momentof inertia I . In that case the angular moment Iω = l (l = 0, 1, 2, ...) and the angularenergy is El = Iω2 /2 = 2 l2 /2I. In 1924, Louis de Broglie (in his Ph.D thesis!) generalizing Sommerfeld’s idea sug-gested that all matter are described by waves. So, just as a photon possesses a momen-tum p = k, thus an electron is characterized by wavevector: k = p/ . The Wilson-Sommerfeld rule is thus equivalent to the requirement that the electron wave in a boundsystem interfers constructively to form a standing wave. As we shall see in the followingde Broglie’s analogy opened the way for the formal development of quantum mechan-ics from its analogy with optics: classical mechanics becoming to quantum mechanicswhat geometrical optics is to electromagnetic waves.
  • 15. 16 QUANTUM MECHANICSF. 2.4. (A) The absorption of radiation by an atom in its ground state. (B) the stimu- lated emission of a photon in presence of radiation by an atom in its excited state: notice that this process is the time reversal of adsorption. (C) The spontaneous emis- sion of a photon (in absence of radiation) by an atom in its excited state.2.1.4 Adsorption, stimulated emission and the laserBased on the Bohr-Sommerfeld model, Einstein proposed in 1917 a simple theory oflight-matter interaction which could account for Planck’s formula and would be (40years later) the basis for the invention of the laser. First he pointed out that since micro-scopic processes are reversible the adsorption of a photon is indistinguishible from theprocess of stimulated emission, see Fig.2.5. In other words in presence of an externalelectro-magnetic field the transition rate B21 to state 2 from 1 should be the same as thetransition rate B12 to state 1 from 2. The transition being due to the interaction betweenradiation and matter, the overall rate T i j (i, j = 1, 2) will depend on the spectral densityof radiation at the transition energy ρ(ν = ∆E/h) and on the density of states 1 and2: n1 = N1 /V and n2 = N2 /V (where Ni is the number of atoms in state i and V thevolume): T i j = Bi j ρN j /V. Einstein also recognized that in abscence of interaction withthe external field, an atom in an excited state 2 could spontaneously return to the groundstate 1 by emission of a photon of energy hν. That process depends on the lifetime τ sof the excited state and occurs at a rate A12 = 1/τ s . To summarize in steady state thetransition rates to and from each state should balance: B21 ρN1 /V = B12 ρN2 /V + A12 N2 /VFrom which since B12 = B21 we derive: A12 /B12 ρ(ν) = N1 /N2 − 1Since at thermal equilibrium the probability of being in the excited state is smaller thanin the ground state by the Boltzmann factor N2 /N1 = exp(−∆E/kB T ) one obtains: A12 1 8πhν3 1 ρ(ν) = = B12 e hν/kB T − 1 c3 ehν/kB T − 1
  • 16. THE PUZZLES OF MATTER AND RADIATION 17F. 2.5. Principle of operation of a laser. An amplifying medium is pumped by an external energy source (e.g. a flash lamp) to generate a higher density of excited states than of ground states (population inversion). The medium is placed in a cavity with reflecting mirrors, one of which lets a small fraction of the light out. The light reflected in the cavity is amplified by the stimulated emission of the excited states. When the threshold for lasing is achieved the losses in the cavity are balanced by the gain from the amplifying medium.with the identification: A12 /B12 = 8πhν3 /c3 or: c3 λ3 B12 = 3τ = 8πhν s 8πhτ s Einstein model could account for Planck’s Black-body radiation if atoms capableof absorbing radiation at all frequencies are uniformly present. It also made possibledecades later the invention of the laser, acronym for Light Amplification by StimulatedEmission of Radiation. A laser consists of a light amplifying medium inside a highlyreflective optical cavity, which usually consists of two mirrors arranged such that lightbounces back and forth, each time passing through the gain medium, see Fig.2.5. Typi-cally one of the two mirrors is partially transparent to let part of the beam exit the cavity.To achieve light amplification the excited state in the medium of a laser cavity emittingat frequency ν0 = ∆E/h has to be more populated than the lower energy state. Since atthermodynamic equilibrium low energy states are always more populated than higherenergy ones, energy must be injected in the medium to ”pump” (i.e. excite) atoms fromthe ground state into the light emitting state. Let us therefore consider light of intensity I(z, ν) = ρ(z, ν)c (0 < z < l) propagatingin a cavity of length l and cross section S . Due to stimulated emission of power dPemitin a volume dV = S dz the increase in the light intensity is:
  • 17. 18 QUANTUM MECHANICS dI = dPemit /dV = hν0 (n2 − n1 )B12 ρ(ν)δ(ν − ν0 ) dz c2 = (n2 − n1 ) I(z, ν)δ(ν − ν0 ) (2.11) 8πν0 τ s 2For various reasons that we shall discuss later, the transition frequency between twostates is never infintely sharp and one replaces the δ-function in the preceeding equationby a function g(ν) which approximates it: g(ν)dν = 1. Quite often the response g(ν)of a resonant oscillator is appropriate (see Appendix on Fourier transforms): γ/π g(ν) = (ν − ν0 )2 + γ2One then obtains: dI c2 = (n2 − n1 ) g(ν)I(z, ν) ≡ γ(ν)I(z, ν) dz 8πν0 τ s 2As argued above, the light intensity grows exponentially when the population of atomsin the medium is inverted, i.e. when n2 > n1 . As light propagates in the cavity it sufferlosses (due to adsorption for example) of magnitude α cm−1 . Since light has to comeout of the cavity there are also losses due to the fact that only a portion R of the intensityis reflected back into the cavity. For a laser to operate the losses must equal the gain:R exp[(γ(ν) − α)l] = 1, which imply that the population inversion at threshold has tosatisfy: 8πτ s 1 (n2 − n1 )t = 2 (α − ln R) λ g(ν) lHence the longer the wavelength the smaller the required population inversion for las-ing. That is one of the reasons that masers (lasers in the microwave range) were thefirst to be invented while X-ray lasers, even though of great utility, have been difficultto develop. During steady-state laser operation the balance of losses and gain imply thatthe population inversion remains at threshold. The more the ground state is pumped, themore the excited state is induced to emit by the increased light density in the cavity, thuskeeping the difference between the density of the two states fixed at its threshold.
  • 18. QUANTUM MECHANICAL FORMALISM 192.2 Quantum Mechanical formalismThe early 20th century investigations by Planck, Einstein, Bohr, Sommerfeld, de Broglie, etc. revealed a picture of matter that was different from the infinitely divisible contin-uum of energy and momentum that prevailed untill then. Many of the properties ofmatter could be explained by assuming that these quantities were discrete rather thancontinuous. Thus could Planck explain the radiation spectrum of black-bodies, Einsteinthe photo-electric effect and Bohr the stability of atoms and the emission spectrum ofhydrogen (though not of other elements). These early efforts suggested that matter andradiation shared similar properties: light came as photons, particles of zero mass butpossessing definite energy and momentum. Similarly electrons had wave-like proper-ties and could interfere with themselves, as in the orbitals of Bohr’s atom. What wasmissing was a conceptual framework that would unite these observations and modelswith classical mechanics. The breakthrough came with the works of Werner Heisenbergand Max Born in 1925 and Erwin Schroedinger in 1926. The later in particular wrote anequation for the probability of finding a particle at a given position that was inspired bythe analogy pointed out by de Broglie between waves and particles. As we shall see laterthe eigenvalues of the famed Schroedinger equation yield the energy levels of a boundsystem, much as one determines the resonant modes of electromagnetic radiation in acavity (in both cases one solves a Helmholtz equation, see Appendix). Heisenberg proposed a matrix formulation of Quantum Mechanics that was latershown by Schroedinger to be equivalent to his own formulation. Heisenberg’s approachhowever inspired the mathematically rigorous and clear formulation of Quantum Me-chanics presented in 1930 by Paul A.M.Dirac in his landmark book (”the Principlesof Quantum Mechanics”, Clarendon press, Oxford). In the following we shall followDirac’s lead. According to Dirac, a physical system is characterized by its state |Ψ > (also calledwave-function by Schroedinger). These states are complex unit vectors in a so-calledHilbert space, defined so that their scalar product < Ψ|Ψ >= 1. When an observableO is measured, the system is perturbed and ends up in an eigenstate |n > of O witheigenvalue On : O|n >= On |n > (in linear algebra the eigenstates are the vectors thatdiagonalize the matrix O). As in linear algebra one can expressing the vector-state |Ψ >in terms of the eigenvectors |n > of O: |Ψ >= αn |n > (2.12) nSince |Ψ > is a complex vector its conjugate is: < Ψ| = n α∗ < n|. The physical ninterpretation of the amplitudes αn is at the core of Dirac’s formulation: the probabilityof observing a system in state |n > after the measurement of O is: |αn |2 . So that theprobability of measuring a value On when the system is in state |Ψ > is: P(On ) = | < n|Ψ > |2 = |αn |2 (2.13)
  • 19. 20 QUANTUM MECHANICSNotice that |Ψ > being a unit vector: n |αn |2 = 1 as it should for |αn |2 to be interpretedas a probability. This is the physical interpretation of the wave-function and the intrinsicindeterminism of QM that so annoyed Einstein. The average value of O in state |Ψ > is: < Ψ|O|Ψ > = (< n|α∗ n )O(αm |m >) n,m = Om α∗ n αm < n|m > (2.14) n,m = |αn |2 On = P(On )On =< O > (2.15) n nwhere we assumed the eigenstates to be orthonormal < n|m >= δnm . Quantum me-chanics in Dirac’s formulation is thus reduced to linear algebra: the states are complexvectors and the observables complex matrices with real eigenvalues, i.e. Hermitian ma-trices satisfying Amn = A∗ . If the eigenstates of one operator (observable) A are also nmthe eigenstates of an other operator B then A and B commute: AB|n >= Abn |n >= an bn |n >= bn an |n >= bn A|n >= BA|n >If the operators commute they can be both measured simultaneously: they are diagonal-ized by (i.e they share) the same eigenstates. If on the other hand the eigenstates of Aand B are not identical, their simultaneous measurement is not possible. Let {|n >} bethe eigenvectors of A, then BA|n >= Ban |n >= an B|n >= an |m >< m|B|n >= an Bmn |m > m mOn the other hand: AB|n >= A|m >< m|B|n >= am Bmn |m > an Bmn |m >= BA|n > m m mhence the operators do not commute [A, B] ≡ AB − BA 0. We shall see later thatthe non-commutability of operators (which is quite common with matrix operations)is at the core of Heisenberg uncertainty principle which states that the position andmomentum of a particle cannot be simultaneously determined. The energy being an important observable in physics, the energy operator (or Hamil-tonian, H) plays an important role in Quantum Mechanics. Its eigenvalues are the pos-sible measured energies of the system (which can be discrete or continuous) and itseigenmodes are like the resonant modes of an oscillator or the specific orbits of theelectron in Bohr’s atom: H|n >= n |n >From Planck and Einstein, we know that there is a relation between energy and fre-quency: n = ωn , so that an eigenstate with given frequency ωn evolves in time asexp(−iωn t) = exp(−i n t/ ). Notice that if the initial state of the system is one of the
  • 20. SIMPLE QM SYSTEMS 21eigenstates |n >, it remains there with probability P = | exp(−iωn t)|2 = 1. If however theinitial state is |Ψ(0) >= n αn |n > then: |Ψ(t) >= αn e−i n t/ t |n > nFrom which one derives Schroedinger’s equation: ∂ 1 1 |Ψ(t) >= αn n e−i n t/ t |n >= H|Ψ(t) > ∂t i n ior in the more common notation: ∂ i |Ψ(t) >= H|Ψ(t) > (2.16) ∂t This is essentially all of Quantum Mechanics: a definition of physical states as vec-tors in a complex space, measurements as matrix operations on these vector states, theoutcome of the measurements as eigenvalues of those matrices and a description of thetime evolution of the physical states by Schroedinger’s equation. The rest is applicationof this linear algebra formalism!2.3 Simple QM systems2.3.1 The chiral amonia moleculeOur first application of the QM formalism described above will be the amonia moleculeand the amonia maser (the microwave equivalent of the laser discussed earlier). We willconsider here the chiral amonia molecule NHDT (D and T stand for the isotopes of Hy-drogen: Deuterium and Tritium), rather than the achiral NH3 considered by Feynman (inVol3 of his ”Lectures on Physics”). This choice allows us not to care about rotationalmotions and it exemplifies the queer nature of QM better than the achiral molecule.NHDT is a tetrahedron with the four atoms sitting at the four apexes. The moleculepossesses distincts enantiomers, i.e. distinct states |1 > and |2 >, which are mirror im-ages of each other depending on whether the Nitrogen atom is on the right or the leftof the HDT plane, see fig.2.6. These states are not eigenstates of the Hamiltonian asthe Nitrogen in state |1 > can end up in state |2 > by passing through the HDT plane,like a left-handed glove can be transformed into a right-handed one by turning it insideout. Even though this energetically costly transition is classically impossible there is inQM always a small probability for such a process to happen (this tunneling through anenergetically forbidden zone (a wall, see below) is one of the oddities of QM). Due to the symmetry of states |1 > and |2 >, the Hamiltonian of this two-statesystem can thus be written as: E0 −A H0 = −A E0
  • 21. 22 QUANTUM MECHANICSF. 2.6. The chiral amonia molecule, NHDT consist of a nitrogen atom bound to the different isotopes of hydrogen(H): deuterium(D) and tritium(T). This molecule exist with different chirality: left-handed or right-handed which are mirror images of each other. Since nitrogen is slightly more electrophilic (negatively charged) than the hydrogen isotopes the molecule possesses a small electric dipole moment µ.Its eigenvalues (eigen-energies) are: E I,II = E0 ± A and its eigenvectors are: 0 1 1 |I >= 1 √ |II >= 1 √ 2 −1 2 1You can check that H0 |I >= E I |I > and H0 |II >= E II |II >. In the eigenvector basis the 0 0Hamiltonian is diagonal: E +A 0 H0 = 0 D 0 E0 − AIt is a general result from linear algebra that the matrix of eigenvectors: 1 1 1 Λ= √ 2 −1 1diagonalizes the original matrix: H0 = ΛT H0 Λ. D Notice that the energy eigenstates for this chiral amonia molecule consist of a co-herent superposition of a left- and a right-handed state with probability 1/2! This is aclassically absurd situation akin to Schroedinger’s famous cat paradox, see Fig.2.7. Inthis gedanken experiment he proposed to couple a cat enclosed in a box to a two-stateQM system: the cat is dead if the system is in state |1 > and alive if in state |2 >. Ac-cording to QM, before one looks into the box the cat exists as a superposition of the twostates: dead and alive; just like our chiral amonia molecule is described as a superpo-sition of left and right-handed states. However once a measurement is made the cat iseither dead or alive; just as the chiral amonia molecule is -when observed- either left-or right-handed.
  • 22. SIMPLE QM SYSTEMS 23F. 2.7. The gedanken (thought) experiment that Schroedinger proposed to test the va- lidity of QM. In a closed box isolated from the external world there is a cat and a radioactive source of say β−particles (electrons). If a particle is emitted and detected by a Geiger counter a poison vial is broken that kills the cat. Otherwise the cat is alive. As the state of the particle is a superposition of bound and emitted particle, one must consider the cat to be in a superposition of a live and dead animal. For Ein- stein that thought experiment demonstrated that QM was incomplete since it gave rise to absurd assumptions. Yet, when isolated from the external world (a tall order for macroscopic systems) all experiments so far are consistent with this ”absurd” superposition of states. So did Einstein ask: how was the cat ”really” before we looked into the box? Heclaimed that the superposition is non-sensical and the cat is either dead or alive. He thenargued that some unknown factors (hidden variables) in the description of the micro-scopic two-state sytem that determines the cat’s observed state results in the QM proba-bilistic and verified predictions. However as mentioned earlier, the experimental viola-tion of Bells’ inequalities suggest that Einstein was wrong and that such ”Schroedingercats” exist as a superposition of dead and alive states, even if we have no clue what thatmeans! We will discuss again that point more quantitatively below. A crucial ingredient enters into the QM picture and it is the measurement process.For a system to exhibit the interference effects resulting from QM state superpositionsit must not interact with the environment. These interactions are like independent mea-surements of the system and they destroy its coherence (e.g. the two-state superposition)by implicating (entangling them with) many external uncontrolled states. Now it is veryeasy for a macroscopic system like a cat to interact with the world outside the box (forexample through the radiation it emits or adsorbs). Hence quantum experiments withlarge objects are notoriously difficult to perform. So far the largest molecule for whichquantum interference effects have been demonstrated is the buckyball: C60 (see below). Since the chiral states are given by:
  • 23. 24 QUANTUM MECHANICS √ |1 > = ( |I > + |II >)/ 2 √ |2 > = (−|I > +|II >)/ 2 (2.17)if the molecule has left-handed chirality to begin with: |Ψ(0) >= |1 >, it will evolve as: 0 0 e−iE I t/ |I > + e−iE II t/ |II > |Ψ(t) >= √ 2The probability of finding the system with a left-handed chirality (state |1 >) after atime t is: 0 0 |e−iE I t/ + e−iE II t/ |2 P1 (t) = | < 1|Ψ(t) > |2 = = cos2 (At/ ) 4and the probability of finding the system with right-handed chirality is P2 (t) = sin2 (At/ ).When a given molecule is measured its chirality is well defined (either left or right), butmeasurements over many molecules yield the oscillating probability distribution justcomputed. Since cos2 (At/ ) = (1 + cos 2At/ )/2 the oscillation frequency is related tothe difference between the energy levels of the eigenstates ω0 = 2A/ . For NH3 thatfrequency ν0 = ω0 /2π = 24 Ghz is in the microwave range. For NHDT it is slighlylower due to the higher mass of Deuterium and Tritium. The essence of Bell’s inequalities may be grasped from this simple example. Imag-ine that a NHDT molecule prepared in a definite chiral state (|1 > or |2 >) is observeda time δt later so that the probability of finding it in the same state is 99%. If it ismeasured again a time δt later it will be observed in the same state as previously withprobability 99%. One may now ask: what is the probability of observing the systemin its initial state if we look at it a time not δt but 2δt later? If as Einstein believedthe system is at δt in a definite state that is only once in a hundred times differentfrom the initial state, then at 2δt the state of the system would be at worst twice ina hundred times different from the initial state, namely the probability of observingthe system in its initial state would be at worst 98%. However the QM prediction is:P(2δt) = cos2 (2Aδt/ ) = 1 − 4(Aδt/ )2 = 0.96 (since we assumed that P(δt) = 0.99,i.e. Aδt/ = 0.1). The QM mechanical prediction violates the lower bound (the Bellinequality) set by ”realistic” theories which assume that the system is in a definite statewhich we have simply no way of determining, not a ”meaningless” superposition ofleft and right-handed molecules. As mentioned earlier the experimental results (mea-sured not on chiral amonia molecules but some other two-state system) vindicate theQM prediction and rule out the ”realistic” theories.2.3.2 The amonia molecule in a constant electric fieldSince nitrogen is more electrophilic than hydrogen, it tends to be slightly more nega-tively charged than the hydrogen isotopes and the molecule ends up with a permanent
  • 24. SIMPLE QM SYSTEMS 25electric dipole moment µ, as shown in Fig.2.6. In presence of an electric field E the en-ergy of a dipole is W = −µ · E, (Eq.??). If the electric field is along the x-axis in Fig.2.6,then we expect state |1 > to have higher energy than state |2 >. The Hamiltonian of themolecule in an external electric field is thus: E0 + W −A H= −A E0 − WNotice that in the eigenbasis representation (where the unperturbed Hamiltonian H0 isdiagonal), the perturbed Hamiltonian can be written as: E0 + A 0 0 W H = ΛT HΛ = H0 + δH = D + (2.18) 0 E0 − A W 0 √ √The eigenvalues of H (and H ) are: E I,II = E0 ± A2 + W 2 . Defining tan θ ≡ ( A2 + W 2 −W)/A the eigenvectors of H are: cos θ sin θ |I >= |II >= − sin θ cos θWhen W = 0 one recovers the previous result. When W A: |I > |1 > and |II >|2 >, in which case the enantiomers are also eigenstates. In practice however W =µE A and the energies vary as: E I,II = E I,II ± µ2 E 2 /2A. State |I > (with its dipole 0essentially opposite to the electric field) has higher energy than state |II >. Notice thatwe can write the energies E I,II in the following form (we will see later that this is ageneral result when the diagonal Hamiltonian is perturbed by an amount δH) δH12 δH21 EI = EI + 0 0 0 (2.19) E I − E II δH21 δH12 E II = E II + 0 0 0 (2.20) E II − E I These results suggest a way to separate the eigenstates by passing a beam of amoniamolecules through a strong electric field gradient. This field gradient generates a forceon the molecules: µ2 F I,II = − E I,II = E2 2Awhich separates them: state |I > is deflected to regions of small electric fields, whilestate |II > is deflected to regions of high electric fields. Thus can a sub-populationinversion be generated where high energy amonia molecules are separated from lowerenergy ones. This high energy sub-population can then be used to amplify microwaveradiaton by stimulated emission. The resulting device, known as a maser, was the firstimplementation of a stimulated radiation amplification device and served as the firstatomic clock.
  • 25. 26 QUANTUM MECHANICSF. 2.8. Principle of operation of a maser. An amonia beam (here the chiral molecule NHDT) is sent trough a slit into a beam splitter that consists of a strong inhomoge- nous electric field. At high field one can separate the different enantiomers which are eigenstates of the energy. The high energy eigenstate (|I >) is sent into a cavity where it goes into the low energy state |II > by emitting stimulated radiation at the resonant frequency ω0 = (E I − E II )/2.3.3 The amonia maser and atomic clocksIn an amonia maser, Fig.2.8, the high energy state |I > selected as described, entersa resonant cavity (tuned to the transition frequency ω0 ). A population inversion of theamonia molecules is thus generated in the cavity and amplification of radiation can beexpected. The emitted radiation stimulated by the presence in the cavity of an externalsource generates a highly coherent microwave beam. To analyse the operation of a maser, let us assume that amonia molecules enter acavity in which they experience a time varying electric field: E = Ee−iωt x. This field ˆcouples with the dipole moment of the molecules to modulate their energy by W(t) =−µEe−iωt . The perturbed Hamitonian in the eigenbasis (|I > and |II >) is then, seeEq.2.18: EI 0 0 W(t) H = H0 + δH(t) = D + (2.21) 0 E II W ∗ (t) 0Looking for a general solution: Ψ(t) >= C I (t)|I > +C II (t)|II >, Eq.2.16 yields: i ∂t C I = E I C I + W(t)C II i ∂t C II = W ∗ (t)C I + E II C II (2.22)Looking for a solution C I,II = αI,II (t) exp(−iE I,II t/ ) yields the following equation forαI,II : i ∂t αI = W(t)e−i(E II −EI )t/ αII = −µEe−i(ω−ω0 )t αII i ∂t αII = W ∗ (t)e−i(E I −E II )t/ αI = −µEei(ω−ω0 )t αI (2.23)At the resonance: ω = ω0 , one obtains:
  • 26. SIMPLE QM SYSTEMS 27 ∂2 αI,II + Ω2 αI,II = 0 t with Ω = µE/which solution is αI (t) = cos Ω(t − t0 ) and αII = sin Ω(t − t0 ). The probability of being instate |I > (α2 ) or II > (α2 ) oscillates with frequency 2Ω: the molecule go periodically I IIfrom stimulated emission in state |I > to adsorption in state II >. These oscillationsare different from the oscillations between the enantiomers |1 > and |2 > observed atfrequency ω0 in absence of electric field. This oscillation in the probability of observinga specific enantiomer is not associated to emission of any radiation, it results from thefact that the enantiomers are not eigenstates of the Hamiltonian: in absence of a timevarying electric field, if the molecule is in eigenstates |I > or |II > it remains there. If the frequency of the electric field is slightly off resonance ω − ω0 0 and if themolecule remains in the cavity for a short time t 1/Ω we may assume that αI 1and integrate the equation for αII to yield: ei(ω−ω0 )t − 1 αII (t) = iΩ i(ω − ω0 )The probability of transition from state |I > to state |II > is then: sin2 (ω − ω0 )t/2 PI→II (t) = |αII |2 = (Ωt)2 [(ω − ω0 )t/2]2The function sinc x ≡ sin x/x decays rapidly for values of |x| > π. Hence the transitionrate is significant only for frequencies which are very close to resonance: |ω − ω0 | <2π/t. If the molecule remains in the cavity for 1 sec, the relative possible detuning :|ν/ν0 − 1| = 1/ν0 t ∼ 4 10−11 . Only frequencies within that very narrow range can beamplified by stimulated emission. This allows for very high Q resonators, Eq.??, i.e.very precise frequency generators and clocks. Because of the sharpness of the functionsinc2 (ω − ω0 )t/2 one often rewrites the previous equation using the limit 2π lim sinc2 (ω − ω0 )t/2 = δ(ω − ω0 ) t→0 tto obtain the transition rate from state |I > to |II >: dPI→II (t) 2π|δH12 |2 BI,II = = 2πΩ2 δ(ω − ω0 ) = δ(E I − E II − ω) (2.24) dtNotice that if we had assumed the amonia beam to enter the cavity in state |II >, i.e.with αII = 1, then we could have similarly integrated the equation for αI to yield thetransition probability from state |II > to |I >, which comes to be: 2π|δH21 |2 BII,I = δ(E I − E II − ω) = BI,IIwhich was the intuitive assumption Einstein made: the rate of stimulated emission BI,IIis equal to the rate of adsorption BII,I . Since the energies of the excited states are never
  • 27. 28 QUANTUM MECHANICSperfectly sharp, due to the Heisenberg uncertainty principle that we shall see below anddue to thermal motion that leads to Doppler broadening of the emission lines (see sec-tion??), the δ-function is replaced by the density of states with the appropriate energy:ρ(E) (with E = E II + ω) and the normalisation ρ(E)dE = 1). The amonia maser was the first atomic (or molecular) clock. The present genera-tion of atomic clocks use a microwave transition in Cesium (Cs) as the reference fre-quency for the clock. The atoms are cooled to very low temperatures (µ◦ K) to reducethe Doppler broadening of their emission lines and keep them for as long as possiblein the cavity, usually a Penning trap (see ????). As a result the record for the frequencyprecision of atomic clock is: |ν/ν0 − 1| ∼ 10−16 .2.3.4 The energy spectrum of aromatic moleculesA simple generalization of the two-state system that we considered previously is then-state system consisting for example of a circular chain of n identical atoms such asbenzene C6 H6 (n = 6) around which electrons can hop. If an electron has energy E0when associated with a particular atom |n > and can hop only between nearest neighborsthe Hamiltonian for this system in the basis of the atom’s position |n > is:  E0 −A 0 . . . −A     −A E −A . . . 0       0  H= .   .      .  (2.25)  .    .  .       −A . . . 0 −A E0  The eigenstates of that system obey: H|Ψ >= E|Ψ >, with |Ψ >= n C n |n > Fromwhich we derive the equations: E − E0 C1 + C2 + Cn = 0 A E − E0 C1 + C2 + C3 = 0 A . . . E − E0 C1 + Cn−1 + Cn = 0 ALooking for a solution: Clm = exp i(2πlm/n) where l = 1, 2, ...., n we get: El − E0 = −(ei2πl/n + e−i2πl/n ) AThus the eigen-energies of the system are: El = E0 − 2A cos 2πl/n (2.26) For benzene (n = 6) we have: E6 = E0 − 2A; E1,5 = E0 − A; E3 = E0 + 2A and E2,4 =E0 + A. Energies E1,5,6 which are smaller that E0 are associated to so-called bonding
  • 28. SIMPLE QM SYSTEMS 29F. 2.9. Band-gap theory of material. (a) If the gap between the valence and conduction band is large (a few electron-Volts) the material is an insulator. (b) Variation of the band-gap energy with wave-vector k. Top curve: energy of electrons Ee ; bottom curve: energy of holes Eh (increases towards the bottom). (c) A semi-conductor which due to thermal excitation or illumination has some electrons in the conduction band and some holes in the valence band. (d) a conductor is a material for which the valence and conduction band overlap or equivalently for which an energy band is not filled.orbitals or wave-functions, whereas energies E2,3,4 > E0 are associated to anti-bondingorbitals, on which we shall have more to say later. Notice that the eigen-state associated √to the maximally bonding orbital (l = 6) is fully symmetric: |Ψ6 >= (1, 1, 1, 1, 1, 1)/ 6,while that associated to the maximally anti-bonding orbital (l = 3)√ antisymmetric to isthe permutation of nearest neighbors: |Ψ3 >= (1, −1, 1, −1, 1, −1)/ Conduction bands in solidsAn interesting generalization of the previous analysis is the case of a long chain ofn atoms a distance a apart. In that case we have: Clm = exp[i(2πl/na)ma]. Since theposition of atom m is x = ma we may write Ck (x) = exp ikx with k = 2πl/na. Like theoscillation modes on a string of length na, the electron’s eigenstates are 1D transversewaves of wavelength λ = 2π/k = na/l. The energy of such a mode is: E(k) = E0 − 2A cos ka (2.27)The energy of the electron is bounded: E0 − 2A < E(k) < E0 + 2A. If each atomcontributes two electrons, these 2n electrons will occuy all the n-energy states (we shallsee later that each state can accomodate 2 electrons) and electron hoping in this energy-band will be impossible. If however the on-site energy E0 can possess discrete values(El as it does indeed, for example in Bohr’s model), the coupling of the n−atoms willgenerate energy bands around each energy value El into which electron may hop. Thisforms the basis of the band-theory of conduction in materials, see Fig.2.9: a materialwill conduct if there are empty states into which electrons can hop. If the low energy
  • 29. 30 QUANTUM MECHANICSF. 2.10. A p-type semiconductor consists of a material (usually Silicon, Si) doped with an element (such as Aluminium) which having less electrons in its outer shell than Si tend to trap electrons from the lattice, leaving a electron vacancy instead: a hole. This depletes the valence bands of electrons allowing conduction through motion of holes. A n-type semiconductor consists of a material doped with an elec- tron donor, an element (such as Phosphate) which has more electrons in its outer shell than Si. This injects electrons into the conduction band allowing the material to conduct electricity. A pn-junction is formed when p-type and n-type semiconduc- tors are brought in contact. Such a junction can be used as a Light Emitting Diode (LED) when electron from the n-side of the junction recombine with holes from the p-side. The wavelength of emitted light is determined by the energy-gap between the conduction and valence and bands.(valence) band is filled with electrons and the next (conduction) band is empty butmany electron-volts (eV) above it, then the electrons have no states in which to go andthe material is an insulator. If the conduction band overlaps with the valence band thenthe electrons have empty states to hop to and current can flow in the material. Theinteresting and technologically important case is the situation where the band-gap Egbetween the conduction and valence bands is small: Eg 1eV (the gap for Silicon(Si) is: Eg = 1.1eV; for Germanium (Ge): Eg = 0.72eV). In that case electrons can betransfered from the valence to the conduction band by thermal agitation or via the photo-electric effect resulting in a material that can behave either as a metal or an insulatordepending on the external conditions (temperature, voltage, illumination wavelengthand intensity, etc.). In particular the introduction of atomic impurities (doping) that donate electrons to(or accept electrons from) a semiconductor lattice creates a situation where few elec-trons occupy an almost empty conduction band (or few holes (electron vacancies) oc-cupy an almost full valence band). As we shall see later the electrons in an atom oc-cupy certain orbitals or shells around the nucleus. Atoms (such as Phosphate or Ar-senic) that have more electrons in their outer shell than the bulk semiconductor (theso-called majority carrier, usually Silicon) will usually donate an electron to the lat-tice, whereas atoms that have less electrons in their outer shell (such as Boron or Alu-minium) will accept an electron from the lattice. The doping creates so called n- and
  • 30. MOMENTUM AND SPACE OPERATORS 31p-type semiconductors that have electrons in their conduction-band (n-type) or holes intheir valence-band (p-type), see Fig.2.10. The energy of the electrons moving near thebottom Emin = E0,c − 2Ac of the conduction band (i.e. when ka << 1 in Eq.2.27) is: E(k) = Emin + k /2mc 2 2where the effective mass of the electron in the conduction band is defined as: mc = /∂k E(k) = 2 /2a2 Ac which can be quite different from the mass of a free electron. 2 2Notice that if, as proposed by de Broglie, we identify the momentum of the electronas p = k , then Eq.2.3.5 represents the kinetic energy of an electron with mass mc atthe bottom of the conduction band. Similarly the energy of the hole near the top of thevalence band is: Eh (k) = Emax − 2 k2 /2mvThe movement of holes in the valence band is similar to that of air bubbles in water: asthe moving water displaces the bubble up, the electrons moving in the opposite directionto the hole minimize their energy by displacing it to the top of the valence band. As aresult holes have minimal energy at the top of the valence band: their energy increasesas k increases. Doped semiconductors are the basic ingredients of all the semi-conductor industry(transistors, diodes, integrated circuits, etc.). For example the coupling of p-type andn-type semiconductors, generate a pn-junction which acts like a diode (current flows inonly one direction). It can also be used as a powerful and efficient light source. Electronsin the n-type part of the junction can recombine with holes in the p-type (i.e. transit tothe valence band) with emission of light (just as in the atomic or molecular transitionsdiscussed earlier in the context of stimulated emission and the laser). The advantageof these Light Emitting Diodes (LED) is that by appropriate tuning of the energy gap(appropriate choice of the semi-conducting material) one can tune the wavelength atwhich the LED will emit light. For example in the red and infrared part of the spectrumGalium-Arsenide (GaAs) is the material of choice for LEDs. The intensity of light iscontrolled by the current flowing through the junction. These LED are used in all kindof electronic displays, in new high efficiency spot-lamps and traffic lights, in the remotecontrol of various electronic device, in the laser diode of DVD players, etc.2.4 Momentum and space operatorsIn the example above we considered the case of a QM system that could occupy onlydiscrete states |n >. It is easy to generalize that to free particles that can be found atcontinuous positions |x >. In that case the general wave-function in position (or real)space can be formally written as: |Ψ >= Ψ(x)|x > x
  • 31. 32 QUANTUM MECHANICSWhere Ψ(x) is the probability amplitude of finding the particle at position |x >, so that: P(x) = Ψ∗ (x)Ψ(x) (2.28)Of course the probability of finding the particle somewhere is one, so that the wave-function Ψ(x) is normalized: dxΨ∗ (x)Ψ(x) = 1 (2.29)The mean position of the particle is: < x >= dxP(x)x = dxΨ∗ (x)xΨ(x) (2.30)For example, the wavefunction can be: 1 e−(x−x0 ) /4σ 2 2 Ψ(x) = (2πσ2 )1/4for which the probability distribution: 1 e−(x−x0 ) /2σ 2 2 P(x) = √ 2πσcorresponds to a gaussian distribution with mean < x >= x0 and standard deviation < (x− < x >)2 > = σ. In Fourier-space (see Appendix) we can write: 1 1 Ψ(x) = √ dkΨ(k)eikx = √ d pΨ(p)eipx/ (2.31) 2π 2πwhere we used de Broglie’s relations: p = k. Conversely: 1 Ψ(p) = √ dxΨ(x)e−ipx/ (2.32) 2πΨ(p) is the wavefunction in the momentum (or Fourier) space: |Ψ >= p Ψ(p)|p >. Forthe example chosen above the momentum wavefunction is: e−ipx0 / ∞ dxe−(x−x0 ) /4σ e−ip(x−x0 )/ 2 2 Ψ(p) = √ 2π (2πσ2 )1/4 −∞ −ipx0 / −p2 σ2 / 2 ∞ e e = √ dxe−(x−x0 +2ipσ/ )2 /4σ2 2π (2πσ2 )1/4 −∞ √ 2σ/ −ipx0 / −p2 σ2 / 2 = e e (2π)1/4Being the Fourier transform of Ψ(x), Ψ(p) satisfies the normalization condition:
  • 32. MOMENTUM AND SPACE OPERATORS 33 Ψ∗ (p )Ψ(p) 1= dxΨ∗ (x)Ψ(x) = d pd p dxei(p−p )x/ 2π = d pΨ∗ (p)Ψ(p) = d pP(p) (2.33)where we used the identity (see Appendix on Fourier transforms): dxeipx/ = 2π δ(p) (2.34)Eq.2.33 is known in the theory of Fourier transforms as Parseval’s theorem. For a Gaus-sian wavefunction Ψ(x) (see above example) P(p) corresponds to a Gaussian centeredon p = 0 with standard deviation < (p− < p >)2 > = /2σ. The mean value of themomentum satisfies: < p >= d pΨ∗ (p)pΨ(p) = d p d pΨ∗ (p )δ(p − p)pΨ(p)Using the identity Eq.2.34 and Eq.2.31 allow us to express the momentum operator inreal space as: 1 <p>= dxd p d pΨ∗ (p )ei(p−p )x/ pΨ(p) 2π 1 ∂ = dx d p Ψ∗ (p )e−ip x/ dp Ψ(p)eipx/ 2π i ∂x ∂ = dxΨ∗ (x) Ψ(x) ≡ dxΨ∗ (x) pΨ(x) ˆ i ∂xwhere we identify the momentum operator in real space as: ∂ p= ˆ ≡ −i ∂ x (2.35) i ∂xSimilarly by writing the mean position < x > in momentum space we come to identifythe position operator in momentum space as: ∂ x=i ˆ ≡ i ∂p (2.36) ∂pNotice that momentum and space-operators do not commute. In real space: < x|[x, p]|x >= dxΨ∗ (x)(x p − px)Ψ(x) = ˆ ˆ dxΨ∗ (x)[x∂ x Ψ − ∂ x (xΨ)] = i iWe would have obtained the same result by computing the commutator in momentumspace: < p|[x, p]|p >= d pΨ∗ (p)( x p − p x)Ψ(p) = i . Since two observable cannot ˆ ˆshare the same eigenstates if they do not commute, the position and momentum of aparticle cannot be simultaneously determined with absolute precision. The same resultholds also for the commutator of time and energy: < x|[H, t]|x >=< x|Ht − tH|x >= dxΨ∗ (x)(i ∂t t − ti ∂t )Ψ(x) = i
  • 33. 34 QUANTUM MECHANICS2.4.1 Heisenberg uncertainty principleFrom the result that momentum and space do not commute we can derive the Heisenberguncertainty principle. Consider the action on state |ψ > of the operator xo +iλpo where λis a number and the operators xo and po are the deviation of the position and momentumoperators from their mean: xo = x− < x > and po = p− < p >: |φ >= (xo + iλpo )|ψ >Since [xo , po ] = [x, p] = i the positiveness of the probability implies that: 0 ≤ < φ|φ > = < ψ|(xo − iλpo )(xo + iλpo )|ψ > = < ψ|xo + iλ[xo , po ] + λ2 p2 |ψ > 2 o = < ψ|xo |ψ > − λ+ < ψ|p2 |ψ > λ2 = P2 (λ) 2 oFor the quadratic polynomial P2 (λ) to be non-negative for any real λ, its determinanthas to satisfy: 2 − 4 < ψ|xo |ψ >< ψ|p2 )|ψ >≤ 0 2 oOr in terms of the position and momentum variables: 2 < ∆x2 >< ∆p2 > ≥ (2.37) 4This is Heisenberg’s principle: it sets a limit on the precision with which one can mea- √sure both the position δx = < ∆x2 > and the momentum δp = < ∆p2 > of a physi-cal system. The smallest uncertainty (the equality in Eq.2.37) is obtained for a Gaussianprobability distribution as can be verified from the example worked out above. Sincethe Hamiltonian and time operators do not commute similar uncertainty relation can beobtained for the energy and time uncertainties: 2 < ∆E 2 >< ∆t2 > ≥ (2.38) 4 Notice that Heisenberg principle is a direct mathematical consequence of the QMdescription of physical systems by a complex wave-function Ψ(x) (Eq.2.13) and of deBroglie’s relation between wavelength and momentum. Heisenberg uncertainty princi-ple is a tautology: a consequence of the definition of Fourier transforms (see Appendix).In the context of communication it has been known for a long time: a very short timesignal is spread over a very large frequency spectrum. In the context of optics we havealso encountered it in the diffraction pattern from a hole which is larger the smaller thehole is.
  • 34. SCHROEDINGER’S EQUATION 352.5 Schroedinger’s equationWe have determined the representation of the position operator in momentum space andof the momentum operator in real space. Note that in its eigenspace the momentumspace p is diagonal, i.e. it is a number. We can now write the representation of the ˆHamiltonian in any of these Hilbert spaces. It is often easier to work in real space, inwhich case the Hamiltonian which is the sum of kinetic and potential energy is writtenas: p2 ˆ 2 2 H= + V(x) = − (∂2 + ∂2 + ∂2 ) + V(x) = − x y z 2 + V(x) (2.39) 2m 2m 2mand Schroedinger’s equation, Eq.2.16 can be recast as: ∂ 2 i Ψ(x, t) = − 2 Ψ(x, t) + V(x)Ψ(x, t) (2.40) ∂t 2mMultiplying Eq.2.40 by Ψ∗ , its complex conjugate by Ψ and subtracting the two yieldsa conservation law for the probability distribution P(x) = |Ψ(x)|2 : 2 i ∂t P = − (Ψ∗ 2 Ψ−Ψ 2 Ψ∗ ) 2m 2 = · (Ψ∗ Ψ − Ψ Ψ∗ ) 2mwhich can be recast in the usual form (see Eq.?? for the charge distribution in EM): Ψ∗ vΨ + c.c. ˆ ∂t P + ·J=0 with J = Ψ∗ Ψ + c.c. = (2.41) 2im 2where c.c stands for the complex conjugate and v = ˆ /im is the velocity operator.Eq.2.41 is a very important self-consistency check of Quantum Mechanics, since forP(x) to be interpreted as a probability distribution it must satisfy a conservation law.This equation expresses the intuitive expectation that the change in the probability offinding a particle at a given position is equal to the particle flux gradient.2.5.1 Diffraction of free particlesIf the potential is null V(x) = 0 then the eigen-solutions of Schroedinger’s equation,Eq.2.40: 2 dΨ i =− 2 Ψ (2.42) dt 2mare plane waves: Ψ(x, t) = eik·x−iEk t/with p = k and Ek = k /2m. These plane-waves are eigenmodes of the momentum 2 2operator: pΨ(x, t) = −i ˆ Ψ(x, t) = p Ψ(x, t)
  • 35. 36 QUANTUM MECHANICSF. 2.11. The double-slit or Young’s experiment. (a) a wave passing through two slits in a screen generates two wave-sources which interference creates on a far-away screen a pattern of interference consisting of alternating minima and maxima of intensity. (b) a particle in state |O > impinging on a double slit generates two states |I > and |II > corresponding to its passage through slit 1 or 2. The phase of these states evolves as exp ikl. If their coherence is maintained they can interfere on a screen a large distance z from the slits, generating an oscillating pattern related to their phase difference: φint = k(l2 − l1 ) = kd sin θ. (c) Observation of the interference pattern of electron passing through a double slit and impinging on a camera. Each electron is observed as a particle (white dot) with a well defined position on the camera. The QM interference pattern is only visible when a sufficiently large number of particles has been observed (A.Tonomura, Proc.Natl.Acad.Sci. 102, 14952 (2005). One of the most striking confirmations of the QM mechanics picture is the obser-vation of a diffraction pattern like the one seen with electro-magnetic radiation whena free particle is passed through one or a few slits (see ??? and Fig.2.11(a)). Let theparticle be in an eigenstate |0 >= Ψ(x, t) = eikz−iEk t/as it impinges on a screen that is absoring except for two apertures of size a a distanced apart. In the far-field, i.e. at distances z d2 /λ the wave amplitude is given byHuygens’s principle, Eq.??: eik(z+(x +y )/2z 2 2 Ψdi f f (x, y, z) = dx dy e−i(kx x +ky y ) (2.43) iλzWhere k x = kx/z and ky = ky/z. Thus the probability of detecting a particle on a screena distance z from the slits is: 4a2 |Ψdi f f |2 = sinc2 k x a cos2 k x d/2 (2.44) (λz)2where sinc x ≡ sin x/x. If the distance between the diffraction slits is much larger thantheir width (d a), the probability oscillates with a period: δx = λz/d. While each
  • 36. SCHROEDINGER’S EQUATION 37particle is observed as such, i.e. localized at a given position, the overall probabilitydistribution of the particles varies in space just as the diffraction pattern of EM radiationdoes. There is an other way to derive that result, see Fig.2.11(b). The wave-function of theparticle on the screen can be written as: |Ψ >= |I > +|II > where state |I >∼ eikl1 |O >corresponds to the particle passing through the first slit and state |II >∼ eikl2 |O > to theparticle passing through the second slit. If the slits are assumed to be thin, i.e. a d,then |Ψdi f f |2 = < Ψ|Ψ >∼ |1 + eik(l2 −l1 ) |2 < O|O >= 4 cos2 kd sin θ/2 = 4 cos2 k x d/2 ≡ 4 cos2 (φint /2)since l2 − l1 d sin θ xd/z. Notice that the interference pattern is observed only if thephase-coherence between states |I > and |II > is maintained. For example if one wantedto see through which slit the particle has passed, one would have to use radiation withwavelength λo d. Such radiation would impart on the detected particle a momentumuncertainty δpo ∼ h/λo . In the far-field d2 /λz 1 the observation imparts on theobserved state a phase uncertainty: δφo = δpo z/ ∼ 2πz/λo > 2πx/d 2πxd/λz ∼ φintwhich is enough to destroy the interference pattern. This is the essence of Heisenberg’sprinciple: if the particle interacts with radiation of small enough wavelength as requiredto determine its position the perturbation to its momentum will be so large as to destroythe coherence of the states. Trying to determine through which slit the particle haspassed, namely to localize its position to better than the distance between the slits d,imparts on the particle a momentum uncertainty large enough to destroy the interferencepattern, replacing it with an image of the two slits.2.5.2 Quantum interference observed with C60In 1999, the group of Anton Zellinger observed QM interference effects in the largestmolecule investigated so far: the fullerene C60 , a molecule that can be seen with aScanning Tunneling Microscope (STM, see below), Fig.2.12(a). A beam of moleculesat ∼ 900◦ C with a velocity distributed around 220 m/sec was shot at slits of sizea ∼ 40nm and inter-distance d ∼ 100nm. At this velocity the wavelength of the parti-cles is λ = 0.025Å! They were observed on a detector positioned a distance z 1.25mfrom the slits (i.e. in the far-field λz d2 ). Since the size of the slits a is only slightlysmaller than d the intensity is strongly damped (see Eq.2.44) by the factor sinc2 k x a whenx > λz/2a. Hence only a few oscillations in the probability of detection can be observedin Fig.2.12(c). In particular the first peaks (fringes) at x = 0 and x = ±λz/d ∼ 30µmcan be seen. The contrast is not as high as expected since the velocity of the molecules(hence their momentum and the distance between fringes) is thermally spread. It is in-teresting however that the coherence of the molecule passing through the slits can be
  • 37. 38 QUANTUM MECHANICSF. 2.12. The double-slit or Young’s experiment performed with the C60 molecule (Na- ture 401, 680 (1999)). (a) Scanning tunneling microscope (STM) images of a C60 cluster and the reconstructed theoretical image of the cluster. (b) Schematic drawing of the double slit experiment with C60 and the expected pattern, which is a cosine wave modulated by the one-slit diffraction pattern (sinc2 k x a). (c) Upper plot: results of the diffraction of the molecule from the slits (the continuous line is a best fit to the theoretical prediction). Notice that the small side peaks are observed at a dis- tance δx = λz/d ∼ 30µm from the central peak as expected for the diffraction from two slits (see text). Bottom plot: the results when no slits are present and hence no diffraction is expected (control experiment).maintained with molecules at such high temperatures. The reason for that coherence isthat the particle is emitting black-body radiation at a wavelength of a few µm which ismuch larger than the distance between the slits d ∼ 0.1µm. Hence the uncertainty in themolecule’s momentum imparted by its black-body radiation is not enough to destroythe coherence of the wave-function passing through contiguous slits.2.5.3 QM tunneling and the Scanning Tunneling MicroscopeA consequence of Heisenberg’s principle and a queer feature of Quantum mechanics isthe possibility for particles to pass through energetically forbidden regions. Imagine aparticle moving with momentum p along the x-axis and encountering a wall (an energybarrier of height V > p2 /2m = E and size x = d, see Fig.2.13(a)). In classical mechanicsthe particle is reflected from such a wall. However in QM, the particle has a finiteprobability of passing through the wall and emerging on its other side! On the left sideof the wall the particle’s wavefunction is the sum of an incident and a reflected wave: Ψl (x) = Aeikx + Be−ikxwith k2 = 2mE/ 2 , while on its right side the transmitted wavefunction is: Ψr (x) = Ceikx
  • 38. SCHROEDINGER’S EQUATION 39In the forbidden region (i.e. within the wall: 0 < x < d) Schroedinger’s equation,Eq.2.40 2 − 2 Ψ + VΨ = EΨ 2mhas solutions of the form: Ψo (x) = De−κx + Eeκx √with κ = 2m(V − E)/ . At the interface with the wall (at x = 0 and x = d) thecontinuity of the wavefunction and its first derivative require that: A+B = D+E ik(A − B) = κ(−D + E) Ceikd = De−κd + Eeκd ikCeikd = κ(−De−κd + Eeκd )The transmission probability is therefore: C 4k2 κ2 T = | |2 = A 4k2 κ2 + (k2 + κ2 ) sinh2 κd 4E(V − E) 16E −d/λt = → e when : E V and κd 1 4E(V − E) + V 2 sinh2 κd VWhere λt = 1/2κ is the typical tunneling length. For an energy barrier V = 1eV: λt1Å which is the typical size of an atom. The transmission probability thus decreasesexponentially with the size d of the gap which can be but a few Å wide. This sensitivity of QM tunneling to atomic size dimension has led to the develop-ment in 1981 of the Scanning Tunneling Microscope (STM) by Gerd Binning and Hein-rich Rohrer. They used it to image surfaces with atomic resolution and were awardedfor it the Nobel prize in Physics in 1986. In a STM a conducting tip is scanned abovea conducting surface, see Fig.2.13(b). The electrons tunnelling between the tip and thesurface (across an insulating gap, usually vacuum) generate a current which amplitudeis exponentially sensitive to the size of the gap and the voltage across it. A piezo-electrictube displaces the tip above the surface while maintaining a constant current, namely afixed distance between the tip and the surface via an appropriate feedback loop. Mea-suring the amplitude of the feedback signal while scanning the tip generates an image ofthe surface with atomic resolution, Fig.2.13(c). This invention revolutionized the studyof matter at the nanoscale. It also led to the development of many other scanning probemicroscopes, such as the Atomic Force Microscope (AFM) which uses the deflectionof a small cantilever, like the needle in a gramophone, to map a surface with atomicresolution.
  • 39. 40 QUANTUM MECHANICSF. 2.13. (a) An electron wave of energy E is incident on a barrier of energy V. Some of the wave is reflected and some tunnels through the barrier to be transmitted. (b) The principle of the Scanning Tunneling Microscope (STM). A conducting tip mounted on a piezo-electric drive is scanned across a surface. The current tunneling between the surface and the tip is kept constant via an appropriate feedback loop that maintains a constant distance between tip and surface. The feedback signal is used to generate an image of the surface with atomic resolution. (c) Image of the surface of a gold crystal taken in vacuum at 77◦ K. To minimize their surface energy the gold atoms deviate from the bulk crystal structure and arrange in columns several atoms wide with regularly-spaced pits between them. The STM also made possible the study of QM problems which for decades wereassumed to be thought experiments, such as the problem of finding the eigen-energiesof a particle confined in a 2D box. With the help of an STM such a situation could beconstructed and studied: consider for example the case of a free particle (an electronon the surface of a conductor (Cu) with effective mass m∗ = 0.38me ) confined in a twodimensional ring of radius a = 71.3Å formed by 48 atoms of iron (Fe) deposited andmanipulated with an STM to form a corral, Fig.2.14. The Schroedinger equation forsuch a particle is: 2 − 2 Ψ = EΨ 2m∗for r < a and Ψ = 0 for r > a. This is a typical Helmholtz equation (see Ap-pendix) which solutions are: |n, l >≡ Ψn,l (r, φ) = Jl (κn,l r)eilφ where the eigen-energiesEn,l = 2 κn,l /2m∗ are determined by the boundary condition Jl (κn,l a) = 0. Fig.2.14 2presents the results from the group of Don Eigler at IBM who observed with an STMthe wavefunction of an electron trapped in such a corral. The measured tunneling cur-rent is proportional to the probability of finding the electron under the scannig tip, which
  • 40. THE CORRESPONDANCE PRINCIPLE 41F. 2.14. Electrons on the surface of copper (Cu) with effective mass m = 0.38me are confined in a corral of radius a = 71.3Å formed by 48 atoms of iron (Fe), see Crommie et al., Science 262, 218 (1993). The probability of finding an electron at a given place within the corral is proportional to the intensity of the current tunneling into the scanning conducting tip of an STM. The observations are remarquably fit by a linear superposition of just three eigen-functions, solutions of Schroedinger’s equation for this problem.is remarquably fit by a linear combination of just three eigenstates: |5, 0 >, |4, 2 > and|2, 7 > with similar energies (since: κ5,0 a = 14.931, κ4,2 a = 14.796 and κ2,7 a = 14.821).In this corral the surface electrons with the highest energy are thus occupying the top ofthe conduction band from which they hop into the tip of the STM.2.6 The correspondance principleOne of the main self-consistency checks of the interpretation of QM is that its equationsyield the classical results when the system is large and its wavefunction is localized sothat one can replace the operators with their mean values. The correspondance principletherefore states that the time evolution of the mean of an observable should satisfy theequation of classical mechanics. ˆ If O is an operator that does not depend explicitely on time, the evolution of its meanvalue is given by: d d d ˆd <O>= ˆ < Ψ|O|Ψ >= ( < Ψ|)O|Ψ > + < Ψ|O |Ψ > ˆ ˆ dt dt dt dt 1 1 = − < Ψ|H O|Ψ > + < Ψ|OH|Ψ > ˆ ˆ ˆ ˆ i i ˆwhere we used Eq.2.16 with H as the Hamiltonian operator. Hence the time evolutionof the mean value of an operator obeys: d i < O >=< Ψ|[O, H]|Ψ >=< [O, H] > ˆ ˆ ˆ ˆ ˆ (2.45) dtConsider for example the problem of a particle moving in a potential V(x). Its Hamilto-nian is H = p2 /2m + V(x). The mean velocity of the particle is ˆ ˆ
  • 41. 42 QUANTUM MECHANICS d < [ x, H] > < [ x, p2 ] > ˆ ˆ ˆ ˆ < v >= < x >= ˆ = =< p > /m dt i 2i mwhich is indeed the classical result p = mv. Similarly the evolution of the velocityoperator yields: d < [ p, H] > < [ p, V(x)] > ˆ ˆ ˆ ∂V m < v >= ˆ = =−< >=< F > dt i i ∂xWhich is Newton’s law: mdv/dt = −∂ x V = F. For a particle in a magnetic field theHamiltonian is: ˆ2 ˆ ( p − (q/c)A) H= ˆ (2.46) 2m If the magnetic field B is constant along the z−axis, the vector potential is: A =(−yB, xB, 0)/2. The mean velocity v x is: d < [ x, H] > < [ x, ( p2 − (q/c)(A p + pA))] > ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ <x>= ˆ = dt i 2i m < [ x, p2 ] > −2(q/c)A x [ x, p x ] > ˆ ˆx ˆ ˆ ˆ = 2i m < p x − (q/c)A x > = m Which is the classical result: mv = p − (q/c)A. Notice that the canonical momentump is different from the kinetic momentum π = mv. We will leave it as an exercise for thereader to derive the Lorentz force Law: dπ/dt = (q/c)v × B, from the time evolution ofthe velocity operator: v = ( p − (q/c)A)/m. ˆ ˆ ˆ2.6.1 Gauge invariance and the Aharonov-Bohm effectThe Hamiltonian of a charged particle in a magnetic field, Eq.2.46, appears to violatethe principle of gauge invariance. This principle implies that Nature is unaffected bya change in the potential fields (Φ, A) that leave the real electromagnetic fields (E, B)unchanged. A is defined up to the addition of the gradient of an arbitrary function f :A → A + f which does not affect the real field B = × A, see Eq.??. But such a gaugetransformation does modify the Hamiltonian: ˆ ( p − (q/c)A − (q/c) f )2 ˆ H = ˆ 2mHowever this transformation can be compensated by a similar gauge transformation onthe wavefunction: Ψ (x) = eiq f / c Ψ(x)In that case:
  • 42. THE CORRESPONDANCE PRINCIPLE 43F. 2.15. The double-slit or Young’s experiment performed in presence of a magnetic field confined in a region isolated from the interfering electrons, see Tonomura et al., Phys.Rev.Lett. 56, 792 (1986). An electron passes through two slits and interferes on a far-field screen. On its way it is incident on a microscopic toroidal ring of ferro- magnetic material (with azymuthal magnetization) enclosed by a superconducting Niobium (Nb) shield to prevent any flux leakage. Notice the phase shift δφ = π in the interference pattern of the electrons that have passed in the center of the ring as compared with those that have passed outside (bright stripes appear instead of dark ones). This phase shift is due to the flux ∆Φ = hc/2e (= 2 10−7 Gauss cm2 = 2 10−15 Tesla m2 ) enclosed by the paths surrounding the ring (the value of this flux quantum is itself a result of the Aharonov-Bohm effect on the superconducting current with charge 2e flowing and interfering constructively in the Niobium shield ). 1 H Ψ (x) = ˆ ˆ ( p − (q/c)A − (q/c) ˆ f )2 eiq f / c Ψ(x) 2m 1 = ˆ ( p − (q/c)A − (q/c) ˆ f )eiq f / c ( p − (q/c)A)Ψ(x) ˆ ˆ 2m 1 = ( p − (q/c)A)2 Ψ(x) = ˆ ˆ ˆ HΨ(x) 2m In general the extra phase q f / c has no effect on the probability of detecting a parti-cle at a given position, since it depends on the absolute value of Ψ(x). There is howevera situation first pointed out by Aharonov and Bohm in 1959 where this gauge transfor-mation leads to detectable effects. This happens when the probability of detecting anelectron in presence of a magnetic field results from the interference between differentpaths the electron can follow, as in the diffraction experiment shown in Fig.2.11 andFig.2.15. Following our earlier derivation, the wave-function of the particle on a screenfar away from the diffraction slits is:
  • 43. 44 QUANTUM MECHANICS |Ψ > = |I > +|II >= eikl1 +i(q/ c) 1 A·dr |O > +eikl2 +(q/ c) 2 A·dr |O > = (1 + e ik(l2 −l1 )+i(q/ c) A·dr )|O > = (1 + eik(l2 −l1 )+iδφ )|O >where the phase difference δφ is related to the magnetic flux Φ enclosed by the interfer-ing paths: δφ = (q/ c) A · dr = (q/ c) × A · dS = (q/ c) B · dS = qΦ/ cHence the enclosed magnetic flux can change the pattern of interference of the elec-tron wavefunction, even if the electron does not (and cannot) pass through the regioncontaining the magnetic field and no force is exerted on it! As shown in Fig.2.15, thiseffect has been clearly evidenced using superconducting materials to prevent leakageof the magnetic field. Notice that the interference pattern is periodic in Φ with a pe-riod ∆Φ = hc/q a property which is used to measure the magnetic field in extremelysensitive Superconducting Quantum Interference Devices (SQUID).2.7 Dirac’s equation: antiparticles and spinWe have seen the power of the QM formalism in predicting effects that are not onlycounter-intuitive but at odds with classical mechanics: molecules that can exist as su-perpositions of enantiomers, particles that behave like waves and diffract and electronsthat sense the magnetic field in regions they cannot accede. The most striking demon-stration of the power and validity of QM has been the prediction of spin (a propertyof electrons which has no classical equivalence) and the existence of anti-particles thatcame out of the successful attempt by P.A.M. Dirac to unify QM with relativity. Sincespin is crucial in understanding the properties of atoms and molecules, we shall see howit arises from Dirac’s equation. The Hamiltonian in Shroedinger’s equation describes a non-relativistic particle forwhich the energy E = p2 /2m mc2 . For a relativistic particle, Einstein showed (seeApendix) that E = p2 c2 + m2 c4 . However writing the relativistic analogue of Eq.2.40: ∂Ψ √ 2 2 i = − c 2 + m2 c4 Ψ ∂tled to insurmountable problems related to the interpretation of the square-root of themomentum operator. Similarly writing H 2 = p2 c2 + m2 c4 to derive the so-called Klein-Gordon equation: ∂2 Ψ = [ 2 − (mc/ )2 ]Ψ (2.47) ∂(ct)2yields an equation which is second order in time (Schroedinger’s equation is first order)and does not allow for the interpretation of |Ψ|2 as a probability density. Dirac searchedfor a relativistic formulation that like Eq.2.16 would be first order in time, yet wouldtreat time and space in a relativistic invariant way. He noticed that he could obtain
  • 44. DIRAC’S EQUATION: ANTIPARTICLES AND SPIN 45such an equation if he rewrote the relativistic invariant Klein-Gordon equation for awavefunction |Ψ(x, t) > that is a vector rather than a scalar: ∂2 [ 2 − ]|Ψ(x, t) >= ( Ak ∂ xk + i(β/c)∂t )( Ak ∂ xk + i(β/c)∂t )|Ψ(x, t) > ∂(ct)2Where Ak , β are hermitian matrices that satisfy the following relations: A2 = β2 = 1 and kAk A j + A j Ak = Ak β + βAk = 0 (for k j). To satisfy these 10 equations the matriceshave to be at least 4 × 4 matrices and the eigenvectors are 4-vectors which satisfy: ( Ak ∂ xk + i(β/c)∂t )|Ψ >= mc/ |Ψ >Further application of the operator ( Ak ∂ xk + i(β/c)∂t ) yields the Klein-Gordon equa-tion, Eq.2.47 which is compatible with relativity. Multiplying the preceding equation onthe left by the matrix cβ yields the Dirac equation: i ∂t |Ψ(x, t) >= (c αk pk + mc2 β)|Ψ(x, t) > (2.48) kwhich is formally equivalent to Eq.2.16 and where: αk = −iβAk and pk = −i ∂ xk . Asmentioned by Dirac (see ”Principles of Quantum Mechanics”, p.253ff) the matrices αk , βdescribe some new internal degree of freedom of the electron which commutes with thespace and time operators. They can be written as: 0 σk I 0 αk = β= (2.49) σk 0 0 −Iwhere I is the 2 × 2 unit matrix and σk are the 2 × 2 Pauli matrices: 01 0 −i 10 σx = σy = σz = (2.50) 10 i 0 0 −1 In Dirac’s equation as in Schroedinger’s equation the wavefunction has a naturalinterpretation as a probability distribution. Multiplying Eq.2.48 by < Ψ(x, t)| and itscomplex conjugate −i (< Ψ(x, t)|∂t ) = [< Ψ(x, t)|(c αk pk + mc2 β)] kby |Ψ(x, t) > and subtracting the two yields: i ∂t < Ψ|Ψ >= c pk < Ψ|αk |Ψ >= −i c ∂ xk < Ψ|αk |Ψ > k kwhich expresses a conservation law for the probability density P(x, t) =< Ψ(x, t)|Ψ(x, t) >with current density Jk = c < Ψ|αk |Ψ >: ∂P + ·J=0 ∂t
  • 45. 46 QUANTUM MECHANICS Let us now investigate Eq.2.48. Let us first consider a free electron with zero mo-mentum. The equation is then: i ∂t |Ψ >= mc2 β|Ψ >which solutions are |Ψ0 >1,2 = e−i(mc / )t (φ0 , 0, 0) and |Ψ0 >3,4 = ei(mc / )t (0, 0, χ0 ) where 2 2 ˜ ˜φ and χ are 2D vectors. The first solution describes an electron with positive energy˜ 0 ˜ 0Ee = mc2 and the second a hole with energy Eh = −mc2 on which we will have more tosay shortly. In the presence of an electro-magnetic field Dirac’s equation is generalized just asSchroedinger’s equation was generalized: by replacing the momentum pk by πk = pk −(q/c)Ak and adding the potential energy qΦ i ∂t |Ψ(x, t) >= [c αk (pk − (q/c)Ak ) + qΦ + mc2 β]|Ψ(x, t) > (2.51) kLet us look for a solution in terms of the two-component vector: |Ψ >= (φ, χ). Dirac’s ˜ ˜equation can then be recast as: φ ˜ χ ˜ φ ˜ φ ˜ i ∂t =c σk πk ˜ + qΦ + mc2 (2.52) χ ˜ k φ χ ˜ −χ˜To obtain the low energy limit of Dirac’s equation we look for a solution which is aperturbation of the previous solution for positive energy Ee = mc2 at zero momentum: φ ˜ φ(x, t) = e−i(mc / 2 |Ψ(x, t) >= )t (2.53) χ ˜ χ(x, t)Where φ and χ are slowly time-varying fields (i.e. slower than ω0 = mc2 / ), solutionsof the coupled equations: φ χ φ 0 i ∂t =c σk πk + qΦ − 2mc2 (2.54) χ k φ χ χIn the low energy limit (when mc2 is larger than any other energy) the field χ is slavedto φ (i.e. all terms in χ are negligible with respect to the 2mc2 χ term) and thus: σk πk φ χ= k 2mcwhich yields the following equation for φ: ( σk πk )2 i ∂t φ = [ k + qΦ]φ 2mUsing the following property of Pauli matrices:
  • 46. DIRAC’S EQUATION: ANTIPARTICLES AND SPIN 47 (σ · A)(σ · B) = A · B + iσ · A × Bone obtains: (σ · π)2 = π2 + iσ · π × π = π2 + iσ · (−i − (q/c)A) × (−i − (q/c)A) (2.55) = π − ( q/c)σ · 2 × A = π − ( q/c)σ · B 2From which one obtains the low energy version of Dirac’s equation: ˆ ( p − (q/c)A)2 ˆ i ∂t φ = [ − ( q/2mc)σ · B + qΦ] φ (2.56) 2mwhich ressembles Schroedinger’s equation except for the additional term on the righthand side which is the energy of a dipole m = ( q/2mc)σ in a magnetic field B. HenceDirac’s equation implies that the electron possesses an intrinsic magnetic moment, afact that was known from the spectrum of atoms in a magnetic field (see below). Exem-plifying yet again the ”unreasonable effectiveness of Mathematics” it is amazing thatthis intrinsic properties of the electron is a mathematical consequence of the unificationof QM and relativity! Moreover Dirac’s was able to relate this magnetic moment to the intrinsic angularmomentum of the particle. Consider the time evolution in abscence of magnetic field ofthe angular momentum L = r × p. As we have seen the time evolution of an observableobeys: i ∂t < L x > = < [L x , H] >=< [ypz − zpy , (c αk pk + mc2 β)] k = c < αy [y, py ]pz − αz [z, pz ]py >= i c < αy pz − αz py > 0Thus in Dirac’s equation < L x > is not a constant of motion, in contrast with theSchroedinger’s equation for a free particle where it is. Now consider the evolution of σ x ˆ(the 4 × 4 matrix which diagonal elements are the Pauli σ x matrices): i ∂t < σ x > = < [σ x , H] >=< [σ x , (c ˆ ˆ ˆ αk pk + mc2 β)] k = c < [σ x , αy ]py + [σ x , αz ]pz >= 2ic < αz py − αy pz > ˆ ˆWhere we used the following property of Pauli’s matrices: [σ x , σy ] = 2iσz , [σ x , σz ] =−2iσy . Therefore L x + σ x /2 commutes with the Hamiltonian, i.e. it is a constant of ˆthe motion. Dirac interpreted this result to mean that the particle has a spin angularmomentum S = σ/2 which must be added to the orbital angular momentum L to getthe total angular momentum, J = L + S which is a constant of the motion. According to Eq.2.56, the magnetic dipole moment of the electron is related to itsspin by: m = ( q/2mc)σ = (qg/2mc)S (2.57)Where g = 2 is known as the gyromagnetic ratio. This is in contrast with the mag-netic moment arising from the angular momentum m = (q/2mc)L (see Eq.??) for which
  • 47. 48 QUANTUM MECHANICSg = 1. The prediction of the gyromagnetic ratio of the electron was an other successof Dirac’s equation. In fact the value of g has become the most stringent test of quan-tum electro-dynamics (QED), the theory that generalized Dirac’s approach and unifiedElectromagnetism and Quantum Mechanics. It is the most precisely known constant ofnature: its value has been tested to 12 decimal places g = 2.0023193043617(!!), makingQM and QED the most precisley tested theories ever proposed. Let us now return to the hole solution with energy Eh = −mc2 . We have already seensuch solutions when studying the band theory of solids. In fact the original interpretationof Dirac was very similar. He assumed that all negative energy levels were occupied by a”sea” of electrons. Out of this sea, electrons could be excited to an energy level Ee = mc2leaving behind a hole (or positron) of energy Eh = mc2 . This process, electron-positroncreation (and its reverse, annihilation) is routinely observed in particle accelerators oncean energy E > 2mc2 ∼ 1MeV is achieved. The dynamic behaviour of the positron issimilar to that of an electron with opposite charge. To derive the positron wave-equationat low energies, we shall follow our previous study of the electron in a similar regime.We look for a low energy solution: φ ˜ φ = ei(mc / 2 |Ψ >= )t (2.58) χ ˜ χWhich leads to the following equation for the positron-wavefunction at low energies: ( σk πk )2 i ∂t χ = [− k + qΦ]χ 2mSince the Pauli matrices are hermitian, the equation for the complex conjugate of χ is: (σk (i ∂ xk − (q/c)Ak )2 k i ∂t χ∗ = [ − qΦ]χ∗ (2.59) 2m ( k σk (−i ∂ xk + (q/c)Ak )2 =[ − qΦ]χ∗ (2.60) 2mThis is the low energy limit of Dirac’s equation for a particle of mass m and charge −q.As seen above in the band-gap theory of conduction in solids, the hole solution describesthe behaviour of a particle of the same mass as the electron and of opposite charge. Thisprediction of Dirac was vindicated by the discovery of the positron three years later byCarl Anderson, for which they both shared the 1933 Nobel prize in Physics. To summarize, the unification in 1929 of relativity and QM by Dirac led to manyunexpected predictions of which we have seen three: 1) the existence of anti-particles (e.g. the positron) of mass identical to the particleand with opposite charge. These are created when an energy 2mc2 is available to createfor example an electron-positron pair (raise the electron from the top of the ”sea” atEh = −mc2 to the electron rest energy Ee = mc2 ).
  • 48. DIRAC’S EQUATION: ANTIPARTICLES AND SPIN 49 2) The association of an intrinsic angular momentum variable, the spin, S = σ/2for both the electron and its anti-particle. 3) The existence of a magnetic dipole moment linked to the spin of the particle insimilitude to the classical relation between magnetic moment and angular momentumbut larger by a factor g = Angular momentum and spinThe Hamiltonian for a free particle is: 2 1 ∂22 1 ∂ ∂ 1 ∂2 H=− 2 =− r+ 2 { sin θ + } (2.61) 2m 2m r ∂r 2 r sin θ ∂θ ∂θ r2 sin θ 2 ∂φ2 2 ∂2 2 ∂ 2 =− { 2 + }− L2 (2.62) 2m ∂r r ∂r 2mr2 The operator : L2 = − 2 L2 is known as the angular momentum operator, since its ˆcontribution to the Hamiltonian, i.e. L2 /2mr2 is the classical angular momentum kinetic ˆenergy: E L = L2 /2I (whereI = d3 rρ(r)r2 is the moment of inertia). As shown inthe Appendix on the Helmholtz equation, the eigenfunctions of L2 are the sphericalharmonics Ylm [θ, φ) with eigen-value −l(l + 1), hence: L2 |l, m > = ˆ 2 l(l + 1)|l, m > (2l + 1)(l − m)! m with :|l, m > = Ylm [θ, φ) = Pl (cos θ)eimφ (2.63) 4π(l + m)! The linear angular momentum operator L = r × p can be written as: ˆ ˆ ˆ ∂ ∂ ∂ ∂ L x = −i ˆ y −z = i sin φ + cot θ cos φ (2.64) ∂z ∂y ∂θ ∂φ ∂ ∂ ∂ ∂ Ly = −i ˆ z −x = i cos φ − cot θ sin φ (2.65) ∂x ∂z ∂θ ∂φ ∂ ∂ ∂ Lz = −i ˆ x −y = −i (2.66) ∂y ∂x ∂φ ˆ ˆThe eigenfunctions of L2 are also eigenfunctions of Lz with eigenvalue m: Lz |l, m >= m|l, m > ˆ ˆ ˆ ˆNotice that Lz does not commute with L x or Ly . In fact these operators satisfy the fol-lowing commutation relations: [L x , Ly ] = i Lz ˆ ˆ ˆ ˆ x , Lz ] = −i Ly [L ˆ ˆ [Ly , Lz ] = i L x ˆ ˆ ˆ or : [Li , L j ] = i ˆ ˆ ˆ i jk Lk (2.67)
  • 49. 50 QUANTUM MECHANICSwhere i, j, k stands for x, y or z and i jk = 1 for an even permutation of the indices and−1 for an odd permutation. As we shall see below, it is usefull to define the raising andlowering operators: ∂ ∂ L± = L x ± Ly = e±iφ ± + i cot θ ˆ ˆ ˆ ∂θ ∂φwhich satisfy the following commutation rules: [L+ , Lz ] = − L+ ˆ ˆ ˆ [L− , Lz ] = L− ˆ ˆ ˆ [L+ , L− ] = 2 Lz ˆ ˆ ˆNotice that L2 = L2 + Ly + Lz = Lz + L+ L− − Lz = Lz + L− L+ + Lz commutes with Lz ˆ ˆ x ˆ2 ˆ2 ˆ2 ˆ ˆ ˆ2 ˆ ˆ ˆ ˆ x , Ly , L± .but also with L ˆ ˆ [L2 , L x ] = [Ly , L x ] + [Lz , L x ] = Ly [Ly , L x ] + [Ly , L x ]Ly + Lz [Lz , L x ] + [Lz , L x ]Lz ˆ ˆ ˆ2 ˆ ˆ2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = −i Ly Lz − i Lz Ly + i Lz Ly + i Ly Lz = 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆThese operators therefore do not mix eigenstates with different values of l but since they ˆdo not commute with Lz they do mix eigenstates with different values of m. Using theircommutation rules one can show that L± |l, m >∼ |l, m ± 1 >. Consider for example the ˆ ˆcommutation relations for L+ : − L+ |l, m > = [L+ , Lz ]|l, m >= (L+ Lz − Lz L+ )|l, m > ˆ ˆ ˆ ˆ ˆ ˆ ˆ = mL ˆ + |l, m > −Lz L+ |l, m > ˆ ˆ namely : Lz (L+ |l, m >) = (m + 1)(L+ |l, m >) ˆ ˆ ˆHence: L± |l, m >= a± |l, m ± 1 >= ˆ lm (l + 1) − m(m ± 1)|l, m ± 1 > (2.68)Where the coefficient a+ is determined by the normalization condition: lm |a+ |2 = |a+ |2 < l, m + 1|l, m + 1 >=< l, m|L+ L+ |l, m >=< l, m|L− L+ |l, m > lm lm ˆ∗ ˆ ˆ ˆ = < l, m|L2 − Lz − Lz |l, m >= 2 (l(l + 1) − m(m + 1)) ˆ ˆ2 Notice that the action of L+ on a state |l, m > is to generate the higher m-state|l, m + 1 > whereas the action of L− on |l, m > generates the lower m-state l, m − 1 >,which justifies their name as raising and lowering operators. Since the spin is also an angular momentum operator the same commutation rela- ˆ ˆ ˆ ˆ ˆtions exist between S 2 , S x , S y , S z and S ± . For spin one-half particles, the eigenstates
  • 50. DIRAC’S EQUATION: ANTIPARTICLES AND SPIN 51|s, sz > of S 2 = 2 σ2 /4 and S z = σz /2 with eigenvalues s(s + 1) ˆ ˆ 2 = 3 2 /4 and sz = (±1/2) are: 1 0 |1/2, 1/2 >= |1/2, −1/2 >= 0 1We have seen that one of the results of Dirac’s equation is that spin and angular mo-mentum operators Lz , S z do not commute with the Hamiltonian and are therefore not ˆ ˆgood (i.e. conserved) quantum numbers. Rather the total angular momentum operator:Jz = Lz + S z with eigenvalue k = m + sz is conserved, together with J 2 , L2 and S 2 . Thus ˆ ˆ ˆ ˆ ˆ ˆthe eigenstate of the total angular momentum is: | j, l, s, k >. Since m = k 1/2 and theeigenstates of the angular momentum are Ylm we can write: 1 0| j, l, s, k >= c1 |l, k−1/2 > |s, 1/2 > +c2 |l, k+1/2 > |s, −1/2 >= c1 Yl,k−1/2 + c2 Yl,k+1/2 0 1For this state to be an eigenstate of J 2 = L2 + S 2 + 2L · S it has to satisfy: ˆ ˆ ˆ ˆ ˆ 3 Lσˆ J 2 | j, l, s, k >= ˆ 2 [l(l + 1) + + ]| j, l, s, k >= 2 γ| j, l, s, k > 4Using the relation: L L Lσ = L x σ x + Ly σy + Lz σz == z − ˆ ˆ ˆ ˆ L+ −Lzand the previously derived equation for the raisng and lowering operators, Eq.2.68, oneobtains the following eigenvalue equation for γ: √ l(l + 1) + 1/4 + k − γ l(l + 1) − (k − 1/2)(k + 1/2) √ =0 l(l + 1) − (k − 1/2)(k + 1/2) l(l + 1) + 1/4 − k − γwhich yields the quadratic equation: [(l + 1/2)2 − γ]2 − (l + 1/2)2 = 0which solution is γ = j( j+1) with j = l±1/2. Hence the eigenvalues of the total angularmomentum operator J 2 are 2 j( j + 1) with eigenfuctions: ˆ √ √ l + k Yl,k−1/2 1 | j, l, s, k > = |l + 1/2, l, 1/2, k >= √ 2l + 1 l − k + 1 Yl,k+1/2 √ 1 l − k + 1 Yl,k−1/2 | j, l, s, k > = |l − 1/2, l, 1/2, k >= √ √ 2l + 1 − l + k Yl,k+1/2
  • 51. 52 QUANTUM MECHANICS2.8 The Hydrogen atom and electronic orbitalsWe have seen above how the formalism of QM could be applied to simple cases (twostate systems, free particle, particle in a box, etc.). However its first and major successwas in solving for the energy level of the hydrogen atom and in explaining the propertiesof the elements as classified in Mendeleev’s periodic table. Schroedinger’s equation fora hydrogen-like atom is: 2 Ze2 − 2 Ψ(x, t) − Ψ(x, t) = En Ψ(x, t) (2.69) 2me rwhere the −Ze2 /r term stands for the Coulomb interaction between the orbiting electron(of mass me ) and a nucleus containing Z protons (in Hydrogen Z = 1). In sphericalcoordinate the differential operator can be written as: ∂2 2 ∂ 1 1 ∂ ∂ 1 ∂2 2 = + + 2[ (sin θ ) + ] ∂r2 r ∂r r sin θ ∂θ ∂θ sin2 θ ∂φ2As for the Helmholtz problem (see Appendix) we look for a eigen-function: Ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ)where the spherical harmonics Ylm (θ, φ) are eigen-functions of the angular operator (seeabove): 1 ∂ ∂ 1 ∂2 L2 Ylm (θ, φ) = [ (sin θ ) + ]Ylm (θ, φ) = −l(l + 1)Ylm (θ, φ) sin θ ∂θ ∂θ sin2 θ ∂φ2The equation for the radial function Rnl (r) then becomes: ∂2 2 2 ∂ 2 l(l + 1) Ze2 [− (+ )+ − ]Rnl = En Rnl 2me ∂r 2 r ∂r 2me r2 rThe first term on the left is the radial part of the kinetic energy, the second term is thecentrifugal potential (the angular part of the kinetic energy) and the third term is theCoulomb repulsion. Writing Rnl (r) = unl (r)/r yields the following equation for u(r): d2 unl 2me Ze2 l(l + 1) 2me En 2 +( 2 − 2 )unl = − 2 unl dr r rDefining a = 2Zme e2 / 2 ≡ 2/r0 and λ2 = −2me En / n 2 one can recast the previousequation as: a l(l + 1) unl + ( − )unl = λ2 unl n r r2In the limit r → ∞, the terms in the bracket on the left hand side are negligible andthe solution in that limit is unl (r) ∼ exp(−λn r). Let us therefore look for a solution:unl (r) = µnl (r)exp(−λn r). The equation that µnl (r) obeys is:
  • 52. THE HYDROGEN ATOM AND ELECTRONIC ORBITALS 53 a l(l + 1) µnl − 2λn µnl + ( − )µnl = 0 r r2Notice that close to r = 0 the solution behaves as µnl (r) ∼ rl+1 . Let us therefore lookfor a polynomial solution: µnl (r) = n c j r j . The coefficient of the term of order j − 1 j=1reads: j( j + 1)c j+1 − 2λn jc j + ac j − l(l + 1)c j+1 = 0which yields: 2λn j − a c j+1 = cj j( j + 1) − l(l + 1)To be defined for all values of j the series {c j } must start at j = l + 1 (the small r limitof µnl (r)) and to be finite it must end at a value l < j = n for which: a = 2nλn , i.e. for avalue of the eigen-energy, En : λn 2 2 2 1 En = − =− 2 2me 2me r0 n2which is the result obtained by Bohr, see Eq.2.8. Notice that while the radial eigen-functions depend on the two quantum numbers n, l, the eigen-energies depend only onn. This is a peculiarity of central potentials (such as the Coulomb potential betweenproton and electron). In non-hydrogen atoms that degeneracy is raised due to the extraCoulomb repulsion between the electrons. It is customary to label the various orbitals(the hydrogen atom’s eigensolutions) by their shell number n = 1, 2, 3, .... and theirmain angular momentum number l, such that l = 0 is called the s-orbital; l = 1 thep−orbital; l = 2 the d−orbital; l = 3 the f −orbital; l = 4 the g−orbital ... etc. (inalphabetical order skipping j). With this nomenclature we can proceed to write downthe radial eigen-functions Rnl (r) = µnl (r) exp(−r/nr0 )/r for the first orbitals: 1s orbital : R10 (r) = 2r0 e−r/r0 −3/2 r −r/2r0 2s orbital : R20 (r) = (2r0 )−3/2 (2 − )e r0 1 r 2p orbital : R21 (r) = √ (2r0 )−3/2 e−r/2r0 3 r0 In abscence of other terms in the Hamiltonian, the orbitals of the electron in theHydrogen atom are degenerate, i.e. for different values of the spin and angular quantumnumbers (l, m) the energy of the electron is the same. Since there are 2 possible spinstates for every eigenfunction Ψnlm and 2l + 1 states with different values of m (andsame value of l: −l ≤ m ≤ l) and since 0 ≤ l ≤ n − 1, the degeneracy of a statewith shell number n is: 2n2 . As we shall see next this degeneracy is partially raised byvarious relativistic effects (for example by the interaction between the electron spin andthe magnetic field the electron generates when orbiting the nucleus, so-called spin-orbitcoupling), by the action of external fields or by the presence of other electrons.
  • 53. 54 QUANTUM MECHANICS2.8.1 Spin-orbit couplingWe have seen that Schroedinger’s equation, Eq.2.56 is only an approximation to themore correct relativistically invariant Dirac’s equation, Eq.2.51. Although Dirac’s equa-tion for the Hydrogen atom can be solved exactly (see Appendix), it is instructive toinvestigate the various terms that were neglected when deriving Eq.2.56. One of these terms is the so called spin-orbit coupling, which is due to the interactionbetween the electron spin and the magnetic field experienced by the orbiting electron.In the rest frame of the nucleus there exists only an electric field E = − Φ = −ˆ∂r Φ = rer/r3 . However in the frame of the electron orbiting at velocity v = p/me , the nucleus ismoving at velocity −v and is generating both electric and magnetic fields. If the electronframe of reference was inertial (i.e. if the electron was moving in a straight line) thenthe magnetic field experienced by the electron would be p×r e e B = −v × E/c = − 3 = L me c r me cr3This magnetic field which is parallel to the angular momentum L = r × p, interacts withthe electron spin S = σ/2 to contribute an energy, see Eq.2.56: e e δH so = σ·B= S ·B 2me c me cSince the electron frame is not inertial the final result differs by a factor 2 from thisestimate: e2 δH so = S · L ≡ g so S · L 2m2 c2 r3 e As was argued by Dirac (see above), Lz and S z are not eigenvalues of the relativisticHamiltonian, rather the projection of the total angular momentum Jz = Lz + S z is.Rewriting 2S · L = J 2 − L2 − S 2 one can write the Spin-orbit contribution as: e2 δH so = (J 2 − L2 − S 2 ) 4m2 c2 r3 eIts contribution is best estimated in the basis which commutes with Dirac’s Hamiltonian,i.e. in terms of eigenstates | j, l, s, k > of J 2 , L2 , S 2 and Jz : δE so =< j, l, s, k|δH so | j, l, s, k >To compute how that spin-orbit coupling modifies the energy spectrum of the Hydrogenatom computed above, first notice that it couples between eigenstates with the samevalue of l but different values of j, i.e. j = l ± 1/2. Hence the term: J 2 − L2 − S 2with eigenvalue: j( j + 1) − l(l + 1) − 3/4 can have two values: 2 l (if j = l + 1/2) and− 2 (l + 1) (if j = l − 1/2). The term in 1/r3 in δH so can be computed directly from theradial solutions Rnl of the Hydrogen atom with the result: e2 e2 gnl ≡< n, l| |n, l >= 4m2 c2 r3 e 2m2 c2 a3 n3 l(2l + 1)(l + 1) e 0
  • 54. THE HYDROGEN ATOM AND ELECTRONIC ORBITALS 55Thus because of the relativistic coupling between the electron’s angular momentum andits spin the energy levels of the Hydrogen atom with same values of n and l are split intwo: e2 2 l δE ± = so 2m2 c2 r n l(2l + 1)(l + 1) −l − 1 e 3 3 0Notice that that the spin-orbit interaction is a very small (yet measurable) perturbationto the electronic energies of the hydrogen atom: |En | e2 δE so ∼ ∼ 7.4 10−4 eV |En | me c2 r0Where we used the fact that e2 /2r0 = |E1 | = 13.6 eV and me c2 = 0.5MeV.2.8.2 Many electron systemsUntill now we have mainly dealt with the problem of finding the energy level of a QMsystem consisting of a single electron (hoping between atoms, confined in a box or orbit-ing the nucleus). When considering a QM system consisting of two (or more) electronsthere arises besides the question of their interaction (which as we shall see below canbe treated perturbatively) and the issue of their identity. In classical mechanics whiletwo particles (e.g. billiard balls) may look similar they are never identical: they can beidentified from their different trajectories. In QM not only do all electrons ”look” sim-ilar (for example all hydrogen atoms have the same spectrum) but they cannot be toldapart: they are identical. As we shall see, this identity of QM particles has profoundconsequences on the properties of matter. ˆ Consider the Hamiltonian of a two particle system H(1, 2). Because of the iden-tity of the particles the Hamiltonian is unchanged upon a permutation of the particles:H(1, 2) = H(2, 1). Hence the permutation operator P commutes with the Hamiltonian: ˆ ˆ ˆ PH(1, 2)|1, 2 >= H(2, 1)|2, 1 >= H(1, 2)P|1, 2 > ˆ ˆ ˆ ˆ ˆand is a constant of motion. Two successive permutations brings us back to the initialstate: P2 = 1 which implies that the eigenvalues of the permutation operator are ±1: ˆ|2, 1 >= P|1, 2 >= ±|1, 2 >. These eigenvalues are constants of motion, namely thewavefunction of a many particles system is either even or odd under permutation of its(identical) particles. Like spin this is a QM property of the particle. Particles knownas fermions (electrons, protons, neutrons, etc.) are odd under permutations: |2, 1 >=−|1, 2 >, while particles which are even under permutation |2, 1 >= |1, 2 > are knownas bosons (e.g. photons). In 1940 Wolfgang Pauli, requiring QM to be invariant under achange of reference frames (relativistic invariance), related this permutation property tothe spin of the particle: particles with half integer spin are fermions while particles withinteger spin are bosons.
  • 55. 56 QUANTUM MECHANICS If the particles can occupy states |ψ > and |φ >, then for two bosons the joint wave-function |1, 2 > (symmetric under exchange of the particles) is: |ψ(1) > |φ(2) > +|ψ(2) > |φ(1) > |1, 2 >= √ = |2, 1 > 2 For two fermions the joint wavefunction is: |ψ(1) > |φ(2) > −|ψ(2) > |φ(1) > |1, 2 >= √ 2which is an eigenfunction of the permutation operator P with eigenvalue -1: |2, 1 >=P|1, 2 >= −|1, 2 >. This wavefunction implies that in contrast to bosons, two fermionscannot share the same eigenfunctions: |ψ > |φ >. If their spatial wavefunction is iden-tical then their spin-state must differ and if their spin-state is identical then their spatialwavefunction must differ. This extremely important property of fermions is known asPauli’s exclusion principle. It implies that the lowest energy level of an electronic sys-tem can be occupied by at most two electrons (with opposite spins). Contrast that withthe situation for bosons where the lowest energy state can be occupied by an arbitrarynumber of particles (for example photons of same energy as in Plank’s black body radi-ation).2.8.3 The periodic tablePauli’s exclusion principle and the solution of Schroedinger’s equation for the hydro-gen atom allows for a classifications of atoms according to the degree to which theirorbitals are successively filled by the orbiting electrons: the so-called Aufbau principle.As noticed above due to electron-electron interactions the near angular degeneracy (ne-glecting spin-orbit coupling) existing in the hydrogen atom is lifted in multi-electronsystems. For the same value of the shell number n, orbitals with smaller values of l(more symmetric angular wavefunctions) have lower energy, see Fig.2.16. These ener-gies cannot be calculated analytically, but numerical schemes have been developped toestimate them with great accuracy, see perturbation method below. Hydrogen (H) has one electron in the lowest orbital: 1s. Its spin S = 1/2 implythat there are 2s + 1 = 2 possible states in a s−orbital. In terms of Dirac’s eigenstates| j, l, s, k > with total angular momentum j = l + 1/2 = 1/2 that state can also be writtenas 2S +1 L J =2 S 1/2 , a notation known as spectroscopic notation. Helium (He) has two electrons occupying the lowest orbital (written as (1s)2 ) withanti-parallel spin, i.e with total spin S = 0. The spectroscopic notation of that state is2S +1 L J =1 S 0 . Lithium (Li) has three electrons, two occupying with opposite spin the lowest or-bital (1s)2 and one electron occupying the next orbital: 2s, hence the electron occupancyis : (1s)2 (2s). Since the chemical properties of atoms are determined by their more outerelectrons (so called valence electrons which can engage in bonding interactions), the
  • 56. THE HYDROGEN ATOM AND ELECTRONIC ORBITALS 57F. 2.16. The orbital energies of many-electron atoms. The energy levels are ordered along the abscissa according to the main shell number n, which is also indicative of the size of the atom (the spread of the radial wavefunction increases roughly as nr0 ). Notice that levels with different angular quantum number l (different colors) have different energies, though they are still degenerate with respect to the azymutal quantum number m: the number of circles indicate the level of degeneracy 2l + 1. Notice that in abscence of an external field there is no preferred axis that would break that azymutal symmetry. To minimize the energy electrons fill these levels from the bottom up while satisfying Pauli’s principle: no more than two electrons (with opposite spin) per energy level. The first level to be filled is 1s, then 2s, 2p, etc... Notice that the more outer orbitals 4s having lower energy is filled before 3d; 5s before 4d and 6s before 4 f and 5d.property of Lithium with one valance electron in the same s−state as the electron inHydrogen imply similar chemical properties, as indeed all the alkalines (the elements inthe first column of Mendeleev’s table, see Fig.2.17: Sodium (Na), Potassium (K), Ru-bidium (Rb), Cesium (Cs), Francium (Fr)) which all have a single electron in an s−state.In spectroscopic notation the state of the electron in the unfilled shell is: 2S +1 L J =2 S 1/2 ,which is the same as Hydrogen. Since (as we shall see below) it is the electronic state inthe outer shell that determines the chemical characteristics of the element the chemicalproperties of Lithium and hydrogen are similar. Beryllium (Be) has four electrons, two occupying with opposite spin the lowestorbital (1s)2 and two electrons occupying the next orbitals (2s) with opposite spin. Itselectron occupancy is thus: (1s)2 (2s)2 which is similar to Helium. Beryllium is howevermore reactive than Helium since its 2p orbital has an energy only slightly higher than
  • 57. 58 QUANTUM MECHANICSF. 2.17. The periodic-table of the elements. The periodic table devised by Mendeleev and ordered according to increasing atomic number (number of electrons) and sim- ilar chemical properties. In comparison the table is redrawn as deduced from the filling up of the orbitals from the bottom up, taking into account Pauli’s exclusion principle and ordering the columns according to the number of valence (outer shell) electrons. The agreement between the two is one of the greatest success of QM.2s which result in the electron easily hoping to the 2p orbital. Its outer shell electron oc-cupancy is identical to that of Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium(Ba) and Radium (Ra) which have similar chemical properties. Boron (B) has 5 electrons, two in orbital 1s, two in orbital 2s and one in orbital 2p:(1s)2 (2s)2 (2p). Its outer shell occupancy and chemical properties are similar to that ofAluminium (Al), Gallium (Ga), Indium (In) and Thalium (Tl). These elements are oftenused as p-type dopants (electron acceptors) in semi-conductors. The value of the totalangular momentum is j = l − s = 1/2 (due to spin-orbit coupling (see above) the state
  • 58. THE CHEMICAL BOND 59 j = l + s = 3/2 has higher energy). Its outer electronic state is thus 2 P1/2 . Carbon (C) has 6 electrons, two in orbital 1s, two in orbital 2s and two in orbital 2p.Since orbital 2p is triply degenerate, the question arises as to where the two electronsin 2p settle. Because of Pauli’s principle, electrons with identical wavefunction (i.e.orbitals) have opposite spin, while electron in different orbitals may have similar spin.Since electrons in different orbitals overlap and repel each other less, that situation isenergetically favorable giving rise to Hund’s rule: ”the state of highest spin has thelowest energy”. Hence in Carbon the electrons in the 2p orbital have similar (parallel)spin and occupy orbitals with different azymutal number m. The outer shell propertiesof Carbon and its chemical properties are similar to Silicon (Si) and Germanium (Ge),i.e. semiconductors. Tin (Sn) and Lead (Pb) have smaller band gaps (see above) and arein fact conductors at room temperatures. Nitrogen (N) has 7 electrons, two in orbital 1s, two in orbital 2s and three in orbital2p. These outer electrons have parallel spin because of Hund’s rule and occupy thethree 2p orbitals with different azymutal number m. The outer shell occupancy andchemical properties of Nitrogen are similar to that of Phosphorus (P), Arsenic (As),Antimony (Sb) and Bismuth (Bi). These elements are often used as n-type dopants(electron donors) in Si and Ge semi-conductors. Oxygen (O) has 8 electrons, two in 1s, two in 2s and four in 2p. Two of those arepaired (they share the same orbital with opposite spin) and two have parallel spins andshare orbitals with different m. Its outer shell electron occupancy is identical to that ofSulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po). Fluorine (F) has nine electrons, the last five of which occupy the three 2p orbitals,four are paired (they share the same orbital with opposite spin) and one occupy thethird orbital. The outer shell occupancy of Fluorine is similar to that of other Halogens:Chlorine (Cl), Bromine (Br), Iodine (I) and Astatine (At). Neon (Ne) has ten electrons which occupy all the 1s, 2s and 2p orbitals. All electronsare paired and therefore not disponible for bonding. Neon like Helium and the otherNoble gases: Argon( Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) have a similar outerelectron shell and share a very low chemical reactivity. An interesting situation occurs when orbitals 4s and 5s fill up (for elements Ca andSr). The next ten elements have their electrons fill the more inner 3d or 4d shells. Theseelements are known as transition metals. Similarly when orbitals 6s fill up (for Ba), thenext shell to fill up is the inner 4f shells which defines a set of 14 rare-earth elements(Lanthanides) with similar chemical properties. Once this shell is filled the next one, 5dis also an inner shell and the next ten elements share similar properties with transitionmetals.2.9 The chemical bondThe role that the Hydrogen atom has played in the understanding of the periodic tableof the elements has been played by the Hydrogen molecule H+ in understanding the 2chemical bond. Setting the protons at positions ±a = ±aˆ, the problem is to solve the zSchroedinger equation for a single electron:
  • 59. 60 QUANTUM MECHANICS 2 e2 e2 e2 HΨ(x, t) = − 2 Ψ(x, t) − ( + − )Ψ(x, t) = En Ψ(x, t) (2.70) 2me |x + a| |x − a| 2a By rewriting the differential operators in prolate spheroidal coordinates: x = a sinh µ sin ν cos φ y = a sinh µ sin ν sin φ z = a cosh µ cos νthis equation can be solved by separation of variables (i.e. looking for a solution Ψ(µ, ν, φ) =M(µ)N(ν)Φ(φ)) as was done for the hydrogen atom (in spherical coordinates). Noticingthat : |x ± a| = a(cosh µ ± cos ν, we leave it as an exercise to the reader to show thatEq.2.70 then yields the following equations: ∂2 Φ + m2 Φ = 0 ∂φ2 ∂2 M ∂M m2 + coth µ + (2α cosh µ − + En sinh2 µ − λ)M = 0 ∂µ2 ∂µ sinh2 µ ∂2 N ∂N m2 + cot ν + (− 2 + En sin2 ν + λ)N = 0 (2.71) ∂ν 2 ∂ν sin νWhere α ≡ 2a/r0 (with r0 = 2 /me e2 , the Bohr radius of Hydrogen), En ≡ (2me a2 / 2 )(En −e2 /2a) and m = 0, ±1, ±2, . . . and λ are quantum numbers (playing the role that m, lplayed in the hydrogen atom). The energy levels En have been solved in 1927 (only oneyear after Schroedinger solved the Hydrogen atom) by O.Burrau who then looked forthe distance 2a between the atoms that would minimize the ground state energy E0 . Since most problems in physics are not exactly soluble and in particular the com-putation of energy spectra of complex molecules, the existence of an exact solution forH+ was very helpfull in the development and testing of approximate methods to com- 2pute eigen-energies and eigen-functions (orbitals) for more complex QM systems. Oneof these approximation is known as the variational method (for details see L.P.Paulingand E.B.Wilson, ”Introduction to quantum mechanics”, Chap.7). The idea is pretty sim-ple. One assumes that in some limit a given problem can be solved exactly, i.e. a setof eigen-functions can be computed for the Hamiltonian in that limit. One then uses alinear superposition of these solutions as a trial function and one estimates the energy ofthat state for the complete Hamiltonian. This energy is then minimized with respect tothe parameters of the superposition (for example the linear coefficients). Since one doesnot expect the trial function to yield the exact solution of the full problem, the estimatedminimal energy Emin is an upper bound to the true energy of the system: Emin > E0 . Ingeneral the more complete the set of functions used as a trial, the better the approxima-tion. However the main problem with the variational approach is that there is no way of
  • 60. THE CHEMICAL BOND 61F. 2.18. The ground state energy level of the Hydrogen molecule ion H+ as a function 2 of the distance R between the two protons. The exact solution E0 (R) is shown in dashed and compared to the results obtained from the variational method for a sym- metric and an anti-symmetric electronic wavefunction. Only the symmetric solution describes a bound state (energy E s smaller than the hydrogen ground state energy E1 ).estimating how good an approximation it is. For that one needs to resort to perturbationexpansions, which we will study later. In the case of the hydrogen molecule ion H+ its Hamiltonian can be written as: 2 H = HH (r1 ) − e2 /r2 + e2 /2a = HH (r2 ) − e2 /r1 + e2 /2awhere HH is the Hamiltonian of the Hydrogen atom. When the two protons are far apart(a r0 ) the solutions for the electron wavefunction are simply those computed for ahydrogen atom centered on either proton, i.e. |Ψnlm (r1 ) > or |Ψnlm (r2 ) > with r1 = r − a;r2 = r + a. When the atoms are brought in proximity, we shall use for simplicity a trialwavefunction |ΨT > which is a superposition of hydrogen 1s ground states (with energyE1 = −e2 /2r0 = −13.6eV) : |Ψ(r) >= R10 (r)Y00 (θ, φ) = exp(−r/r0 )/ πr0 : 3 |ΨT >= c1 |Ψ(r1 ) > +c2 |Ψ(r2 ) >Where the coefficients c1 , c2 are parameters to be determined such as to minimize theenergy E of that state. Inserting the trial wavefunction into the full Schroedinger equa-tion yields:
  • 61. 62 QUANTUM MECHANICS < ΨT |H|ΨT >= E < ΨT |ΨT >which defining H11 =< Ψ(r1 )|H|Ψ(r1 ) >; H12 =< Ψ(r1 )|H|Ψ(r2 ) >= H21 ; H22 =<Ψ(r2 )|H|Ψ(r2 ) > and the overlap function: ∆ ≡< Ψ(r2 )|Ψ(r1 ) > is equivalent to: c2 (H11 − E) + 2c1 c2 (H12 − ∆E) + c2 (H22 − E) = 0 1 2 (2.72)We now have to determine the coefficients c1 , c2 that minimize E, i.e. for which ∂E/∂c1 =∂E/∂c2 = 0. Taking the derivative of the previous equation with respect to c1 and c2yields at the minimum of the energy a set of coupled homogenous equations: (H11 − E)c1 + (H12 − ∆E)c2 = 0 (H12 − ∆E)c1 + (H22 − E)c2 = 0Which has a solution when the determinant: H11 − E H12 − ∆E =0 H12 − ∆E H22 − ESince H11 = H22 and H12 = H21 we obtain: H11 + H12 Es = 1+∆ H11 − H12 Ea = 1−∆The first solution corresponds to a symmetric trial wavefunction: c1 = c2 : 1 |Ψ s >= √ (|Ψ(r1 ) > +|Ψ(r2 ) >) 2 + 2∆The second solution corresponds to an anti-symmetric wavefunction: c2 = −c1 : 1 |Ψa >= √ (|Ψ(r1 ) > −|Ψ(r2 ) >) 2 − 2∆The various integrals can then be evaluated numerically or in this case even exactly toyield: r0 H11 = E1 (1 − )+J a r0 H12 = ∆E1 (1 − ) + K a (2.73)where J known as the Coulomb integral, K the resonance integral and ∆ the overlapintegral are:
  • 62. THE CHEMICAL BOND 63 r0 r0 J= d3 rΨ(r1 )(−e2 /r2 )Ψ(r1 ) = −2E1 − + e−4a/r0 (1 + 2a 2a 2a K= d3 rΨ(r1 )(−e2 /r2 )Ψ(r2 ) = 2E1 e−2a/r0 (1 + ) r0 2a 4a2 −2a/r0 ∆= d3 rΨ(r1 )Ψ(r2 ) = (1 + + 2 )e r0 3r0The energies of the symmetric and antisymmetric wavefunctions can then be recast as: r0 J+K E s = E1 (1 − )+ (2.74) a 1+∆ r0 J−K Ea = E1 (1 − ) + (2.75) a 1−∆Only the symmetric solution has a minimal energy E s = −15.36eV lower than E1 =−13.6eV at an internuclear distance 2a ∼ 1.3Å. The anti-symmetric solution has anenergy Ea always larger than E1 and thus cannot represent a bound state of the H+ 2ion. Comparing these values to the exact solution (E0 = −16.4eV at 2a = 1.06Å), seeFig.2.18, suggests that in this case the simple variational approach yields a reasonableapproximation to the ground state of the system. Just as we constructed the orbitals in atoms, by filling up the (perturbed) Hydro-gen orbitals from the bottom up taking into consideration Pauli’s exclusion principle wemay arrange the two electrons in the H2 molecule by filling up the molecular orbitalsjust determined taking into account Pauli’s principle. The problem in that case is thatthe electron-electron repulsion (e2 /|r1 − r2 |) may not be negligible (as in the atomicconstruction where it is somewhat neutralized by the higher positive charge of the nu-cleus). However because the electron charges are spread over a region of size ∼ r0 , theirrepulsive interaction turns out to alter only slightly the ground state energy which isstill stable for the symmetric wavefunction and unstable for the antisymmetric one. Thetwo electrons in H2 thus share the same ground state orbital with opposite spin (theyare paired). This is a general result: a chemical bond between two elements is formedby electron pairs whose spatial wavefunction is symmetric with respect to particle ex-change and of opposite spin (to satisfy Pauli’s exclusion principle). ¨2.9.1 Huckel’s molecular orbital theoryThe variational approach used to estimate the ground state of H+ has been used by 2Eric H¨ ckel in 1930 to estimate the eigen-energies and orbitals of hydro-carbon chains uand rings of n carbons. As we shall now show this H¨ ckel Molecular Orbital (HMO) utheory justifies a-posteriori our initial approach to multi-state systems (the amonia andaromatic molecules). H¨ ckel’s idea was to use a trial molecular wavefunction |Ψ MO > which is a super- uposition of atomic wavefunctions |φ > (p-orbitals) centered on the individual Carbonatoms:
  • 63. 64 QUANTUM MECHANICSF. 2.19. The band gap ∆E of a few alkenes with n = 7, 9, 11 measured by the adsorp- tion of photons of energy hν = ∆E. Notice that the measured energies are propori- onal to sin π/(n + 1) as predicted from H¨ ckel’ molecular orbital theory. u |Ψ MO >= ci |φi > iJust as we have done for H+ the idea is to determine the coefficients {ci } that minimize 2the energy of the full Hamiltonian: < Ψ MO |H|Ψ MO >= E < Ψ MO |Ψ MO >Neglecting the overlap between neighboring atomic orbitals, i.e. assuming: ∆ =< φi |φ j >0 one gets: ci c j Hi j − Ec2 δi j = 0 i ijWhere Hi j =< φi |H|φ j >. Hii and Hi j are respectively the Coulomb and resonance inte-grals we saw above. If only the overlap between nearest neighbours j = i ± 1 contributesignificantly to the resonance integral, minimization of the energy with respect to thevector of coefficients {ci } yields:  E0 −A 0 . . . 0   c   c1     −A E −A . . .   0  1   .  .  = E .        0     .   .    . .   .  .  (2.76) .      . .               cn   cn    0 . . . 0 −A   E0
  • 64. THE CHEMICAL BOND 65Where we defined: E0 ≡ Hii and A ≡ Hi,i+1 = Hi−1,i . Notice the similarity with themodel studied earlier for aromatic molecules, Eq.2.25. The eigen-energies can then becomputed as follows: 2 cos θ 1 0 . . . 0 1 2 cos θ 1 . . . 0 ∆n = . =0 . . 0 ... 0 1 2 cos θwith 2 cos θ ≡ (E − E0 )/A. Computing the determinant yields a recursion relation: ∆n =2 cos θ∆n−1 − ∆n−2 with ∆0 = 1 and ∆1 = 2 cos θ. Its solution is ∆n = sin(n + 1)θ/ sin θand the eigenvalues are consequently determined by: ∆n = 0, i.e. θ = mπ/(n + 1)(m = 1, 2, ..., n): mπ lπ E = E0 + 2A cos = E0 − 2A cos (2.77) n+1 n+1with l = n + 1 − m, compare with Eq.2.26. The energy gap betwen the highest occupiedmolecular orbital(HOMO, the top of the valence band) with lhomo = n/2 (or (n + 1)/2 forodd n) and the lowest unoccupied molecular orbital (LUMO, bottom of the conductionband) with llumo = lhomo + 1 is: ∆E = −4A sin π/(n + 1). H¨ ckel’s computation of the uenergy gap of alkene chains has been measured for chains of various lengths and com-pares nicely with this prediction, see Fig.2.19. The filling up of the valence orbitals issimilar to the filling up of atomic orbitals, each orbit accomodating an electron pair withopposite spin. The small gap in long alkene chains such as poly-acetylene and progressin their doping (with electron donors or acceptors) have led to the development of con-ducting polymers for which Alan J. Heeger, Alan MacDiarmid and Hideki Shirakawawere awarded in 2000 the Nobel prize.2.9.2 Molecular vibrational spectrumWa have seen that the nuclei of diatomic molecules such as H2 are stabilized by the ex-change of paired electrons at the bottom of an effective potential V(r), see Fig.2.18. TheHamiltonian describing a system of two bound atoms of mass m1 , m2 at an equilibriumdistance d = 2a is thus: 2 ∂2 2 ∂2 H = −( + ) + V(|r1 − r2 |) 2m1 ∂r1 2m2 ∂r2 2 2Defining the center of mass (CM) coordinate : RCM = (m1 r1 + m2 r2 )/(m1 + m2 )| andr12 = r1 − r2 allows to rewrite the previous Hamiltonian as: H = HCM + Hosc , whereHCM is the Hamiltonian of the center of mass (free particle of mass m1 + m2 ) and Hoscis the Hamiltonian describing the two bound atoms in the CM coordinate system. Nearequilibrium the binding potential V(r12 ) can be expanded in a Taylor series: 1 ∂2 V V(r12 ) = V(d) − | (r − d)2 + . . . 2 r =d 12 2 ∂r12 12
  • 65. 66 QUANTUM MECHANICSF. 2.20. (a) The first eigen-functions of the harmonic oscillator. Notice the inversion symmetry of the even function and the antisymmetry of the odd ones. (b) The com- puted and measured eigenenergies of the vibrational spectrum of C6 H++ (Vanoshi et 4 al., Theor. Chem. Account 120,45 (2008)). Notice the equal spacing of the spectral lines and their value (∼ 0.1eV) typical of vibrational energies in molecules.Defining k ≡ ∂2 V/∂r12 |r12 =d and looking at the deviations from equilibrium: r = r12 − d 2yields the Hamiltonian of an harmonic oscillator: 2 ∂2 1 Hosc = − + kr2 (2.78) 2m ∂r2 2with m = m1 m2 /(m1 + m2 ). The determination of the eigen-energies and eigen-functionsof a harmonic oscillator turns out to be one of those rare examples of an exactly solubleproblem in QM (the hydrogen atom is an other). As done earlier, we shall look for a separable solution Ψ(x, y, z) = X(x)Y(y)Z(z) ofSchroedinger’s equation: Hosc Ψ(x, y, z) = EΨ(x, y, z)Since r2 = x2 + y2 + z2 we obtain: ∂2 X(x) + (λ x − α2 x2 )X(x) = 0 (2.79) ∂x2 √with α = mω/ , ω = k/m and similar equations for Y(y) and Z(z). The eigenenergy isgiven by: (λ x + λy + λz ) 2 E= 2mIn the limit x → ∞, Eq.2.79, simplifies to X = α2 x2 X, for which the asymptoticbounded solution is X∞ ∼ exp(−αx2 /2). As we have done for the solution of the radialequation of the Hydrogen atom, let us then look for a solution: X(x) = f (x) exp(−αx2 /2)where f (x) is a finite polynomial. Inserting that Ansatz into Eq.2.79 yields:
  • 66. THE CHEMICAL BOND 67 f − 2αx f + (λ x − α) f = 0Looking for a polynomial solution: f (x) = n an xn yields the recursion: n(n + 1)an+2 + ((λ x − (2n + 1)α)an = 0which ends at a value n = 0, 1, 2, .... such that: λ x = (2n + 1)α. Thus for a one dimen-sional oscillator the energy levels are: 2 λx En = 1D = ( n + 1/2 ) ω (2.80) 2mwhich (except for the 1/2) is the result obtained using the Bohr-Sommerfeld rule,Eq.2.10. The energy levels in the harmonic oscillator are equally spaced by ω. Formolecules this spacing is typically of order 0.1eV (see Fig.2.20), i.e. in the infrared. √The polynomials f (x) are known as Hermite polynomials of ξ = αx: dn −ξ2 Hn (ξ) = (−1)n eξ 2 e dξnso that the eigenfunctions read: 2 /2 √ α/π1/4 Xn (x) = An e−αx Hn ( αx) with : An = √ (2.81) 2n n!The first Hermite polynomials describing the lowest eigenstates are: H0 (ξ) = 1 (2.82) H1 (ξ) = 2ξ (2.83) H2 (ξ) = 4ξ2 − 2 (2.84) H3 (ξ) = 8ξ3 − 12ξ (2.85)The first eigenfunctions of the 1D-harmonic oscillator are shown in Fig.2.20. Noticethat that the even functions in n are symmetric with respect to inversion x → −x,whereas the odd functions are antisymmetric. For identical di-atoms the spin of thenuclei in the ground state (n = 0) must thus be opposite (Pauli’s principle). The 3D-harmonic oscillator can be decomposed into three components with resulting energylevels: En1 ,n2 ,n3 = (n1 + n2 + n3 + 3/2) ω.2.9.3 Molecular rotational spectrumA diatomic molecule may not only vibrate but also rotate in space. In the CM coordinatesystem the rotation Hamiltonian is: 2 2 J 2 1 ∂ ∂ 1 ∂2 Hrot = =− [ (sin θ ) + ] (2.86) 2I 2md2 sin θ ∂θ ∂θ sin2 θ ∂φ2where I = mr2 is the moment of inertia. Notice the similarity with the angular part ofthe Hydrogen atom Hamiltonian. The eigen-energies of the rotation Hamiltonian aretherefore:
  • 67. 68 QUANTUM MECHANICS 2 j( j + 1) Ej = (2.87) 2IThe eigenfunctions are the spherical harmonics (see Appendix): Y jm (θ, φ) and as in theHydrogen atom, each eigenstate is 2 j+1 -fold degenerate (since m = − j, − j+1, ..., j). Ifthe molecule both vibrates and rotates its energy levels are only approximately a sum ofrotational and vibrational levels (in fact the system does not separates nicely into purevibrational and pure rotational modes). For more complex molecules with different moments of inertia, the analysis be-comes more complicated. For a symmetric top, i.e. if two of the moments of inertia areequal (I1 = I2 I3 , the Hamiltonian can still be written as: 2 2 2 2 2 2 2 2 2 J1 J2 J3 J 1 1 2 H= + + = + ( − )J3 2I1 2I2 2I3 2I1 2 I3 I1where the body-fixed angular momentum operators J1 , J2 , J3 , J 2 are defined in therotating frame of the top (and not with respect to a fixed frame as is the case in theHydrogen atom for the angular momentum operator J). In particular the commutationrelations of Ji are anomalous: [J1 , J2 ] = −iJ3Notice the different sign as compared to the commutation relations of J ([J1 , J2 ] = iJ3 ).These different angular momentum operators commute: [Ji , J j ] = 0 and satisfy J 2 =J 2 . Hence the eigenstates of the symmetric top Hamiltonian are characterized by threequantum numbers that define the eigen-values of J 2 (or J 2 ), J3 and J3 . For a giveneigenstate j, k of J 2 and J3 the eigen-energies are: 2 2 1 1 E j,k,m = j( j + 1) + ( − )k2 2I1 2 I3 I1When k > 0, the energy levels are 2(2 j + 1)-fold degenerate (the factor 2 j + 1 becauseof the degeneracy in m (m = − j, ...., j) and the factor 2 because of the degeneracyk → −k). The typical energies associated with molecular rotational levels correspondsto microwave radiation energies (meV).2.10 Time independent perturbation theoryMost of the problems in physics and in QM in particular are not exactly soluble. There-fore the development of methods that yield approximate solutions is of particular im-portance. We have seen earlier one such method (the variational approach) and havepointed out its main drawback: the abscence of a systematic way of estimating the errormade. In the following we will study a systematic method to get an approximate solu-tion and to estimate the error made within this approximation. The idea is to express the
  • 68. TIME INDEPENDENT PERTURBATION THEORY 69Hamiltonian H as the sum of a known exactly soluble one H0 plus a perturbation λH1 ,where λ is a small parameter (λ 1): H = H0 + λH1One seeks the eigenstates |Ψn > of the full Schroedinger’s equation: H|Ψn >= En |Ψn >knowing the eigenstates |φn > of the soluble problem: H0 |φn >= 0 n |φn >If these eigenstates span the Hilbert space of H (the space of its solutions) we may lookfor an eigenstate of H as: |Ψn >= anm |φm > mInserting that Ansatz into the full Schroedinger equation yields: anm H|φm >= anm (H0 + λH1 )|φm >= anm ( 0 m + λH1 )|φm >= anm En |φm > m m m mmultiplying this equation on the left by < φl | yields: anm (En − 0 m) < φl |φm >= λ anm < φl |H1 |φm > m mSince < φl |φm >= δlm we obtain: (En − 0 l )anl =λ anm < φl |H1 |φm > (2.88) mWhen λ = 0: En = n and |Ψn >= |φn >, i.e. anl = δnl . When λ 0 1 we will thereforelook for a solution as an expansion in λ: anl = δnl + λa(1) + λ2 a(2) + . . . nl nl En = 0 n + λEn + λ2 En + . . . (1) (2)Inserting that Ansatz into Eq.2.88 yields:( n − l0 +λEn +λ2 En +. . . )(δnl +λa(1) +λ2 a(2) +. . . ) = λ 0 (1) (2) nl nl < φl |H1 |φm > (δnm +λa(1) +λ2 a(2) +. . . ) nm nm mEquating terms order by order in λ yields: λ: En δnl + ( (1) 0 n − l )anl =< φl |H1 |φn 0 (1) > (2.89) λ2 : En δnl + (2) En a(1) + ( n − l0 )a(2) = (1) nl 0 nl < φl |H1 |φm > a(1) nm m . . .
  • 69. 70 QUANTUM MECHANICSAt order λ we obtain: when n=l: En =< φn |H1 |φn > (1) < φl |H1 |φn > when n l: a(1) = nl 0 0 (2.90) n − lNotice that a(1) is determined by the normalization condition of nn |Ψn >= (1 + λa(1) )|φn > +λ nn a(1) |φm > nm m n< Ψn |Ψn >= 1 at order λ, which implies: a(1) + a(1)∗ nn nn = 0. Hence a(1) is purely imaginary: nna(1) = iα(1) , so that to first order in λ: nn |Ψn >= eλα |φn > +λ (1) a(1) |φm > nm m nThe phase-shift by λα(1) is of no physical consequences (at this order it can be adsorbedinto a phase shift of |Ψn >) and can be chosen as zero. To second order in λ, Eq.2.90yields: when n=l: En + En a(1) = (2) (1) nn < φn |H1 |φm > a(1) nm m 0 (2) when n l: ( 0 n − l )anl = < φl |H1 |φm > a(1) − En a(1) nm (1) nl mUsing the first-order results,i.e. Eq.2.90, yields:when n = l : En = (2) < φn |H1 |φm > a(1) nm m n < φn |H1 |φm >< φm |H1 |φn > | < φn |H1 |φm > |2 = 0 0 = 0 0 (2.91) m n n − m m n n − m 1 < φl |H1 |φm >< φm |H1 |φn > < φn |H1 |φn >< φl |H1 |φn >when n l: a(2) = nl 0 0 0 0 − 0 0 n − l m n n − m n − lThe coefficient a(2) is determined as before from the normalization condition up to sec- nnond order in λ of: |Ψn >= (1 + λ2 a(2) )|φn > + nn (λa(1) + λ2 a(2) |φm > nm nm m nwhich yields: 1 | < φm |H1 |φn > |2 a(2) = − nn 0 0 2m n ( n − m )2 One may compute higher order terms in this expansion, but usually only the firstnon-zero contribution is computed. The advantage of this perturbation expansion over
  • 70. TIME INDEPENDENT PERTURBATION THEORY 71the variational approach seen earlier is that the error in a perturbation expansion up toorder n is of O(λn+1 ). Example: Energy levels in non-Hydrogen atoms We have seen that in Hydrogen, the energy levels have an accidental degeneracy inthe angular momentum quantum number l. While the wavefunctions Rnl depend on bothquantum numbers, the energy levels themselves depends only on n. To see how this de-generacy is lifted in non-Hydrogen atoms consider the alkalines which have their innershells filled and one electron on a new shell. The electrostatic potential experienced bythe outer electron is: e2 (Z − 1)e2 e2 V(r) = − − (1 − p(r)) ≡ − − δV(r) r r rThe term δV(r) represents the Coulomb interaction of the outer electron with the Z − 1partially screened charges of the nucleus (p(r) represent the effective negative chargewithin a sphere of radius r due to the inner Z − 1 electrons: p(0) = 0 and limr→∞ p(r) =1). The shift in the first order energy levels is given by Eq.2.90: ∆Enl = − < Ψnlm |δV|Ψnlm > (1)Since δV is maximal at r = 0 (the nuclear charge is less screened the closer one gets tothe nucleus) , the energy shift is larger the larger the probability of finding the electronnear the nucleus at r = 0. When solving the Schroedinger equation for the Hydrogenatom, we found that near r = 0: Rnl ∼ (r/rn )l with rn = nr0 . Assume for simplicity thatδV = δV0 exp(−r/r s )/r. Then: ∞ δV0 ∞ ∆Enl = − (1) drr2 R2 (r)δV(r) = − nl dxx2l+1 e−rn x/rs 0 rn 0 δV0 =− (2l + 1)!(r s /rn )2l+2 rnSince the inner electrons are spread on a distance r s < rn the negative energy shift ismaximal for l = 0 and decreases as l increases. This justifies the lift in the accidentaldegeneracy of the Hydrogen energy levels in multi-electron atoms which we used whenfilling-up their electronic orbitals (see Aufbau principle above and Fig.2.17).2.10.1 The polarizability of atoms in an electric fieldTo study the polarizability of an atom in a constant electric field E, we shall estimate itschange in energy when subjected to this field. We shall deduce its electric susceptibilityχ (see Chapter on electrostatics) from the electrostatic energy E p of an induced dipoleP = χE: E p = −P · E/2 = −χE2 /2 (2.92) Without loss of generality we will assume that the direction of E defines the z-axis.Thus to the unperturbed Hamiltonian H0 of the atom one needs to add the perturbationdue to the electric field:
  • 71. 72 QUANTUM MECHANICS H = H0 + ezETo first order the energy shift, Eq.2.90: ∆Enl =< Ψnlm |eEr cos θ|Ψnlm >= 0 (1)Indeed, using the mathematical identities: l+1−m m l+m mcos θ Pm (cos θ) = Pl+1 (cos θ) + P (cos θ) (2.93) l 2l + 1 2l + 1 l−1 (l + 1 + m)(l + 1 − m) (l + m)(l − m) cos θ Ylm (θ, φ) = Yl+1,m (θ, φ) + Yl−1,m (θ, φ) (2l + 3)(2l + 1) (2l + 1)(2l − 1) ≡ Cl+1,m Yl+1,m (θ, φ) + Cl−1,m Yl−1,m (θ, φ)we may compute the first order energy shift (with |Ψnlm >= Rnl (r)Ylm (θ, φ)) as follows: ∆Enl = (1) r2 drdΩRnl (r)Ylm (θ, φ)eEr cos θRnl (r)Ylm (θ, φ) ∗ = eE r3 drdΩR2 (r)Ylm (θ, φ)[Cl+1,m Yl+1,m (θ, φ) + Cl−1,m Yl−1,m (θ, φ)] = 0 nl ∗where we used the orthogonality of the spherical harmonics (see Appendix). The secondorder correction in the energy, Eq.2.91 is then: | < Ψnlm |eEr cos θ|Ψn l m > |2 Enl = (2) 0 0 (2.94) n ,l n,l nl − n ,lNotice that this sum is well defined only for non-Hydrogen atoms where the denom-inator is non-zero (as argued above). In Hydrogen the degeneracy of the energy withrespect to the angular quantum number l necessitates a perturbation analysis specific fordegenerate eigenstates which we will present below. In view of Eq.2.93, we see that inthe sum the only terms that contribute have l = l ± 1. In fact since the lifting of thel−degeneracy is a small effect with respect to the variation of the energy with the majorquantum number n, the main contribution in the sum appearing in Eq.2.94 will comefrom levels n = n. | < Ψnlm |r cos θ|Ψnl m > |2 Enl = e2 E2 (2) 0 0 l =l±1 nl − nlSince the shift in the angular l-degeneracy is larger the smaller l is: nl − n,l−1 > n,l+1 − 0 0 0 nl > 0 and the energy shift E nl is in general negative, see Fig.2.21. This raising of the 0 (2)degeneracy in l with an applied electric field is known as the Stark effect. Assuming the outer electrons to be in their ground states so that only levels withl = l + 1 contribute in the prevuious equation, we may now compare the change inenergy of this atom in an electric field with the energy of an induced dipole, Eq.2.92and identify the electric susceptibility χ of a non-Hydrogen atom as: | < Ψnlm |r cos θ|Ψn,l+1,m > |2 χ = 2e2 0 0 n,l+1 − nl
  • 72. TIME INDEPENDENT PERTURBATION THEORY 73F. 2.21. The shift in the spectral line of an alkaline, Rubidium excited to a high n level (35s state). Notice the negative parabolic shift as a function of electric field, from V.Bendkowsky et al., Nature, 458,1005 (2009).F. 2.22. (a-b) Strong magnetic field lines pierce the surface of the Sun at its spots. These are regions on the Sun’s surface that are cooler by about 1000◦ K than their surrounding due to the damping of the underlying convective motion by the mag- netic field. (c) This field of a few thousand Gauss causes the splitting of the spectral emission lines measured along the vertical black line passing through a Sun’s spot shown in (b).2.10.2 Atom in a constant magnetic field: the Zeeman effectJust as we have studied the lifting of the angular l−degeneracy resulting from the ap-plication of an electric field, we shall now investigate the response of a Hydrogen-likeatom to a constant magnetic field. We have seen that the low energy approximation of
  • 73. 74 QUANTUM MECHANICSDirac’s equation yields the Schroedinger equation with a magnetic correction due to theelectron spin S with magnetic moment m = −( e/2me c)σ = −(eg/2me c)S , see Eq.2.57:    (p + (e/c)A)2 Ze2 eg  + § · B  Ψ = EΨ   −   2me r 2me c     In the Coulomb gauge: · A = 0 (i.e. A = ( B × r)/2) this equation becomes: 2 2 Ze2 e e2 eg − − + A· p+ 2 A2 + §·B Ψ = EΨ 2me r me c 2me c 2me cIn order to know what type of perturbation we need to consider let us estimate theorder of magnitude of the various terms on the left. The first two terms (the kineticand Coulomb energies) are of order of the ionization energy of the atom (typically 1-10eV). The third term is of order eB/me c 10−8 B eV/Gauss, which for a strongmagnetic field B ∼ 104 Gauss is of O(10−4 eV). The fourth term on the left is of order(e2 r0 /2me c2 )B2 ∼ 10−9 eV and is thus negligible with respect to the third one. The 2Hamiltonian can thus be written as: e egB H = H0 + ( B × r) · p + Sz 2mc 2me cSince: ( B × r) · p = B · (r × p) = B · L, one obtains: eB H = H0 + (Lz + gS z ) 2me cSince H0 commutes with Lz (which eigenvalues are m; m = −l, ..., l) and with S z (witheigenvalues ± /2), the energy shift is: e B ∆E = (m ± 1) = ωL (m ± 1) (2.95) 2mcwhere we set: g = 2 (see Eq.2.57). The frequency ωL = eB/2mc is known as the Larmorfrequency: it is the frequency of precession of a magnetic dipole about the direction ofthe magnetic field. Notice that the magnetic field lifts the degeneracy in the azymuthalquantum number m. It should also be noted that the present description of the Zeemaneffect is only an approximation valid for strong magnetic fields. Indeed as we have seenwhen studying Dirac’s equation Lz and S z do not commute with the full Hamiltonian.The reason that they do commute with H0 is that this Hamiltonian which we derivedfrom Dirac’s equation is only the lowest term in a low energy expansion of Dirac’srelativistic Hamiltonian. In a strong magnetic field however, the magnetic perturbationterm is larger than the next (relativistic) terms in Dirac’s expansion which can thusbe neglected (for example the spin-orbit coupling discussed previously). The Zeemaneffect has been used to estimate the strong magnetic field coming out of the Sun’s spotsthrough the spliting of the Zeeman lines of Hydrogen atoms at the Sun’s surface, seeFig.2.22
  • 74. TIME INDEPENDENT PERTURBATION THEORY 75F. 2.23. A Magnetic Resonance Imaging machine consists of an apparatus in which a strong permanent magnet and various coils allow for the generation of strong mag- netic field gradients along a pre-determined direction. This field gradient results in a splitting of the proton energy levels with a gap that varies with space. By measuring the adsorption of radio-waves emitted by a transmitter at given (Larmor) frequency (corresponding to a given gap) one can deduce the concentration of Hydrogen atoms at the corresponding location. By measuring the lifetime of the excited state, one can further distinguish betwen different tissues. The result is a very detailed image of various cuts through the body, here the head. Nuclear magnetic resonance imaging (MRI). One of the main applications of the Zeeman effect is in nuclear Magnetic ResonanceImaging which functioning principle is based on the use of a magnetic field to align thenuclear spin of atoms (mainly Hydrogen), in other words to split their nuclear energy p plevels according to Eq.2.95: ∆E = ωL σz (where ωL is the Larmor frequency of the pro-ton which for the typical fields of MRI, i.e. 1 Tesla=104 Gauss is in the radio-frequencyrange: 42Mhz). The low energy state (spin aligned with the field) will be more occupiedthan the high energy one. As seen earlier in the study of the Amonia molecule, radiation pat the resonant frequency (ω = ωL ) will be absorbed by the protons. As a result theywill be excited into their high energy state (spin anti-parallel to the magnetic field) fromwhich they return to the ground state by spontaneous emission. In MRI one applies astrong magnetic field gradient on a patient. As a result the protons (e.g. the nuclei ofhydrogen in H2 O)) display a Larmor frequency that varies with distance along the gra-dient. By measuring the adsorption of radio-waves at a specific frequency, one can thusdeduce the concentration of protons at the corresponding location. The Fourier trans-form of the received radio-signal (i.e. the intensity of the signal at varying frequencies)therefore provides a map of proton density at varying locations along the field gradient.Moreover by measuring the lifetime of the excited state (which depends on the chemicalenvironment of the Hydrogen nucleus, e.g. whether it is bound to Oxygen or Carbon)one can distinguish between different tissues (e.g. richer in fat or water). The result is ahigh resolution image of various cuts through the patient’s body, see Fig.2.23.
  • 75. 76 QUANTUM MECHANICS2.10.3 Degenerate eigenstatesWhen eigenstates have the same energy the perturbation expansion we described abovediverges at second order since the denominator in Eq.2.91 is zero for degenerate states(i.e. when n = m ). The way to deal with degenerate eigenstates is to neglect all tran- 0 0sitions to non-degenerate eigenstates and to solve the equations exactly taking into ac-count only the transitions between states of the same energy. This will usually raise thedegeneracy with the new eigenstates expressed as a superposition of the original un-perturbed eigenstates. One may then proceed using non-degenerate perturbation theorywith these new eigenstates as zeroth order approximation. As above we assume that the Hamiltonian is H = H0 + λH1And we want to solve for it taking into account only the N degenerate eigenstates |φn >(n = 1, . . . , N) of H0 with energy 0 : H0 |φn >= 0 |φn >One seeks the new eigenstates |ψn > of H as a superposition of these degenerate eigen-states: |ψn >= anm |φm > mwhich yields: H|ψn >= En |ψn > or: anm (( 0 − En ) + λH1 )|φm >= 0 mDefining Vnm ≡ λ < φn |H1 |φm > yields the following equation: 0 + V11 − En V12 ... V1N V21 0 + V22 − En ... V2N . . . . = 0 (2.96) . . . . . . . . VN1 ... VN,N−1 0 + VNN − EnThe solution of this N th order equation for En will usually yield N eigenvalues and Northogonal eigenstates ψn > which can then be used instead of the |φn > in a perturbationexpansion together with the other non-degenerate eigenstates of H0 . Example: The Stark effect in Hydrogen. When previously studying the response of atoms to a constant electric field E wementioned that the case of Hydrogen was particular since its energy levels are degen-erate with respect to the quantum numbers l (and m) and thus the second order non-degenerate perturbation expansion breaks down. We shall now see how the degeneracyof the electronic energy levels in hydrogen are affected by the presence of an electricfield. The Hamiltonian is: H = H0 + ezEConsider the case of the n = 2 level of hydrogen for which the states: l = 0; m = 0and l = 1; m = 0, ±1 all have the same energy: E2 = E1 /4 (with E1 = −13.6eV).
  • 76. TIME DEPENDENT PERTURBATION THEORY 77Following our previous perturbation analysis for degenerate energy levels, we needto diagonalize the Hamiltonian matrix: < 2, l , m |H|2, l, m >. While the unperturbedHamiltonian H0 is diagonal with energy E2 , using Eq.2.93 we deduce that the dipolecoupling < 2, l , m |ezE|2, l, m > is non-zero only if m = m and l = l ± 1, i.e. only whencoupling states |2, 0, 0 > and |2, 1, 0 >. In that case: < 2, 1, 0|ezE|2, 0, 0 > = eE r3 drR21 (r)R20 (r) dΩY10 (θ, φ) cos θY00 (θ, φ) = −3er0 E ≡ −∆E2The eigen-energies E are thus determined by the equation: E2 − E −∆E2 0 0 −∆E2 E2 − E 0 0 = 0 0 0 E2 − E 0 0 0 0 E2 − Ewhich solutions are: E so = E2 + ∆E2 = E2 + 3er0 E and E so = E2 − ∆E2 with eigenstates: I I |2, 0, 0 > −|2, 1, 0 > |2, I > = √ 2 |2, II >= |2, 0, 0 > +|2, 1, 0 > √ 2 III,IVand E so = E2 with eigenstates |2, 1, 1 > and |2, 1, −1 >. Notice that for the hydrogenatom the Stark effect, i.e. the shift in the energy level, is linear with electric field andnot quadratic as it is for the other atoms for which the electronic energy levels are non-degenerate, see Eq. Time dependent perturbation theoryUntill now, we have dealt mostly with time independent problems (the one exceptionwas our investigation of the amonia maser), namely systems for which the Hamiltoniandid not depend explicitly on time and for which the goal was to determine the eigen-states and eigen-energies. Very often though to investigate these states, one drives thesystem with a time dependent external perturbation, such as an electro-magnetic field,and studies the induced transitions between the various eigenstates. In such a situationthe Hamiltonian of the system can be written as: H(t) = H0 + λH1 (t), where H0 is thetime independent unperturbed Hamiltonian for which the eigenstates and eigen-energiesare known: H0 |φn >= n |φn > 0To solve the time dependent Schroedinger equation: i ∂t |Ψ(t) >= H|Ψ(t) >= H0 |Ψ(t) > +λH1 (t)|Ψ(t) > (2.97)with initial condition: |Ψ(0) >= |φi >, we shall look for a solution as an expansion interms of the eigenstates of H0 :
  • 77. 78 QUANTUM MECHANICS |Ψ(t) >= an (t)|φn > nInserting that Ansatz into Eq.2.97 yields: dan i |φn >= [ 0 n + λH1 (t)]an (t)|φn > n dt nMultiplying this equation on the left by < φm | and recalling that the eigenstates areorthonormal: < φm |φn >= δmn yields: dam (t) i = 0 m am (t) +λ < φm |H1 (t)|φn > an (t) dt n 0Define: am (t) = µm (t)e−i m t/ to obtain the following equation for µm (t): dµm (t) 0 0 i =λ < φm |H1 (t)|φn > ei( m − n )t/ dt nIntegrating yields: λ t µm (t) − δmi = dt < φm |H1 (t)|φn > eiωmn t µn i n 0with ωmn = ( m − n )/ and the initial condition µm (0) = δmi . For short enough times, 0 0the system remains essentially in its initial state |φi > and for states m i we may write: λ t µm (t) dt < φm |H1 (t)|φi > eiωmi t i 0Now consider a periodic perturbation, such as an electro-magnetic wave, so that: λH1 (t) =(Ve−iωt + V ∗ eiωt )/2 = |V| cos(ωt + θ): 1 t < φm |V|φi > i[ωmi −ω]t < φm |V ∗ |φi > i[ωmi +ω]t µm (t) dt e + e i 0 2 2 = −[Vmi F(ω) + Vmi F(−ω)] ∗with Vmi =< φm |V|φi > /2 and : t ei[ωmi −ω]t − 1 F(ω) = dt ei[ωmi −ω]t = m− i − ω 0 0 0The probability that the system is in state |φm > at time t is: Pm (t) = |am (t)|2 = |µm (t)|2 = |Vmi F(ω) + Vmi F(−ω)|2 ∗This transition probability will be large when either F(ω) or F(−ω) are large, whichoccurs when m − i0 = ω or m − i0 = − ω. The former case corresponds to absorption 0 0
  • 78. TIME DEPENDENT PERTURBATION THEORY 79of energy from the perturbing field ( 0 m > 0 i ), the latter to emission of radiation fromthe initial state ( i0 > m ). 0 In case of adsorption the probability of finding the system in state |φm > at times t isthus: 2π|Vmi |2 Pm (t) = |Vmi |2 |F(ω)|2 = |Vmi |2 (t/ )2 sinc2 [(ωmi − ω)t/2] → 2 tδ(ωmi − ω) (2.98)where we used the mathematical identity: lima→0 sinc2 ax = (π/a)δ(x), which is a validapproximation as long as (ωmi − ω)t < 1. Hence the transition rate from state |φi > intostate |φm >: T mi = dPm /dt is: 2π|Vmi |2 2π|Vmi |2 T mi = 2 δ(ωmi − ω) = δ( 0 m − 0 i − ω)Since the frequency of the initial beam is not infinitely sharp and if there exists a dis-tribution of final state energies (which is usually the case due to Doppler broadening ofthe final state energy, etc.) then the δ-function in the above equation has to be replace bythe energy density ρ( m ) of the final state (such that 1 = d δ( − i0 − ω) = d ρ( )): 2π|Vmi |2 T mi = ρ( m ) 0 (2.99)This equation is known as Fermi’s Golden rule. It relates the transition rate from theground state |φi > to the final state |φm > to this state’s energy density and to the couplingbetween initial and final states. This is the result we obtained when studying the amoniamaser, see Eq.2.24. In case of emission, the probability of finding the system in a lower energy state|φm > after a time t is computed exactly as before: ∗ 2π|Vmi |2 2π|Vmi |2 T mi = δ( 0 m − 0 i + ω) = ρ( i0 )which is the same result as that for adsorption (notice that here it is the energy density ofthe initial state that appears in Fermi’s Golden rule, as the ground state being infinitelylived is infinitely sharp). This is also in agreement with Einstein treatment of radiation,see above. Let us know investigate in more details the factors affecting the coupling element:Vmi . Let the perturbation be a plane electromagnetic wave: A = A0 ei(kx−ωt) + A∗ e−i(kx−ωt) 0propagating along the x-axis and let us choose the z-axis as parallel to A0 (which isalways possible unless some external field exist which defines the z-axis). The electric
  • 79. 80 QUANTUM MECHANICSfield is : E = −(1/c)∂t A and B = × A. The average intensity (Poynting vector) of thiselectromagnetic wave is, see Eq.??: I = (c/4π) < E × B >= ω2 |A0 |2 /2πc In presence of this EM-wave, the momentum p in the Hamiltonian is replaced byp − qA(t)/c), see Eq.2.46. The perturbation to the Hamiltonian is: q q2 2 λH1 (t) = − (p · A + A · p) + A 2mc 2mc2In the following we shall neglect the second term on the right, which is usually smallerthan the first (if however the matrix element computed with the first term vanishes, thesecond term must be retained). One obtains: qA0 Vmi =< φm |λH1 |φi >= − d3 xφm (x) pz φi (x)eikx ˆ mcSince the electromagnetic wavelength λ is usually much larger than atomic scales wemay assume that kx 1 and thus approximate eikx 1, hence: i qA0 Vmi d3 xφm ∂z φi mcThe integral can be evaluated by studying the evolution of the following transition ma-trix element,see Eq.2.45, : 2 d i i < φm |z|φi > = < φm |[z, H]|φi >= < φm |pz |φi >= d3 xφm ∂z φi dt m m d d dbut : i < φm |z|φi > = i < φm | z |φi > + < φm | z i |φi > = −( m − 0 0 i) < φm |z|φi > dt dt dtSince by energy conservation: ωmi = ( 0 m − 0 i )/ = ω: iqA0 ωzmi Vmi = cAnd the transition rate is then: 2π|Vmi |2 2πq2 |A0 |2 ω2 T mi = 2 δ(ωmi − ω) = 2 |zmi |2 δ(ωmi − ω) c2 2πq 2 =( ) cI(ω) |zmi |2 δ(ωmi − ω) (2.100) cThe transition rate depends on the dipole moment between the initial and final states:< φ f |qz|φi >. It is therefore known as the dipole transition rate. When investigating thepolarizability of atoms in a constant electric field by time independent perturbation the-ory, see above, we have already been led to compute the transition element z f i between
  • 80. TIME DEPENDENT PERTURBATION THEORY 81various states |n, l, m > of the hydrogen atom, see Eq.2.94. We have seen there that thiselement is zero unless m = m and l = l ± 1. If an other field (e.g. a constant mag-netic field) exist, then the polarization of the EM-wave will in general be along an otheraxis (say y) and the transition element will be non-zero if m = m ± 1. Therefore onlytransitions between states differing in their angular momentum l by ±1 and in m by 0or ±1 are possible (at leading order), for example transitions between s and p states arepossible, but not between two s or two p states. Example: Rotational-vibrational transitions in diatomic molecules. Since the rotational eigenstates of a diatomic molecule are the same as the Hydrogenatom angular wavefunctions, the same selection rules hold for dipole transitions in adiatomic molecule as in the Hydrogen atom. With respect to the initial state, the finalstates differ in their angular quantum numbers j by ±1 and m by 0 or ±1. If vibrational states of the diatomic molecule are excited, then the transition ratedepend on the dipole moment between initial and final eigenstates, |n1 ,2 , n3 >, of theharmonic oscillator. < n1 , n2 , n3 |z|n1 , n2 , n3 >= dzXn3 (z)zXn3 (z)Using the results of our previous investigations, √ Eq.2.81, the eigenfunctions of the see1D-harmonic oscillator are: Xn (z) ∼ e−αx /2 Hn ( αz) (for the 3D-oscillator the eigen- 2functions are products of 1D-eigenstates). Using the property of the Hermite polynomi-als: 2zHn = Hn+1 + 2nHn−1It is easy to check that the dipole moment between initial and final states will be non-zero only if: n = n ± 1. Diatomic molecules can both vibrate and rotate. Their vibrational energy levels aregiven by: En1 ,n2 ,n3 = (n1 + n2 + n3 + 3/2) ωThey are separated by an energy gap: ∆Evib = ω of O(0.1eV). The diatomic rotationalenergy spectrum is: 2 j( j + 1) Ej = 2Iwith a splitting between levels: ∆Erot = 2 (2 j + 1)/2I of O(1meV) which is muchsmaller than the energy gap between successive vibrational energy level. Electromag-netically, i.e. light-induced transitions between vibrational-rotational energy levels indiatomic molecules occur between vibrational levels with ∆n = ±1 and ∆ j = ±1.Hence the emission spectrum from the first excited vibrational level to the ground statewill exhibit two bands known as R- and P-bands corresponding to transitions betweenrotational energy levels differing by ∆ j = ±1, see Fig.2.24.
  • 81. 82 QUANTUM MECHANICSF. 2.24. (a) Transitions between vibrational energy level n = 0 and n = 1 and rota- tional energy levels differing by ∆ j = ±1 in a diatomic molecule such as CO. These transitions define two characteristic spectral bands known as P-band when ∆ j = −1 and R-band when ∆ j = 1.