Quantum physics

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Quantum physics

  1. 1. DeBroglie HypothesisProblem with Bohr Theory: WHY L = n ?• have integers with standing waves: n(λ/2) = Lconsider circular path for standing wave:nλ = 2πr• from Bohr theory: L = mvr = nh/2π,Re-arrange to get 2πr = nh/mv = nλ,Therefore, nλ = 2πr = nh/mv, which leads toλ = h/mv = h/p .
  2. 2. DeBroglie HypothesisλDeBroglie = h/mv = h/pIn this case, we are considering theelectron to be a WAVE, andthe electron wave will “fit” around theorbitif the momentum (and energy) is just right(as in the above relation). But this will happenonly for specific cases - and those are thespecific allowed orbits (rn) and energies (En)that are allowed in the Bohr Theory!
  3. 3. DeBroglie HypothesisThe Introduction to Computer Homework onthe Hydrogen Atom (Vol. 5, number 5)shows this electron wave fitting around theorbit for n=1 and n=2.What we now have is a wave/particle dualityfor light (E&M vs photon), AND awave/particle duality for electrons!
  4. 4. DeBroglie HypothesisIf the electron behaves as a wave, withλ = h/mv, then we should be able to test thiswave behavior via interference anddiffraction.In fact, experiments show that electrons DOEXHIBIT INTERFERENCE when theygo through multiple slits, just as theDeBroglie Hypothesis indicates.
  5. 5. DeBroglie HypothesisEven neutrons have shown interferencephenomena when they are diffracted from acrystal structure according to the DeBroglieHypothesis: λ = h/p .Note that h is very small, so that normally λwill also be very small (unless the mv isalso very small). A small λ means verylittle diffraction effects [1.22 λ = D sin(θ)].
  6. 6. Quantum TheoryWhat we are now dealing with is theQuantum Theory:• atoms are quantized (you can have 2 or 3,but not 2.5 atoms)• light is quantized (you can have 2 or 3photons, but not 2.5)• in addition, we have quantum numbers(L = n , where n is an integer)
  7. 7. Heisenberg Uncertainty PrincipleThere is a major problem with the wave/particleduality:a) a wave with a definite frequency andwavelength (e.g., a nice sine wave) does nothave a definite location.[At a definite location at a specific time the wavewould have a definite phase, but the wavewould not be said to be located there.][ a nice travelling sine wave = A sin(kx-ωt) ]
  8. 8. Heisenberg Uncertainty Principleb) A particle does have a definite location ata specific time, but it does not have afrequency or wavelength.c) Inbetween case: a group of sine waves canadd together (via Fourier analysis) to give asemi-definite location: a result of Fourieranalysis is this: the more the group showsup as a spike, the more waves it takes tomake the group.
  9. 9. Heisenberg Uncertainty PrincipleA rough drawing of a sample inbetween case,where the wave is somewhat localized, andmade up of several frequencies.
  10. 10. Heisenberg Uncertainty PrincipleA formal statement of this (from Fourieranalysis) is: ∆x * ∆k = 1/2(where k = 2π/λ, and ∆ indicates theuncertainty in the value).But from the DeBroglie Hypothesis, λ = h/p,this uncertainty relation becomes:∆x * ∆(2π/λ) = ∆x * ∆(2πp/h) = 1/2 , or∆x * ∆p = /2.
  11. 11. Heisenberg Uncertainty Principle∆x * ∆p = /2The above is the BEST we can do, since thereis always some experimental uncertainty.Thus the Heisenberg Uncertainty Principlesays: ∆x * ∆p > /2 .
  12. 12. Heisenberg Uncertainty PrincipleA similar relation from Fourier analysis fortime and frequency: ∆t * ∆ω = 1/2 leads toanother part of the Uncertainty Principle(using E = hf): ∆t * ∆E > /2 .There is a third part: ∆θ * ∆L > /2 (whereL is the angular momentum value).All of this is a direct result of thewave/particle duality of light and matter.
  13. 13. Heisenberg Uncertainty PrincipleLet’s look at how this works in practice.Consider trying to locate an electronsomewhere in space. You might try to“see” the electron by hitting it with aphoton. The next slide will show anidealized diagram, that is, it will show adiagram assuming a definite position for theelectron.
  14. 14. Heisenberg Uncertainty PrincipleWe fire an incoming photon at the electron,have the photon hit and bounce, then tracethe path of the outgoing photon back to seewhere the electron was.electronincomingphoton
  15. 15. Heisenberg Uncertainty Principleelectronoutgoingphotonslit so we candetermine directionof the outgoingphotonscreen
  16. 16. Heisenberg Uncertainty Principle• Here the wave-particle duality creates aproblem in determining where the electronwas.electronslit so we candetermine directionof the outgoingphotonphoton hits here
  17. 17. Heisenberg Uncertainty Principle• If we make the slit narrower to betterdetermine the direction of the photon (andhence the location of the electron, the wavenature of light will cause the light to bediffracted. This diffraction pattern willcause some uncertainty in where the photonactually came from, and hence someuncertainty in where the electron was .
  18. 18. Heisenberg Uncertainty PrincipleWe can reduce the diffraction angle if wereduce the wavelength (and hence increasethe frequency and the energy of thephoton).But if we do increase the energy of thephoton, the photon will hit the electronharder and make it move more from itslocation, which will increase the uncertaintyin the momentum of the electron.
  19. 19. Heisenberg Uncertainty PrincipleThus, we can decrease the ∆x of the electrononly at the expense of increasing theuncertainty in ∆p of the electron.
  20. 20. Heisenberg Uncertainty PrincipleLet’s consider a second example: trying tolocate an electron’s y position by making itgo through a narrow slit: only electronsthat make it through the narrow slit willhave the y value determined within theuncertainty of the slit width.
  21. 21. Heisenberg Uncertainty PrincipleBut the more we narrow the slit (decrease ∆y),the more the diffraction effects (wave aspect),and the more we are uncertain of the y motion(increase ∆py) of the electron.
  22. 22. Heisenberg Uncertainty PrincipleLet’s take a look at how much uncertaintythere is: ∆x * ∆p > /2 .Note that /2 is a very small number(5.3 x 10-35J-sec).
  23. 23. Heisenberg Uncertainty PrincipleIf we were to apply this to a steel ball ofmass .002 kg +/- .00002 kg, rolling at aspeed of 2 m/s +/- .02 m/s, the uncertaintyin momentum would be 4 x 10-7kg*m/s .From the H.U.P, then, the best we could besure of the position of the steel ball wouldbe: ∆x = 5.3 x 10-35J*s / 4 x 10-7kg*m/s= 1.3 x 10-28m !
  24. 24. Heisenberg Uncertainty PrincipleAs we have just demonstrated, the H.U.P.comes into play only when we are dealingwith very small particles (like individualelectrons or photons), not when we aredealing with normal size objects!
  25. 25. Heisenberg Uncertainty PrincipleIf we apply this principle to the electron goingaround the atom, then we know the electronis somewhere near the atom,(∆x = 2r = 1 x 10-10m)then there should be at least some uncertaintyin the momentum of the atom:∆px > 5 x 10-35J*s / 1 x 10-10m = 5 x 10-25m/s
  26. 26. Heisenberg Uncertainty PrincipleSolving for p = mv from the Bohr theory[KE + PE = Etotal, (1/2)mv2- ke2/r = -13.6 eVgives v = 2.2 x 106m/s ] givesp = (9.1 x 10-31kg) * (2.2 x 106m/s)= 2 x 10-24kg*m/s;this means px is between -2 x 10-24kg*m/s and2 x 10-24kg*m/s, with the minimum ∆pxbeing 5 x 10-25kg*m/s, or 25% of p.
  27. 27. Heisenberg Uncertainty PrincipleThus the H.U.P. says that we cannot reallyknow exactly where and how fast theelectron is going around the atom at anyparticular time.This is consistent with the idea that theelectron is actually a wave as it movesaround the electron.
  28. 28. Quantum TheoryBut if an electron acts as a wave when it ismoving, WHAT IS WAVING?When light acts as a wave when it is moving,we have identified theELECTROMAGNETIC FIELD as waving.But try to recall: what is the electric field?Can we directly measure it?
  29. 29. Quantum TheoryRecall that by definition, E = F/q. We canonly determine that a field exists bymeasuring an electric force! We havebecome so used to working with the electricand magnetic fields, that we tend to taketheir existence for granted. They certainlyare a useful construct even if they don’texist.
  30. 30. Quantum TheoryWe have four LAWS governing the electricand magnetic fields: MAXWELL’SEQUATIONS. By combining these lawswe can get a WAVE EQUATION forE&M fields, and from this wave equationwe can get the speed of the E&M wave andeven more (such as reflection coefficients,etc.).
  31. 31. Quantum TheoryBut what do we do for electron waves?What laws or new law can we find that willwork to give us the wealth of predictivepower that MAXWELL’S EQUATIONShave given us?
  32. 32. Quantum TheoryThe way you get laws is try to explainsomething you already know about, andthen see if you can generalize. A successfullaw will explain what you already knowabout, and predict things to look for thatyou may not know about. This is where thevalidity (or at least usefulness) of the lawcan be confirmed.
  33. 33. Quantum TheorySchrodinger started with the idea ofConservation of Energy: KE + PE = Etotal .He noted that• KE = (1/2)mv2= p2/2m, and that λ=h/p, sothat p = h/λ = (h/2π)*(2π/λ) = k = p, soKE = 2k2/2m• Etotal = hf = (h/2π)*(2πf) = ω.
  34. 34. Quantum TheoryHe then took a nice sine wave, and calledwhatever was waving, Ψ:Ψ(x,t) = A sin(kx-ωt) = Aei(kx-ωt) .He noted that both k and ω were in theexponent, and could be gotten down bydifferentiating. So he tried operators:
  35. 35. Quantum TheoryΨ(x,t) = A sin(kx-ωt) = Aei(kx-ωt) .popΨ = i[dΨ/dx] = i[-ikAe-i(kx-ωt)] = kΨ= (h/2π)*(2π/λ)*Ψ = (h/λ)Ψ = pΨ .similary:EopΨ = i[dΨ/dt] = i[-iωAei(kx-ωt)] = ωΨ= ((h/2π)*(2πf)*Ψ = (hf)Ψ = EΨ .
  36. 36. Quantum TheoryConservation of Energy: KE + PE = Etotalbecomes with the momentum and energyoperators:-(2/2m)*(d2Ψ/dx2) + PE*Ψ = i(dΨ/dt)which is called SCHRODINGER’SEQUATION. If it works for more than thefree electron, then we can call it a LAW.
  37. 37. Quantum TheoryWhat is waving here? ΨWhat do we call Ψ? the wavefunctionSchrodinger’s Equation allows us to solvefor the wavefunction. The operators thenallow us to find out information about theelectron, such as its energy and itsmomentum.
  38. 38. Quantum TheoryTo get a better handle on Ψ, let’s considerlight: how did the E&M wave relate to thephoton?
  39. 39. Quantum TheoryThe photon was the basic unit of energy forthe light. The energy in the wave dependedon the field strength squared.[Recall energy in capacitor, Energy = (1/2)CV2,where for parallel plates, Efield = V/d andC// = KεoA/d, so thatEnergy = (1/2)*(KεoA/d)*(Efieldd)2= KεoEfield2* Vol, or Energyα Efield2.]
  40. 40. Quantum TheorySince Energy is proportional to field strengthsquared, AND energy is proportional to thenumber of photons, THEN that implies thatthe number of photons is proportional to thesquare of the field strength.This then can be interpreted to mean that thesquare of the field strength is related tothe probability of finding a photon.
  41. 41. Quantum TheoryIn the same way, the square of thewavefunction is related to the probability offind the electron!Since the wavefunction is a function of both xand t, then the probability of finding theelectron is also a function of x and t!Prob(x,t) = Ψ(x,t)2
  42. 42. Quantum TheoryDifferent situations for the electron, like beingin the hydrogen atom, will show up inSchrodinger’s Equation in the PE part.Different PE functions (like PE = -ke2/r for thehydrogen atom) will cause the solution toSchrodinger’s equation to be different, justlike different PE functions in the normalConservation of Energy will cause differentspeeds to result for the particles.
  43. 43. Schrodinger’s EquationLet’s look at the case of an electron confinedto a length, L, but otherwise free. We willfirst consider the 1-D case. Since theelectron is free, PE=0. However, since it isconfined to the length, L, we haveboundary conditions:Ψ(x=0,t) = 0, and Ψ(x=L,t) = 0.( / )( / ) ( / )− + = 2 2 22 0m x i t∂ ∂ ∂ ∂Ψ Ψ
  44. 44. Schrodinger’s EquationTo solve this differential equation, we use thetechnique of separation of variables:Ψ(x,t) = X(x)*T(t) .The partial derivative with respect to x doesnot affect T(t), and the partial derivativewith respect to t does not affect X(x).We can also divide both sides of the equationby X(x)*T(t) to get:
  45. 45. Schrodinger’s EquationNote that the left side depends solely on x,and the right side solely on t. This meansthat as far as x is concerned, the right side isa constant, and as far as t is concerned, theleft side is a constant (which turns out to bethe Energy).( / )( / ) / ( / ) /− = = 2 2 22m X x X E i T t T∂ ∂ ∂ ∂
  46. 46. Schrodinger’s EquationWe now have two ordinary differentialequations instead of one partial differentialequation:-(2/2m)d2X/dx2= E*X for X(x)i dT/dt = E*T for T(t).In fact, any situation where the PE does notexplicitly depend on time will have thissolution.
  47. 47. Schrodinger’s Equationi dT/dt = E*T for T(t)The above 1st order ordinary differentialequation is easily solved if we recall that thedifferential of an exponential gives anexponential back again. So try T(t) = Aeat:i(aAeat) = E(Aeat) or ia = E, or a = E/i.But we know that E = ω, so this means that:a = ω/i = -iω, and T(t) = A e-iωt.
  48. 48. Schrodinger’s Equation• Note that the probability depends on Ψ2(actually, since Ψ can be a complexquantity, Prob(x,t) = Ψ∗∗Ψ, where Ψ∗is thecomplex conjugate of Ψ).• Note also that T**T = Ae-iωt* Ae+iωt= A2, butA can be incorporated as a constant intoX(x). Thus the T part of Ψ can effectivelybe ignored if the PE does not depend on t.
  49. 49. Schrodinger’s EquationThis means that when the PE does not dependon time, Schrodinger’s Equation can be re-written as (since idΨ/dt = E*Ψ) :(-2/2m)*d2Ψ/dx2+ PE*Ψ = E*Ψ .Note that Ψ can be complex (just like E couldbe written in complex form), but theProbability must be real (since it is ameasurable quantity).Now we go on to the X equation.
  50. 50. Schrodinger’s Equation-(2/2m)d2X/dx2= E*X for X(x)This equation can again be solved by inspectionsince we know that the second derivative of a sine(or cosine) function gives itself back again with aminus sign, so we’ll try X(x) = B sin(kx+φo) :-(2/2m)(-Bk2sin(kx+φo)) = E B sin(kx+φo) , or2k2/2m = E (since p = k, E = p2/2m = KEbut this is correct since PE = 0 in this case!)
  51. 51. Schrodinger’s EquationX(x) = B sin(kx +φo)We now need to apply the Boundaryconditions:X(x=0) = 0 works fine for the sine function,but not the cosine function so φo= 0.X(x=L) = 0 demands that kL = nπ, or thatk = nπ/L (note that k is quantized, so p isquantized, and E is quantized!).
  52. 52. Schrodinger’s EquationNote that Schrodinger’s Equation GIVES USA QUANTUM NUMBER. It comes whenwe apply the boundary conditions! We didnot have to put the quantum number into thetheory as Bohr did.Note that the Schrodinger’s equation we havewritten down is for 1-D only (x). We nowneed to see about a 3-D form.
  53. 53. Schrodinger’s EquationMomentum is a vector, so the DeBroglierelation: p = k is really a componentrelation: px = kx, py = ky, pz = kz , wherethe kx simply describes how the E&M fieldvaries in the x-direction.Thus the pop = i d/dx must now becomep i x p i yp i zx op x opz op− −−= ; == ∂ ∂ ∂ ∂∂ ∂/ / ;/
  54. 54. Schrodinger’s EquationKE = p2/2m = (px2+ py2+ pz2)/2m.This can be used in the Schrodinger’sequation with the component operators forp, and using the symbol∇2 2 2 2 2 2 2= + +∂ ∂ ∂ ∂ ∂ ∂/ / /x y z
  55. 55. Schrodinger’s EquationThe three dimensional Scrodinger’sEquation becomes (for PE not dependenton time):( / )− ∇i m2 22 + PE = EΨ Ψ Ψ
  56. 56. Schrodinger’s EquationThis equation can also usually be solved byseparation of variables, and when applyingthe boundary conditions (for x, y and z) weusually get THREE QUANTUMNUMBERS (just like we got 1 quantumnumber in the 1-D case).These quantum numbers come out of thetheory rather than being put into the theoryas Bohr did in his.
  57. 57. Quantum TheoryFrom Chemistry, you may recall that FOURquantum numbers were required, insteadof the three that the Schrodinger’s Equationpredicts.The fourth quantum number (spin) can bepredicted from a relativistic quantumtheory, where the equation is called theDIRAC EQUATION. (This also predictsanti-matter as well!)
  58. 58. Quantum TheoryIn the hydrogen atom we do not userectangular coordinates (x,y,z); instead weuse spherical coordinates (r,θ,ϕ) becauseof the spherical symmetry of the potentialenergy (PE = -ke2/r).The solution of the Schrodinger’s equationgives us three quantum numbers: n, l, andml . The fourth number is due to spin, ms .
  59. 59. Quantum TheoryThe n quantum number is related to the r equationand is related to energy.The l quantum number is related to the θ equationand is related to angular momentum.The m quantum number is related to the ϕ equationand is related to the z-component of angularmomentum, which is related to the magneticproperties of the state.
  60. 60. Quantum TheoryThe fourth quantum number, ms, does nothave a classical explanation - it is arelativistic quantum phenomenon. It isrelated to magnetic behavior, and hence hasthe m name. The closest thing classicallywe can relate it to is to the case of theelectron “spinning”, so that its spinningcharge creates a magnetic field. But this doesnot work out according to classical calculations.
  61. 61. Pauli Exclusion PrincipleHow do we extend the quantum theory tosystems beyond the hydrogen atom?For systems of 2 electrons, we simply have aΨ that depends on time, and the coordinatesof each of the two electrons:Ψ(x1,y1,z1,x2,y2,z2,t)and the Schrodinger’s equation has twokinetic energies instead of one.
  62. 62. Pauli Exclusion PrincipleIt turns out that the Schodinger’s Equationcan again be separated:Ψ = Xa(x1,y1,z1) * Xb(x2,y2,z2) * T(t) .This is like having electron one in state a, andhaving electron two in state b.
  63. 63. Pauli Exclusion PrincipleHowever, from the Heisenberg UncertaintyPrinciple (from wave/particle duality), weare not really sure which electron is electronnumber 1 and which is number 2. Thismeans that the wavefunction must alsoreflect this uncertainty.
  64. 64. Pauli Exclusion PrincipleThere are two ways of making thewavefunction reflect the indistinguishabilityof the two electrons:Ψsym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)andΨanti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t) .
  65. 65. Pauli Exclusion PrincipleΨsym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)Ψanti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t) .Which (if either) possibility agrees withexperiment?It turns out that some particles are explainednicely by the symmetric, and some areexplained by the antisymmetric.
  66. 66. BOSONSΨsym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)Those particles that work with the symmetricform are called BOSONS. All of theseparticles have integer spin as well. Notethat if electron 1 and electron 2 both havethe same state, Ψ > 0. This means that bothparticles CAN be in the same state at thesame location at the same time.
  67. 67. FERMIONSΨanti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t)Those particles that work with the anti-symmetricwavefunction are called FERMIONS. All of theseparticles have half-integer spin. Note that ifelectron 1 and electron 2 both have the samestate, Ψ = 0. This means that both particles CANNOT be in the same state at the same location atthe same time.
  68. 68. Pauli Exclusion PrincipleBOSONS. Photons and alpha particles (2neutrons + 2 protons) are bosons. Theseparticles can be in the same location withthe same energy state at the same time.This occcurs in a laser beam, where all thephotons are at the same energy(monochromatic).
  69. 69. Pauli Exclusion PrincipleFERMIONS. Electrons, protons and neutronsare fermions. These particles can NOT be inthe same location with the same energy stateat the same time.This means that two electrons going aroundthe same nucleus CAN NOT both be inthe exact same state at the same time!This is known as the Pauli ExclusionPrinciple!
  70. 70. Pauli Exclusion PrincipleSince no two electrons can be in the sameenergy state in the same atom at the sametime, chemistry is possible (and so isbiology, psychology, sociology, politics,and religion)!Thus, the possibility of chemistry is explainedby the wave/particle duality of light andmatter, and electrons acting as fermions!

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