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PART II - Continuation of PART I - Quantum Mechanics.

PART II - Continuation of PART I - Quantum Mechanics.

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Linked in slides - Quantum Fields Linked in slides - Quantum Fields Presentation Transcript

  • From First Principles PART II – QUANTUM FIELDS August 2014 – R1.7 Maurice R. TREMBLAY
  • “A poet once said, ‘The whole universe is in a glass of wine.’ We will probably never know in what sense he meant that, for poets do not write to be understood. But it is true that if we look at a glass of wine closely enough we see the entire universe. There are the things of physics: the twisting liquid which evaporates depending on the wind and weather, the reflections in the glass, and our imagination adds the atoms. The glass is a distillation of the earth’s rocks, and in its composition we see the secrets of the universe’s age, and the evolution of stars. What strange array of chemicals are in the wine? How did they come to be? There are the ferments, the enzymes, the substrates, and the products. There in wine is found the great generalization: all life is fermentation. Nobody can discover the chemistry of wine without discovering, as did Louis Pasteur, the cause of much disease. How vivid is the claret, pressing its existence into the consciousness that watches it! If our small minds, for some convenience, divide this glass of wine, this universe, into parts – physics, biology, geology, astronomy, psychology, and so on – remember that nature does not know it! So let us put it all back together, not forgetting ultimately what it is for. Let it give us one more final pleasure: drink it and forget it all! ” Richard Feynman Epicatechin TARTARIC ACID (C4H6O6) 2,3-dihydroxybutanedioic acid Tartaric acid is, from a winemaking perspective, the most important in wine due to the prominent role it plays in maintaining the chemical stability of the wine and its color and finally in influencing the taste of the finished wine. [Wikipedia] MALIC ACID (C4H6O5) hydroxybutanedioic acid Malic acid, along with tartaric acid, is one of the principal organic acids found in wine grapes. In the grape vine, malic acid is involved in several processes which are essential for the health and sustainability of the vine. [Wikipedia] CITRIC ACID (C6H8O7) 2-hydroxypropane-1,2,3-tricarboxylic acid The citric acid most commonly found in wine is commercially produced acid supplements derived from fermenting sucrose solutions. [Wikipedia] Three primary acids are found in wine grapes: RESVERATROL DERIVATIVES trans cis Malvidin-3-glucoeide Procyanidin B1 Quercetin R = H; resueratrol R = glucose; p Ice Id TYPICAL WINE FLAVONOIDS Resveratrol (3,5,4'-trihydroxy- trans-stilbene) is a stilbenoid, a type of natural phenol, and a phytoalexin produced naturally by several plants. [Wikipedia] In red wine, up to 90% of the wine's phenolic content falls under the classification of flavonoids. These phenols, mainly derived from the stems, seeds and skins are often leached out of the grape during the maceration period of winemaking. These compounds contribute to the astringency, color and mouthfeel of the wine. [Wikipedia] Prolog
  • Contents PART II – QUANTUM FIELDS Review of Quantum Mechanics Galilean Invariance Lorentz Invariance The Relativity Principle Poincaré Transformations The Poincaré Algebra Lorentz Transformations Lorentz Invariant Scalar Klein-Gordon & Dirac One-Particle States Wigner’s Little Group Normalization Factor Mass Positive-Definite Boosts & Rotations Mass Zero The Klein-Gordon Equation The Dirac Equation “It is more important to have beauty in one’s equations than to have them fit experiment … because the discrepency may be due to minor features that are not properly taken into account and that will get cleared up with further development of the theory….” Paul Dirac, Scientific American, May 1963. 2014 MRT Determining the structure of the proton: a Feynman diagram for deep inelastic scattering process. The diagram shows the flow of momentum when a high energy electron e (•) scatters (hense the exchange of a photon γ with momentum q) from a quark (•) taken from the wavefunction of the proton p (•). This is a simple case called the Parton Model invented by Richard Feynman. We assume that the parton () has negligeable (i.e., a small fraction ξ of) transverse momentum with respect to the proton p, so the parton momentum ξ pis in the same direction as the proton momentum p, that is, the parton has momentum ξ p µ , where 0 ≤ ξ ≤ 1. Finally, momentum conservation forces us to have the equality p′=ξ pµ + q given vertex couplings of the form ±ieγ µ where the gamma matrices satisfy γ µ γ ν + γ ν γ µ = 2gµν.
  • 2014 MRT PART III – THE HYDROGEN ATOM What happens at 10−10 m The Hydrogen Atom Spin-Orbit Coupling Other Interactions Magnetic & Electric Fields Hyperfine Interactions Molecules Appendix The Harmonic Oscillator Electromagnetic Interactions Quantization of the Radiation Field Transition Probabilities Einstein’s Coefficients Planck’s Law A Note on Line Broadening The Photoelectric Effect Higher Order Electromagnetic Interactions References “Quantum field theory is the way it is because […] this is the only way to reconcile the principles of quantum mechanics […] with those of special relativity. […] The reason that quantum field theory describes physics at accessible energies is that any relativistic quantum theory will look at sufficiently low energy like a quantum field theory.” Steven Weinberg, Preface to The Quantum Theory of Fields, Vol. I.PART 0 – PHYSICAL MATHEMATICS PART I – QUANTUM MECHANICS Introduction Symmetries Angular Momentum Quantum Behavior Postulates Quantum Angular Momentum Spherical Harmonics Spin Angular Momentum Total Angular Momentum Momentum Coupling General Propagator Free Particle Propagator Wave Packets Non-Relativistic Particle
  • Review of Quantum Mechanics We will provide only the briefest of summaries of PART I - Quantum Mechanics, in the generalized version of Dirac. This will also strengthen our mathematical conventions. It has a norm; for any pair of vectors there is a complex number 〈Φ |Ψ 〉, such that: ΨΦΨΦΨΦΦ ΨΦΨΦΨΨΦ ΦΨΨΦ 2 * 21 * 12211 22112211 * ηηηη ξξξξ +=+ +=+ = POSTULATE #1: Physical states are represented by rays in Hilbert space. where the asterisk (∗) indicates that the complex conjugate is taken. The norm 〈Ψ |Ψ 〉 also satisfies a positivity condition: A ray is a set of normalized vectors: A Hilbert space is a kind of complex vector space; that is, if |Φ 〉 and |Ψ 〉 are vectors in the space (often called ‘state-vectors’ or ‘kets’) then so is η|Φ 〉+ξ|Ψ 〉, for arbitrary complex numbers η and ξ. 0≥ΨΨ and vanishes if and only if the state-vector (or ket) is null: |Ψ 〉=0. with |Ψ 〉 and |Ψ 〉 belonging to the same ray if |Ψ 〉=ξ|Ψ 〉, where ξ is (as above) an arbitrary complex number with the extra condition that it’s magnitude is unity: |ξ |=1. 1=ΨΨ 2014 MRT
  • POSTULATE #2: Observables are represented by Hermitian operators. The above relation satisfies the reality condition: ΦΨΦΨ AAA ηξηξ +=+ These are mappings |Ψ 〉→A|Ψ 〉 of Hilbert space into itself, linear in the sense that: *† ΦΨΨΦΨΦ AAA == AA =† ( )Rinfor ΨΨΨ α=A The state represented by a ray R has a definite value α for the observable represented by an operator A if vectors |Ψ 〉 belonging to this ray are eigenvalues of A with eigenvalue α: An elementary theorem tells us that for A Hermitian, α is real, and eigenvalues with different αs are orthogonal. 2014 MRT where the daggar (†) definies the Hermitian Operator and it indicates that the complex conjugate (i.e., replacing i → −i) and transpose (i.e., matrix elements are ‘transposed’ on either side of the diagonal, Aij →Aji ), A† =A*T , is taken. For any linear operator A the adjoint A† is defined by:
  • POSTULATE #3: If a system is in a state represented by a ray R, and an experiment is done to test whether it is in any one of the different states represented by mutually orthogonal rays R1, R2, …, then the probability of finding it in the state represented by Rn is: where |Ψ 〉 and |Ψn 〉 are any vectors belonging to rays R and Rn, respectively. 2 )( nnP ΨΨ=→ RR Another elementary theorem gives a total probability unity (i.e., they add up to 100%): if the state-vectors |Ψn 〉 form a complete set. 1)( =→∑n nP RR 2014 MRT
  • Let us now review symmetries: A symmetry transformation is a change in our point of view that does not change the result of possible experiments. For any such transformation R → R, of rays we may define an operator U on Hilbert space H , such that if |Ψ 〉 is in ray R then U|Ψ 〉 is in the ray R, with U either unitary and linear: or else antiunitary and antilinear: If an inertial observer O (i.e., he ain’t movin’) sees a system in a state represented by a ray R or R1 or R2, …, then an equivalent observer O (i.e., he’s moving away!) who looks at the same system will observe it in a different state, represented by R or R1 or R2, …, respectively, but the two observers must find the same probabilities: )()( nOnO PP RRRR →=→ ΦΨΦΨ ΨΦΨΦ UUU UU βαβα +=+ = ΦΨΦΨ ΨΦΨΦ UUU UU ** * βαβα +=+ = 2014 MRT This is a fundamental theorem from Wigner (1931): Any symmetry transformation can be represented on the Hilbert space of physical states by an operator that is either linear and unitary or antilinear and antiunitary.
  • As mentionned in the condition 〈Φ |A† Ψ 〉=〈AΦ |Ψ 〉=〈Ψ |A† Φ 〉∗ above, the adjoint of a linear operator L is defined by: With this definition, the conditions for unitarity or antiunitarity both take the form: ΨΦΨΦ LL ≡† ΦΨΨΦΨΦ AAA =≡ *† 1=UU† and there is always a trivial symmetry transformation R → R, represented by the identity operator U=1. This operator is, of course, unitary and linear. Continuity then demands that any symmetry (e.g., a rotation or translation or Lorentz transformation) that can be made trivial by a continuous change of some parameters (i.e., like angles or distances or velocities) must be represented by a linear unitary operator U (rather than one that is antilinear and antiunitary). This condition cannot be satisfied for an antilinear operator, because in this case the right-hand side of the above equation would be linear in |Φ 〉, while the left-hand side is antilinear in |Φ 〉. Instead, the adjoint of an antilinear operator A is defined by: 2014 MRT 1† − =UU Multiplying this unitary operator by it’s inverse U−1 wegetU† (UU−1 )=1⊗U−1 =U−1 . Thus:
  • A symmetry transformation that is infinitesimally close to being trivial can be re- presented by a linear ‘unitary operator’ that is infinitesimally close to the identity: The set of symmetry transformations has certain properties that define it as a group. If T1 is a transformation that takes rays Rn into Rn, and T2 is another transformation that takes Rn into Rn, then the result of performing both transformations is another symmetry transformation, which we write T2T1 (T1 then T2) that takes Rn into Rn. Also, a symmetry transformation T which takes rays Rn into Rn has an inverse, written T−1 , which takes Rn into Rn, and there is an identity transformation T =1, which leaves rays unchanged. with ε a real infinitesimal (e.g., an infinitesimal change in the coordinates dxµ or an angle dϕ). For this to be unitary and trivial, T must be Hermitian and linear, so it is a can- didate for an observable. Indeed, most (and perhaps all) of the observables of physics (e.g.,angularmomentumormomentum) arise in this way from symmetry transformations. For φ =0, U(T) furnishes a representation of the group of symmetry transformations. 2014 MRT The unitary operators U(T) act on vectors in the Hilbert space, rather than on rays. If T1 takes Rn into Rn, then acting on a vector |Ψn 〉 in the ray Rn, U(T1) must yield a vector U(T1)|Ψn 〉 in the ray Rn, and if T2 takes this ray into Rn, then acting on U(T1)|Ψn 〉 it must yield a vector U(T2)U(T1)|Ψn 〉 (again U(T1) then U(T2) but this time on |Ψ 〉) in the ray Rn. But U(T2T1)|Ψn 〉 is also in this ray, so these vectors can differ only by a phase φn(T2,T1) : )(e)()()(e)()( 12 ),( 1212 ),( 12 1212 TTUTUTUTTUTUTU TTi n TTi n nn φφ =⇒= ΨΨ )( 2 εε OTiU ++= 1
  • where Λµ ν is a constant matrix (a function of the velocity v of a ‘moving’ frame). Under an infinitesimal transformation of the variable θ, the coordinate differential dxµ is given by:         ==≡+Λ=→ ∑= c xfaxxx v v βζµµ ν ν ν µµµ tanh);()]([ 3 0 θ θ θ θ µ dxd 0= ∂ ∂ ≡ all );( θµ xf so that the state vector |ψ 〉 will transform according to (i.e., by using Taylor’s expansion): As an example of symmetry,consideratransformation (parametrizedbythevariableθ, e.g., an angle ϕ,a translation a or aLorentzboostζ )onthespace-timecoordinates xµ : in which the real infinitesimal is ε =dθ and the generator for the parameter θ is given by: ∑= = ∂ ∂ ∂ ∂ −= 3 0 0 );( µ µ θ µ θ θ x xf iT all ψεεψθψ θ θ θψ ψ θ θ θψψψψψψ µµ µ µ θ µ µ µ µ µ θ µ µµ µ µ µµµµµ )]([)()()( );( )( )( );( )()()()()( 2 0 0 OTixTdix x xf idix x x xf dxx x dxxxdxx T ++=+=         ∂ ∂ ∂ ∂ −+= ∂ ∂ ∂ ∂ +≡ ∂ ∂ +≈+=→ ∑ ∑∑ = = 111    )(Generator all all 2014 MRT
  • with f a (θ,θ) a function of the θs and θs. Taking θa =0 as the coordinate of the identity, we must have: )),(()()( θθθθ fTTT = aaa ff θθθ == ),0()0,( As mentionned above, the transformation of such continuous groups must be represen- ted on the physical Hilbert space by unitary operators U[T(θ )]. For a Lie group these op- erators can be represented by a power series (e.g., in the neighborhood of the identity): A finite set of real continuous parameters θ a describe a group of transformations T(θ ) with each element of the group connected to the identity by a path (i.e., UU−1 =1) within the group. The group multiplication law U(T2)U(T1)=U(T2T1) thus takes on the form (i.e., a connected Lie group): According to f a (θ,0)= f a (0 ,θ)=θ a above, the expansion of f a (θ,θ) to second-order must take the form: 2014 MRT +++= ∑∑∑ c b bc cb a a a TTiTU θθθθ 2 1 )]([ 1 where Ta, Tbc =Tcb, &c. are Hermitian operators independent of the θs. Suppose that the U[T(θ )] form an ordinary (i.e., φ =0) representation of this group of transformations, i.e.: ))],(([)]([)]([ θθθθ fTUTUTU = ...),( +⊕+= ∑∑c b cba bc aaa ff θθθθθθ with real coefficients f a bc. The addition of the second-order term was highlighted by⊕.
  • (The Σ were also omitted to get space.) The terms of order 1, θ, θ, and θ2 automatically match on both sides of this equation – from the θθ terms we get a non-trivial condition:  ++++++++=+++×+++ bc ccbb a cba bc aa bc cb a a bc cb a a TTfiTTitti ))(()(][][ 2 1 2 1 2 1 θθθθθθθθθθθθθθ 111 ∑−−= a a a bccbbc TfiTTT Since we are following Weinberg’s development, he points out that: This shows that if we are given the structure of the group, i.e., the function f a (θ,θ), and hence its quadratic coefficient f a bc, we can calculate the second-order terms (i.e., Tbc) inU[T(θ)] from the generators Ta appearing in the first-order terms. (A pretty amazing fact I would say!) Applying the multiplication rule U[T(θ)]U[T(θ)]=U[T( f (θ ,θ )] and using the series U[T(θ)]=1+iθ a Ta +½θ b θ c Tbc +…above withθ → f a (θ,θ)=θ a +θ a + f a bc θ b θ c ,we get: where Ca bc are a set of real constants known as structure constants: ∑=−≡ a a a bcbccbcb TCiTTTTTT ],[ 2014 MRT However, as he points out: There is a consistency condition: the operator Tbc, must be symmetricinbandc(becauseitisthesecondderivativeof U[T(θ)]withrespect toθb andθc ) so the equation Tbc =−Tb Tc −iΣa f a bc Ta above requires that the commutation relations be: a cb a bc a bc ffC +−= Such a set of commutation relations is known as a Lie algebra.
  • This is the case for instance for ‘translations’ in spacetime, or for ‘rotations’ about any one fixed axis (though not both together). Then the coefficients f a bc in the function f a (θ,θ)=θa +θa +Σbc f a bc θb θc vanish,and so do the structure constants Ca bc = − f a bc + f a cb. The generators then all commute: aaa f θθθθ +=),( N N TUTU                   = θ θ)]([ Such a group is called Abelian. In this case, it is easy to calculate U[T(θ)] for all θa . Again, from the group multiplication rule U[T(θ)]U[T(θ)]=U[T( f (θ ,θ ))] and the function f a (θ,θ)= θa +θa above, and taking ε=θ /N, we have for any integer N: As a special case of importance, suppose that the function f a (θ,θ) is simply additive: and hence: 0],[ =cb TT 2014 MRT Letting N→∞, and keeping only the first-order term in U[T(θ/N)], we have then: N a a a N T N iTU                 += ∑∞→ θ θ 1lim)]([ Ti Ti UTU a a a θ θ θθ e)(e)]([ =⇒= ∑
  • Lorentz invariance is needed to replace the principle of Galilean invariance and the discovery of the non-conservation of parity in weak interactions (1956) has reemphasized that an invariance principle and its consequences must be experimentally verified. One key invariance principle in quantum mechanics and quantum field theory is that: Different equivalent observers make the same predictions as to the outcome of an experiment carried out on a system. The vectors |φO〉 and |ψO〉 seen by observer O and vectors |φO〉 and |ψO〉 seen by observer O. A unitary transformation U(L) (a function of the Lorentz transformation) relates both systems. |φO〉 |ψO〉 We shall call the vector |ψO 〉 the translation of the vector |ψO〉. Stated mathematically, the postulate above asserts that if |ψO〉 and |φO 〉 are two states and |ψO〉 and |φO 〉 their translations, then: OO LU ψψ )(= where U depends on the coordinate systems between which it affects the correspondence and U(L =1) =1 if L is the identity transformation 1, i.e. if O and O are the same coordinate system. If all rays in Hilbert space are distinguishable, it the follows from the above equation – as a mathematical theorem (Wigner, 1931) – that the correspondence |ψO〉 → |ψO 〉 is effected by a unitary or anti-unitary operator, U(O,O), the operator U is completely determined up to a factor of modulus 1 by the transformation L which carries O in O. We write: 22 OOOO ψφψφ = |φO〉 |ψO〉 O O U(L) OOOO OO LU LU ψφψφ ψψ )( )( = = 2014 MRT This statement means that observer O will attribute the vector |ψO〉 to the state of the system, whereas observer O will describe the state of this same system by a vector |ψO〉. 〈φO |U(L)|ψO〉 Lorentz Invariance
  • v For special relativity, as an example, we consider the inhomogeneous Lorentz transformations. A relativity invariance requires the vector space describing the possible states of a quantum mechanical system to be invariant under all relativity transformations, i.e., it must contain together with every |ψ 〉 all transformations U(L)|ψ 〉 where L is any special relativity transformation. The transformed states can always be obtained from the original state by an actual physical operation on the system. Consider for example a Lorentz transformation along the z-axis with velocity v. The transformed state, which arises from the momentum eigenstate |ΨO (p,σ )〉, is given by U(v)|ΨO (p,σ )〉. This is the state of the system as seen by observer O. It is, however, also a possible state of the system as seen by O and which can be realized by giving the system a velocity −v along the z-axis. The state vector |ΨO〉 seen by observer O and the vector |ΨO 〉 seen by observer O moving away from O at velocity v. A unitary trans- formation U(v) brings state |Ψ 〉 into state |Ψ 〉. OO z,z U(v) |ΨO 〉 |ΨO 〉 Here are two typical problems*: 1.Suppose that observer O sees a W-boson (spin one and mass m≠0) with momentum p in the y-direction and spin z- component σ . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the W state? 2.Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? sees sees 2014 MRT * S. Weinberg, Quantum Theory of Fields, Vol. I – Foundations, 1995 – P. 104-105. y x p p σ m Solving these two problems is the goal of the following slides and they involve modern concepts starting with particle definitions as unitary representations. x
  • The Relativity Principle Einstein’s principle of relativity states the equivalence of certain ‘inertial’ frames of reference. It is distinguished from the Galilean principle of relativity, obeyed by Newtonian mechanics, by the transformation connecting coordinate systems in different inertial frames. 2014 MRT Here ηµν is the diagonal 4×4 matrix, with elements (i.e., the Minkowski metric): η00 =+1, η11 =η22 =η33 =−1 and ηµν ≡0 for µ ≠ν. These transformations have the special property that the speed of light c is the same in all inertial frames (e.g., a light wave traveling at speed c satisfies |dr/dt|=c or in other words Σµνηµν dxµ dxν =c2 dt2 − dr2 =0, from which it follows that Σµνηµν dxµ dxν =0, and hence |dr /dt|=c.) If the contravariant vector xµ =(ct,r) are the coordinates in one inertial frame (x1 , x2 , x3 ) i.e., r = x i (i=1,2,3), as Cartesian space coordinates, and x0 =ct a time coordinate, the speed of light being c) then in any other inertial frame, the coordinates xµ must satisfy: OO xdxdxdxd FrameFrame ∑∑∑∑ = == = ≡ 3 0 3 0 3 0 3 0 µ ν νµ µν µ ν νµ µν ηη or equivalently stated as the Principle of covariance: ∑∑ ∂ ∂ ∂ ∂ = µ ν µνσ µ ρ µ σρ ηη x x x x The covariant vector can be given as xµ=Σνηµν xν =(ct,−r). The norm of the vector Σ x xµ =(x0 )−Σ (xi )2 =c2 t2 −|r|2 is a Lorentz invariant term.
  • Poincaré Transformations with aµ arbitraryconstants(e.g.,‘leaps’),and Λµ ν aconstantmatrixsatisfyingthecondition: Any coordinate transformation xµ → xµ that satisfies Σµνηµν (∂xµ /∂xρ )(∂xν /∂xσ )=ηρσ is linear and allows us to define the Poincaré Transformations: 2014 MRT µµµµµµ ν ν ν µµ axxxxaxx +Λ+Λ+Λ+Λ=+Λ= ∑= 3 3 2 2 1 1 0 0 3 0 σρ µ ν σ ν ρ µ µν ηη =ΛΛ∑∑ The matrix ηµν has an inverse, written ηµν , which happens to have the same components: it is also diagonal matrix, with elements: η00 =+1,η11 =η22 =η33 =−1 and ηµν ≡0 for µ ≠ν. To save on using summation signs (Σ), we introduce the Summation Convention: We sum over any space-time index like µ and ν (or i, j or k in three dimensions) which appears twice in the same term, if they appear only once ‘up’ and also only once ‘down’. As an example, and while also enforcing this tricky summation convention, we now multiply ηµ ν Λµ ρ Λν σ =ηρσ above with ηστ Λκ τ and when inserting parentheses for show: ρ κ ρ κκ ρ κ ρ κ ρ κ ρ ηηηηηηηηη µν µν σ σ τσ τσ τσ τσ νµ µν µν τσ τσ τσ νµ µν Λ=Λ=Λ=Λ=ΛΛΛ=ΛΛΛ∑ ∑ ][)(])[(])[(   and now, when multiplying with the inverse of the matrix ηµν Λµ ρ, we get from this: τσ τσ ηη κνκν ΛΛ= ( )3,2,1,0=µ
  • 2014 MRT ( )νρaxaxx andoversumρν ν ρ ρ ρ ρ i.e.)( +ΛΛ=+Λ= µµµµ These transformations do form a group. If we first perform a Poincaré transformation xµ → Λµ ν xν +aµ, and then a second Poincaré transformation xµ → xµ, with: )()( µρ ρ µν ν ρ ρ µµ aaxx +Λ+ΛΛ= then the effect is the same as the Poincaré transformation xµ → xµ, with: Taking the determinant of ηµν Λµ ρ Λν σ =ηρσ gives: so Λµ ν has an inverse, (Λ−1 )ν σ , which we see from ηµν Λµ ρ Λν σ =ηρσ and takes the form:  σ µσρ µν ρ νν ρ ηη Λ=Λ=Λ− Note )( 1 1)Det( 2 =Λ The transformation T =T(Λ,a) induced on physical states therefore satisfy the group composition rule: ),(),(),( aaTaTaT +ΛΛΛ=ΛΛ and the inverse of this T(Λ,a) transformation is also obtained from T(Λ,a)T(Λ,a)= T(ΛΛ,Λa+a) above to be T(Λ−1 ,−Λ−1 a) such that: The whole group of transformations T(Λ,a) is properly known as the inhomogeneous Lorentz group, or Poincaré group. It has a number of important subgroups – notably T(Λ,0) which we will look at in greater detail in a little while. 1),(),( 11 =Λ−ΛΛ −− aTaT
  • and this unitary operator U satisfies the same group composition rule as T(Λ,a): So, in accordance with the discussion in the previous slides, the transformations T(Λ,a) induced a unitary linear transformation on a state vector in the Hilbert space: For example, we will soon discuss the wave function in its momentum representation: for which the same composition rule applies: 2014 MRT )]()([)]([)]([),( 1 pLpLUpLUpLUpU ΛΛΛ=Λ≡Λ − ),(),(),(),;( jjj mppUmpmjp ΨΨΨ Λ=→µ )(),()()( xaUxx ψψψ µ Λ=→ ),(),(),( aaUaUaU +ΛΛΛ=ΛΛ
  • Now it is time to study three-dimensional rotations and add relativity to the overall description. To this effect we will exploit pretty much all the group symmetry properties! 2014 MRT ζζβγ ζζβγ sinhcosh)( sinhcosh)( 03033 22 11 30300 xxxxx xx xx xxxxx −=−= = = −=−= and v=|v| is the relative velocity of the two frames. or β ζ ζ ζβγβζ β ζγ ==      == − == cosh sinh tanhsinh 1 1 cosh 2 aswellaswithand c v                           − − =               3 2 1 0 3 2 1 0 cosh00sinh 0100 0010 sinh00cosh x x x x x x x x ζζ ζζ that is, assuming propagation in the direction of the x3 -axis. This can be represented in matrix form as: Let us recall a few facts about homogeneous Lorentz transformations: where ν ν µνµµ ζ xxxx )],-([Λ=
  • The explicit matrix representation of a restricted homogeneous Lorentz transformation in the x1 -direction (rotation in the x0 -x1 plane) is given by: Similarly, the infinitesimal generators M20 and M30 for rotations in the x2 -x0 and x3 -x0 planes respectively, are given by: and the infinitesimal generator M10 for this rotation is defined as:             − − = 1000 0100 00coshsinh 00sinhcosh ),-(Λ 01 ζζ ζζ ζxx             ==≡ = 0000 0000 0001 0010 ),-(Λ 0 01 10 1 ζ ζ ζ d xxd MK             ==≡             ==≡ == 0001 0000 0000 1000 ),-(Λ 0000 0001 0000 0100 ),-(Λ 0 03 30 3 0 02 20 2 ζζ ζ ζ ζ ζ d xxd MK d xxd MK , 2014 MRT
  • The infinitesimal generators in the xi -xj plane, i.e. spatial rotations, are: 2014 MRT             − =≡             − =≡             − =≡ 0000 0010 0100 0000 0010 0000 1000 0000 0100 1000 0000 0000 12 3 31 2 23 1 MJMJMJ and, where we define Mµν =−Mνµ . In matrix form these terms come together as:             −− − =               −− − = 0 0 0 0 0 0 0 0 123 132 231 321 231303 231202 131201 030201 JJK JJK JJK KKK MMM MMM MMM MMM M µν
  • The general result for Mµν can now be written alternatively as: 2014 MRT where we can use ηµν Λµ ρ Λν σ =ηρσ to show that ωµν (or ωµ ν ) is antisymmetric ωµν =−ωνµ : ∑+= µν µν µν M i ω 2 1)ω(Λ  which implies ωρσ +ωσρ =0 QED. An arbitrary infinitesimal Lorentz transformation, by expanding according to Λ(ω)= exp(−½iMµν ωµν /) in a power series, can be written as: KζSω •−•=L and KζSω •−• = eA where ω and ζ are constant 3-vectors. ρσσρσρ ρ µ σµσ ν νρσρ σ ν ρ µ σ ν ρ µ σ ν ρ µ µν σ ν σ ν ρ µ ρ µ µνσρ η ηηη δδδδη δδηη ωω ωω )ωω( )ω)(ω( ++= ++= ++= ++=
  • A finite rotation in the µ -ν plane (in the sense µ toν ), is again obtained by exponentiation: 2014 MRT ζνµ µν ζ M xx e),-(Λ = µσνρνρµσµρνσνσµρρσµν ηηηη MMMMMM −−+=],[ One verifies that the infinitesimal generators, Mµν ,satisfy the following commutation rules:
  •         ∂ ∂ = ∂ ∂ −= ∂ ∂ −= ∂ ∂ ∂ ∂ −=• = kj kjii kji kji k k k x xi x xi x i x xf i εε ζ ζ ϕ µ )ˆ()ˆ()ˆ( );(ˆˆ 0 nnrn Jn × hence: kj kjii x xiJ ∂ ∂ = ε For the rotation acting on the space-time coordinates, note that the time coordinate is unaffected (hence only latin indices):    •−= −•−+= = +−+==≡ ζζ ζζ ϕϕϕϕµµ sinhˆcosh ]sinhˆ)1[(coshˆ ]ˆ[sin)]ˆ(ˆ)[cos1()],ˆ([);( rn nrnrr rnrnnn cttc ct xxRxfx kkki i k ××× in which we made use of the relation xj =ηij xj =−x j , η being the Minkowski metric. It can be shown that the generators for rotation are equivalent to the generators for the SO(3) Special Orthogonal group (which are Hermitian). Thus, the representation for a finite rotation acting on the wavefunction is unitary and it is given by U(R)=exp(−iϕ n•J/).ˆ 2014 MRT where v=ctanhζ n. So, we get:ˆ
  • The general Lorentz transformations for a simple spatial rotation, ΛR, is given by: in which the three-dimensional spatial rotations, R(n,ϕ), are elements in the simple orthogonal group SO(3). However, this is not relavant for evaluating the Wigner coefficients since it is trivial to show that both the Wigner transformation and Lorentz rotation, ΛR, belong to the same little group H(Λ,k). µν µν ω 2 1 e)ω( M i  − =Λ Note that the parameters associated with the Lorentz transformation are given by the anti-symmetric tensor ωµν . (In addition, the matrix representations of the Lorentz generators in the four-vector coordinates are given by:       − =      = iab i ai bi i i J i i K εδ δ 0 00 0 0 and where the indices a, b represent the rows and columns, respectively.) The convenional way of characterizing the Lorentz transformation, Λ, is described by the generators for boosts with Ki =M0i =−Mi0 and rotations Ji =½ε ijk Mjk, that is: ˆ             =Λ 0 )],ˆ([0 0 0001 )( j iR R ϕ ν µ n 2014 MRT
  • in which we recall that coshζ =γ and sinhζ =βγ where γ =1/√(1−β2 ) and β =v/c. In four- vector notation we have:               + + =                           =               ζζ ζζ ζζ ζζ coshsinh sinhcosh cosh00sinh 0100 0010 sinh00cosh 30 2 1 30 3 2 1 0 3 2 1 0 xx x x xx x x x x x x x x );()]([ ζζ µν ν µµ xfxx ≡Λ= and the associated generator is given as: Now, consider a Lorentz boost along the x3 -axis: since Λ0 0|ζ =0 =coshζ |ζ =0=1, &c.       ∂ ∂ − ∂ ∂ =      ∂ ∂ + ∂ ∂ −=         ∂ ∂ ∂ Λ∂ + ∂ ∂ ∂ Λ∂ + ∂ ∂ ∂ Λ∂ + ∂ ∂ ∂ Λ∂ −= ∂ ∂ ∂ Λ∂ −= ∂ ∂ ∂ ∂ −= ==== == 30033 0 0 3 3 3 0 3 3 3 0 0 0 3 0 3 0 3 0 0 0 0 0 0 00 3 );( x x x xi x x x xi x x x x x x x xi x xi x xf iK ζζζζ µ ν ζ ν µ µ ζ µ ζζζζ ζζ ζ 2014 MRT
  • Similarly: )( 0000 iiiii xxi x x x xiK ∂−∂=      ∂ ∂ − ∂ ∂ = † 00 ])([)()()( xKx x xx x xixK iiii ψψψψ ≠      ∂ ∂ − ∂ ∂ = Consider that the state-vector of the system is given by |ψ 〉, hence the action of Ki on the state is: This implies that the generator for the Lorentz boost, Ki, is not Hermitian and hence the exponentiation of the generator (i.e., exp(−iζ i Ki /)) will not be unitary. The representation of the Lorentz boost acting on the wavefunction is not unitary and hence is not trace-preserving. We can summarize the effects of the rotations and the Lorentz boost into one second- rank covariant tensor: )( µννµµν ∂−∂= xxiM in which Ji =½ε ijk Mjk and Ki =Mi0. 2014 MRT
  • These generators (i.e., Pµ and Mµν ) obey the following commutation relations, which characterize the Lie algebra of the Poincaré group (and adding  and c for reference): kkjijikkjijikkjiji JiKKKiKJJiJJ εεε  −=== ],[],[],[ and; 0],[ )(],[ )(],[ = −−= −+−−= λµ νµλµνλλµν νρµσµρνσµσνρνσµρµνρσ ηη ηηηη PP PPiPM MMMMiMM   The rotation Ji and boost Kj generators can be written in covariant notation Mµν and the commutation relations are then re-written as: 2014 MRT 00 ],[],[],[ P c iPKcPiPKPiPJ jijiiikkjiji δε   === and; 0],[0],[0],[ 00 === jiii PPPPPJ and; The first of these is the usual set of commutators for angular momentum, the second says that the boost K transforms as a three-vector under rotations, and the third implies that a series of boosts can be equivalent to a rotation. Next we have: where P0 c=H, the Hamiltonian, and finally all components of Pµ should commute with each other: Together, these equations above form the Lie algebra of the Poincaré group.
  • we get: µµµµµ axaxfx +=≡ );(    ≠= == = ∂ ∂ = νµ νµ δ ν µ ν µ µ when when 0 1);( 0a a axf and using: ννµ ν µ ν δ ∂−= ∂ ∂ −= ∂ ∂ −=  i x i x iP For a simple translation aµ of the coordinates xµ: i.e., the momentum operator. For a finite translation aµ , the state-vector |ψ 〉 of a relativistic system (expressed by the Dirac wavefunction ψ ) will transform as: )(e)()( xxx aP i ψψψ µ µ µ  − =→ Note that the generators P0 =−i∂0 and Pk = −i∂k are known as the Hamiltonian and momentum operators and they are Hermitian since their associated eigenvalues are defined to be real. This implies that the representation for a finite translation acting on the wavefunction (e.g., as you will see soon it is given by U(1,a)=exp(−iPµ aµ /)) is actually unitary. 2014 MRT ψεψεψψ µν µ ν       +=         ∂ ∂ ∂ ∂ −+= = P i xa axf i i a    11 0 );( all we obtain:
  • µ µ µ aP i aIT  − × = e),( 44 00 ωωω iiiji kjik −=== ζεϕ and 0 2 1 iikj kjii MKMJ == andε The contravariant generators for the space-time translations aµ are defined by Pµ, in which the time translation, P0 and the spatial translation, Pi are the Hamiltonian and momentum operators, respectively, of a free particle. The finite translations acting on the space-time coordinates are well-defined (with no Lorentz transformation): µν µν ω 2 1 e)ω( M i  − =Λ The Lorentz transformation can be described by its generators (without translation): Here ζ i is the Lorentz boost along the i-th axis, and θk are the parameters involved in the rotation along the axial vector. The generators for these Lorentz transformations are given by Mµν have been explicitely derived previously. Here the generators for the spatial rotation Ji and Lorentz boost Ki are given by: with the corresponding covariant generators Pµ =ηµν Pν . The Lorentz transformation Λ can be described by an antisymmetric second-rank tensor ωµν which is defined by the parameters in Λ: 2014 MRT
  • and it produces the following space-time coordinate transformation: In general, the elements of the Poincaré group are given as: µ µ µν µν aP i M i aT  −− =Λ ee),( ω 2 1 in which Pµ is the momentum of the particle in the new coordinate frame. µν ν µµµ axxx +Λ=→ µν ν µµµ PPPP +Λ=′→ Similarly, the momentum of a free particle also transforms according to: By definition, a momentum contravariant four-vector is given by Pµ Pµ =(P0 /c)2 −(Pi )2 = (E/c)2 −|p|2 =mo 2 c2 is another Lorentz invariant term as well. In this context, mo is defined as the rest mass of the particle. 2014 MRT
  • For the inhomogeneous Lorentz group, the identityis thetransformation Λµ ν =δ µ ν , aµ =0, so we want to study those transformations with: µµ ν µ ν µ ν µ εδ =+=Λ aandω where ωσρ≡ηµσ ωµ ρ and ωµ ρ≡ηµσ ωσρ. Keeping only the terms of first order in ω in the Lorents condition ηµν Λµ ρ Λν σ =ηρσ , we see that this condition now reduces to the antisymmetry of ωσρ : both ωµ ν and ε µ being taken as infinitesimal. The Lorentz condition ηµν Λµ ρ Λν σ =ηρσ reads here: An antisymmetric second-rank tensor in four dimensions has (4×3)/2=6 independent components, so including the four components of em, an inhomogeneous Lorentz transformation is described by 6+4=10 parameters. )ω(ωω )ω()ω( 2 O+++= ++= ΛΛ= σρρσρσ σ ν σ ν ρ µ ρ µ µν σ ν ρ µ µνσρ η δδη ηη ρσσρ ωω −= 2014 MRT
  • Since U(1,0) carries any ray into itself, it must be proportional to the unit operator, and by a choice of phase may be made equal to it. For an infinitesimal Lorentz transformation Λµ ν =δ µ ν +ωµ ν and aµ =εµ, U(1 + ω,ε) must then equal 1 plus terms linear in ωρσ and ερ . We write this as: ...ω 2 1 1),ω1( +−+=+ ρ ρ σρ σρ εε P i M i U  σρσρρρ MMPP == †† )()( and Since ωρσ is antisymmetric, we can take its coefficients Mρσ to be antisymmetric also: Here Pρ and Mρσ are ε- and ω-independent operators, respectively, and the dots denote terms of higher order in ε and/or ω. In order for U(1 + ω,ε) to be unitary, the operators Pρ and Mρσ must be Hermitian: 2014 MRT As we shall see, P1 , P2 , and P3 are the components of the momentum operator, M23 , M31 , and M12 are the components of the angular momentum vector, and P0 is the energy operator, or Hamiltonian. ρσσρ MM −=
  • ]ω,ω)1([ ]ω)(,ω)1([ ]ω)1(,ω)1([ ]ω)1(,ω)1[(),( ])ω)(1(,ω)1[(),(),(),ω1(),(),(),ω1(),( 11 111 11 ω)1(&ω)1(;; 11 11 &;ω;1 111 11 11 aU aaaU aaU aUaU aUaUaUUaUaUUaU aaaa aaa −− −−− −− +Λ+−=Λ+=Λ=Λ=Λ −− −− Λ−=Λ=Λ=+=Λ −−− ΛΛ−ΛΛ+Λ≡ ΛΛ+Λ+ΛΛ−Λ+Λ= +Λ+Λ+Λ−Λ+Λ= +Λ+−Λ+Λ= +Λ−+Λ+Λ=Λ−Λ+Λ=Λ+Λ −− −− ε ε ε ε εεε ε ε       2014 MRT where Λµ ν and aµ are here the parameters of a new transformation, unrelated to ω and ε. Let us consider the Lorentz transformation properties of Pρ and Mρσ. We consider the product: ),(),( 1 aUaU Λ+Λ − ),(1 εωU In the end the transformation rule is given by: ],1[),(),1(),( 111 aUaUUaU −−− ΛΛ−ΛΛΛ+≡Λ+Λ ωωω εε since Λ1Λ−1 = 1. According to the composition rule T(Λ,a)T(Λ,a)= T(ΛΛ,Λa+a) with T =U(Λ,a) the product U(Λ−1 ,−Λ−1 a) U(Λ,a) equals U(1,0), so U(Λ−1 ,−Λ−1 a) = U−1 (Λ,a), i.e., U(Λ−1 ,−Λ−1 a) is the inverse of U(Λ,a). It follows from U(Λ,a)U(Λ,a)= U(ΛΛ,Λa+a) that, in sufficient detail to show these important group operations to that they be understood, we have:
  • 2014 MRT Using U(1+ω,ε)=1+½(i/)ωρσ Mρσ −(i/)ερ Pρ to first order in ω and ε we have then: µρ µρ µννµµνσ ν ρ µρσ µρ µρ µννµνµµνσ ν ρ µρσ µνσ ν ρ µρσ µν ν σ ρσ ρ µ µ ρ ρ µ µν ν σ ρσ ρ µ µ µ µν µν ρ ρ σρ σρ ε ε ε εε P i PaPaM i P i PaPaPaPa i M i PaP i M i Pa i M i aUP i M i aU Λ−−−ΛΛ+= Λ−       ++−ΛΛ+ ΛΛ+= ΛΛ−Λ− ΛΛ+= ΛΛ−Λ−ΛΛ+=Λ      −+Λ − − −−−        )(ω 2 1 1 )( 2 1 )( 2 1 ω ω 2 1 1 ])(ω[ )(ω 2 1 1 )ω()ω( 2 1 1),(ω 2 1 1),( 1 1 111 Equating coefficients of ωρσ and ερ on both sides of this equation we find: )(),(),( ),(),( 1 1 µννµµνσ ν ρ µ σρ µρ µ ρ PaPaMaUMaU PaUPaU −−ΛΛ=ΛΛ Λ=ΛΛ − − where we have exploited the antisymmetry of ωρσ , i.e., ωρσ =−ωσρ , and we have used the inverse (Λ−1 )ν σ =Λν ρ =ηνµ ηρσ Λµ ρ .
  • Next, let’s apply these rules rules to a transformation that is itself infinitesimal, i.e., Λµ ν = δ µ ν +ωµ ν and aµ =εµ, with infinitesimals ωµ ν andεµ unrelated to the previous ω and ε. 2014 MRT Equating coefficients of ωρσ and ερ on both sides of these equations, we would find these commutation rules: µρνσρνµσσνρµµσρνρσµν ρσµσρµρσµ ρµ ηηηη ηη MMMMMM i PPMP i PP +−+= −= = ],[ ],[ 0],[   This is the Lie algebra of the Poincaré group. µρ µ ρµ µ µν µν ε PPPM i ω,ω 2 1 =      −  ρσσρνρσ ν µσρ µ µνµ µ µν µν εεε PPMMMPM i +−+=      − ωω,ω 2 1  and By using U(1+ω,ε)=1+½(i/)ωρσ Mρσ −(i/)ερ Pρ and keeping only terms of first order in ωµ ν andεµ , our equations for Pρ and Mρσ become:
  • In quantum mechanics a special role is played by those operators that are conserved, i.e., that commute with the energy operator H=P0 . We just saw that [Pµ,Pρ]=0 and (i/) [Pµ,Mρσ ]=ηµρPσ –ηµσPρ shows that these are the momentum three-vectors: 2014 MRT and the angular momentum three-vector: These are not conserved, which is why we do not use the eigenvalues of K to label physical states. In a three-dimensional notation, the commutation relations may be written: where i, j, k, &c. run over the values 1, 2, and 3, and εijk is the totally antisymmetric quantity with ε123 =+1. The commutation relation [Ji ,Jj ]=iεijk Jk is the angular- ],,[ 321 PPP=P ],,[],,[ 211332123123 JJJJJJ −−−==J and the energy P0 itself. The remaining generators form what is called the ‘boost’ three- vector: ],,[],,[ 030201302010 JJJJJJ −−−==K jijikjkiji kjkijikjkijikjkiji iiii HiPKPiPJ JiKKKiKJJiJJ PiHKHHHPHJ δε εεε == −=== ==== ],[],[ ],[],[],[ ],[0],[],[],[ and and, and
  • Now,* there is one peculiar consequence to one of these commutators – the two boost generators are: 2014 MRT This commutator means that two boosts Bi and Bj in different directions (i.e., the indices i and j can’t equal each other at the same time) are not equivalent to a single boost B: where B is some boost. The reason things aren’t equal is the factor Wn× m(Ω), the Wigner Rotation where Ω is the Wigner Angle (i.e. a true space-time rotation although to be realistic, for practical reason it is usually an infinitesimal one.) BWBB )(ˆˆˆˆ Ω= mnmn × kjkiji JiKK ε−=],[ )()()()( ˆˆˆˆ 1 ˆˆˆˆˆˆ Ω=ΩΩΩ= − mnmnmnmnmn ×××× WBWBWWBB * Credit for developing this in the way it is shown here and in the next few slides and example is given pretty much as it is in Entanglement in Relativistic Quantum Mechanics, E. Yakaboylu, arxiv:1005.0846v2, August 2010. By using B=WBW−1 , the expression Bn Bm =Wn× m (Ω)B above can be re-written as:
  • For example, let us use Lorentz transformations as boost matrices along the x- and y- direction, respectively, as defined by: 2014 MRT Notice that the result is non-symmetric. Now we can write this matrice for By Bx as: ff BWBWBB )()( ˆˆˆˆˆ Ω=Ω= −zxyxy × where the ‘arbitrary’ Wigner Rotation matrix here is given by:             − − −− =             − −             − − = 1000 0 00 0 1000 0100 00 00 1000 00 0010 00 ˆˆˆˆˆˆˆˆ ˆˆˆ ˆˆˆˆˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆ yxyxyyxy xxx yyxxyxy xxx xxx yyy yyy xy γββγγβγγ γβγ βγβγγγγ γβγ βγγ γβγ βγγ BB         Ω=             ΩΩ− ΩΩ =Ω − − )(0 01 1000 0cossin0 0sincos0 0001 )( ˆ ˆ z z RW The ordered (i.e., reading from right-to-left as the are applied) product gives:             − − =             − − = 1000 00 0010 00 1000 0100 00 00 ˆˆˆ ˆˆˆ ˆ ˆˆˆ ˆˆˆ ˆ yyy yyy y xxx xxx x γβγ βγγ γβγ βγγ BB and ˆ ˆ
  • The result of the (group) matrix multiplication is (remember that M−1 =η MT η): 2014 MRT The ‘Wigner Angle’ Ω can be obtained by demanding that the Bf matrix be symmetric in, say, its M23 and M32 components: and after solving for the ratio sinΩ/cosΩ=tanΩ we get:             ΩΩ+ΩΩ−Ω− Ω−Ω−ΩΩ+Ω− −− =             − − −−             ΩΩ Ω−Ω =Ω= − − 1000 0coscossincossin 0sinsincossincos 0 1000 0 00 0 1000 0cossin0 0sincos0 0001 )( ˆˆˆˆˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆˆ ˆˆˆ ˆˆˆˆˆˆˆ ˆˆ 1 ˆ yxyxyxyxyxx yxyxyxyxyxx yyxxyxy yxyxyyxy xxx yyxxyxy xyz γββγγγβγγβγ γββγγγβγγβγ βγβγγγγ γββγγβγγ γβγ βγβγγγγ BBWBf Ω+Ω=Ω− cossinsin ˆˆˆˆˆˆ xyxyxy ββγγγγ xy xyxy ˆˆ ˆˆˆˆ tan γγ ββγγ + −=Ω or 1 cos 1 sin ˆˆ ˆˆ ˆˆ ˆˆˆˆ + + −=Ω + −=Ω xy xy xy xyxy γγ γγ γγ ββγγ and
  • By replacing sinΩ and cosΩ in the boost matrix Bf one gets: 2014 MRT Notice now that this Bf matrix is symmetric. So, as a result in this case (i.e., a boost along the x-direction followed by a boost in the y-direction), we obtain a boost along ‘some’ direction given by Ω=tan−1 [− βx γx βyγ y /(γx +γy)] in the x-y plane.                     + + + − ++ +− −− =Ω= − − 1000 0 1 )( 1 0 11 1 0 )( ˆˆ ˆˆˆ ˆˆ 2 ˆˆˆˆ ˆˆ ˆˆ 2 ˆˆˆˆ ˆˆ 2 ˆ 2 ˆ 2 ˆ ˆˆˆ ˆˆˆˆˆˆˆ ˆˆ 1 ˆ yx yxy yx yyxx yy xy yyxx yx yxx yxx yyyxxyx xyz γγ γγγ γγ γβγβ βγ γγ γβγβ γγ γβγ γβγ βγγβγγγ BBWBf mnmn ˆˆ 1 ˆˆ )( BBWB Ω= − × Note that we can read Bn Bm =Wn× m (Ω)Bf (e.g., By Bx =Wy× x (Ω)Bf =W−z (Ω)Bf in the example above) backward to note that any boost B in the n-m (e.g., the x-y plane in the example above) can be decomposed into two mutually perpendicular boosts (in order) followed by a Wigner rotation (using group algebra – we mean it’s inverse)*: ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ * Generic composition of boosts: An elementary derivation of the Wigner rotation, R. Ferraro and M. Thibeault, Eur. J. Phys. 20 (1999) 143-151. This result will be used later when we discuss particle representation in quantum field theory using Wigner basis states and especially when we calculate one first hand… ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
  • Since we will be using this soon let us look at a general example. Suppose that a parti- cle of mass mo is seen from system O with momentum p along the +z-axis. A second ob- server sees the same particle from a system O moving with velocity v along the +x-axis:             ΩΩ Ω−Ω−Ω+Ω−− Ω+ΩΩ−Ω =             ΩΩ Ω−Ω ΩΩ             − − =             ΩΩ Ω−Ω             + +             − − =ΛΛ=Λ − −−= cos0sin 0100 sincos0cossin sincos0cossin cos0sin 0100 sin0cos0 cos0sin 1000 0100 00 00 cos0sin0 0100 sin0cos0 0001 00 0100 0010 00 1000 0100 00 00 ),()()( ˆˆˆˆ ˆˆˆˆˆˆˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆˆˆˆˆˆ ˆˆˆˆ ˆˆˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ 1 ˆˆˆˆˆˆ zzzz xzzxxxzzxxzxx xxzzxxxzzxzx zzzz zzzzz xxx xxx zzz zzz xxx xxx yzxyzx γγγβ γγβγβγγβγβγγβ γβγβγγβγβγγγ γγγβ γβγβγ γγβ γβγ γγβ γβγ γγβ γβγ pWpLpL × 2014 MRT      + −=Ω⇒ + −=Ω= Ω Ω Ω=Ω+−⇒Ω=Ω−Ω− zx zzxx zx zzxx zzxxzxzxzzxx ˆˆ ˆˆˆˆ ˆˆ ˆˆˆˆ ˆˆˆˆˆˆˆˆˆˆˆˆ arctantan cos sin cossin)(sinsincos γγ γβγβ γγ γβγβ γβγβγγγγγβγβ Since Lx×z(Λp) is symmetric we can extract the [Lx×z(Λp)]3 2 =[Lx×z(Λp)]2 3 components:ˆ ˆ ˆ ˆ ˆ ˆ
  • px W−y(Λ,p) Now, since: This provides us with the three cartesian values for the boosted momentum: 2014 MRT y mo Suppose that observer O sees a particle (mass mo ≠0) with momentum pz in the z- direction. A second observer O moves relative to the first with velocity v in the x-direction. How does O describe the particle’s motion? Λp )ˆˆ(ˆˆ ˆˆˆˆˆoˆˆ kikip zzzxxzx γβγγβ +−=+−=Λ cmpp pz Λ(v) y x v ↑             −− − =                         − − =Λ zzz zzxxxzxx zzxxxzx zzz zzz xxx xxx zx ˆˆˆ ˆˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆ 00 0100 0 0 00 0100 0010 00 1000 0100 00 00 )( γγβ γβγβγγγβ γβγγβγγ γγβ γβγ γγβ γβγ pL             − =                         −− − =Λ cm cm cmcm kpL oˆˆ oˆˆˆ oˆˆo ˆˆˆ ˆˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆ ˆˆ 0 0 0 0 00 0100 0 0 )( zz zxx zx zzz zzxxxzxx zzxxxzx zx γβ γγβ γγ γγβ γβγβγγγβ γβγγβγγ µ we then have (with kµ =[moc,0,0,0]T a standard rest momentum for massive particles):          − =           Λ Λ Λ zz zxx ˆˆo ˆˆˆo 3 2 1 0 γβ γγβ cm cm p p p )ˆˆ(ˆˆ ˆˆˆˆˆooˆˆoˆˆˆ kikip zzzxxzzzxx γβγγβγβγγβ +−=+−=Λ cmcmcm x z z Lz(p) which when applied in the Figure will look like:
  • The Poincaré Algebra and is a space-time translation, i.e., as the product operation of a translation by a real vector aµ =(τ,a) and a homogeneous Lorentz transformation, Λµ ν (the translation being performed after the homogeneous Lorentz transformation.) It can conveniently be represented by the following matrix equation: The inhomogeneous Lorentz transformation (or Poincaré group), L={Λ,a}, is defined by: 2014 MRT                                 ΛΛΛΛ ΛΛΛΛ ΛΛΛΛ ΛΛΛΛ =                 1100001 3 2 1 0 33 3 3 2 3 1 3 0 22 3 2 2 2 1 2 0 11 3 1 2 1 1 1 0 00 3 0 2 0 1 0 0 3 2 1 0 x x x x a a a a x x x x The commutation rules of these generators with themselves are the Poincaré Algebra: µν ν µµµ axx +Λ== )( xL ρµσνρνσµµσµρµσνρρσµνρµ ηηηη MMMMMMiPP +−−== ],[0],[ and ρµσσµρσρµ ηη PPMPi −=],[ where the last coordinate, i.e., 1, has no physical significance and is left invariant by the transformation. The generators for infinitesimal translations are the Hermitian operators Pµ , and their commutative relations with the Hermitian generators for ‘rotations’ in the xµ -xν plane, Mµν =−Mνµ are as expressed in contravariant form:
  • Lorentz Transformations We note from (DetΛ)2 =1 above that either DetΛ=+1 or DetΛ=−1; those transformations with DetΛ=+1 form a subgroup of either the homogeneous or the inhomogeneous Lorentz group. Furthermore, from the 00-components of ηµν Λµ ρ Λν σ =ηρσ and Λν σ Λκ τ ηστ =ηνκ , we have: These transformations also form a group. Now, those transformations with aµ =0 form a subgroup with to the Poincaré group: with i summed over the values 1, 2, and 3. We see that either Λ0 0 ≥+1 or Λ0 0 ≤−1. Those transformations with Λ0 0 ≥+1 form a subgroup. Note that if Λµ ν and Λµ ν are two such Λs, then: ii ii 00 00 2 0 0 11)( ΛΛ+=ΛΛ+=Λ 0 3 3 0 0 2 2 0 0 1 1 0 0 0 0 0 0 0 0 0 )( ΛΛ+ΛΛ+ΛΛ+ΛΛ=ΛΛ≡ΛΛ µ µ )0,()0,()0,( ΛΛ=ΛΛ TTT 2014 MRT Taking the determinant of ηµν Λµ ρ Λν σ =ηρσ gives: so Λµ ν has an inverse, (Λ−1 )ν σ , which we see from ηµν Λµ ρ Λν σ =ηρσ and takes the form: σ µσρ µν ρ νν ρ ηη Λ=Λ=Λ− )( 1 These transformations also form a group. known as the homogeneous Lorentz group. If we first perform a Lorentz transformation Λ: ν ν ρ ρ µρ ρ µµ xxx ΛΛ=Λ= 1)Det( 2 =Λ
  • The problem of classifying all the irreducible unitary representations of the inhomo- geneous Lorentz group (i.e., the Poincaré group) can again be formulated in terms of finding all the representations of the commutation rules above by self-adjoint operators. commute with all the infinitesimal generators, Mµν and Pµ , and constitute the invariants of the group. 2014 MRT ν µνσ µσ σ σ µν µνµ µ µ µ PPMMPPMMwwWPPP −=== 2 122 and where ρσν µνρσ µ ε MPw 2 1 = They are therefore multiples of the identity for every irreducible representation of the inhomogeneous Lorentz group and their eigenvalues can be used to classify the irreducible representations. The set of all four-dimensional translations is a commutative subgroup of the inhomogeneous Lorentz group. Since it is commutative, the irreducible unitary representations of this subgroup are all one-dimensional and are obtained by exponentiation. Believe it or not, the following scalar operators:
  • and so: The subgroup of Lorentz transformations with DetΛ=+1 and Λ0 0 ≥+1 is known as the proper orthochronous Lorentz group. 1)(1)( 2 0 02 0 0 0 3 3 0 0 2 2 0 0 1 1 0 0 0 −Λ−Λ≤ΛΛ+ΛΛ+ΛΛ≡ΛΛ i i But (Λ0 0)2 =1+Λi 0Λi 0 =1+Λ0 iΛ0 i shows that the three-vector [Λ1 0,Λ2 0,Λ3 0] has length √[(Λ0 0)2 −1], and similarly the three-vector [Λ0 1,Λ0 2,Λ0 3] has length √[(Λ0 0)2 −1], so the scalar product of these two three-vector is bounded by: 1)(1)()( 2 0 02 0 0 0 0 0 0 0 0 −Λ−Λ−ΛΛ≥ΛΛ 2014 MRT
  • If we look over the Lorentz transformation properties of Pρ and Mρσ is the case for homogeneous Lorentz transformations (i.e., aµ =0), we get: 2014 MRT µνσ ν ρ µ σρ µρ µ ρ MaUMaU PaUPaU ΛΛ=ΛΛ Λ=ΛΛ − − ),(),( ),(),( 1 1 These transformation rules simply say that Mρσ is a tensor and Pρ is a vector. For pure translations (i.e., with Λµ ν =δ µ ν ), they tell us that Pρ is translation-invariant, but Mρσ is not.
  • The group of Lorentz transformations contains a subgroup which is isomorphic to the familiar three-dimensional rotation group. This subgroup consists of all Λµ ν of the form: 2014 MRT             = RR 0 01 Λ where ΛR(R1) and ΛR(R2) are spatial rotations and Λ(L1) a Lorentz transformation in the x1 - direction. If we set µ =ν =0 in the equation Λλµ ηλρ Λρν =ηµν, we then obtain: so that Λ0 0 ≥1 or Λ0 0 ≤−1. where R is a 3×3 matrix with RRT =RT R=1. We call such a ΛR a ‘spatial’ rotation. Every homogeneous Lorentz transformation can be decomposed as follows: 1)()()(1)(1)( 2 0 32 0 22 0 1 3 1 2 0 2 0 0 ≥Λ+Λ+Λ+=Λ+=Λ ∑=i i )(Λ)(Λ)(ΛΛ 112 RLR RR= Remember that a Lorentz transformation for which Λ0 0 ≥1 is called an orthochronous Lorentz transformation. A Lorentz transformation is orthochronous if and only if it transforms every positive time-like vector into a positive time-like vector. The set of all orthochronous Lorentz transformations forms a group: the orthochronous Lorentz group.
  • The problem of finding the representation of the ‘restricted’ Lorentz group is equivalent to finding all the representations of the commutation rules above. The finite dimensional irreducible representation of the restricted group can be labeled by two discrete indices which can take on a values the positive integers, the positive half-integers, and zero. To show this, let us define the operators: 2014 MRT and their commutation rules are: From these operators, we can construct these invariants of the group: ( ) ( ) ( )Boost momentumAngular Momentum ],,[ ],,[ ],,[ 030201 211323 321 MMM MMM PPP = = = K J P kijkji kijkji kijkji KKJ JKK JJJ ε ε ε = −= = ],[ ],[ ],[ which commute with all the Ji and Ki. They are therefore the invariants of the group and they are multiples of the identity inanyirreduciblerepresentation.The representations can thus be labeled by the values of these operators in the given representation. and µν µν MM2 122 =− KJ ρσµν µνρσε MM8 1=•− KJ
  • where E(p)=ωp =√(p2 c2 +mo 2 c4 ) is the one-particle energy. As seen earlier, a rotation Rθ by an angleθ =|θ | about the direction of θ =θp is representedontheHilbert space by: 2014 MRT ),(e),(e),(),1( j aP i j aP i j mpmpmpaU ΨΨΨ µ µ µ µ  −− == hence ),(e),()0,( j i j mpmpU ΨΨ θJ θ •− =  A boost K in the direction of the momentum ζ: The operator corresponding to the translation by the four-vector aµ is given by: ),(e),()0,( j ci j mpmpU ΨΨ Kζ ζ •− =          =         = −− cmcm i i o 1 o 1 sinˆsin p p p p p ζ and similarly for evolution: ),(e),(e ω j t i j tH i mpmp p ΨΨ  = In an irreducible representation, the operation of translation by a thus corresponds to multiplying each basis vector |Ψ(p,mj)〉 by exp(−iPµ aµ /): µ µ aP i aU  − = e),1( ˆ
  • which satisfy the following commutation rules: To make the range of values of the label more transparent, let us introduce the following generators: 2014 MRT )()( 2 1 2 1 iiiiii KiJiKKiJiJ −=+= and and It follows from the commutation rules that a finite dimensional irreducible representation space, Vjj' can be spanned by a set of (2j +1)(2j' +1) basis vectors | jmj, j'm'j 〉 where j, mj, j' and m'j are integers or half-odd integers, −j ≤ mj ≤ j, −j' ≤ m'j ≤ j' and in terms of which the J and K operators have the following representation: 0],[ ],[ ],[ = = = ji kijkji kijkji KJ JiKK JiJJ ε ε jjjjj jjjjjjjj mjmjmmjmjJ mjmjmjmjmjmjJiJmjmjJ ′′=′′ ′′±+±=′′±=′′± ,;,,;, ,;1,)1)((,;,)(,;, 3 21   jjjjj jjjjjjjj mjmjmmjmjK mjmjmjmjmjmjKiKmjmjK ′′′=′′ ±′′+′±′′′=′′±=′′± ,;,,;, 1,;,)1)((,;,)(,;, 3 21  
  • These are thus a denumerable infinity of non equivalent finite-dimensional (in general they are non-unitary) irreducible representations. These can be labeled by two non-ne- gative indices ( j, j') where j, j' = 0, 1/2, 1, 3/2,…. The dimension of the representation is (2j +1)(2j' +1) and D( j, j') is single-valued if j+j' is an integer and double-valued. A quantity which transforms under D(0,0) is called a scalar, one which transforms under D(1/2,1/2) a four-component vector, one which transforms under the (1/2,0) representation a two-component spinor. A quantity which transforms under (0,1/2) is called a conjugate spinor. For the D(0,1/2) and the D(1/2,0) representations, an explicit matrix representation of the infinitesimal generators can be given in terms of the Pauli matrices with: 2014 MRT iiii iiii KK iJiJ σσ σσ 2 1)2/1,0( 2 1)0,2/1( 2 1)2/1,0( 2 1)0,2/1( +=−= −=−= Note also that the quantity ξ∗ ξ is not a scalar. A two-component spinor, ξ, transforms under spatial rotation as in the three- dimensional situation. For example, under an infinitesimal rotation ε about the i-th axis: and under an infinitesimal Lorentz transformation in the xi -direction, this spinor transforms according to: ( )32 1 )1( Ri i ξξξ σε+=→ ( )42 1 Λ)1( ξξξ εσi+=→
  • ∑∫∑∫ == + = jj m jj j m jj P pd mpψmp P pd xψmpmpψ 0 * 12 1 0 ),(),()(),(),( φφφ ΨΨ Lorentz Invariant Scalar For the last case, Pµ = 0, the complete system of (infinite dimensional) unitary representations coincides with the complete system of (infinite dimensional) unitary representations of the homogeneous Lorentz group which we studied earlier. The irreducible representations of the inhomogeneous Lorentz group can now be classi- fied according to whether Pµ is space-like, time-like, or null vector, or Pµ is equal to zero. The representations of principle interest for physical applications are those for which P2 =mo 2 c2 =positive constant, and those for which P2 =0. 2014 MRT Defining a Lorentz invariant scalar product within the vector space by integrating over a set of p (with P2 =mo 2 c2 , P0 =+√(p2 + mo 2 c2 )=E/c) and summing over the index mj: where j =0,1/2,1,3/2,2,…. So, we have hinted that an irreducible representation of the type P2 >0, P0 >0 is labeled by two indices (mo, j), where mo is a positive number and j is an integer or half-integer. The index mo characterizes the mass of the elementary system, the index j the angular momentum in its rest frame, i.e., the spin of the elementary system. The fact that the irreducible representation is infinite dimensional is just the expression of the fact that each elementary system is capable of assuming Let us first discuss the case P2 =mo 2 c2 . In that case, P0 /|P0 |, the sign of the energy, commutes with all the infinitesimal generators and is therefore an invariant of the group.
  • Klein-Gordon & Dirac For j = 0, the representation space is spanned by the positive energy solutions of the relativistically covariant equation for a spin-0 particle – the Klein-Gordon equation: For each (mo , j) – and a given sign of the energy – there is one and only one irreducible representation of the inhomogeneous Lorentz group to within unitary equivalence. For j half-integral, the representation is double-valued. 2014 MRT For j = 1/2 by the positive energy solution of the Dirac equation (free particle case): For j = 1 by the positive energy solutions of the Proca equation (not discussed). ),()( ),( 42 o 222 2 2 2 tcmc t t r r ϕ ϕ +∇−= ∂ ∂ −  ),()( ),( 2 o tcmci t t i rα r ψβ ψ +•−= ∂ ∂ ∇ We will discuss both the Klein-Gordon and Dirac equations later. )( 2 1 4 1)( 42 o µν µ ν µ µ ν ν µ µ ν µ φφφ φφφφφ xcm xxxxt x i         +         ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂ −= ∂ ∂  0)( 2 o =+∂− ψγ µ µ xcmc )(),()()(),( tdmpxxψt j ΨΨΨ ⋅==≡ ∫ ∞+ ∞− ppprr ψµ or such that in what follows,it applies also to states that get acted on by the Dirac equation:
  • One-Particle States The physical states of particles are described by the Wigner basis states |Ψkmo ( j,mj)〉 (which are equivalent to the states|Ψ(k,mj)〉) for a unitary irreducible representation of the inhomogeneous Lorentz group (Poincaré group) with pµ pµ =p02 –p2 =(moc)2 . Thesestatesformthe Hilbert space of the theory and the momentum states |Ψ(p,mj)〉 can be obtained from the standard state |Ψpmo ( j,mj)〉≡|Ψ( p,mj)〉 by a unitary transformation: ∑ + −=′ Λ′ −•− Λ Λ =≡ j jm jEp j mm tE i j jj mjpW pE pE xx ),()],([e )( )( )π2( 1 )()( )( )( 2/3 ΨD rp   ψψ µ Our goal is to find eigenkets of |Ψpmo ( j,mj)〉 as they appear following an homogeneous Lorentz transformation group U(Λ,a) product on a state-vector |Ψpmo ( j,mj)〉 is as follows: and L( p) is some standard Lorentz transformation matrix that depends on p=pµ and the momentum states are normalized over intermediate states andweget the coordinate ket: ),()(),( jj mkpLmp ΨΨ = ),()]([)]([)]([)]([ )( ),( ),()]([)]([ )( ),(),(),( o oo 1o o jmk jmkjmp mjpUpLUUpLU pE m aU mjpUU pE m aUmjaU Ψ ΨΨ LvΛ1 LvΛ1Λ ΛΛ= = − 2014 MRT The spin j corresponds to the eigenvalues J 2 = j( j +1)2 of J2 and J3 =mj  (mj = j, j –1,…, −j ). Also, U(1,a)|Ψkmo ( j,mj)〉 meansthesamethingas exp(−ikµ aµ )|Ψkmo ( j,mj)〉(pµ =kµ ).
  • The Lorentz transformations associated with the Poincaré group can be constructed in which the quantum state differs by only a mixture of the internal spin indices mj (i.e., run- ning over the discrete values j, j –1,…,−j) but possess the same physical observables. The effects of the inhomogeneous Lorentz transformations (e.g., acting on the Dirac fields) can be elucidated by considering the particle states obtained from the irreducible unitary representations of the Poincaré group. The unitary operation representing the Lorentz transformations acting on the Poincaré generators are given by the equations U(Λ,a)Pρ U−1 (Λ,a)=[Λ−1 ]ρ µ Pµ and U(Λ,a)MρσU−1 (Λ,a)=[Λ−1 ]ρ µ [Λ−1 ]σ ν (Mµν −aµ Pν −aν Pµ ). and the state transforms accordingly for a space-time translation: ),(),( jj mppmpP ΨΨ µµ = ),(e),(e),(),( j ap i j aP i j mpmpmpaU ΨΨΨ µ µ µ µ  −− =≡1 Thus, the one-particle state is an eigenvector of the momentum operator: Since the momentum four-vector commutes among each other according to [Pµ ,Pν ]= 0, the particle states can be characterized by the four-momentum Pµ together with additional internal degrees of freedom mj. The internal degrees of freedom pertain to the spin vector which can be affected by transformations in the space-time coordinates. 2014 MRT
  • Under space-time translations, the states |Ψ (p,mj)〉 transforms as: 2014 MRT It is thus natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible. Hence U(Λ)|Ψ (p,mj)〉 must be a linear combination of the state vectors |Ψ (Λp,mj)〉: We must now consider how these states transform under homogeneous Lorentz transformations U(Λ,0)≡U(Λ) is to produce eigenvectors of the four-momentum with eigenvalues Λp (where we use U−1 (Λ,a)Pµ U(Λ,a)=[Λ−1 ]µ ν Pν ): ),()()(),()( ),()]([)(),()]()([)(),()( 11 jj jjj mpUpmpUp mpPUmpUPUUmpUP ΨΨ ΨΨΨ ΛΛ ΛΛΛΛΛ µν ν µ ν ν µµµ Λ=Λ= Λ== −− ),(),( jj mppmpP ΨΨ µµ = The components of the energy-momentum four-vector operator Pµ all commute with each other (i.e., [Pµ,Pν]=0), so it is natural to express physical state-vectors in terms of eigenvectors of the four-momentum. This four-momentum Pµ is a trusted observable! ),(e),(),( j api j mpmpaU ΨΨ µ µ − =1 ∑′ ′ ′ΛΛ= j jj m jmmj mppCmpU ),(),(),()( ΨΨΛ We now introduce a label mj to denote all other degrees of freedom (i.e., all other total angular momentum orientations), and thus consider state-vectors |Ψ (p,mj)〉 with:
  • In other words, if a system is confronted with a homogeneous Lorentz transformation Λ, the momentum p is changed to Λp. According to Pµ |Ψ(p,mj)〉=pµ |Ψ(p,mj)〉, the one- particle state must possess an eigenvalue of Λp as well: in which U(Λ,a)Pµ U−1 (Λ,a)=[Λ−1 ]ν µ Pν has been used for the Lorentz transformed momentum generator. It can be seen then that U(Λ)|Ψ(p,mj)〉 is a linear combination of the states |Ψ(Λp,m′j)〉, where: and this, believe it or not, does leave the momenta of all the particle states invariant. ∑ + −=′ ′ ′ΛΛ=Λ j jm j j mmj j jj mppmpU ),(),(),()( )( ΨΨ D ),()(),(),( 2 o jjj mpcmmpppmpPP ΨΨΨ == µ µ µ µ ),()()( ),(])[(),()]()()[(),()0,( 1 j jjj mpUp mpPUmpUPUUmpUP Ψ ΨΨΨ ΛΛ= ΛΛ=ΛΛΛ=Λ − µ ν ν µµµ in which Dm′jmj ( j) (Λ,p) is termed the Wigner coefficient and, as said previously, they depend on the irreducible representations of the Poincaré group. As a special example, the Casimir operator Pµ Pµ cannot change the value of the momentum (i.e., its an invariant – as also stated previously): 2014 MRT
  • Hence, to distinguish each state, the standard four-momentum given by kµ =[moc,0,0,0] is chosen, from which all momenta can be achieved by means of a pure Lorentz boost: which implies that there exists a subgroup of elements consisting of some arbitrary Wigner rotations, W, and this subgroup is called the little group. µ ν ν ν µ kkW =∑ Notice that a simple three-dimensional rotation, W (which is an element of the Poincaré group), will render the standard four-momentum invariant: It is important to distinguish that this little group is not unique in the Lorentz group but it is actually isomorphic to other bubgroups under a similarity transformation. This is because there is no well-defined frame for the standard momentum kµ due to the equivalence principle in special relativity.Thedefinitionofthelittle group is dependent on the choice of the standard momentum as well as the Lorentz transformation Λ. Wigner’s Little Group (N.B., the standard four-momentum is non-unique and it also depends on the charac- teristics of the particle, e.g., whether it is a massive or a massless particle). kpLp )(= 2014 MRT ),(),(),(),(),( 2 o jjjjjj mkmmkmkmkcmmkH ΨΨΨΨΨ === J0p and& In this momentum rest frame, the state |Ψ(k,mj)〉 is specified in terms of the eigen- values of the Hamiltonian H=p0 , the momentum p, and the z-component of the total angular momentum operator J as:
  • When L(p) is applied to k we get the momentum p. When Λ is applied next to p it becomes Λp. When L−1 (Λp) is applied to Λp we recover k! So, the only functions of pµ that are left invariant by all proper orthochronous Lorentz transformations Λµ ν are the invariant square p2 =Σµν ηµν pµ pν, and p2 ≤ 0, also the sign of p 0 . Hence, for each value of p2 , and (e.g., for p2 ≤0) each sign of p0 , we can choose a ‘standard’ four-momentum, kµ , and express any pµ of this class as: where Lµ ν is some standard Lorentz transformation that dependsonpµ ,andalso implicitly on our choice of the standard kµ . We can define the states |Ψ (p,mj)〉 of momentum p by: ∑= ν ν ν µµ kpLp )( Wigner’s little group is the Lorentz transformation L−1 (Λp)Λ L(p) that takes k to L(p)k= p, and then to Λp, and then back to k, so it belongs to the subgroup of the homogeneous Lorentz group. Operating on this equation with an arbitrary homogeneous Lorentz transformation U(Λ), we find: k µ =[moc,0,0,0] pµ =Σν Lµ ν ( p)kν Λ p=Σν Λ0 ν pν L−1 (Λ p) L(p) Λ ),()]([)(),( jj mkpUpNmp ΨΨ L= N(p) is a numerical normalization factor to be chosen on the next slideandwhere U[L(p)] is a unitary operator associated with the pure Lorentz ‘boost’ thattakes[moc,0] into[p0 ,p]. The transformation Wµ ν (i.e., L−1 (Λp) Λ L(p) = W(Λ,p)) thus leaves kµ invariant: Σν Wµ ν kν = kµ . For any Wµ ν satisfying this relationship, we have: ),()]()([)]([)( ),()]([)(),()( 1 1 j jj mkppUpUpN mkpUpNmpU Ψ ΨΨ LΛLL LΛΛ    = − ΛΛ= = ∑ + −=′ ′ ′Λ=Λ j jm j j mmj j jj mkpmkpU ),()],([),()],([ )( ΨΨ WW D where the coefficients Dm′j mj ( j) [W(Λ,p)] furnish the 2014 MRT
  • The Wigner coefficient for the standard four-momentum can be evaluated as: which implies directly that: )()()( WWUWUWU ′≡′ Note that the Wigner coefficients form a representation of the little group, which means that for any little group elements W, W′, the group multiplication property holds: ∑∑ ′ ′ + −=′ ′ ′≡′= j jj j jj m jmm j jm j j mmj mkWmkkWmpWU ),()(),(),(),()( )( ΨΨΨ DD thus: ∑ ∑ ′ ′′ ′ ′ ′′= ′≡′′=′ σ σσ σ σσ σ σσ , ),()()( ),()()(),()(),()( j jj m mm j kWW kWUWUkWWmkWWU Ψ ΨΨΨ DD D ∑ ′=′ ′′ j jj m mm WWWW )()()( σσσσ DDD However, this choice for the normalization condition leads to problems with the subsequent transformation equations relating to the particle states which involve some tedious momentum-dependent constants. 2014 MRT
  • Normalization Factor Using W=L−1 Λ L and inserting into U(Λ)|Ψ(p,mj)〉=NU[L(Λp)]U[L−1 (Λp)Λ L(p)]|Ψ(p,ms) 〉 and using U(W)|Ψ(k,mj)〉=Σm′j Dm′j mj (W)|Ψ(k,m′j)〉 we get: or, recalling the definition |Ψ 〉=NU[L(p)]|Ψ 〉, we finally get: 2014 MRT for which The normalization factor N(p) is sometimes chosen to be N(p)=1 but then we would need to keep track of the p0 /k0 factor in scalar products. Instead, the convention is that: The normalization condition is achieved using the scalar product: ∑′ ′ ′ΛΛ Λ = j jj m jmmj mpp pN pN mpU ),()],([ )( )( ),()( ΨΨ WΛ D ∑′ ′ ′ΛΛ= j jj m jmmj mkpUppNmpU ),()]([)],([)(),()( ΨΨ LWΛ D 0 0 )( p k pN = jj mmjj mpmp ′−′=′′ δδ )(),(),( ppΨΨ )()(),(),( 2 kk −′=′′ ′ δδ jj mmjj pNmpmp ΨΨ )( )()( )( )( )( )( 0 0 0 0 pE pE p p pN pN p k pN Λ = Λ = Λ ⇒ Λ =Λ and
  • We choose the following normalization condition (c.f. Weinberg QTF I): which implies that: This implies that the Wigner coefficients cannot sum up to unity, but instead up to a phase factor that depends on the momentum of the particle, p, and the Lorentz transformation,Λ: jj mmjj pmkmp ′′−=′ δδ )()2(),(),( )3(0 ppΨΨ σσσσ δ ′′ Λ =ΛΛ∑ 0 0 * )( ),(),( p p pp j jj m mm DD jj jj jj jj jj mm mm mm mm jjmm pppppp mpmppppUUp ′ ′ ′′′ ′ ′′ ∑ ∑ ΛΛΛ==′ ′ΛΛΛΛ=′ΛΛ δδσσ σσ σσσσ σσ , 0*0 , *† )2(),(),(2),(),( ),(),(),(),(),()()(),( DD DD ΨΨ ΨΨΨΨ and finally: 220 0 † )( ),(),( × Λ =ΛΛ I p p pp DD The transformations mentionned above whereby the representations of the Poincaré group is being derived from the little group is termed as the method of induced representations. 2014 MRT
  • The explicit form for W(Λ,p) is: where the angle θ is defined by tanhθ =√[(p0 )2 –mo 2 c4 ]/p0 and the components of the particle momentum are given by pi (i=1,2,3). The Lorentz boost L(p) transforms the standard momentum kµ to momentum pµ . In addition, the general Lorentz transformation ΛL for ‘pure boost’ only is written as: ( )ppkpLpLpW pL Λ→ →ΛΛ=Λ Λ− )(1 )(),(),(         −+ = ji ji i j ppp p pL ˆˆ)1(coshsinhˆ sinhˆcosh )( θδθ θθ         −+ =Λ ji ji i j L nnn n ˆˆ)1(coshsinhˆ sinhˆcosh ξδξ ξξ and it can be shown that this is an element of the little group associated with the standard vector kµ . (Note that L(Λp) is the Lorentz boost of the momentum (Λp)µ from standard momentum vector kµ ). Then the Wigner transformation corresponding to the present choice of kµ is a SO(3) rotation. Furthermore, the Lorentz boost L(p) can be computed from the boost generator Ki =Mi0=i(xi ∂0 −x0∂i) where L(p)=exp(−icθ p•K/). The explicit form of L(p) is given by: ˆ where n is a unit vector along the direction of the boost. This transforms the momentum pµ to a Lorentz-transformed momentum (Λp)µ . ˆ 2014 MRT
  • Considering now a Taylor expansion of the Wigner transformation W(Λ,p)= L−1 (Λp)ΛL(p) as well as the explicit form of the matrix for L(p) just obtained for an arbitrary Lorentz transformation Λ: in which the Lorentz boosts, L(p) and L(Λ,p) and the arbitrary Lorentz transformation, Λ, are parametrized in terms of its generators. In addition, the spatial component of the Lorentz-transformed momentum is given by pΛ=Λp (Note that the corresponding Lorentz boost for the Lorentz transformed momentum starting from the standard vector, kµ , is given by tanhθ′=|pΛ|/(Λp)0 , in which L(Λp) k=Λp). +−=Λ µν µν Mω 2 1 1)ω( thus: KpKp •−−•′ − ≡ ΛΛ=Λ ˆω 2 1 ˆ 1 eee )()(),( Λ θθ µν µν  ci M ci pLpLpW βα β αβα βµν µναα pppM i pp ]ω[]ω[ 2 )( +≡−≅Λ Consider the inifinitesimal Lorentz transformation being parametrized by the anti- symmetric tensor ωµν (expressed to first order terms of ω): 2014 MRT
  • A Taylor series expansion of the Wigner transformation is considered as fallows: thus: In the ‘second term’ of this last equation the expression can be re-written as:   + ∂ ΛΛ∂ += + ∂ Λ∂ +Λ=Λ = − = = 0ω 1 0ω 0ω ω )]()([ ω1 ω ),( ω),(),( µν µν µν µν pLpL pW pWpW   +            −Λ+         ∂ Λ∂ += +         ∂ Λ∂ Λ+         Λ ∂ Λ∂ +≅Λ = − = − = − = − )( 2 )(ω)( ω )( ω1 )( ω )ω( )(ω)()ω( ω )( ω1),( 0ω 1 0ω 1 0ω 1 0ω 1 pLM i pLpL pL pLpLpL pL pW µν µν µν µν µν µν µν µν in which the substitution Λp=p′ has been made and, in general, L−1 (p′)=η LT (p′)η.                       −+               ′ ′′         − ′ + ′ − ′ − ′ ∂ ∂ +=      ∂ Λ∂ = = − 22 o 0 o o 2 o 0 0ω 22 o 0 o o 2 o 0 0ω 1 11 ω ω1)( ω )( ω p pp cm p cm p cm p cm p p pp cm p cm p cm p cm p pL pL kj kj j k ji ji j i δδ µν µν µν µν 2014 MRT
  • Now, since (Λp)α ≅pα −(i/2)[ωµν Mµν]α β pβ ≡pα +[ω]α β pβ, we get (=c=1): and since ξ i =ω0i =−ωi0 and θ k =εijk ωij as well as Ji =εijk Mjk and Ki =Mi0: βα β βα βµν µν µν α µν ppM ip ]ω[][ω 2ω )( ω 0ω ≡−≅         ∂ Λ∂ = in which the axial vector is given by θ =[θ 1 ,θ 2 ,θ 3 ]† and mi =(p× θ)i as well as:         − • ≡         − • =         − =                 − = +−=++−= ii ii k jkjii jj j k kjii j k ki i ji ji i i i i p p pp p p p pJK i pMMM i p mp ξpp θpp ξpp ξ ξ θεξ ξ θεξ ξ θξ βββ β α 0 0 0 0 0 0 0 0 )ˆ( )ˆ( )ˆ( 0 ]22[ 2 ]ωωω[ 2 ]ω[ ×         −−−=      ∂ ′∂ ∑ = nm nnmiii i pppp p ][ˆˆ][ 1 ω )ˆ( ω 00 0ω mpmp p ξξµν µν At this junction, the sum Σmn pm [p0 ξ n −|p|mn ] describes the relation between the components of the momentum and its Lorentz boost. ˆ 2014 MRT ˆ
  • Assuming that any sum over the cross-terms is zero for m≠n, that is: )ˆ(][ˆ 00 ξpmp •=−∑ ppp nm nnm ξ in which the boost vector is given by ξ =[ξ 1 ,ξ 2 ,ξ 3 ]† . With the axial vector, θ =[θ 1 ,θ 2 ,θ 3 ]† , θ and ξ become the parameter vectors associated with the infinitesimal Lorentz transformation Λ. In addition, the vector normal to both the axial and momentum vectors is given by m=p× θ. Jmpξp p Km p pξpξ pmξp p m p ξp m p ξp ξp p m p m p ξp p •         −        −+•         −•        −−≡                       −         −−+•         − −+•         − =                       −+                       •         − −−        −+− +−• =         ∂ Λ∂ = − )×× ×× ˆ(2)ˆ(1ˆ)ˆ(1 )ˆ(2)ˆ(1ˆ)ˆ(1 ˆ)ˆ(10 ˆˆ1ˆˆ)ˆ(]ω[ˆ]ω[ˆ1 )ˆ( )( ω )( ω 0 o 0 oo 0 o 0 0 o 0 o 0 oo 0 o 0 oo 0 o 0 o oo 0 o 0 o 0 oo 0 oo 0 o 0ω 1 p m p i mm p m p i p m p m p m p m p m p m p m p pp m p m p m p m p pp mp pppp m p mm p mm p m pL pL nkinniii kkk kj kj j k jiijjiii jj εξ ξ δξ ξ β β β β µν µν ˆ Our equation for ωµν [∂L−1 (Λp)/∂ωµν ]|ω=0L(p) above (i.e. the ‘second term’) can be simplified by using the Lorentz boosts from the matrices for L(p) and ΛL further up, and this can be worked out as: 2014 MRT
  • The third term in our Taylor expansion for W(Λ,p) above (i.e. the term ωµν [L−1 (Λp)(−½i Mµν )]|ω=0L(p)) can also be simplified in terms of its matrix elements (Note that the expression (−½i ωµν Mµν ) can be re-written as the matrix [ω]): Jmpθξp p Km p pξpξ mp p ξm p ξp m p ξp m p ξp m p ξp •                 −+−−+•         −•        −−−≡                                 −+−        −•         − − −+•         − =               −         − +−         −−•         − − −•         − − =                       −+        −               ′′         −+ ′ − ′ − ′ =′− )×× )×× ˆ(1)ˆ( 2 ˆ)ˆ(1 ˆ(1ˆ)ˆ( ˆ)ˆ(10 )ˆˆˆˆ(ˆ)ˆ( )(ˆ)ˆ(0 ˆˆ1 0 ˆˆ1 )](][ω)][([ o 00 oo 0 o 0 o 0 ooo o 0 o 0 o 0 oo 0 o o 0 oooo o 0 o 0 oo o 0 o 0 o 0 o oo 0 o 0 o oo 0 1 m pp i mm p m p i m p mm p m mp m p m p m p m p pmmp m mp m p m p m p m mp m p m p m mp m p pp m p m p m p m p pp m p m p m p m p pLpL mlimm m iii kkk lilimmlil il iiii lll lk lk k l mmkjj k ji ji i j εθξ ξ θεξξξ ξ δ θεξ ξ δ in which ωµν [∂( p′)α /∂ωµν ]|ω=0 above has been used for the matrix reprentation of the arbitrary infinitesimal Lorentz transformation Λ(ω). 2014 MRT
  • After some careful re-grouping of the terms from all the matrix elements given in ωµν [∂L−1 (Λp)/∂ωµν]|ω=0L(p) and [L−1 (p′)][ω][L(p)] just obtained, the boost terms in our expansion for W(Λ,p) has cancelled out completely and leaving behind the rotation terms: where Ω( p)k =−ε ijk [ωij −(pi ω0j −pj ω0i )/( p0 +mo)] is the Wigner angle and Jk =½ε ijk Mij is the rotation generator for the Poincaré group. Thus, we obtain: JΩ Jθξp p •+=         − − −+=         −− − −= •         + − −≡Λ )(1 )ωω( )( 1 ω1 )( )( 1 1 )( )( 1),( 00 o 0 o 0 o 0 pi Mpp mp i Mpp mp i mp ipW ji ijjiji jin njiijji n θεξξ ×                   − − −+=Λ njinnmppW εθ)ˆ( )( 0 00 1)],([ o 0 ξp p × 2014 MRT
  • The Wigner angle can be re-written as a sum of the contributions from the rotation and the boost: Here the angle of rotation are represented by the Euler angles θk =ε ijk ωij and the boost parameter ξi =ξ ni =ω0i. The finite Wigner transformation is given by: )ˆˆ( )( o 0 o 0 pn p θ ξp θΩ × × mp mp p + −≡ + −≡ ξ JΩ• ∞→ =                     Λ=Λ i N N p N WpW e , ω lim]),ω([ ˆ 2014 MRT
  • The representation matrix Dmsσ ( j) [W(Λ,p)] can be constructed from the angular momentum generators Ji explicitly, depending on the angular momentum of the particle. For example, spin-½ particles will be considered to have appropriate generators as given by J=½σ according to the isomorphism between the proper Lorentz group and the SU(2)⊗SU(2) algebra. JΩ• =Λ i pW e),( The Wigner transformation corresponding to an arbitrary Lorentz transformation Λ is given by: where Ω( p)k =−½ε ijk [(piωj 0 −pj ωi 0 )/(p0 +mo)] and Jk =½ε ijk Mjk in the absence of rotation. This can also be written as: o 0 o 0 ˆ )( mpmp p + = + −≡ pnξp Ω ×× ξ Here the boost parameters are represented by ξi =ξni =ωi 0 . For an infinitesimal variation ωµν, the transformation matrix is (e.g., spin-½ case described above): ˆ                 + − +        •+                 + −=                 −         •+                 −≅                •+        =Λ × ×× 3 o 0 o 0 2 o 022 32 2222 )2/1( )(2 ˆ !3 1 )(2 ˆ )(2 ˆ !2 1 1 2!3 1 22!2 1 1 2 sin 2 cos)],([ mpmp i mp I iIiIpW p pnpn σ Ω Ωpn ΩΩ σ Ω ΩΩΩ σ Ω ΩΩ ××× ξξξ D 2014 MRT
  • The representation matrix becomes (i.e., spin-½ case): For the sake of convenience, I2×2 is assigned as the 2×2 identity matrix, and σ as the ‘vector’ that is comprized of the Pauli 2×2 matrices [σ1,σ2,σ3]. (Note that the Wigner angle |Ω|=Ω is dependent on both the rotation and boost parameters for an arbitrary Lorentz transformation Λ. In addition, notice that an additional normalization factor of √[p0 /(Λp)0 ] has been appended to the matrix so as to maintain the condition given D† (Λ,p) D(Λ,p)=p0 /(Λp)0 I ). by using the well known Pauli spin vector σ relation (σ •a)(σ •b)=(a•b)+iσ •(a× b) and remembering that Ω =Ω /|Ω |. Thus, with the well known sinξ=ξ−ξ 3 /3!+ξ 5 /5!−… and cosξ =1–ξ 2 /2!+ξ 4 /4!−… identities:                 •+        Λ =Λ 2 sin)ˆ( 2 cos )( )],([ 0 0 )2/1( Ω σΩ Ω 1 i p p pWD                 +         −•+         +         − Λ =         +      •+      •+ Λ = Λ =Λ × × •   32 220 0 2 220 0 2 0 0 )2/1( 2!3 1 2 )ˆ( 2!2 1 1 )( 2!2 1 2!1 1 )( e )( )],([ ΩΩ σΩ Ω σ Ω σ Ω σ Ω iI p p iiI p p p p pW i D 2014 MRT ˆ
  • In the absence of rotation, the arbitrary Lorentz boost can be parametrized according to the matrix defined by ΛL above. We can construct the matrix by the associative property of the representation matrix: in which: )]([][)]([ )]()([)],([ )2/1()2/1(1)2/1( 1)2/1()2/1( pLpL pLpLpW DDD DD ΛΛ≡ ΛΛ=Λ − − The representation matrix D(1/2) [W(Λ,p)], in the absence of rotation, is this given by:             Ω •+      Ω Λ =             •−      •+      + +Λ+ Λ =Λ × ×× 2 sin)ˆ( 2 cos )( 2 sinh)]ˆ([ 2 sinh)ˆ( 2 cosh)( ]))[(( )/( )],([ 220 0 22o 0 22 o 0 o 0 00 )2/1( mσ npσnp iI p p iImpI mpmp pp pW ξξξ ×D )ˆˆ(sinhsinh 2 1 coshcosh 2 1 2 1 )ˆˆ( 2 sinh 2 sinh ˆ 2 Ω sin )ˆˆ(sinhsinh 2 1 coshcosh 2 1 2 1 )ˆˆ( 2 sinh 2 sinh 2 cosh 2 cosh 2 Ω cos pn pn m pn pn •++             =      •++ •            +            =      θξθξ θξ θξθξ θξθξ × & where coshθ =p0 /mo and m=n× p represents the axis of rotation of the equivalent Wigner transformation (e.g., of the Dirac spinors in the spin-½ case). Here the parameters ξ and n are defined in the same manner as the general Lorentz boost as given by the ΛL ˆ ˆˆ 2014 MRT ˆ
  • ∑−=′ ′ ′ΛΛ Λ = j jm j j mmj j jj mppW p p mpU ),()],([ )( ),()( )( 0 0 ΨΨ DΛ Our transformation thus becomes: where Wα µ (Λ,p) = (L−1 )α σ (Λp)⊗ Λρ ν Lν µ (p) is the Wigner Rotation and E is the relativistic energy and since U(W)|Ψ(k,mj)〉=Σm′j Dm′j mj ( j) [W(Λ,p)]|Ψ(k,m′j)〉: 2014 MRT since N(Λp)U[L(Λp)]|Ψ(k,m′j)〉=|Ψ(Λp,m′j)〉⇒U[L(Λp)]|Ψ(k,m′j)〉=N−1 (Λp)|Ψ(Λp,m′j)〉 where the normalization constant is N(Λp)=√[k0 /(Λp)0 ]. Finally we have: ),()],([)]([),(])()([)]([),()( 0 0 1 0 0 jjj mkpUpLU p k mkpLpLUpLU p k mpU ΨΨΨ ΛΛ=ΛΛ= − WΛΛ       ),()()],([ ),()],([)(),()( )( 0 0 )( 0 0 j m j mm j m j mmj mkpLpW p k mkpWpL p k mpU j jj j jj ′ΛΛ= ′ΛΛ= ∑ ∑ ′ ′ ′ ′ Ψ ΨΨ D DΛ
  • Mass Positive-Definite To define the mass positive-definite, we must recall a few definitions such as the unitary quantum transformation for coordinate and angular momenta: where P={P1 ,P2 ,P3 }, J={M23 ,M31 ,M12 } and K={M10 ,M20 ,M30 } and their properties are P ρ† =P ρ , M ρσ † =M ρσ and M ρσ =−M σρ . Now for the definition of invariance under a unitary transformation of a homogeneous Lorentz Transformation Λ and a uniform translation a: ...ω1),ω1( 2 1 ++−=+ ρσ ρσ ρ ρεε MPU ii  ),(),(),(),ω1(),( 1 jj mpmpaUUaU ΨΨ =Λ+Λ − ε which also applies to both momenta: )(),(),(),(),( 11 µννµµνσ ν ρ µ µνµρ µ µ PaPaMaUMaUPaUPaU +−ΛΛ=ΛΛΛ=ΛΛ −− & For the momentum, they commute with each other: 0],[ =νµ PP but for the angular momentum they do not: ρµσσµρρσµ ηη PPMPi −=],[ nor do the momenta together commute: ρµσνρνσµνσµρµσνρρσµν ηηηη MMMMMMi +−−=],[ As comparison, we show the case when Λ=1 then a= 0 and a rotation around the 3-axis: θµ µ pJ ˆ 3 e)0,(e),1( •−− ==  i aP i RUaU and ( ))()0,( Λ=Λ UU 2014 MRT
  • j i jj i jki j mm mjmjmjmjR jj ,e,,e,)]([ ˆ )( Ω•−•− ′ ′=′=Ω pJΩJ D ∑∑ ′ ′ ′ ′Ω=′Ω′=Ω j jj j m j j mm m jjjj mRmmjRmjmjR )]([,)]([,,)]([ )( DDD θϕ θϕ j jjjj j jj mi kmmkmmj mi k jjjj jjjjkj mm kmmkmjkmjk mjmjmjmj e 2 sin 2 cose )!()!()!(! )!()!()!()!( )1(),,( 222 )( +−′−′−+ ′       Ω       Ω +−′−′−−+ ′−′+−+ −=Ω ∑D ∑∑∑ Ω=′⇒Ω= ′          m mm m mmmmR zpzp ˆ,),,(ˆ,ˆ,,)]([ˆ )( ϕθDD 2014 MRT Recall also that your mathematical toolbox consists of these fundamental relations: and with Finally: jjjjjj mm j kikimm j mm M i ′′′ Ω+=Ω+ ][ 2 )1( )()( δD In this instance, we take (i.e., Weinberg’s QFT definition – he also uses Θ for our Ω): where Rik =δik +Ωik with Ωik =−Ωki infinitesimal and for ms running over j, j–1,…, −j. Also: jjjjjj jjjjjj mmjmm j mm j z mmjjmm jj mm jj mJMJ mjmjJiJMiMJ ′′′ ±′′′± ==⇒ +±=±=±⇒ δ δ   )()( )1)(()()( )( 3 )( 12 1 )( 2 )( 1 )( 31 )( 23
  • Now, let us expand |Ψ(Λp;j,mj)〉=|Ψ(Λp)〉|,s; j,mj 〉 in the case of an electron (s=±½): 2014 MRT ∑ +−−++−−−−+−−+ −+−+−+− × +++ +−+−+−+ +=≡ k ss jjss k sjjs j mm kmsjkmjkmskmkjsk mjmjmsmsmm sj jsjsjjs mmmmjsmms s )!()!()!()!()!(! )!()!()!()!()!()!()1( )!1( )12()!()!()!( ),(,;,,;,          δC Using Jz|,s;j,mj 〉=mj|,s;j,mj 〉 we found out that that |,s;j,mj 〉 is an eigenfunction of Jz and since Jz commutes with J2 the eigenfunction of Jz is simultaneously an eigenfunction of J2 which means that |,s;m,ms 〉 is not an eigenfunction of J2 . To get around this, we constructed a linear combination of |,s;m,ms 〉 instead: such that |,s; j,mj 〉 is simultaneously an eigenfunction of Jz and J2 . The quantities Cj mms = 〈,s;m,ms |,s; j,mj 〉 are numerical coefficients which are known as Clebsch-Gordan (CG) coefficients – and for our benefit j=+s and mj =m +ms. A general formula for these coefficients is due to Wigner: with =0,1,2,…, s=±½, j =0,1/2,1,3/2,…, m =,−1,…,−, ms =±s & mj = j, j –1,…,−j. ∑ ∑′ ′ Λ′Λ Λ =ΛΛ j s sj m mm sj j mmmsj pmmsmjspWmms p p mjpU     )(,;,,;,)],([,;, )( ),;()( )( )(0 0 ΨΨ D ∑∑ ±= += ±= == )21( 21 ,;,,;,,;,,;,,;, s mmm s j mm mm jssj sj s s mmsmjsmmsmmsmjs        C
  • a•+ ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ ∇ t a z a y a x a t a a tzyxt µµ Point P at the tip of the distance vector xµ ={ct,r} is given in 4-dimensional Rectangular (Cartesian) Coordinates by the intersection of constant x, constant y and constant z planes and time t. The speed of light c is a constant of motion (the same everywhere!)       ∂ ∂ =      ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =∇ φ φφφφφ φµ ∇,,,, tzyxt x r O The Laplacian ∇ •∇ ≡ ∇2 = ∂2 /∂x2 + ∂2 /∂y2 +∂2 /∂z2 leads to the D’Alembertian: The divergence ∇ •A= ∂Ax /∂x+∂Ay /∂y+∂Az/∂z gives: P 2014 MRT A ],[],,,[ atzyxt aaaaaa ==µ y φ z If the scalar product is A•B=Ax Bx +Ay By +Az Bz, then: ba•−=−−−= ttzzyyxxtt babababababa µµ       − ∂ ∂ =      ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ ∇ ∇,,,, tzyxt µ The gradient ∇ψ = ∂ψ/∂xi + ∂ψ/∂yj +∂ψ/∂zk of a scalar function ψ : ˆ ˆ ˆ Del ∇ = ∂/∂xi +∂/∂yj+∂/∂zk= [∂/∂x,∂/∂y,∂/∂z] is now: ˆ ˆ ˆ τ = ct Boosts & Rotations Electrodynamics provides the differential equations for the potentials:  2 φ =ρ /εo and  2 A= j/εo and the continuity equation: ∂φ/∂t + ∇ •A= 0. In this new in four-dimensional notation: Aµ ={φ,A} we get  2 Aµ = jµ /εo and ∇µ Aµ =0. y z x y z x x0 = ct v The transformation laws which give φ and A in a moving system in terms of φ and A in a stationary system. Since Aµ = {φ,A} is a four-vector, the equation must just look like t =γ(t − |v|z/c2 ) and z= γ(z −|v|t) with γ = (1 −|v|2 /c2 )−1/2 except that t is replaced by φ, and r is replaced by A. Thus: v 22 2 )(1)(1 c A AAAAA c cA z zxxyy z v v v v − − === − − = φφ φ ,,, If the vector A= Ax i + Ay j+Az k =[ Ax , Ay , Az ], then we have the definition of the 4-vector aµ : ˆ ˆ ˆ 22 2 2 2 2 2 2 2 2 2 2 =− ∂ ∂ = ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =∇∇ ∇ tzyxt µµ So = (1/c2 )∂t −∇ 2 =ηµν∂µ ∂ν =∂µ ∂µ with ∂µ=∂/∂µ .
  • ρ ρ ρσ ρσ P i M i UaU εω 2 1 1)εω,1(),(  −+=+=Λ ]ωinTerms[)ωω()ωδ)(ωδ( 2 O+++=++=ΛΛ= ρσσρσρσ ν σ ν ρ µ ρ µ µνσ ν ρ µ µνρσ ηηηη µµ ν µ ν µ ν µ εω =+=Λ aδ and               − − =                           −+− − =Λ==             )( )( ˆˆ)1(100 0100 0010 00 2 ctz y x zt z y x t c c xx z y x t v v vvv v γ γ γγ γγ νµ ν µ The Lorentz transformation transformsframe xν (say x,y,z,t) intoframe x µ (say x, y,z,t): assuming the inertial frameis going in the x3 (+z) direction.If both ω µ ν and εµ are taken to be an infinitesimal Lorentz transformation and an infinitesimal translation, respectively: This allows us to study the transformation: The linear unitary operator U=1+i ε T was constructed: where the generators G of translation (Pµ ) and rotation (Mµν ) are given by P1 , P2 , and P3 (the components of the momentum operator), M23 , M31 , and M12 (the components of the angular momentum vector), and P0 is the energy operator. 2014 MRT x2 = y x3 = z x1 = x x2 = y x3 = z x1 = x x0 = ct β=|v|/c
  • 2 o 2 o 2 22 o 2 222 2 1 1 1 1 1 1 1 1 1 1 β γ γ γ β γ − = +      = − =⇒ − =         − = • − = m m c cmcc p pvvv and 2 22 )()( c tt vv t vr vvr r vvr rrrrrrvrr • −= • = • −=−=== || ||||⊥⊥⊥|||| γγ γγ andorand− ⊥|| += rrr For a boost in an arbitrary direction with velocity v, it is convenient to decompose the spatial vector r into components perpendicular and parallel to the velocity v: . Then only the component in the direction of is ‘warped’ by the gamma factor: where the gamma factor γ (which is a function of the rapidity β =|v|/c) is:       • −=−+=      •− = ||⊥ 22 )( ))(1( c tttt v vr vrrv vr rr γγγ γ and−+ Now r can also be written as: and the relativistic momentum p can be written as: vpp p o 2 o 2 22 o 2 0 )( )()( mcm c cmc c E p γ=−= − == and 2014 MRT
  • These equations can be expressed in matrix form as: ν ν µ ν µµ γ γ γ γ x x x x ctx ct c c tc x x x tcx x Λ=               = Λ=                         −+− − =             =               = = 3 2 1 0 T T 3 2 1 0 ˆˆ)1( rvv1 v v r where 1 is the identity matrix, v is velocity written as a ‘column vector’, vT is its transpose (i.e., the equivalent of a ‘row vector’) and v is its unit vector.                       −+−−− −−+−− −−−+− −−− =             −+− − =Λ 2 2 ˆ 2 ˆˆ 2 ˆˆ ˆ 2 ˆˆ 2 2 ˆ 2 ˆˆ ˆ 2 ˆˆ 2 ˆˆ 2 2 ˆ ˆ ˆˆˆ T T )1(1)1()1( )1()1(1)1( )1()1()1(1 ˆˆ)1( )( β β γ β ββ γ β ββ γγβ β ββ γ β β γ β ββ γγβ β ββ γ β ββ γ β β γγβ γβγβγβγ γ γ γ γ ν µ zyzxz z zyyxy y zxyxx x zyx vv1 v v v c c More generally for a boost in any arbitrary direction [β x =vx/c,β y =vy/c,β z =vz/c]: 2014 MRT ˆ ˆ ˆ ˆ ˆ ˆ
  •                 −+−−− −−+−− −−−+− −−− =           −+− − = 2 32313 2 3 32 2 212 2 2 3121 2 1 2 1 2 3 2 2 2 1 T2 T2 ˆ)1(1ˆˆ)1(ˆˆ)1(1ˆ ˆˆ)1(ˆ)1(1ˆˆ)1(1ˆ ˆˆ)1(ˆˆ)1(ˆ)1(11ˆ 1ˆ1ˆ1ˆ ˆˆ)1(ˆ1 ˆ1 )( pppppp pppppp pppppp ppp ppp p γγγγ γγγγ γγγγ γγγγ γδγ γγ µ ν jijii i pL Now, since we are using a momentum p represention we generally have for a boost in any arbitrary direction kµ =[k0 =moc,ki =0] with pν =[p0 =E/c,pi ]: 2014 MRT where 1 is the unit/identity matrix, p is the momentum written as a column vector, pT is its transpose (another row vector) and p is the momentum direction unit vector and since γ =1/√(1–β 2 )=1/√(1–|v|2 /c2 ), we have γβ =√(γ 2 –1) and p=γ mov with |p|=γ mo|v|. With these, we also have a unit vector p parallel to the direction of v: ˆ ˆ ˆ o 42 o 22 ˆˆ m cmci iii + ≡≡⇒= p p p pppp γand The result is:                 = = = === = 333 33 322 33 311 330 0 3 o 3 323 22 222 22 211 220 0 2 o 2 313 11 212 11 111 110 0 1 o 1 3 3 0 o 02 2 0 o 01 1 0 o 00 0 0 o 0 ˆˆ)(ˆˆ)(ˆˆ)(0)( ˆˆ)(ˆˆ)(ˆˆ)(0)( ˆˆ)(ˆˆ)(ˆˆ)(0)( 0)(0)(0)()( )( pppppp pppppp pppppp pLppLppLpppLmp pLppLppLpppLmp pLppLppLpppLmp ppLmpppLmpppLmpppLmp pL µ ν
  • 2014 MRT Since the maximum momentum can be moc and the unit time vector labelled ê t (which allows us to set the standard momentum as p0 =E/c):                  ⊗ ⊗= = z y x t=cvp k k e e e eo ˆ ˆ ˆ ][ ]ˆ[][ zyx kkk cm µ             =               = => κ κ µµ 0 0 0 0 0 0 0 0 oo mm k k k and and ki ⊗êj is the input standard momentum direction p=k. This gives us the column kets: ˆ ˆ
  • y x z Pr rˆ ϕˆ ϕ θ Constant ϕ plane θˆ Constant r sphere Constant θ cone x z y O ϕ θ • r rdθ dr dθ d rsinθdϕ dϕ ϕθθ θ θ ϕθ ∂ ∂ + ∂ ∂ + ∂ ∂ =• F r F rr Fr r r sin 1)sin( sin 1)(1 2 2 F∇ ϕˆsinˆˆ ϕθθ drrddrd ++= θr ϕθθ dddrrdV sin2 = θθ ϕθθϕθθ 222222 2222222222 sin sin)sin()( +=ΩΩ+= ++=++= dddrdr drdrdrdrdrdrd where  • Given the geometry involved, a quick review is required. Point P at the tip of the distance vector r is determined in Spherical Coordinates by the intersection of a constant θ (i.e., a cone), constant r (i.e., a sphere) and constant ϕ (i.e., a half-plane) surfaces. y x z r rrr ˆ),,( rr == ϕθ P d is an infinitesimal differential increment of length: ϕ∇ ˆ sin 1ˆ1 ˆ ϕθθ ∂ ∂ + ∂ ∂ + ∂ ∂ = f r f rr f f θr ϕ ×∇ ˆ )(1 ˆ)( sin 11 ˆ )sin( sin 1       ∂ ∂ − ∂ ∂ + +      ∂ ∂ − ∂ ∂ + +      ∂ ∂ − ∂ ∂ = θ ϕθ ϕθ θ θ θ ϕ θϕ r r F r rF r r rFF r FF r θ rF )2 FFF ×(∇×∇∇∇ −)•(=∇ ˆ j i O kˆ dV is an infinitesimal differential increment of volume: 2014 MRT 2 2 22 2 2 2 2 sin 1 sin sin 11 ϕθ θ θ θθ ∂ ∂ + +      ∂ ∂ ∂ ∂ +      ∂ ∂ ∂ ∂ =∇ f r f rr f r rr f Laplacian of a vector function F= F(r,θ ,ϕ ) (identity) The Laplacian of a scalar function f = f (r,θ ,ϕ ) The vector product of a vector function F = F(r,θ ,ϕ ) The scalar product of a vector function F = F(r,θ ,ϕ ) The gradient of a scalar function f = f (r,θ ,ϕ ) Produces another vector perpendicular to the plane formed by ∇ and F with unit vectors given. We can make the following geometric objects into physical realities if we substitute the speed v for f = f (r,θ,ϕ ) and the momentum p for the vector F = F(r,θ,ϕ ). Manifold ˆ p
  • J To calculate this rotation, we need to choose a ‘standard boost’ L(p) which carries the four-momentum from kµ = [moc,0,0,0] to pµ. From above, this is conveniently chosen as: jijijijij i i i ii i pL cm pLpL cm p pL pppp p p p ˆˆ)1()1(coshˆˆ)( ˆ1sinhˆ)()( cosh)( 2 o 0 0 o 0 0 0 −+=−+= −==== === γδθδ γθ γθ where θ =p/moc. Here pi is the same unit vector pi/|p|, and γ =√(|p| 2 +mo 2 c2 )/moc=p0 /moc.ˆ 2014 MRT   jzjjjz jj mJmjmmjJ jjmjjjmj =⇒= +⋅=⇒+⋅= ,, )1(,)]1([, 222 JJ |ΨO ( p,mj )〉 mo Jz = mj  p Suppose that observer O sees a particle (spin-mj and mass mo ≠0) with momentum p in the y-direction and spin z-component mj . A second observer O moves relative to the first with velocity v in the z-direction. How does O Note that for a massive particle (i.e., of mass mo and four-mo- mentum pµ =[p0 c,p]) the standard boost L(p) may be written as:                                   −+        −        −         −        −+        −         −        −        −+ = 2 2 3 2 o 0 2 23 2 o 0 2 13 2 o 0 o 3 2 32 2 o 0 2 2 2 2 o 0 2 12 2 o 0 o 2 2 31 2 o 0 2 21 2 o 0 2 2 1 2 o 0 o 1 o 3 o 2 o 1 2 o 0 1111 1111 1111 )( p p cm p p pp cm p p pp cm p cm p p pp cm p p p cm p p pp cm p cm p p pp cm p p pp cm p p p cm p cm p cm p cm p cm p cm p pL
  • J ]cos,sinsin,cossin[ˆ θϕθϕθ=p where 0≤θ ≤π and 0≤ϕ ≤2π. θϕ θϕ 23 ee)]ˆ([ )()(0 01 )ˆ()ˆ()ˆ( 23 1 J i J i RU RR RRR =⇒      ==− pppp RR Now, suppose we take p to have polar (θ ) and azimutal (ϕ) angles:ˆ 2014 MRT In this case, for a particle of mass mo = 0 and helicity-σ, we have: with the little-group element W(Λ,p) (the Wigner rotation) given by: )ˆ()()ˆ()ˆ()ˆ()ˆ()()(),( 1111 ppBpppBpW −−−− == RRRRpRpRp RR RRRR Then we can take R(p) as a rotation by angle θ around the 2-axis, which takes col[0,0,1] into col[sinθ ,0,cosθ], followed by a rotation by the angle ϕ around the 3-axis: ˆ Suppose that observer O sees a massless particle (helicity-σ and mass mo = 0) with unit momentum p in the z-direction with basis ket |k〉 =col[0,0,1]. R2(p) rotates |p〉 by angle θ around the y-axis into col[sinθ ,0,cosθ ]. Then R3(p) rotates |p〉 by the angle ϕ around the z-axis. ˆ           1 0 0 )( pB1. col[sinθ , 0, cosθ ] y x zϕ First Rotation θ Second Rotation |p〉ˆ           1 0 0 )()(2 pBR θ2. 3.           = 1 0 0 )()()()( 23 pBRRpRB θϕ |k 〉= col[0,0,1] U[R(p)]|Ψ(p,mj)〉 U[L(Rp)R(p)B(|p|)]|Ψ(k,mj)〉ˆ |Ψ( p,mj)〉 = |p〉× |mj 〉ˆ ˆ pˆ ∑ ∑ ∑ ′ −−− ′ + −=′ ′ + −=′ − ′ ′= ′= ′= = − j jj j jj j jj m j j mm j jm j j mmj j jm j j mm jj mkRBRRBRB p B mBB p B mpBU mpB p B mpUB p B mpBU ),(])ˆ()()ˆ()ˆ()()ˆ([)( )( ),()],([)( )( ),()( ),(])()([)( )( ),()()( )( ),()( )ˆ( 1 )ˆ( 11)( 0 0 )( 0 0 1)( 0 0 0 0 1 Ψ ΨΨ Ψ ΨΨ      pLpL ppppppp p ppWp p pLpLp p p p R R R R RRD D RD R ∑ + −=′ ′′= j jm jjjj j mpmjRmjB p B mpBU ),(,)]([,)( )( ),()( 0 0 ΨΨ θp p since Rp = R(p)p and B(|p|) is a Lorentz boost. We finally get:ˆ ˆ ˆ ˆ ˆ σE/c ≈
  • 2014 MRT               − − = γγ γγ 001 0100 0010 100 )( 2 2 pB As just shown, it is very important to note that when Λµ ν is an arbitrary three- dimensional rotation R(3) , the Wigner rotation W(Λ,p) is the same as R(3) for all p. To see this, note that the boost Lµ ν (p) above may be expressed as L(p)=R(p)B(|p|)R−1 (p), where R(p) is a rotation that takes the 3-axis into the direction of p, and: ˆ ˆ ˆ )ˆ( 001 0100 0010 100 )ˆ( 1000 0cossin0 0sincos0 0001 )ˆ( 001 0100 0010 100 )ˆ( )]ˆ()()ˆ([)]ˆ()()ˆ([ )()(),( 1 2 2 1 2 2 1 111 1 pppp ppBpppBp LLW − − − −−− −               − −             ΩΩ− ΩΩ               − − = = = RRRR RRRR ppp γγ γγ γγ γγ RR RRR RR R Then, for a three-dimensional rotation R, we have the following Wigner rotation about the 3-axis with L(p)=R(p)B(|p|)R−1 (p) and L(Bp)=R(Rp)B−1 (|p|)R−1 (Rp):ˆ ˆ ˆ ˆ
  • 2014 MRT The state of a moving massive particle can be shown to have exactly the same trans- formation under rotation as in non-relativistic quantum mechanics (i.e., W(Λ= R,p)= R). )( 100 0cossin 0sincos 100 0cossin 0sincos cos0sin 010 sin0cos 100 0cossin 0sincos )ˆ()ˆ( 3 1 ϕϕϕ ϕϕ αα αα ββ ββ γγ γγ RRR =           −=           −           −           −=− pp RR ϕ ϕ 3 e)]ˆ([ )(0 01 )ˆ()ˆ()ˆ( 3 1 J i RU R RRR =⇒      ==− pppp RR Even in this general case, the rotation R−1 ( Rp) R R(p) takes the 3-axis into the direction p, and then into the direction Rp, and then back to the 3-axis, so it must be just a rotation by some azimutal angle ϕ around the 3-axis: ˆ ˆˆ ˆ      )( 1 )ˆ( 11 )ˆ()()ˆ()ˆ()()ˆ(),( 1 p RBRRBRp LpL ppppppW −−− − = RRRR R and: In conclusion, the Wigner’s formula (J3 ∝ϕ)for the d-matrixelementsdm′jmj ( j) (ϕ) still form a Unitary Group U(ϕ) and most of all, the whole of the dictionary we have learnt already for the Spherical Harmonics, the Clebsch-Gordan coefficients and the Winger-Eckart Theorem apply in the same exact way in relativistic field theory as it does in quantum mechanics – as it does for any problem in which D ( j) [W(Λ,p)] for all spin-j! Then, for an arbitrary rotation R we see that R−1 ( Rp) RR(p) since it is given by:ˆˆ
  • ]0,,0,[]0,ˆ,0[ˆ 22 0 ppp cEpcEp =⇒== µ and ppp Λ=           Λ⇒Λ=Λ⇒= 0 0 0 )(])([)()( LkLpkpLp µµµ µ νν Now, for the W-boson problem (see Figure). This first problem requires us to take p as:ˆ 2014 MRT For a particle of rest mass MW >0 and spin σ = 1 we have the one-particle state which is mass positive-definite for the W- boson: with the Wigner Rotation W(Λ,p), which is given by the following group multiplication: [ ])1,()( )0,()( )1,()( )( ),()( )( )1,()]( )1( 11 )1( 10 )1( 110 0 1,0,1 )1( 10 0 +ΛΩ+ ΛΩ+ −ΛΩ Λ = ′ΛΩ Λ =+=Λ ++ + +− +−=′ +′∑ p p p p p p p p pU Ψ Ψ Ψ ΨΨ D D D D σ σ σσ             0 0 0 )( cM pL Wy z Ω p2 |k 〉 = [MW c,0, 0, 0]T Λ L(p)ˆ We can take [0,0,0]T into p2 using L(p) as a boost along p2 followed by a boost Λ(β3 ) with rapidity β3 along the 3-axis, which takes [0,p2,0]T into something like [0, p2,−p3]T : ˆ ˆ )()(),( 1 ppp LΛLW Λ=Λ − and the unit momentum is, by definition for massive particles: Quantum Fields – Problem 1. T ]0,0,0,[ cMk W=µ Suppose that observer O sees a W boson (spin-1 and mass MW ≠0) with momentum p2 in the y-direction and spin z-component Jz = σ  = . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the |Ψ(p,σ)〉 state? = [0, p2 , − p3 ]T|Λp 〉 ˆ ˆ MW ˆ ˆ ˆ ˆ ˆ
  • where the gamma factor is γ3 =√(1−β3 2 )(which is a function of the rapidity β3 =|v3|/c along the 3-axis). 2014 MRT First we calculate the Wigner Rotation:                         −− + + − =                       + +              − − =Λ 3 2 332 0 33 202 22 22 33 2 32 0 3 202 22 22 2 2 0 333 333 23 0 0 )( 10 0010 0 1000 0 )( 10 0010 00 00 0100 0010 00 )( γγβγβ γβγγ γγβ γβγ cM p cM p cMpcM cp cM p cM p cM p cMpcM cp cM p cM p cM p pL WW WWW WW WWW WW as anticlockwize around the positive 1-axis. We will need this later. Then we have: )()(),( 23 1 23123 pLpLpW ΛΛ=Λ − −= ××
  • 2014 MRT                   +− + =                         + − + =                         + − + =                       − =                         ⋅− ⋅ ⋅ =                                           −− + + − =Λ 222 233 2 222 23 4222 2 33 2 4222 2 3 4222 33 2 4222 3 0 33 2 0 3 2 0 33 2 2 0 3 3 2 332 0 33 202 22 22 33 2 32 0 3 23 0000 0 0 0 0 0 0 )( 10 0010 0 )( cMp p cMp c cMcp p c cMcp c cMc p c cMc c p p c p cM cM p cM cM p cM cM p cM cM p cM p cMpcM cp cM p cM p cM p kpL W W W W W W W W W W W W W WW WWW WW γβ γ γβ γ γβ γ γβ γ γβ γ γγβγβ γβγγ µ p p Then we have with the definition of the basis momentum kµ and since p=p2:           +− =           Λ Λ Λ +=Λ 222 233 2 3 2 1 222 230 0 cMp p p p p cMpp W W γβ γ and and thus:
  • k v v jkjkjip ˆ 1 ˆˆ 1 ˆˆˆˆ0 2 222 2 2 2 3 222 23 2 222 2332       − + −= − + −=+−+=Λ c cMp c p cMp pcMpp W W W β β γβ Finally we get the eigenvector for the boosted momentum Λp: k v v jk v v jkjpp ˆ 1 1 ˆˆ 1 ˆˆˆ 2 2 2 2 2 222 2 232 cM c cM p c p c cMp c ppp W W W       − +      −=       − + −=−=Λ≡Λ so that: x Λ3(v) L2(p) W−1(Ω) 2014 MRT y z | k 〉col[MW c,0, 0, 0] = We see that the effect of a Lorentz boost L2(p) on the initial unit momentum p2 followed by a Lorentz transformation Λ3(v) which has the effect of creating (as seen from O) a momentum vector p3 in the negative 3-axis direction: Jz = Suppose that observer O sees a W boson (spin-1 and mass MW ≠0) with momentum p2 in the y-direction and spin z-component Jz = σ  = . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the |Ψ(p,σ)〉 state? p2 p3 cM c cM p c p W W 2 2 2 3 1 1       − +      −= v v The effect of the Wigner rotation W(Λ,p)will produce a rotation (negative in this case) around the 1-axis which results in an angle Ω between the 2-axis and the boost vector Λp ≡p2 j–p3 k.* J MW ˆ ˆ ˆ ˆ *It is sad to say but in many texts authors suggest that to find the answer to this problem you only have to calculate the Wigner rotation W(Λ, p)=L–1 (Λp)ΛL(p), then extract the angle Ω and calculate D( j) m′ m =exp(i Ω J/). The derivation that follows will be a little more practical. ˆ
  • 2014 MRT Using the group properties described earlier, with the vector product of the 3- and 2-axis: )()(),( 23 1 23123 pLpLpW ΛΛ=Λ − −= ××                       ΩΩ Ω      + +−Ω      + + ΩΩ               − − =             ΩΩ Ω−Ω                       + +              − − =ΛΛ=Λ − − cossin00 sin )( 1cos )( 10 0010 sincos0 00 0100 0010 00 cossin00 sincos00 0010 0001 1000 0 )( 10 0010 00 00 0100 0010 00 ),()()( 202 22 2 202 22 22 22 2 0 333 333 202 22 22 2 2 0 333 333 1 12323 cMpcM cp cMpcM cp cM p cM p cM p cM p cMpcM cp cM p cM p cM p pWpLpL WWWWW WWW WWW WW γγβ γβγ γγβ γβγ × and since L(Λp)L−1 (Λp)=1 we can then go through the matrix multiplicationwiththe order: )()()()()( 1 123 1 23 1 1123 ΩΛΛ=ΩΩ − − −− −−= WpLpLWW ××    1 we get: )()()]()([)( 1 123 1 232323 ΩΛΛΛ=⋅Λ − − − WpLpLpLpL ××× 1 then:
  • 2014 MRT We finally obtain for the boost matrix L3× 2 (Λp) as a function of the deflection angle Ω:                       Ω+Ω−Ω+Ω− Ω      + +−Ω      + + Ω−ΩΩ+Ω =Λ cossinsincos0 sin )( 1cos )( 10 0010 cossinsincos0 )( 3 2 333 2 332 0 33 202 22 2 202 22 22 33 2 33 2 32 0 3 23 γγβγγβγβ γβγβγγ cM p cM p cM p cMpcM cp cMpcM cp cM p cM p cM p cM p pL WWW WWWWW WWW × Ω=Ω               + ++− Ω+Ω=Ω      + +− cossin )( 1 sincossin )( 1 2 33202 22 2 3 3 2 33202 22 2 cM p cMpcM cp cM p cMpcM cp WWW WWW γβγ γγβ       + ++ −=Ω )( 1 tan 202 22 2 3 2 33 cMpcM cp cM p WW W γ γβ Since L3× 2 (Λp) is symmetric we can ‘extract’ the [L3× 2 (Λp)]3 2 =[L3× 2 (Λp)]2 3 components: and at first we obtain: which we can reduce further with the goal of having only to calculate the ratio p2 /MW c.
  • 2014 MRT Going through patiently with the algebra we obtain:     +      ++−+      ++       ++ −=Ω 22 2 24222 2 22 3 24222 2 2 2 24222 23 1 tan cpcMcMcpcMcMcMcpcM cpcMcMcp WWWWWW WW β β                 ++      +−++      +           +      + −=Ω 2 2 2 22 3 2 2 2 2 2 3 11111 11 tan cM p cM p cM p cM p cM p WWW WW β β                                 +                 +      +       +        −−      −≅Ω ... 11 111arctan 2 2 2 2 2 2 cM p cM p ccM p c W W W vv Reducing ruther and deviding both the numerator and denominator by MW 2 c4 we get: By dividing by 1+√[( p2/MW c)2 +1] and using the approximation (1+x)−1 =1–x+… where x=√[1–(|v|/c)2 ]×[1+…/(1+√…)] we finally get the Wigner Angle Ω which is given by: So with only the ratio p2/MW c and |v| we can calculate this relativistic deflection angle.
  • Ω− Ω ±= Ω Ω− ±=Ω 2 2 sin1 sin cos cos1 tan Finally, by using the following trigonometric identities relating tanΩ to cosΩ and/or sinΩ: Ω Ω+ = Ω Ω+ =Ω Ω+ = Ω+ =Ω tan tan1 tan tan1 sin tan1 1 tan1 1 cos 2 2 2 22 and we find that: such that the boost matrix becomes:                                         Ω+− + Ω Ω+ − − Ω Ω+ − + Ω+− +      − − Ω Ω+                 +      +       +− Ω+                 +      +       + Ω+− − Ω Ω+ Ω Ω+ − + Ω+− +      − =Λ 22 3 2 2 2 3 3 2 2 3 2 2 2 3 3 2 2 2 3 3 2 2 2 2 2 22 2 2 2 2 22 3 3 2 2 2 2 3 3 2 2 2 3 2 2 2 3 23 tan1 1 1 1 tan tan1 1tan tan1 1 1 tan11 01 1 tan tan1 11 1 tan1 1 11 10 0010 tan1 1 1tan tan1 tan tan1 1tan11 1 01 1 1 )( ββ β ββ β β β β β β β ββ cM pcM p cM p cM p cM p cM p cM p cM p cM pcM p cM p pL W W W W W W W W W W W × 2014 MRT Sparing you the beast of a representation this will give… but now you have a few ways of calculating it: either by using the relation Ω=arctan{–(|v|/c)(p2/MW c)[1–…]} or the one above by using tanΩ and its square, with p /M c and β =|v|/c given beforehand.
  • σ = +1 σ = 0 σ =−1 σ′= +1 σ′ =0 σ′ =−1 We then obtain for the relativistic one-particle state of spin-1 as seen from observer O: ])1,()()0,()()1,()([ ),()( )0)(())(0()0)(0())(( ),()],([),()],([)1,()( )1( 11 )1( 10 )1( 113 1 1 )1( 10 332 0 3 1 1 )1( 10 30 3 20 2 10 1 00 0 1 1 )1( 10 0 2 +ΛΩ+ΛΩ+−ΛΩ= ′ΛΩ −+++ = ′ΛΛ Λ+Λ+Λ+Λ =′ΛΛ Λ =+=Λ ++++− + −=′ +′ + −=′ +′ + −=′ +′ ∑ ∑∑ ppp p p pp ppW p pppp ppW p p pU ΨΨΨ Ψ ΨΨΨ DDD D DD γ σ γβγ σσσ σ σ σ σ σ σ µ µ )cos1( 2 1 Ω+ σσ′ 0 )cos1( 2 1 Ω− 0 0 0 0Ωsin 2 2 0 The D( j =1) σ′+1(Ω) were provided by the following representation: ])1,()cos1()0,(sin2)1,()cos1[(1 2 1 )1,()( 4/1 2 2 +ΛΩ++ΛΩ+−ΛΩ−                 −=+Λ − ppp c pU ΨΨΨΨ v as well as the final representation (with cosΩ and sinΩ or Ω=arctan(…) as given above): 2014 MRT y z z Suppose that observer O sees a W boson (spin-1 and mass MW ≠0) with momentum p2 in the y-direction and spin z-component Jz = σ  = +. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the |Ψ(p,σ)〉 state? |ΨO (p,σ = +1)〉 |ΨO (Λp,σ ′)〉 ∑′ = +′ Λ σ σ ][ )( )1( 10 0 Wj p p D  2)1(ˆ1ˆˆˆ 222 2 1 2 232 =+=+               −−=−=Λ − jjcMp cc ppp W Jk vv jkjp & p2 Λ(v) y x v →MW Jz =J ˆ Λpˆ
  • 2014 MRT Mass Zero Weinberg suggests that we first have to ‘work out’ the structure of the little group and tells us to consider an arbitrary little-group element Wµ ν , with Wµ ν kν =kµ, where kµ is light- like four-vector, the standard four-momentum for this case, i.e., kµ =[0,0,1,1] (c =1). Acting on a time-like four-vector tµ =[0,0,0,1], such that a Lorentz transformation must yeild a four-vector Wµ ν tν whose length and scalar product with Wµ ν kν =kµ are the same as those of tµ : 1)( 1))(( −== −== µ µ µ ν ν µ µ µ ρµ ρν ν µ ktktW tttWtW Any four-vector that satisfies the second condition (i.e., (Wµ ν tν )kµ =−1) may be written: ]1,,,[ ζζβαν ν µ +=tW and the first condition (i.e., (Wµ ν tν )(Wρ µ tρ )=−1) then yeilds the relation: )( 22 2 1 βαζ += It follows that the effect of Wµ ν ontν is the same as that of the Lorentz transformation:             − − − −+ = ζβαζ ββ αα ζβαζ βαν µ 1 10 01 1 ),(S This does not mean that W equals S(α,β), but it does mean that S−1 (α,β)W is a Lorentz transformation that leaves tµ invariant (i.e., it is therefore a pure rotation!)
  • 2014 MRT Also, Sµ ν (α,β) like Wµ ν leaves kµ invariant, so S−1 (α,β)W must be a rotation by some angle ϕ around the 3-axis:             − = 1000 0cossin0 0sincos0 0001 )( ϕϕ ϕϕ ϕν µ R )(),(1 ϕβα RWS =− where: The most general element of the little group is therefore of the form: )(),(),(),( 1 ϕβαβαβα RSWSS =− When we apply matrix multiplication we get:             −+− −− − −+−+ =             −             − − − −+ = ζϕβϕαϕβϕαζ βϕϕβ αϕϕα ζϕβϕαϕβϕαζ ϕϕ ϕϕ ζβαζ ββ αα ζβαζ βαϕ 1cossinsincos cossin sincos cossinsincos1 1000 0cossin0 0sincos0 0001 1 10 01 1 ),,(W hence: )(),(),,( ϕβαβαϕ RSW =
  • 2014 MRT For ϕ, α, and β infinitesimal, the general group element is:             −− − −− =⇒             − −− − −−− +             = → = = = 00 0 0 00 ω cossin sincos 1000 0100 0010 0001 0 )0( 1sin 0cos βα βϕβ αϕα βα ζβαζ βϕϕβ αϕϕα ζβαζ ν µ ϕ ζ ϕ ϕ ν µ  W ν µ ν µ ν µ δβαϕ ω),,( +=W where (note that we are using the result for W(ϕ ,α,β) but we use the antisymmetric properties of ωµ ν and the equality –α =αcosϕ –βsinϕ and –β =αsinϕ +βcosϕ): From U(1+ω)=1+½(i/)ωρσ Mρσ (we consider no translation so that ε =0) we see then that the corresponding Hilbert space operator is decoupled to 1 followed by 16 terms: )]ωωωω( )ωωωω( )ωωωω( )ωωωω[( 2 1 )ωωωω( 2 1ω 2 1)],,([ 33 33 23 23 13 13 03 03 32 32 22 22 12 12 02 02 31 31 21 21 11 11 01 01 30 30 20 20 10 10 00 00 3 3 2 2 1 1 0 0 MMMM MMMM MMMM MMMM i MMMM i M i WU ++++ ++++ ++++ +++++= ++++=+= ∑∑∑   σ σ σ σ σ σ σ σ σ σ ρ ρσ ρσβαϕ
  • Hence, by inserting the components of the ωρσ matrix and factoring for α, β and ϕ : )]()()([ 2 1 )]()()[( 2 1 )( 2 1 )000 000( 2 1)],,([ 12212332022013310110 12212332022013310110 23133212023121012010 3323130332221202 3121110130201000 MMMMMMMMMM i MMMMMMMMMM i MMMMMMMMMM i MMMMMMMM MMMMMMMM i WU +−+−+−+−+−+= +−+−+−+−+−+= −−++−+−−++= ⋅+⋅−⋅−⋅+⋅+⋅+⋅+⋅− ⋅+⋅−⋅+⋅⋅+⋅+⋅+⋅+= ϕβα ϕϕββββαααα βαβϕβαϕαβα βαβϕβ αϕαβαβαϕ     ]2)(2)(2[ 2 1 )]2()22()22([ 2 1 )]()()([ 2 1)],,([ 1223021301 1223021301 12122323020213130101 MMMMM i MMMMM i MMMMMMMMMM i WU ϕβα ϕβα ϕβαβαϕ +−−+−−+= +−−+−−+= ++−−−−+−−−−+=    In essence this is an acceptable solution: Recalling that Mµν =−Mνµ, we decompose further: ])()([1)],,([ 1223021301 MMMMM i WU ϕβαβαϕ +−−+−−+=  2014 MRT
  • But for the sake of driving things contrary to Weinberg’s notation again, we use temporal-spacial ordering of indices: 31221 1223021301 )()(1 )()(1)],,([ J i JK i JK i M i MM i MM i WU ϕβα ϕβαβαϕ   ++−++−+= +−−+−−+= )(11)],,([ 33 ϕβαϕβαβαϕ JBA i J i B i A i WU +++=+++=  So, finally, we have: where: 2014 MRT             −− − =               −− − = 0 0 0 0 0 0 0 0 123 132 231 321 231303 231202 131201 030201 JJK JJK JJK KKK MMM MMM MMM MMM M µν where we used the Mµν matrix representation arrived at earlier (especially M12 ≡ J3 ): 12 2302 21 1301 JKMMBJKMMA +−=−−=+−=−−= and Weinberg defines his A and B as −J2 + K1 and −J1 + K2, respectively.
  • Massless particles are not ‘observed’ (e.g., in any laboratory experiment) to have any continuous degree of freedon like an infinitesimal angle ϕ so this means that states are distinguished by the eigenvalue of the remaining generator: ),(),(3 εεε kkJ ΨΨ = Since the momentum k is in the 3-direction, ε gives the component of angular momentum in the direction of motion, or helicity. We are now in a position to calculate the Lorentz transformation properties of general massless particles states. The relation U(W)=1+(i/)Aα +(i/)Bβ +(i/)J3ϕ generalises for finite α and β to: βα βα B i A i SU  + = e)],([ ϕ ϕ 3 e)]([ J i RU = and for finite ϕ to: 2014 MRT
  • An arbitraryelementW of the little groupcan be put in the form W(ϕ,α,β)=S(α ,β)R(ϕ ), so that: 2014 MRT and therefore the previously developped equation U(W)|Ψ(k,ε)〉=Σε' Dε' ε (W)|Ψ(k,ε′)〉 gives: ),(e),(ee),()( 3 εεε ϕεϕβα kkkWU i J i B i A i ΨΨΨ  == + εε ϕε εε δ ′′ =  i W e)(D where ϕ is the angle defined by expressing W as in the equation W(ϕ,α,β)=S(α ,β)R(ϕ ). The Lorentz transformation rule for a massless particle of arbitrary helicity is now given by the U(Λ)|Ψ(p, ε )〉=[N(p)/N(Λp)]Σε' Dε' ε [W(Λ,p)]|Ψ(Λp, ε′)〉 and N(p)=√(k0 /p0 ) equations (adapted from the Mass Positive-Definite case and replacing mj by ε) as: ),(e )( ),()( ),( 0 0 εε ϕε p p p pU p i Λ Λ =Λ Λ ΨΨ  with ϕ(Λ,p) defined by: )],([)],(),,([)()(),( 1 pRppSpLpLpW ΛΛΛ≡ΛΛ≡Λ − ϕβα
  • So, for massless particles, the most general element of the little group is of the form: Massless particles are not observed to have any continuous degree of freedom like θ; to avoid such a continuum of states, we must require that physical states (called |Ψ(k,ε )〉) are eigenvectors of A and B. Since the momentum k is in the 3-direction, ε gives the component of angular momentum in the direction in the direction of motion, or helicity. For finite α, β and ϕ:  ϕβα ϕβα 3 e)]([e)],([ )( JiBiAi RUSU == + and             − =             −− − − + == 1000 0cossin0 0sincos0 0001 )( 1 10 01 1 ),()(),(),,( ϕϕ ϕϕ ϕ γβαγ ββ αα γβαγ βαϕβαβαϕ ν µ ν µ RSRSW andwhere Suppose that observer O sees a massless particle with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same relativistic particle’s state? An arbitrary element W of the little group can be put in the form W(ϕ,α,β)= S(α,β)R(ϕ), and: where ϕ is the angle defined by expressing W as in W=S ⊗R above. The Lorentz transformation rule for a massless particle of arbitrary helicity ε is given by (with the basis k=kµ =[κ, 0,0,κ]): |ΨO ( p,ε )〉 |ΨO (Λp,ε )〉 2014 MRT p ε≈ | Ψ≈ 〉 ϕε  i p p e )( 0 0 Λ ),(e),(ee),()( 3 εεε ϕεϕβα kkkU iJiBiAi ΨΨΨ  == + W And therefore, with U(W)|Ψ 〉 =Σε' Dε' ε (W)|Ψ 〉, we get: εε ϕε εε δ ′ Λ ′ = ),( e)( pi WD ),(e )( ),()( ),( 0 0 εε ϕε p p p pU p i Λ Λ = Λ ΨΨ Λ since (Λp)0 = Λ0 µ pµ and can be found with pµ = [p0 ,0]= [|p|2 , 0]. E/c O Λ(v)
  • 2014 MRT To calculate the little-group element                         +− −+ =        = u u u u u u u u uB 2 1 00 2 1 0100 0010 2 1 00 2 1 22 22 κ p ]),,([)],(),,([)],(),,(),,([)()(),( 1 ppRppSpppWpLpLpW ΛΛΛ=ΛΛΛ=ΛΛ≡Λ − ϕβαβαϕ for a given Λ and p we need to fix a convention for the standard Lorentz transformation that take us from k=kµ = [κ,0,0,κ] to p=pµ . This may conveniently be chosen to have the form:         = κ p p BRpL )ˆ()( where B(u=|p|/κ ) is a pure boost along the 3-axis (e.g., z-direction): and R(p) is a pure rotation that carries the 3-axis (e.g., z-direction) into the direction of the unit vector p. ˆ ˆ
  • 2014 MRT Because electromagnetic and gravitational forces obey space inversion symmetry, the massless particles of helicity ±1 associated with electromagnetic phenomena are both called photons. The massless particles of helicity ±2 that are believed to be associated with gravitation are both called gravitons. On the other hand, the supposedly massless particles of helicity ±1/2 that are emitted in nuclear beta decay have no interactions (apart from gravitation) that respect the symmetry of space inversion, so these particles are given different names: neutrino for helicity +1/2 and antineutrino for helicity −1/2. Note that the helicity is Lorentz invariant; a massless particle of a given helicity σ looks the same (aside from its momentum) in all inertial frames. Even though the helicity of a massless particle is Lorentz invariant, the state itself is not. In particular, because of the helicity-dependent phase factor exp(iεϕ/) in: a state formed as a linear superposition of one-particle states with opposite helicities will be changes by Lorentz transformation into a different superposition. ),(e )( ),()( ),(0 εε ϕε p cE p pU p i Λ Λ = Λ ΨΨ Λ
  • 2014 MRT The overall phase of ε + and ε − has no physical significance, and for linear polarization may be ajusted so that ε − =ε + * , but the relative phase is still important. Indeed, for linear polarizations with ε − =ε + * , the phase of ε + may be identified as the angle between the plane of polarization and some fixed reference direction perpendicular to p. The equation: The generic case is one of elliptical polarization, with |ε ±| both non-sero and unequal. Circular polarization is the limiting case where either ε + or ε − valishes, and linear polarization is the opposite extreme, with |ε +|=|ε −|. shows that under a Lorentz transformation Λµ ν , this angle ϕ rotates by an amount ϕ(Λ,p). and has the consequence that a Lorentz transformation Λ rotates the plane of polarization by an angle 2ϕ(Λ,p). ),(e )( ),()( ),(0 εε ϕε p cE p pU p i Λ Λ = Λ ΨΨ Λ Plane polarized gravitons can be defined in a similar way, and here the equation above becomes: ),(e )( ),()( ),(20 εε ϕε p cE p pU p i Λ Λ = Λ ΨΨ Λ For instance, a general one-photon state of four-momenta may be written |Ψ(p;ε)〉= ε +|Ψ(p,+1)〉+ε −|Ψ(p,−1)〉, where |ε +|2 +|ε −|2 =1.
  • Again, now for the calculation based on the data provided: Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? | ΨO (p,σ )〉 |ΨO (Λp,σ )〉 2014 MRT σ | Ψ≈ 〉 ϕσ  i p p e )( 0 0 Λ E/c ν =0 ν =1 ν =2 ν =3 µ =0 µ =1 µ =2 µ =3 κ κ p p 2 122 + ν µ In this case, the Βµ ν are given by (with γ =1/√(1 –β2 ) and β = v/c): κ κ p p 2 122 − κ κ p p 2 122 − 1 1 κ κ p p 2 122 + 00 00 0 0 0 0 00 Quantum Fields – Problem 2. O ),(e )( ),()( ),( 0 0 σσ ϕσ p p p pU pB i Λ Λ =Λ ΨΨ  or, by expansion: where the normalization constant is the same as found in Problem 1, i.e., √γ3.         −Λ++Λ=Λ Λ =Λ ΛΩ−ΛΩΛΩ )1,(e)1,(e),(e )( ),()( ),(),( 3 ),( 0 0 ppp p p pU p i p i p i ΨΨΨΨ  γεε ε |k 〉 ≈p B(|p|/κ ) R L(p)ˆ col[κ ,0, κ , 0] = Λ(v)
  • 2014 MRT First we calculate the Lorentz transformation along the 2-axis:                           − − + −+ =                             − + − − + −+ =                         +− −+             − =         = 1000 001 2 1 0 2 1 00 0 2 1 0 2 1 10090cos 2 1 090cos 2 1 90sin90sin 2 1 090sin 2 1 90cos0 0 2 1 0 2 1 1000 0 2 1 0 2 1 0010 0 2 1 0 2 1 1000 0cossin0 0sincos0 0001 )ˆ()( 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 2332 κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ κ ϕϕ ϕϕ κ p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p    BRpL i.e., boosting along the photon’s the 2-axis direction and carring it along the direction of p in the 3-axis direction. The result is:         = κ 2 2332 )ˆ()( p p BRpL
  • where the gamma factor is γ3 =√(1−β3 2 )(which is a function of the rapidity β3 =|v3|/c along the 3-axis). Then we calculate the Wigner Rotation:                             − − + − − − + − −+ =                           − − + −+               − − =Λ 3 2 22 2 33 2 22 2 33 2 22 2 2 22 2 33 2 22 2 3 2 22 2 3 2 22 2 2 22 2 2 22 2 2 22 2 333 333 23 2 1 0 2 1 001 2 1 0 2 1 00 2 1 0 2 1 1000 001 2 1 0 2 1 00 0 2 1 0 2 1 00 0100 0010 00 )( γ κ κ γβ κ κ γβ κ κ κ κ γβ κ κ γ κ κ γ κ κ κ κ κ κ κ κ γγβ γβγ p p p p p p p p p p p p p p p p p p p p pL as anticlockwize around the positive 1-axis. We will need this again later. Then we have: )()(),( 23 1 23123 pLpLpW ΛΛ=Λ − −= ×× 2014 MRT
  •                     −         −         + =                                     −−         +−         −         +         −+         + =                         ⋅ − −⋅ + − ⋅ − ⋅ + ⋅ − +⋅ + =               = =                         − − + − − − + − −+ =Λ 233 2 2 2 2 2 2 2 2 23 2 2 2 2 2 33 2 2 2 2 2 33 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 3 2 22 2 33 2 22 2 33 2 22 2 2 22 2 2 22 2 3 2 22 2 3 2 0 3 2 22 2 33 2 22 2 33 2 22 2 2 22 2 33 2 22 2 3 2 22 2 3 23 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 0 0 2 1 0 2 1 001 2 1 0 2 1 00 2 1 0 2 1 )( p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p k k p p p p p p p p p p p p kpL γβ κ κ γ κ κ γβ κ κ γβ κ κ κ κ κ κ γ κ κ γ κ κ κ γβκ κ κ γβ κ κ κ κ κ κ κ κ κ γκ κ κ γ κ κ γ κ κ γβ κ κ γβ κ κ κ κ γβ κ κ γ κ κ γ µ Then we have with the definition of the basis momentum kµ and since |p2 |=p2:                     −         −         + =                   Λ Λ Λ =Λ 233 2 2 2 2 2 2 2 2 3 2 1 230 1 2 1 2 p p p p p p p p pp γβ κ κ γ and and thus: 2014 MRT
  • kjikjip ˆ 1 ˆ1 2 ˆ1 2 ˆˆ1 2 ˆ1 2 2 3 23 2 2 2 2 2 2 2 2 2332 2 2 2 2 2 2 2 β βκκ γβ κκ − −         −+         +=−         −+         +=Λ p p p p p p p p p p Finally we get the eigenvector for the boosted momentum Λp: k v v jipp v ˆ 1 ˆ1 1 2 1ˆ1 1 2 1 1 2 2 22 2 22 2 2 2 0 p c c p p p p c p p       − −                     −+                     −=Λ≡       − ≡Λ Λ κκ and so that the four-vector (Λp)µ is given by: x Λ3(v) L2(p) W−1(Ω) y z We see that the effect of a Lorentz boost L2(p) on the initial unit momentum p2 followed by a Lorentz transformation Λ3(v) which has the effect of creating (as seen from O) two component vectors p1 and p2, magnitudes being equal to ½[1– 1/(p2 /κ)2 ]p2, in the positive 1- and 2-axis directions and a momentum vector p3 in the negative 3-axis direction: p2 2 2 3 1 p c c p       − −= v v The effect of the Wigner rotation W(Λ,p)will produce a rotation (negative in this case) around the 1-axis which results in an angle Ω between the 2-axis but also some dispersion resulting in the overall boost vector Λp≡p1 i +p2 j–p3 k. ˆ ˆ ˆ ˆ p3 ˆ ˆ Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? ˆ ˆˆ 2014 MRT Λpˆ ≈ p2 ˆ p1 ˆ Ω
  • 2014 MRT We can then go through the matrix multiplicationwiththe order described in Problem 1:                       ΩΩ − − Ω + −Ω + Ω − −Ω −+               − − =             ΩΩ Ω−Ω                       − − + −+               − − =ΛΛ=Λ − − cossin00 001 2 1)( sin 2 1)( cos 2 1)( 00 sin 2 1)( cos 2 1)( 0 2 1)( 00 0100 0010 00 cossin00 sincos00 0010 0001 1000 001 2 1)( 0 2 1)( 00 0 2 1)( 0 2 1)( 00 0100 0010 00 ),()()( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 333 333 2 2 2 2 2 2 2 2 2 2 2 2 333 333 1 12323 κ κ κ κ κ κ κ κ κ κ κ κ γγβ γβγ κ κ κ κ κ κ κ κ γγβ γβγ p p p p p p p p p p p p p p p p p p p p pWpLpL ×
  • 2014 MRT We finally obtain for the boost matrix L3× 2 (Λp) as a function of the deflection angle Ω:                         Ω+Ω − Ω+Ω − − + − − − Ω−Ω + Ω − −Ω−Ω −+ =Λ cossin 2 1)( sincos 2 1)( 0 2 1)( 001 2 1)( coscos 2 1)( 00 sin 2 1)( sincos 2 1)( 0 2 1)( )( 3 2 2 2 333 2 2 2 33 2 2 2 33 2 2 2 33 2 2 2 2 2 2 333 2 2 2 3 2 2 2 3 23 γ κ κ γβγ κ κ γβ κ κ γβ κ κ γβ κ κ κ κ γγβ κ κ γ κ κ γ p p p p p p p p p p p p p p p p pL × Ω − −=Ω− Ω+Ω − −= cos 2 1)( sin sincos 2 1)( 0 2 2 2 333 3 2 2 2 33 κ κ γβγ γ κ κ γβ p p p p                             −      =Ω⇒               −      =Ω κ κ κ κ 2 2 2 2 2 2 1 2 1 arctan 1 2 1 tan p p cp p c vv Since L3× 2 (Λp) is symmetric we can ‘extract’ the [L3× 2 (Λp)]3 2 =[L3× 2 (Λp)]2 3 components: and we obtain: expressed only by calculating the ratio p2 /κ.
  • Finally, by using the same trigonometric identities relating tanΩ to cosΩ and/or sinΩ: Ω Ω+ =Ω Ω+ =Ω tan tan1 sin tan1 1 cos 2 2 and such that the boost matrix becomes:                             Ω+ + Ω Ω+− Ω Ω+ + Ω+ − − + − − − Ω+ − Ω+ + Ω Ω+− − Ω Ω+ − Ω+ −+ =Λ 2 3 2 2 2 2 33 2 3 2 2 2 2 33 2 2 2 33 2 2 2 2 33 2 2 2 2 2 2 2 2 3 2 33 2 2 2 2 3 2 2 2 3 23 tan1 1 tan tan1 2 1)( tan tan1 tan1 1 2 1)( 0 2 1)( 001 2 1)( tan1 1 tan1 1 2 1)( 00 tan tan1 2 1)( tan tan1 tan1 1 2 1)( 0 2 1)( )( γ κ κ γβγ κ κ γβ κ κ γβ κ κ γβ κ κ κ κ γγβ κ κ γ κ κ γ p p p p p p p p p p p p p p p p pL × 2014 MRT We can now either use the relation Ω=arctan{½(|v|/c)[(p2/κ)2 – 1]/(p2/κ)} or the one above by using tanΩ and its square, with p2/κ and β3 =|v|/c given beforehand to calculate things.
  • The Klein-Gordon Equation When Schrödinger wrote down the non-relativistic equation (i∂t|Ψ 〉=(2 /2m)|Ψ 〉) now bearing his name, he formulated the corresponding relativistic equation. The equation is derived by inserting the operator substitution E → i∂t and p →−i∇r (with p=[E/c,p]= i∂µ where ∂µ =∂/∂xµ ={∂/∂ct,∇} and ∂µ =∂/∂xµ ={∂/∂ct,−∇}) into the relativistic relation between energy and momentum for a free particle: where mo is the rest mass of the particle. We then obtain the Klein-Gordon equation: 2014 MRT 42222 cmcE o+= p where the D’Alembertian operator is defined as = (1/c2 )∂t −∇2 =ηµν ∂µ ∂ν = ∂µ ∂µ . The amplitude ϕ is a one-component scalar quantity which under an inhomogeneous Lorentz transformation, xµ = Λµ ν xν +aµ , transforms as ϕ (x)=ϕ (x) or equivalently as ϕ (x) =ϕ [Λ−1 (x−a)]. In natural units (i.e. c =  =1) and the Dirac notation ϕ (x)=〈x|ϕ〉 with x =xµ =[t,r], we get: ),()( ),( 422222 tcmc t t o r r ϕ ϕ +∇−= ∂ ∂ −  0)( 2 o =+ ϕxm
  • In order to give a physical interpretation to the Klein-Gordon equation, by analogy with the non-relativistic equation, one might try to define a probability density, ρ, and a probability current, j, in such a way that a continuity equation holds between them. One is then led to the following expressions for ρ and j: which by virtue of the Klein-Gordon equation, satisfy: 2014 MRT )( 2 )( 22 ** o * 00 * * * 2 o ϕϕϕϕϕϕϕϕϕ ϕϕ ϕρ ⋅∂−∂⋅=⋅∂−∂⋅=         ∂ ∂ − ∂ ∂ = iii o im j cm i ttcm i  and which for E≈moc2 indeed reduces to the expression for the probability density in non- relativistic quantum mechanics. It is, however, to be noted that in general ρ may assume negative as well as positive values because the Klein-Gordonequationis of second order in the time variable and therefore ϕ and ∂t ϕ can be prescribed arbitrarily at some time to. The constants appearing in the density and current have been so determined that these expressions reduce to the usual expressions for the Schrödinger theory in the non- relativistic limit. If in the expression for ρ we substitute for i∂t ϕ, Eϕ we obtain: 0= ∂ ∂ +• t ρ j∇ ϕϕρ * 2 ocm E = In 1934, Pauli and Weisskopf re-interpreted the Klein-Gordon equation as a quan- tum field equation analogous to Maxwell’s equations for the electromagnetic field.
  • The Dirac Equation The Dirac equation has special importance because it describes particles of spin-½, and both electrons and protons have spin-½. Many other elementary particles, including the neutron, the µ meson, &c., have spin-½. In 1928, Dirac discovered the relativistic equation which now bears his name while trying to overcome the difficulties of negative probability densities of the Klein-Gordon equation. 2014 MRT (where =∂/∂t +∇2 ) if it is to describe a free particle of rest mass mo, since this equation implies that the energy momentum relation for a free particle p2 =mo 2 c2 is satisfied, and that in the correspondence limit, classical relativity still remains a valid concept. The reasoning which led Dirac to the Dirac equation was as follows: If we wish to prevent the occurrence of negative probabilities densities, we must then avoid time derivatives in the expression for the probability density ρ. The wave equation must therefore not contain time derivatives higher than first order. Relativistic covariance, furthermore, requires that there be essentially complete symmetry in the treatment of the spatial and time components. We must therefore also require that only first-order spatial derivatives appear in the wave equation. Thus the Dirac wave function must satisfy a first-order linear differential equation in all four x≡xµ =[t,r]coordinates. The linearity is required in order that the superposition principle of quantum mechanics hold. Finally, we must also require ψ(x) that obey the equation: 0)(2 22 o =        + x cm ψ 
  • Let us therefore assume that ψ consists of N components ψα (α = 1, 2, …, N) where the number N is as yet unspecified; it will turn out to be four. The most general first-order linear equation is then one which expresses the time derivative of one component as a linear combination of all the components as well as their spatial derivatives. Inserting the appropriate dimensional factors (i.e.,c and=h/2π),themostgeneralequationpossibleis: 2014 MRT ( )4,,...,2,10 1 4 1 o 3 1 4 1 i.e.N cmi xtc N k i N ik ik ==+ ∂ ∂ + ∂ ∂ ∑∑∑ = == = = αψβ ψ α ψ α αα α α α  In this equation, ψ is a column matrix of N rows, and α =α i ={α1 ,α2 ,α3 } and β are both Hermitian matrices of N rows and columns. This is the Dirac equation in itssimplestform! The Dirac equation can also be written in Hamiltonian (with H also Hermitian) form: Assuming the homogeneity of space-time, the α i kα and βkα are dimensionless constants, independent of the space-time coordinates x0 , x1 , x2 , x3 . A natural way to simplify these equations is to use matrix notation which reduces them to the following equation:       ΛΨ==+ ∂ ∂ + =∂ ∂ Λ = ∑ ),(0 )( 0 0 )(o 3 1 0 jp p i i i mppx cmi xtcx ψψβ ψ α ψ µ µµ  ψβψ ψ )( 2 ocmciH t i +•−== ∂ ∂ ∇α The probability density, ρ, and a probability current, j, are thus spelled out as: ψαψψψρ ii cj ** == and
  • In order to derive the properties of the α and β matrices, we must multiply the equation (1/c)(∂ψ/∂t)+Σi αi (∂ψ/∂xi )+(imo c/)βψ =0 by the operator: which has the effect of introducing second derivatives. The terms with ∂t or mixed derivatives between space and time cancel and we obtain: 2014 MRT We have symmetrized the α i α j term, which is permissible since ∂/∂xi and ∂/∂x j commute. To agree with the Klein-Gordon equation, the right-hand side of the equation above must reduce to: This imposes the following conditions: βα  cmi xtc i i i o 3 1 1 − ∂ ∂ − ∂ ∂ ∑= ∑∑∑ == = ∂ ∂ ++− ∂∂ ∂ += ∂ ∂ 3 1 o2 2 22 o 3 1 3 1 2 2 1 2 2 2 )()( 1 i i ii i j ji ijji x cmicm xxtc ψ βαβαψβ ψ αααα ψ  ψψ 2 22 o2  cm −∇ Ii ii jiijji == =+ =+ βα βαβα δαααα 2 2 1 )( 0 )( i.e., that the α s as well as any αi and β anticommute, and that the square of all four matrices is unity.
  • Without going into the derivations, if I denotes the unit 2×2 matrix, and σ =σ i are the Pauli matrices, then the 4×4 matrices are given by: satisfy all our conditions: they are Hermitian and can be seen to anticommute by using the anticommutative properties of the three Pauli matrices σ s (i.e.,σiσj =δij +iεijkσk ). 2014 MRT       − =      = I I i i i 0 0 0 0 β σ σ α and We can make this equation look even more symmetrical by introducing the matrices γ µ : 0o 3 1 0 =+∂−∂− ∑= ψψαβψβ cmii i i i      = = = ii αβγ βγ γ µ 0 With these definitions, γ 0 is Hermitian, with (γ 0 )2 =+1, and the γ s are anti-Hermitian, i.e., (γ i )* =−1with (γ i )2 =−1so that the γ matrices satisfy the following commutation rule: ( )AlgebraCliffordTheµνµννµνµ ηγγγγγγ 2},{ =+= Let us finally put the Dirac equation in covariant form. When the Dirac equation is written as (1/c)(∂ψ/∂t)+Σi αi (∂ψ/∂xi )+ (imo c/)βψ =0, the spatial derivatives are multiplied by a matrix – whereas the time derivatives is not. To eliminate this distinction, let us multiply (1/c)(∂ψ/∂t)+Σi αi (∂ψ/∂xi )+(imo c/)βψ =0 by β on the left to obtain:
  • The Dirac matrices γµ are written (N.B., this is one of the many basis available): 2014 MRT or in a more abridged way:             − − =             − − =             =             − − = 0010 0001 1000 0100 000 000 000 000 0001 0010 0100 1000 0010 0001 1000 0100 3210 γγγγ and,, i i i i       − ==      − == 0 0 0 0 0 0 σ σ i i I I γγγγ and By extension, a fifth matrix is defined by:      − ==      − ==−= I I ii I I i 0 0 0 0 3210532105 5 γγγγγγγγγγγ or We list a few identities for these gamma matrices: )(4tr4tr νλµσνσµλλσµνσλνµµννµ ηηηηηηγγγγηγγ +−== and where tr is the trace – or the sum of diagonal components – andη00 =−1,η11 =η22 =η33 =+1 and ηµν ≡0 for µ ≠ν. And the following identities follow from the basic Clifford identity: qprrqpqpqppp ///−=///⋅=///−=/ 242: µ µ µ µ µ µ γγγγγγµ and,overSum which are not really used presently but more often in quantum electrodynamics.
  • In terms of the γ matrices, the equation −iβ∂0ψ −iΣi βαi ∂iψ +mo cψ =0 now reads: 2014 MRT pγ •−===⋅=/ 00 ppppp γγγγ µ µµ µ where γµ is defined by γµ =ηµν γν . With this notation, natural units (i.e., c==1) and the Dirac notation ψ (x)=〈x|ψ〉 with x ≡xµ =[t,r], we get: 0oo =      +•−=      +∂− ψψγ µ µ  cm i cm i ∂γ Feynman has introduced the so-called ‘slash’ notation to simplify the equation even further. For example, he denoted by p the ‘reduced’ quantity: 0)( o =+∂/− ψxcmi where ∂=γ µ ∂µ =γ 0 ∂0 +γ •∇. where our summation convention has been reintroduced. With this last equation, we have written the Dirac equation in a covariant form where space and time derivatives are treated alike.
  • The wave function is now a bi-spinor ψ that holds 4 components: 2014 MRT0)()( 0)()( 0)()( 0)()( 4 2 o21 3 2 o12 2 2 o43 1 2 o34 =+−−+ =+−+− =−+−+ =−++− ψψψ ψψψ ψψψ ψψψ Ecmcpippc Ecmcpippc Ecmcpippc Ecmcpippc zyx zyx zyx zyx The matrices γµ sum-up the notion of spin since they correspond to a generalization of the Pauli spin matrices. For this reason, the Dirac equation is convenient to the description of fermions (i.e., half-integer spin particles). In fact, of the 4 degrees of liberty of the bi-spinor, 2 are used to represent the particle in the states of spin ±½ and the 2 other ones, the antiparticle with states of spin ±½. Under a more explicit form where the matrices are developed, i.e., a system of 4 coupled equations, we have: The Dirac equation describes 4 linear coupled differential equations.                           = 4 3 2 1 ψ ψ ψ ψ ψ (Primus, u or spin-±½ ‘Matter’) (Secundus, v or spin-±½ ‘Anti-matter’)
  • The Dirac equation (cα •p+βmo c2 )ψ =Eψ may also be written in two-component form by means of the Pauli spin matrices: 2014 MRT where so that in place of the 4 coupled equations we have: With σ =[σx,σy,σz] it is seen that and the 2×2 identity matrix:       − =      − =      = 10 01 0 0 01 10 zyx i i σσσ and,       = 10 01 I       − =      = I I 0 0 0 0 βand σ σ α 0)( 0)( 2 o 2 o =+−• =−+• vu uv pσ pσ ψψ ψψ Ecmc Ecmc       =      = 4 3 2 1 ψ ψ ψ ψ ψ ψ vu and
  • The equations cσ •pψv +(moc2 –E)ψu =0 and cσ •pψu −(moc2 +E)ψv =0 provide a useful starting point for the non-relativistic approximation. From cσ •pψv +(moc2 –E)ψu =0, 2014 MRT v c vm vmc 2 2 2 1 =≈ v u ψ ψ Consequently, for v/c<<1, the component of interest is ψu, also known as the “large” component. Eliminating ψv between the equations cσ •pψv +(mo c2 –E)ψu =0 and cσ •pψu −(moc2 +E)ψv =0, ψu satisfies: At non-relativistic energies, we have: uupσ ψψ ))(()( 2 o 2 o 22 cmEcmEc +−=• which enables us to make an estimate of the relative magnitude of ψu and ψv. Approx- imating σ •p≈mv and E–moc2 ≈½mv2 , we get: vu pσ ψψ 2 ocmE c − • = 2 o 2 o cmcmEE <<−≡′ or Ecm cm E EcmcmEcmE ′≈        ′ +′=+− 2 o2 o 2 o 2 o 2 o 2 2 12))(( uu pσpσ ψψ o2 ))(( m E •• =′ The equation c2 (σ •p)2 ψu =(E–moc2 )(E+moc2 )ψu =0 then becomes: u
  • Further ψu =[(σ •p)(σ •p)/2mo]ψu reduction of equation is possible with the aid of the identity (σ •A)(σ •B)=(A•B)+iσ •(A× B) (which is easily verified from the definitions of the Pauli matrices – See Landau-Lifshitz Quantum Mechanics). Therefore: 2014 MRT 2 ))(( p=•• pσpσ or, with p=−i∇: This equation is still in two-component form since ψu is a two-component function. But each component of ψu satisfies this equation too so that we can drop all subscripts and write: and, substituting in ψu =[(σ •p)(σ •p)/2mo]ψu we get: uu ψψ o 2 2m p E =′ u 2 o 2 u 2 ψψ ∇−=′ m E  ψψ 2 o 2 2 ∇−=′ m E  which is the time-independent Schrödinger equation for a free particle.
  • The states of positive energy and spin ±½ are represented by the solutions u1 = col[1 0 0 0] and u2 =col[0 1 0 0] whereas the states of negative energy and spin ±½ are represented by the solutions u3 =col[0 0 1 0] and u4 =col[0 0 0 1]. 2014 MRT 2 o 0 3 2 o 0 3 1 2 0 1 1 0 0 0 1 0 u 0 0 0 1 u cmp Σ cmp Σ = −= = +=             =             = γγ and which shows that antifermions (i.e., positrons) have negative parity. Using a parity transformation such that γ 0 ψ (t,−r)=ψ (t,r), we get: 2 o 0 3 2 o 0 3 1 4 0 1 3 0 1 0 0 0 u 0 1 0 0 u cmp Σ cmp Σ −= −= −= +=             − =             − = γγ and which shows that fermions (i.e., electrons, protons, neutrons) have positive parity whereas:
  • We are now in a position to investigate the Lorentz invariance of the Dirac equation and to establish its connection with the representation of the inhomogeneous Lorentz group. Under an inhomogeneous Lorentz transformation xµ =Λµ ν xν +aµ with ηµν Λµ ρ Λν σ = ηρσ the Dirac equation will be form-invariant if we define: and if S(Λ) satisfies the condition S(Λ)−1 γ λ S(Λ)=Λλ µ γ µ ; S(Λ) is a 4×4 matrix which operates on the components of ψ, i.e., the equation above written explicitly reads: 2014 MRT Let us now exhibit the actual form of S for a given Lorentz transformation. For the infinitesimal Lorentz transformation Λ=1+ελ, xµ =xµ +ελ µ ν xν (λ µν =−λνµ ), we write S(Λ) to first order in ε as follows: S(1 +ελ)=1 +εT and [S(1+ελ)]−1 =S−1 (1+ελ)=1 −εT. For the infinitesimal situations, the equation S(Λ)−1 γ λ S(Λ)=Λλ µ γ µ can be rewritten as: where ∂µ =∂/∂xµ . Note that the γ matrices remainunalteredunderLorentz transformation. The form-invariance of the Dirac equation is the statement that in the new frame ψ obeys the equation: )]([)()()()( 1 axSxSx −ΛΛ=Λ= − ψψψ ∑= − −ΛΛ= 4 1 1 )]([)()( β βαβα ψψ axSx 0)()( o =+∂− xcmi ψγ µ µ ν ν µµν ν µµµµµµ γλεγγγγεγεγεγ +=Λ=−+=+−=− )()1()1(1 TTTTSS where T must be such that γ µ T− Tγ µ =λµ ν γν and is given by T =(1/8)λµν (γµ γν − γν γµ ).
  • As an example of a transformation of single-particle states*, consider a free positive- energy electron (of mass mo) of helicity +1 and momentum p along the positive z- direction. We choose a barred system in such a way that it will coincide with the rest system of the electron (see Figure). In the primed system the electron wave function can be written as: vO O z 2014 MRT * J.J. Sakurai, Advanced Quantum Mechanics, 1967 – P. 101. y pmo E c c 3pv = The question is: What is the wave function for the same physical situation in the unbarred system? tcm i tcm i V x 2 o 2 o e 0 0 0 1 1 e)()(  −−             == xΨψ )()( 1 Lorentz xSx ψψ − = The answer Sakurai derives starts with: where (Majorana representation for gamma matrices and x4 =ict) 2 sinh 0 0 2 cosh 2 sinh 2 cosh 3 3 4443 1 Lorentz θ σ σθθ γγ θ       +=−= × − IiS and with the angle θ given by: cmcm E o 3 2 o 2 sinh 2 cosh p == θθ and and the Pauli spin matrix:       − = 10 01 3σ
  • Sakurai obtains: Then, since This result is in complete agreement with u(1) (p), with p1 =p2 =0 obtained by solving directly the Dirac equation. )(u 0 0 1 2 0 0 0 1 2)(2 )(22 0 0 0 1 )1( 2 o 32 o 2 o 2 o 2 o 2 o 2 o 33 2 o 2 o 33 2 2 o 1 p p p p =               + + =                             + + + + =             − cmE c cm cmE cm cmE cmEcm c cmEcm c mc cmE SLorentz σ σ )(2 1 2 cosh 2 sinh 22 1cosh 2 cosh 2 o 2 o 32 2 o cmEcm ccmE + =−      = + = + = pθθθθ and 2014 MRT
  • and finds: As for the spacetime dependence of the wave function, Sakurai points out that: where he usedV=(mc2 /E)V that follows from the Lorentz contraction of the normalization volume along the direction of motion. In his own words: Once we know the form of the wave function for a particle at rest, the correct wave function for a moving particle of definite momentum can be cons- tructed just be applying S−1 . This operation is sometimes known as the Lorentz boost. tE ii tE i x i tEx i tcm i Lorentz tcm i LorentzLorentz EV cm V cmE c Vcm cmE S V SxSx    −• − −− − − −− = =               + + =             === xp p p p p px e)(u e)(u 1 e 0 0 1 2 e 0 0 0 1 1 e)()()( )1( 2 o )1( )( 2 o 32 o 2 o111 33 33 2 o 2 o Ψψψ 32 o 3 2 o 3 sinhcosh x cm t cm E c x tt p −= −= θθ 2014 MRT
  • For an infinitesimal rotation through an angle ε about the 1-axis, λ23 =−λ32 =+1 with all other λµν =0. The generator for such a transformation is given by the equation T =(1/8)λµν (γµ γν − γν γµ ), and is T =½γ2 γ3 . With the representation of the α matrix: we get: 2014 MRT 2 sinΣ 2 cos ee)( 1 Σ)( 1 121 θθ θθ i RS i RT += == + with The S(θ ) corresponding to a rotation through the angleθ about the 1-axis is therefore: The mapping |ψ〉→ |ψ〉 for an arbitrary Lorentz transformation is induced by a unitary operator U(Λ,a): )]([)( )(),(),( 1 axS xaUxaUx −ΛΛ= Λ==Λ − ψ ψψψ ψψ =Λ ),( aU       = 0 0 i i i σ σ α 1 1 1 322 1 1 Σ 20 0 )( ii RT +=      +=−= σ σ αα 
  • where D, the infinitesimal generator, is determined by using equations xµ =xµ +ελ µ ν xν and S(1+ελ)=1 +εT by equation 〈x|U(Λ,a)|ψ〉=〈x |ψ〉=U(Λ,a)ψ(x)=S(Λ)ψ [Λ−1 (x–a)]: For an infinitesimal quantum transformation, we write: 2014 MRT D i aU ε  +=Λ 1),( For an infinitesimal rotation about the 3-axis, T=½iΣ3 and only λ21 =−λ12 =+1 are different from zero, so that: Thus, a Dirac particle has in addition to its orbital momentum, r× p, an intrinsic angular momentum Σ of magnitude /2. It is to be noted that the spin operator ½Σ is not a constant of motion, since [H,Σ]≠0.The same is true for the orbital angular momentum; however, the total angular momentum J=(/2)Σ +(r× p) is a constant of the motion. so that 3323 )(Σ pr ×+= D )( ρ σ σ ρ λ ∂−−= xTiD     +∂−+=       + ∂ ∂ −+= −+=      + )()()( )( )()()1( )()1(1 xxTx x x xxT xxTD i ψλεψ ψ λεψε λεψεε µ ν ν µ µ µ
  • We now consider the Dirac equation with electromagnetic coupling. in which v is the velocity of a particle with positive charge q and rest mass mo, A and ϕ are the vector and scalar potentials, respectively. The fields are then given by: 2014 MRT ϕq c q vmL −•+= Av2 o 2 1 It will now be assumed that the same modification can be introduced into the free- particle Dirac equation (cα •p+βmo c2 )ψ =Eψ so that the proper equation for a particle in a field with vector potential A and scalar potential ϕ is: ϕ∇×∇ − ∂ ∂ −== tc A EAB 1 and ψϕβψ ])([ 2 o qcmqcE ++−•= Apα In two-component form, by analogy to the equations cσ •pψv +(moc2 –E)ψu =0 and cσ •pψu −(moc2 +E)ψv =0, we set: vvu uuv Apσ Apσ ψψϕψ ψψϕψ Eqcmqc Eqcmqc =−−−• =++−• )()( )()( 2 o 2 o The free-particle Dirac equation (cα •p+βmo c2 )ψ =Eψ must now be modified to include effects due to external fields. Classical considerations suggest how this may be accomplished. In the presence of external fields a possible Lagrangian for the system is:
  • 2014 MRT The identity (σ •A)(σ •B)=(A•B)+iσ •(A× B) gives (σ •A)2 =π2 +iσ •(π × π)=[p–(q/c)A]2 – i(q/c)σ •(p× A+A× p). But p× A=−i∇ × A−A× p, therefore we get: As in the free particle case, we shall be interested mainly in the “large” component ψu . From the equations σ •(cp–qA)ψv +(moc2 +qϕ)ψu =Eψu and σ •(cp–qA)ψu −(moc2 −qϕ)ψv = Eψv, we set: where Our objective now is to obtain an approximation to the equation above to order v2 /c2 . Apπ c q cmqEqcmE cm KcmEE −= −′+ = −+′ =−=′ and, ]2/)[(1 1 2 2 2 o 2 o 2 o2 o ϕϕ uu πσπσ ψψϕ )()( 2 1 )( o ••=−′ K m qE uu AσAp ψψϕ         •−      −=−′ ×∇ cm q c q m qE o 2 o 22 1 )(  and the approximation to (E′– qϕ)ψu=(1/2mo)(σ •π)K(σ •π)ψu with K=1 is: AσApπσ ×∇•−      −=• cm q c q o 2 2 2 )( 
  • 2014 MRT To order v2 /c2 , we get: The next higher approximation is obtained by setting: So that the equation (E′– qϕ)ψu=(1/2mo)(σ •π)K(σ •π)ψu now becomes: Let ψ =Ωψu where Ω=1+(σ •π)2 /8mo 2 c2 and, to order v2 /c2 , Ω−1 =1−(σ •π)2 /8mo 2 c2 . Multiplying the equation above on the left by Ω−1 and replacing ψu by Ω−1 ψu one obtains: 2 o2 1 cm qE K ϕ−′ −= uu πσπσπσ ψ ϕ ψϕ         • −′ •−•=−′ )( 2 )()( 2 1 )( 2 o 2 o cm qE m qE ψ ϕ ψϕ 1 2 o 21 o 11 )( 2 )()( 2 1 )( −−−− Ω      • −′ •−•Ω=Ω−′Ω πσπσπσ cm qE m qE ψ ϕ ψ ϕ ψψψ ψϕψϕψϕψϕ )( 4 )()( 2 )( 2 1 8 )( )( 2 1 )( 2 1 8 )( )()( 8 )( )()( 22 o 1 2 o 1 o 23 o 4 2 o 121 o 22 o 2 22 o 2 11 πσπσπσπσ πσ πσπσ πσπσ • −′ •=Ω• −′ •Ω • −•=Ω•Ω • −′−−′ • −−′=Ω−′Ω −− −− −− cm qE cm qE m cmmm cm qEqE cm qEqE
  • 2014 MRT Substitution of the above simplification into the equation (E′−qϕ)ψ = […]ψ above gives: Therefore: Hence, the transformed equation Ω−1 (E′−qϕ)Ω−1 ψu, to order v2 /c2 , becomes: A number of simplifications are possible with the aid of: ψϕϕ ϕψϕ   • −′+•−′•−    −′ • +•−•=−′ 22 o 2 22 o 22 o 2 4 23 o 2 o 8 )( )())()(( 4 1 )( 8 )( )( 8 1 )( 2 1 )( cm qEqE cm qE cmcmm qE πσ πσπσ πσ πσπσ ϕϕϕϕ ϕϕϕϕϕ ∇×∇∇∇∇ ∇ πσπσσσπσ σpσpσπσπσ •+•−=••−•• •=•−•=•−′−−′• ii qiqqqEqE 2))(())(( )())(())((   ))()((22 ))()((2))]()(())([( )])(())()[(())(()()( 2 22 πσπσπσ πσπσπσπσπσ πσπσπσπσπσ •−′•+•−•= •−′•+••−′−−′•− −•−′−−′••=•−′+−′• ϕϕϕ ϕϕϕ ϕϕϕϕ qEqq qEqEqE qEqEqEqE ∇×∇∇  ψϕϕψϕ       •−•+•−•=−′ ∇×∇∇ πσπσπσ 22 o 22 o 2 4 23 o 2 o 48 )( 8 1 )( 2 1 )( cm q cm q cmm qE 
  • Also, from the relation (σ •A)2 =[p–(q/c)A]2 −(q/c)σ •∇ × A obtained earlier, we get: and, to order v2 /c2 , we finally have: 2014 MRT pppπ πσ ×∇)×∇∇×∇∇×∇× ϕϕϕϕϕ 2222 4 4 4 2 1 ( 111 )( 1 c i ccc c p c −=−−== =•  AσApπσ ×∇•−      −=• cm q c q mm o 2 o 2 o 22 1 )( 2 1  where E′=E–moc2 . This equation, which may be regarded as the Schrödinger equation for an electron interacting with fields describable by the potentials A and ϕ, is the starting point for discussions of atomic and molecular properties. With q =−e, where e is the electronic charge, we obtain for an electron to order v2 /c2 : ψϕϕ ψϕ    •−•−−     −•+      +=+′ pσ AσAp ×∇∇∇ ×∇ 22 o 22 o 2 23 o 4 o 2 o 488 22 1 )( cm e cm e cm p cm e c e m eE  
  • This is the energy associated with the scalar potential energy, ϕ (105 cm−1 ). This significance of the various terms and their energies, indicated to within an order of magnitude, are (the ‘cm−1 ’ scale is given by the wave number 1/λ ≅ 8000 cm−1 ): This contains the kinetic energy (i.e., p2 /2mo) and interaction term (i.e., (e/2moc)(p • A + A • p) + e2 A2 /2moc2 ) with a field represented by a potential vector A (105 cm−1 ). The interaction terms are responsible or contribute to numerous physical processes among which are absorption, emission and scattering of electromagnetic waves, diamagnetism, and the Zeeman effect. 2014 MRT ϕe The spin-orbit interaction (10-103 cm−1 ). More precisely, it is (e /8moc2 ){σ • [p − (e/c)A] × E − σ • E× [p − (e/c)A]} and it arises from the fact that the motion of the magnetic moment gives rise to an electric moment for the particle which then interacts with the electric field. This term appears in the expression of the relativistic energy: and is therefore a relativistic correction to the kinetic energy (i.e., p2 /2m) (0.1 cm−1 ). The interaction of the spin magnetic moment (i.e., µS = 2⋅e/2mo⋅ /2) with a magnetic field B = ∇ × A (1 cm−1 ). Thus, it is the magnetic moment of one Bohr magneton, e/2moc (i.e., µB = 9.2741×10−24 A⋅m2 or J/T) with the magnetic field. 2 o2 1       + Ap c e m Aσ ×∇• cm e o2  23 o 4 8 cm p +−+≅+ 23 o 2 2 o 2222 o 82 )( cm p m p cmcpcm 4 ϕ∇∇ •− 22 o 2 8 cm e This term produces an energy shift in s-states and is known as the Darwin (1887-1962) term (< 0.1 cm−1 ). It is thus a correction to the direct point charge interaction due to the fact that in the representation (Foldy-Wouthuysen), the particle is not concentrated at a point but is spread out over a volume with radius whose magnitude is roughly that of a Compton wavelength, /moc. pσ ×∇ϕ•− 22 o4 cm e ϕe c e m −      + 2 o2 1 Ap As a combination, this term represents the interaction of a point charge with the electromagnetic field.
  • The solution in the Coulomb field has important applications, particularly to the energy levels of the hydrogen atom and to the calculation of x-ray spectra due to the K and L electrons of the heavy elements. 2014 MRT In the general case of hydrogen-like atoms (e.g., the Hydrogen atom H1 has Z=1), the energy levels Enj are described by the total angular momentum quantum number j with: ( ) 2 222 2 1 22 2 o 1     −++′ += Zjn Z cmE jn α α where n′=0,1,2,….; j =1/2,3/2,…; α =e2 /4πc≅1/137 and can be approximated by:         +         − + +−≈  njn Z n Z cmcmE jn 4 31 1 2 1 2 1 2 2 22 2 o 22 o α α and where the principle quantum number n = n′+j +½ =1,2,…. The observed fine structure of the levels of hydrogen and hydrogen-like atoms, particularly Helium He+ , is in good agreement with the Dirac theory. FIN The atomic number is the number of protons found in the nucleus of an atom and therefore identical to the charge number of the nucleus. It is conventionally represented by the symbol Z. The atomic number uniquely identifies a chemical element. In an atom of neutral charge, the atomic number is also equal to the number of electrons.
  • References / Study Guide C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976. California State University, Haywood This is my favorite go-to reference for mathematical physics. Most of the complex variable and matrix definitions, &c. are succinctly spelled out and the wave packet discussions served as a primer followed by more in-depth discussion reference of Appendix A of Sakurai’s book below and also the solutions to the wave function (i.e. Radial, Azimutal and Angular) are very well presented and treated in this very readable 300 page volume. J.J. Sakurai, Modern Quantum Mechanics, Revised Edition, Addison-Wesley, 1994. University of California at Los Angeles Only half (Chapters 1-3 or about the first 200 pages) of this book makes for quite a thorough introduction to modern quantum mechanics. Most of the mathematical and angular momentum framework (i.e. orbital, spin and total as well as spherical harmonics discussed earlier) is based on Sakurai’s post-humanous presentation of the subject. S.S. Schweber, An Introduction to Relativistic Quantum Field Theory, Dover, 1989. Brandeis University Quite a voluminous 900 page tome written by a colleague of Hans Bethe. Most of the postulates and rotation invariance discussion, which was really the base or primer for the rest, is based on the first 50 pages of this book. The other 50 pages formed the basis of the relativistic material as well as the Klein-Gordon and Dirac equations. S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press, 2014. Josey Regental Chair in Science at the University of Texas at Austin Soon to become the seminal work on Quantum Mechanics, this ‘course’ from Weinberg fills-in all the gaps in learning and exposition required to fully understand the theory. Including the Historical Introduction, I would definitely suggest Chapter 2-4 and do the problems! Once you do, you will have truly mastered the theory to handle the next set of books from Weinberg – especially QFT I & II. S. Weinberg, The Quantum Theory of Fields, Volume I, Cambridge University Press, 1995. Josey Regental Chair in Science at the University of Texas at Austin Volume 1 (of 3) introduces quantum fields in which Chapter 1 serves as a historical introduction whereas the first 5 sections of Chapter 2 make for quite an elevated introduction to the relativistic quantum mechanics discussed here. Weinberg’s book was crucial in setting up the notation and the Lorentz and Poincaré transformations and bridging the gap for the relativistic one-particle and mass zero states using Wigner’s little group. Very high level reading! M. Weissbluth, Atoms and Molecules, Academic Press, 1978. Professor Emeritus of Applied Physics at Stanford Weissbluth’s book starts by knocking you off your chair with Angular Momentum, Rotations and Elements of Group Theory! The text is mathematical physics throughout but where more than one sentence occurs, the physics (or quantum chemistry) abounds. Chapter 15 starts Part III – One-electron Atoms with Dirac’s equation and in Chapter 16, it offers most of the 2014 MRT