A Brief Introduction to Relativistic Quantum Mechanics, Slides of Quantum Mechanics

Download A Brief Introduction to Relativistic Quantum Mechanics and more Slides Quantum Mechanics in PDF only on Docsity! A Brief Introduction to Relativistic Quantum Mechanics Hsin-Chia Cheng, U.C. Davis 1 Introduction In Physics 215AB, you learned non-relativistic quantum mechanics, e.g., Schrödinger equation, E = p2 2m + V, E → i~ ∂ ∂t , p → −i~∇, i~ ∂ ∂t Ψ = ~2 2m ∇2Ψ + VΨ. (1) Now we would like to extend quantum mechanics to the relativistic domain. The natural thing at first is to search for a relativistic single-particle wave equation to replace the Schrödinger equation. It turns out that the form of the relativistic equation depends on the spin of the particle, spin-0 Klein-Gordon equation spin-1/2 Dirac equation spin-1 Proca equation etc It is useful to study these one-particle equations and their solutions for certain problems. However, at certain point these one-particle relativistic quantum theory encounter fatal inconsistencies and break down. Essentially, this is because while energy is conserved in special relativity but mass is not. Particles with mass can be created and destroyed in real physical processes. For example, pair annihilation e+e− → 2γ, muon decay µ− → e−ν̄eνµ. They cannot be described by single-particle theory. At that stage we are forced to abandon single-particle relativistic wave equations and go to a many-particle theory in which particles can be created and destroyed, that is, quantum field theory, which is the subject of the course. 1 2 Summary of Special Relativity An event occurs at a single point in space-time and is defined by its coordinates xµ, µ = 0, 1, 2, 3, x0 = ct, x1 = x, x2 = y, x3 = z, (2) in any given frame. The interval between 2 events xµ and x̄µ is called s, s2 = c2(t− t̄)2 − (x− x̄)2 − (y − ȳ)2 − (z − z̄)2 = (x0 − x̄0)2 − (x1 − x̄1)2 − (x2 − x̄2)2 − (x3 − x̄3)2. (3) We define the metric gµν =  1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1  , (4) then we can write s2 = ∑ µ,ν gµν(x µ − x̄µ)(xν − x̄ν) = gµν∆x µ∆xν , (5) where we have used the Einstein convention: repeated indices (1 upper + 1 lower) are summed except when otherwise indicated. Lorentz transformations The postulates of Special Relativity tell us that the speed of light is the same in any inertial frame. s2 is invariant under transformations from one inertial frame to any other. Such transformations are called Lorentz transformations. We will only need to discuss the homogeneous Lorentz transformations (under which the origin is not shifted) here, x′µ = Λµ νx ν . (6) gµνx µxν = gµνx ′µx′ν = gµνΛ µ ρxρΛν σx σ = gρσx ρxσ ⇒ gρσ = gµνΛ µ ρΛ ν σ. (7) It’s convenient to use a matrix notation, xµ :  x0 x1 x2 x3  = x. (8) 2 Substituting it into the third equation, we have ∇× ( E + 1 c ∂A ∂t ) = 0, (27) then we can define a potential φ, such that E = −∇φ− 1 c ∂A ∂t . (28) Gauge invariance: E, B are not changed under the following transformation, A → A−∇χ φ → φ+ 1 c ∂ ∂t χ. (29) (cρ,J) form a 4-vector Jµ. Charge conservation can be written in the Lorentz covariant form, ∂µJ µ = 0, (φ,A) from a 4-vector Aµ (Aµ = (φ,−A)), from which one can derive an antisym- metric electromagnetic field tensor, F µν = ∂µAν − ∂νAµ (note: ∂i = −∂i = − ∂ ∂xi , i = 1, 2, 3). (30) F µν =  0 −Ex −Ey −Ez Ex 0 −Bz By Ey Bz 0 −Bx Ez −By Bx 0  , Fµν =  0 Ex Ey Ez −Ex 0 −Bz By −Ey Bz 0 −Bx −Ez −By Bx 0  (31) Maxwell’s equations in the covariant form: ∂µF µν = 1 c Jν (32) ∂µF̃ µν = ∂µFλν + ∂λFνµ + ∂νFµλ = 0 (33) where F̃ µν ≡ 1 2 εµνλρFλρ, (34) ε0123 and its even permutation = +1, its odd permutation = −1. Gauge invariance: Aµ → Aµ + ∂µχ. One can check F µν is invariant under this transformation. 5 3 Klein-Gordon Equation In non-relativistic mechanics, the energy for a free particle is E = p2 2m . (35) To get quantum mechanics, we make the following substitutions: E → i~ ∂ ∂t , p → −i~∇, (36) and the Schródinger equation for a free particle is − ~2 2m ∇2Ψ = i~ ∂Ψ ∂t . (37) In relativistic mechanics, the energy of a free particle is E = √ p2c2 +m2c4. (38) Making the same substitution we obtain √ −~2c2∇2 +m2c2Ψ = i~ ∂Ψ ∂t . (39) It’s difficult to interpret the operator on the left hand side, so instead we try E2 = p2c2 +m2c4 (40) ⇒ ( i~ ∂ ∂t )2 Ψ = −~2c2∇2 +m2c4Ψ, (41) or 1 c2 ( ∂ ∂t )2 Ψ−∇2Ψ ≡ 2Ψ = −m 2c2 ~2 Ψ, (42) where 2 = 1 c2 ( ∂ ∂t )2 −∇2 = ∂µ∂ µ. (43) Plane-wave solutions are readily found by inspection, Ψ = 1√ V exp ( i ~ p · x ) exp ( − i ~ Et ) , (44) where E2 = p2c2 + m2c4 and thus E = ± √ p2c2 +m2c4. Note that there is a negative energy solution as well as a positive energy solution for each value of p. Näıvely one should just discard the negative energy solution. For a free particle in a positive energy state, there is no mechanism for it to make a transition to 6 the negative energy state. However, if there is some external potential, the Klein- Gordon equation is then altered by the usual replacements, E → E − eφ, p → p− e c A, (45) (i~∂t − eφ)2Ψ = c2(−i~∇− e c A)2Ψ +m2c4Ψ. (46) The solution Ψ can always be expressed as a superposition of free particle solutions, provided that the latter form a complete set. They from a complete set only if the negative energy components are retained, so they cannot be simply discarded. Recall the probability density and current in Schródinger equation. If we multiply the Schródinger equation by Ψ∗ on the left and multiply the conjugate of the Schrödinger equation by Ψ, and then take the difference, we obtain − ~2 2m (Ψ∗∇2Ψ−Ψ∇2Ψ∗) = i~(Ψ∗Ψ̇ + ΨΨ̇∗) ⇒ − ~2 2m ∇(Ψ∗∇Ψ−Ψ∇Ψ∗) = i~ ∂ ∂t (Ψ∗Ψ) (47) Using ρs = Ψ∗Ψ, js = ~ 2mi (Ψ∗∇Ψ − Ψ∇Ψ∗), we then obtain the equation of continuity, ∂ρs ∂t + ∇ · js = 0 (48) Now we can carry out the same procedure for the free-particle Klein-Gordon equa- tion: Ψ∗2Ψ = −m 2c2 ~ Ψ∗Ψ Ψ2Ψ∗ = −m 2c2 ~ ΨΨ∗ (49) Taking the difference, we obtain Ψ∗2Ψ−Ψ2Ψ∗ = ∂µ(Ψ∗∂µΨ−Ψ∂µΨ∗) = 0. (50) This suggests that we can define a probability 4-current, jµ = α(Ψ∂µΨ−Ψ∂µΨ∗), where α is a constant (51) and it’s conserved, ∂µj µ = 0, jµ = (j0, j). To make j agree with js, α is chosen to be α = − ~ 2mi . So, ρ = j0 c = i~ 2mc2 ( Ψ∗∂Ψ ∂t −Ψ ∂Ψ∗ ∂t ) . (52) ρ does reduce to ρs = Ψ∗Ψ in the non-relativistic limit. However, ρ is not positive -definite and hence can not describe a probability density for a single particle. Pauli and Weisskopf in 1934 showed that Klein-Gordon equation describes a spin-0 (scalar) field. ρ and j are interpreted as charge and current density of the particles in the field. 7 Covariant form of the Dirac equation Define γ0 = β, γj = βαj, j = 1, 2, 3 γµ = (γ0, γ1, γ2, γ3), γµ = gµνγ ν (70) Multiply Eq. (55) by iβ, iβ × ( 1 c ∂ ∂t + α ·∇ + imc ~ β ) ψ = 0 ⇒ ( iγ0 ∂ ∂x0 + iγj ∂ ∂xj − mc ~ ) ψ = ( iγµ∂µ − mc ~ ) ψ = 0 (71) Using the short-hand notation: γµ∂µ ≡6∂, γµAµ ≡6A,( i 6∂ − mc ~ ) ψ = 0 (72) From the properties of the αj and β matrices, we can derive γ0† = γ0, (hermitian) (73) γj† = (βαj)† = αj†β† = αjβ = −βαj = −γj, (anti-hermitian)(74) 㵆 = γ0γµγ0, (75) γµγν + γνγµ = 2gµνI. (Clifford algebra). (76) Conjugate of the Dirac equation is given by −i∂µψ †γµ† − mc ~ ψ† = 0 ⇒ −i∂µψ †γ0γµγ0 − mc ~ ψ† = 0 (77) We will define the Dirac adjoint spinor ψ by ψ ≡ ψ†γ0. Then i∂µψγ µ + mc ~ ψ = 0. (78) The four-current is jµ c = ψγµψ = ( ρ, j c ) , ∂µj µ = 0. (79) 10 Properties of the γµ matrices We may form new matrices by multiplying γ matrices together. Because different γ matrices anticommute, we only need to consider products of different γ’s and the order is not important. We can combine them in 24−1 ways. Plus the identity we have 16 different matrices, I γ0, iγ1, iγ2, iγ3 γ0γ1, γ0γ2, γ0γ3, iγ2γ3, iγ3γ1, iγ1γ2 iγ0γ2γ3, iγ0γ3γ1, iγ0γ1γ2, γ1γ2γ3 iγ0γ1γ2γ3 ≡ γ5(= γ5). (80) Denoting them by Γl, l = 1, 2, · · · , 16, we can derive the following relations. (a) ΓlΓm = almΓn, alm = ±1 or ± i. (b) ΓlΓm = I if and only if l = m. (c) ΓlΓm = ±ΓmΓl. (d) If Γl 6= I, there always exists a Γk, such that ΓkΓlΓk = −Γl. (e) Tr(Γl) = 0 for Γl 6= I. Proof: Tr(−Γl) = Tr(ΓkΓlΓk) = Tr(ΓlΓkΓk) = Tr(Γl) . (f) Γl are linearly independent: ∑16 k=1 xkΓk = 0 only if xk = 0, k = 1, 2, · · · , 16. Proof:( 16∑ k=1 xkΓk ) Γm = xmI + ∑ k 6=m xkΓkΓm = xmI + ∑ k 6=m xkakmΓn = 0 (Γn 6= I) . Taking the trace, xmTr(I) = − ∑ k 6=m xkakmTr(Γn) = 0 ⇒ xm = 0. for any m. This implies that Γk’s cannot be represented by matrices smaller than 4 × 4. In fact, the smallest representations of Γk’s are 4 × 4 matrices. (Note that this 4 is not the dimension of the space-time. the equality is accidental.) (g) Corollary: any 4×4 matrix X can be written uniquely as a linear combination of the Γk’s. X = 16∑ k=1 xkΓk Tr(XΓm) = xmTr(ΓmΓm) + ∑ k 6=m xkTr(ΓkΓm) = xmTr(I) = 4xm xm = 1 4 Tr(xΓm) 11 (h) Stronger corollary: ΓlΓm = almΓn where Γn is a different Γn for each m, given a fixed l. Proof: If it were not true and one can find two different Γm, Γm′ such that ΓlΓm = almΓn, ΓlΓm′ = alm′Γn, then we have Γm = almΓlΓn, Γm′ = alm′ΓlΓn ⇒ Γm = alm alm′ Γm′ , which contradicts that γk’s are linearly independent. (i) Any matrix X that commutes with γµ (for all µ) is a multiple of the identity. Proof: Assume X is not a multiple of the identity. If X commutes with all γµ then it commutes with all Γl’s, i.e., X = ΓlXΓl. We can express X in terms of the Ga matrices, X = xmΓm + ∑ k 6=m xkΓk, Γm 6= I. There exists a Γi such that ΓiΓmΓi = −Γm. By the hypothesis that X commutes with this Γi, we have X = xmΓm + ∑ k 6=m xkΓk = ΓiXΓi = xmΓiΓmΓi + ∑ k 6=m xkΓiΓkΓi = −xmΓm + ∑ k 6=m ±xkΓk. Since the expansion is unique, we must have xm = −xm. Γm was arbitrary except that Γm 6= I. This implies that all xm = 0 for Γm 6= I and hence X = aI. (j) Pauli’s fundamental theorem: Given two sets of 4×4 matrices γµ and γ′µ which both satisfy {γµ, γν} = 2gµνI, there exists a nonsingular matrix S such that γ′µ = SγµS−1. Proof: F is an arbitrary 4 × 4 matrix, set Γi is constructed from γµ and Γ′ i is constructed from γ′µ. Let S = 16∑ i=1 Γ′ iFΓi. ΓiΓj = aijΓk ΓiΓjΓiΓj = a2 ijΓ 2 k = a2 ij ΓiΓiΓjΓiΓjΓj = ΓjΓi = a2 ijΓiΓj = a3 ijΓk Γ′ iΓ ′ j = aijΓ ′ k 12 and rewrite the Dirac equation in terms of φ1 and φ2, i ∂ ∂x0 φ1 − iσ ·∇φ1 = mc ~ φ2, i ∂ ∂x0 φ2 + iσ ·∇φ2 = mc ~ φ1. (90) On can see that φ1 and φ2 are coupled only via the mass term. In ultra-relativistic limit (or for nearly massless particle such as neutrinos), rest mass is negligible, then φ1 and φ2 decouple, i ∂ ∂x0 φ1 − iσ ·∇φ1 = 0, i ∂ ∂x0 φ2 + iσ ·∇φ2 = 0, (91) The 4-component wavefunction in the Weyl representation is written as ψWeyl = ( φ1 φ2 ) . (92) Let’s imagine that a massless spin-1/2 neutrino is described by φ1, a plane wave state of a definite momentum p with energy E = |p|c, φ1 ∝ e i ~ (p·x−Et). (93) i ∂ ∂x0 φ1 = i 1 c ∂ ∂t φ1 = E ~c φ1, iσ ·∇φ1 = −1 ~ σ · pφ1 ⇒ Eφ1 = |p|cφ1 = −cσ · pφ1 or σ · p |p| φ1 = −φ1. (94) The operator h = σ · p/|p| is called the “helicity.” Physically it refers to the component of spin in the direction of motion. φ1 describes a neutrino with helicity −1 (“left-handed”). Similarly, σ · p |p| φ2 = φ2, (h = +1, “right-handed”). (95) The γµ’s in the Weyl representation are γ0 = ( 0 I I 0 ) , γi = ( 0 σi −σi 0 ) , γ5 = ( −I 0 0 I ) . (96) Exercise: Find the S matrix which transform between the Pauli-Dirac represen- tation and the Weyl representation and verify that the gaµ matrices in the Weyl representation are correct. 15 5 Lorentz Covariance of the Dirac Equation We will set ~ = c = 1 from now on. In E&M, we write down Maxwell’s equations in a given inertial frame, x, t, with the electric and magnetic fields E, B. Maxwell’s equations are covariant with respect to Lorentz transformations, i.e., in a new Lorentz frame, x′, t′, the equations have the same form, but the fields E′(x′, t′), B′(x′, t′) are different. Similarly, Dirac equation is Lorentz covariant, but the wavefunction will change when we make a Lorentz transformation. Consider a frame F with an observer O and coordinates xµ. O describes a particle by the wavefunction ψ(xµ) which obeys( iγµ ∂ ∂xµ −m ) ψ(xµ). (97) In another inertial frame F ′ with an observer O′ and coordinates x′ν given by x′ν = Λν µx µ, (98) O′ describes the same particle by ψ′(x′ν) and ψ′(x′ν) satisfies( iγν ∂ ∂x′ν −m ) ψ′(x′ν). (99) Lorentz covariance of the Dirac equation means that the γ matrices are the same in both frames. What is the transformation matrix S which takes ψ to ψ′ under the Lorentz trans- formation? ψ′(Λx) = Sψ(x). (100) Applying S to Eq. (97), iSγµS−1 ∂ ∂xµ Sψ(xµ)−mSψ(xµ) = 0 ⇒ iSγµS−1 ∂ ∂xµ ψ′(x′ν)−mψ′(x′ν) = 0. (101) Using ∂ ∂xµ = ∂ ∂x′ν ∂x′ν ∂xµ = Λν µ ∂ ∂x′ν , (102) we obtain iSγµS−1Λν µ ∂ ∂x′ν ψ′(x′ν)−mψ′(x′ν) = 0. (103) Comparing it with Eq. (99), we need SγµS−1Λν µ = γν or equivalently SγµS−1 = ( Λ−1 )µ ν γν . (104) 16 We will write down the form of the S matrix without proof. You are encouraged to read the derivation in Shulten’s notes Chapter 10, p.319-321 and verify it by yourself. For an infinitesimal Lorentz transformation, Λµ ν = δµ ν + εµν . Multiplied by gνλ it can be written as Λµλ = gµλ + εµλ, (105) where εµλ is antisymmetric in µ and λ. Then the corresponding Lorentz transfor- mation on the spinor wavefunction is given by S(εµν) = I − i 4 σµνε µν , (106) where σµν = i 2 (γµγν − γνγµ) = i 2 [γµ, γν ]. (107) For finite Lorentz transformation, S = exp ( − i 4 σµνε µν ) . (108) Note that one can use either the active transformation (which transforms the ob- ject) or the passive transformation (which transforms the coordinates), but care should be taken to maintain consistency. We will mostly use passive transforma- tions unless explicitly noted otherwise. Example: Rotation about z-axis by θ angle (passive). −ε12 = +ε21 = θ, (109) σ12 = i 2 [γ1, γ2] = iγ1γ2 = i ( −iσ3 0 0 −iσ3 ) = ( σ3 0 0 σ3 ) ≡ Σ3 (110) S = exp ( + i 2 θ ( σ3 0 0 σ3 )) = I cos θ 2 + i ( σ3 0 0 σ3 ) sin θ 2 . (111) We can see that ψ transforms under rotations like an spin-1/2 object. For a rotation around a general direction n̂, S = I cos θ 2 + in̂ ·Σ sin θ 2 . (112) 17 ψ′ 2(x ′) has the same form except that ( 1 0 ) is replaced by ( 0 1 ) . For the negative energy solutions E ′ − = −E ′ = − √ |p|2 +m2 and p′ − = vn̂E ′ − = −vE ′n̂ = −p′ +. So we have ψ′ 3(x ′) = 1√ V √ m+ E ′ − σ·p′− E′+m ( 1 0 ) ( 1 0 )  ei(p′−·x′+E′t′), (124) and ψ′ 4(x ′) is obtained by the replacement ( 1 0 ) → ( 0 1 ) . Now we can drop the primes and the ± subscripts, ψ1,2 = 1√ V √ E +m ( χ+,− σ·p E+m χ+,− ) ei(p·x−Et) = 1√ V u1,2e i(p·x−Et), ψ3,4 = 1√ V √ E +m ( − σ·p E+m χ+,− χ+,− ) ei(p·x+Et) = 1√ V u3,4e i(p·x+Et), (125) where χ+ = ( 1 0 ) , χ− = ( 0 1 ) (126) (V is the proper volume in the frame where the particle is at rest.) Properties of spinors u1, · · ·u4 u†rus = 0 for r 6= s. (127) u†1u1 = (E +m) ( χ†+ χ†+ σ·p E+m )( χ+ σ·p E+m χ+ ) = (E +m)χ†+ ( 1 + (σ · p)(σ · p) (E +m)2 ) χ+. (128) Using the following identity: (σ · a)(σ · b) = σiaiσjbj = (δij + iεijkσk)aibj = a · b + iσ · (a× b), (129) we have u†1u1 = (E +m)χ†+ ( 1 + |p|2 (E +m)2 ) χ+ = (E +m)χ†+ E2 + 2Em+m2 + |p|2 (E +m)2 χ+ = χ+ 2E2 + 2Em E +m χ+ = 2Eχ†+χ+ = 2E. (130) 20 Similarly for other ur we have u†rus = δrs2E, which reflects that ρ = ψ†ψ is the zeroth component of a 4-vector. One can also check that urus = ±2mδrs (131) where + for r = 1, 2 and − for r = 3, 4. u1u1 = u†1γ 0u1 γ0 = ( I 0 0 −I ) = (E +m)χ†+ ( 1− |p|2 (E +m)2 ) χ+ = (E +m)χ†+ E2 + 2Em+m2 − |p|2 (E +m)2 χ+ = χ+ 2m2 + 2Em E +m χ+ = 2mχ†+χ+ = 2m (132) is invariant under Lorentz transformation. Orbital angular momentum and spin Orbital angular momentum L = r × p or Li = εijkrjpk. (133) (We don’t distinguish upper and lower indices when dealing with space dimensions only.) dLi dt = i[H, Li] = i[cα · p + βmc2, Li] = icαn[pn, εijkrjpk] = icαnεijk[pn, rj]pk = icαnεijk(−iδnj~)pk = c~εijkαjpk = c~(α× p)i 6= 0. (134) We find that the orbital angular momentum of a free particle is not a constant of the motion. 21 Consider the spin 1 2 Σ = 1 2 ( σi 0 0 σi ) , dΣi dt = i[H, Σi] = i[cαjpj + βmc2, Σi] = ic[αj, Σi]pj [ using Σiγ5 = ( σi 0 0 σi )( 0 I I 0 ) = ( 0 σi σi 0 ) = αi = γ5Σi ] = ic[γ5Σj, Σi]pj = icγ5[Σj, Σi]pj = icγ5(−2iεijkΣk)pj = 2cγ5εijkΣkpj = 2cεijkαkpj = −2c(α× p)i. (135) Comparing it with Eq. (134), we find d(Li + 1 2 ~Σi) dt = 0, (136) so the total angular momentum J = L + 1 2 ~Σ is conserved. 7 Interactions of a Relativistic Electron with an External Electromagnetic Field We make the usual replacement in the presence of external potential: E → E − eφ = i~ ∂ ∂t − eφ, e < 0 for electron p → p− e c A = −i~∇− e c A. (137) In covariant form, ∂µ → ∂µ + ie ~c Aµ → ∂µ + ieAµ ~ = c = 1. (138) Dirac equation in external potential: iγµ(∂µ + ieAµ)ψ −mψ = 0. (139) Two component reduction of Dirac equation in Pauli-Dirac basis:( I 0 0 −I ) (E − eφ) ( ψA ψB ) − ( 0 σ σ 0 ) (p− eA) ( ψA ψB ) −m ( ψA ψB ) = 0, ⇒ (E − eφ)ψA − σ · (p− eA)ψB −mψA = 0 (140) −(E − eφ)ψB + σ · (p− eA)ψA −mψB = 0 (141) 22 (σ · π − 2ikσ ·E)ψB = (W − qφ− 2kσ ·B)ψA, (163) (σ · π + 2ikσ ·E)ψA = (2m+W − qφ+ 2kσ ·B)ψB. (164) Again taking the non-relativistic limit, ψB ' 1 2m (σ · π + 2ikσ ·E)ψA, (165) we obtain (W − qφ− 2kσ ·B)ψA = 1 2m (σ · π − 2ikσ ·E)(σ · π + 2ikσ ·E)ψA. (166) Let’s consider two special cases. (a) φ = 0, E = 0 (W − 2kσ ·B)ψA = 1 2m (σ · π)2ψA ⇒ WψA = 1 2m π2ψA − q 2m σ ·BψA + 2kσ ·BψA ⇒ µ = q 2m − 2k. (167) (b) B = 0, E 6= 0 for the neutron (q = 0) WψA = 1 2m σ · (p + iµnE) σ · (p− iµnE)ψA = 1 2m [(p + iµnE) · (p + iµnE) + iσ · (p + iµnE)× (p− iµnE)]ψA = 1 2m [ p2 + µ2 nE 2 + iµnE · p− iµnp ·E + iσ · (iµnp×E − iµnE × p) ] ψA = 1 2m [ p2 + µ2 nE 2 − µn(∇ ·E) + 2µnσ · (E × p) + iµnσ · (∇×E) ] ψA = 1 2m [ p2 + µ2 nE 2 − µnρ+ 2µnσ · (E × p) ] ψA. (168) The last term is the spin-orbit interaction, σ · (E × p) = −1 r dφ dr σ · (r × p) = −1 r dφ dr σ ·L. (169) The second to last term gives an effective potential for a slow neutron moving in the electric field of an electron, V = −µnρ 2m = µn 2m (−e)δ3(r). (170) It’s called “Foldy” potential and does exist experimentally. 25 8 Foldy-Wouthuysen Transformation We now have the Dirac equation with interactions. For a given problem we can solve for the spectrum and wavefunctions (ignoring the negative energy solutions for a moment), for instance, the hydrogen atom, We can compare the solutions to those of the schrödinger equation and find out the relativistic corrections to the spectrum and the wavefunctions. In fact, the problem of hydrogen atom can be solved exactly. However, the exact solutions are problem-specific and involve unfamiliar special functions, hence they not very illuminating. You can find the exact solutions in many textbooks and also in Shulten’s notes. Instead, in this section we will develop a systematic approximation method to solve a system in the non-relativistic regime (E−m m). It corresponds to take the approximation we discussed in the previous section to higher orders in a systematic way. This allows a physical interpretation for each term in the approximation and tells us the relative importance of various effects. Such a method has more general applications for different problems. In Foldy-Wouthuysen transformation, we look for a unitary transformation UF removes operators which couple the large to the small components. Odd operators (off-diagonal in Pauli-Dirac basis): αi, γi, γ5, · · · Even operators (diagonal in Pauli-Dirac basis): 1, β, Σ, · · · ψ′ = UFψ = eiSψ, S = hermitian (171) First consider the case of a free particle, H = α · p + βm not time-dependent. i ∂ψ′ ∂t = eiSHψ = eiSHe−iSψ′ = H ′ψ′ (172) We want to find S such that H ′ contains no odd operators. We can try eiS = eβα·p̂θ = cos θ + βα · p̂ sin θ, where p̂ = p/|p|. (173) H ′ = (cos θ + βα · p̂ sin θ) (α · p + βm) (cos θ − βα · p̂ sin θ) = (α · p + βm) (cos θ − βα · p̂ sin θ)2 = (α · p + βm) exp (−2βα · p̂θ) = (α · p) ( cos 2θ − m |p| sin 2θ ) + β (m cos 2θ + |p| sin 2θ) . (174) To eliminate (α · p) term we choose tan 2θ = |p|/m, then H ′ = β √ m2 + |p|2. (175) This is the same as the first Hamiltonian we tried except for the β factor which also gives rise to negative energy solutions. In practice, we need to expand the Hamiltonian for |p|  m. 26 General case: H = α · (p− eA) + βm+ eΦ = βm+O + E , (176) O = α · (p− eA), E = eΦ, βO = −Oβ, βE = Eβ (177) H time-dependent ⇒ S time-dependent We can only construct S with a non-relativistic expansion of the transformed Hamiltonian H ′ in a power series in 1/m. We’ll expand to p4 m3 and p×(E, B) m2 . Hψ = i ∂ ∂t ( e−iSψ′) = e−iSi ∂ψ′ ∂t + ( i ∂ ∂t e−iS ) ψ′ ⇒ i ∂ψ′ ∂t = [ eiS ( H − i ∂ ∂t ) e−iS ] ψ′ = H ′ψ′ (178) S is expanded in powers of 1/m and is “small” in the non-relativistic limit. eiSHe−iS = H + i[S,H] + i2 2! [S, [S,H]] + · · ·+ in n! [S, [S, · · · [S,H]]]. (179) S = O( 1 m ) to the desired order of accuracy H ′ = H + i[S,H]− 1 2 [S, [S,H]]− i 6 [S, [S, [S,H]]] + 1 24 [S, [S, [S, [S, βm]]]] −Ṡ − i 2 [S, Ṡ] + 1 6 [S, [S, Ṡ]] (180) We will eliminate the odd operators order by order in 1/m and repeat until the desired order is reached. First order [O(1)]: H ′ = βm+ E +O + i[S, β]m. (181) To cancel O, we choose S = − iβO 2m , i[S,H] = −O + β 2m [O, E ] + 1 m βO2 (182) i2 2 [S, [S,H]] = −βO 2 2m − 1 8m2 [O, [O, E ]]− 1 2m2 O3 (183) i3 3! [S, [S, [S,H]]] = O3 6m2 − 1 6m3 βO4 (184) i4 4! [S, [S, [S, [S,H]]]] = βO4 24m3 (185) −Ṡ = iβȮ 2m (186) − i 2 [S, Ṡ] = − i 8m2 [O, Ȯ] (187) 27 Relativistic corrections:〈 − p4 8m3 〉 = Z4α4m 2n4 ( 3 4 − n l + 1 2 ) . (207) We find ∆E(l = 0) = Z4α4m 2n4 ( 3 4 − n ) (208) = ∆E(l = 1, j = 1 2 ), (209) so 2S1/2 and 2P1/2 remain degenerate at this level. They are split by Lamb shift (2S1/2 > 2P1/2) which can be calculated after you learn radiative corrections in QED. The 2P1/2 and 2P3/2 are split by the spin-orbit interaction (fine structure) which you should have seen before. ∆E(l = 1, j = 3 2 )−∆E(l = 1, j = 1 2 ) = Z4α4m 4n3 (210) 9 Klein Paradox and the Hole Theory So far we have ignored the negative solutions. However, the negative energy solu- tions are required together with the positive energy solutions to form a complete set. If we try to localize an electron by forming a wave packet, the wavefunc- tion will be composed of some negative energy components. There will be more negative energy components if the electron is more localized by the uncertainty relation ∆x∆p ∼ ~. The negative energy components can not be ignored if the electron is localized to distances comparable to its Compton wavelength ~/mc, and we will encounter many paradoxes and dilemmas. An example is the Klein paradox described below. In order to localize electrons, we must introduce strong external forces confining them to the desired region. Let’s consider a simplified situation that we want to confine a free electron of energy E to the region z < 0 by a one-dimensional step- function potential of height V as shown in Fig. 1. Now in the z < 0 half space there is an incident positive energy plan wave of momentum k > 0 along the z axis, ψinc(z) = eikz  1 0 k E+m 0  , (spin-up). (211) 30 Figure 1: Electrostatic potential idealized with a sharp boundary, with an incident free electron wave moving to the right in region I. The reflected wave in z < 0 region has the form ψref(z) = a e−ikz  1 0 −k E+m 0 + b e−ikz  0 1 0 k E+m  , (212) and the transmitted wave in the z > 0 region (in the presence of the constant potential V ) has a similar form ψtrans(z) = c eiqz  1 0 q E−V +m 0 + d eiqz  0 1 0 −q E−V +m  , (213) with an effective momentum q of q = √ (E − V )2 −m2. (214) The total wavefunction is ψ(z) = θ(−z)[ψinc(z) + ψref(z)] + θ(z)ψtrans(z). (215) Requiring the continuity of ψ(z) at z = 0, ψinc(0) + ψref(0) = ψtrans(0), we obtain 1 + a = c (216) b = d (217) (1− a) k E +m = c q E − V +m (218) b k E +m = d −q E − V +m (219) 31 From these equations we can see b = d = 0 (no spin-flip) (220) 1 + a = c (221) 1− a = rc where r = q k E +m E − V +m (222) ⇒ c = 2 1 + r , a = 1− r 1 + r . (223) As long as |E−V | < m, q is imaginary and the transmitted wave decays exponen- tially. However, when V ≥ E+m the transmitted wave becomes oscillatory again. The probability currents j = ψ†αψ = ψ†α3ψẑ, for the incident, transmitted, and reflected waves are jinc = 2 k E +M , jtrans = 2c2 q E − V +m , jref = 2a2 k E +m . (224) we find jtrans jinc = c2r = 4r (1 + r)2 (< 0 for V ≥ E +m), jref jinc = a2 = ( 1− r 1 + r )2 (> 1 for V ≥ E +m). (225) Although the conservation of the probabilities looks satisfied: jinc = jtrans + jref , but we get the paradox that the reflected flux is larger than the incident one! There is also a problem of causality violation of the single particle theory which you can read in Prof. Gunion’s notes, p.14–p.15. Hole Theory In spite of the success of the Dirac equation, we must face the difficulties from the negative energy solutions. By their very existence they require a massive reinterpretation of the Dirac theory in order to prevent atomic electrons from making radiative transitions into negative-energy states. The transition rate for an electron in the ground state of a hydrogen atom to fall into a negative-energy state may be calculated by applying semi-classical radiation theory. The rate for the electron to make a transition into the energy interval −mc2 to −2mc2 is ∼ 2α6 π mc2 ~ ' 108sec−1 (226) and it blows up if all the negative-energy states are included, which clearly makes no sense. 32