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 respecct to Lorentz transformations, i.e., in a new Lorentz frame, x0, t0, the equations have the same form, but the fields E0(x0, t0), B0(x0, t0) 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 m ψ(xµ). (97) ∂xµ − In another inertial frame F 0 with an observer O0 and coordinates x0ν given by

0ν ν µ x = Λ µx , (98)

O0 describes the same particle by ψ0(x0ν ) and ψ0(x0ν ) satisﬁes

∂ iγν m ψ0(x0ν ). (99) ∂x0ν − 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 ψ0 under the Lorentz transformation? ψ0(Λx) = Sψ(x). (100) Applying S to Eq. (97),

∂ iSγµS−1 Sψ(xµ) mSψ(xµ) = 0 ∂xµ − ∂ iSγµS−1 ψ0(x0ν) mψ0(x0ν) = 0. (101) ⇒ ∂xµ − Using ∂ ∂ ∂x0ν ∂ = = Λν , (102) ∂xµ ∂x0ν ∂xµ µ ∂x0ν we obtain ∂ iSγµS−1Λν ψ0(x0ν ) mψ0(x0ν ) = 0. (103) µ ∂x0ν − Comparing it with Eq. (99), we need

µ −1 ν ν µ −1 −1 µ ν Sγ S Λ µ = γ or equivalently Sγ S = Λ ν γ . (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.

µ µ µ νλ For an inﬁnitesimal Lorentz transformation, Λ ν = δ ν + ν. Multiplied by g it can be written as Λµλ = gµλ + µλ, (105) where µλ is antisymmetric in µ and λ. Then the corresponding Lorentz transformation on the spinor wavefunction is given by i S(µν ) = I σ µν , (106) − 4 µν where i i σ = (γ γ γ γ ) = [γ , γ ]. (107) µν 2 µ ν − ν µ 2 µ ν For ﬁnite Lorentz transformation,

i S = exp σ µν . (108) −4 µν Note that one can use either the active transformation (which transforms the object) or the passive transformation (which transforms the coordinates), but care should be taken to maintain consistency. We will mostly use passive transformations unless explicitly noted otherwise. Example: Rotation about z-axis by θ angle (passive).

12 = +21 = θ, (109) − i σ = [γ , γ ] = iγ γ 12 2 1 2 1 2 iσ3 0 = i −0 iσ3 − σ3 0 = Σ (110) 0 σ3 ≡ 3 i σ3 0 θ σ3 0 θ S = exp + θ = I cos + i sin . (111) 2 0 σ3 2 0 σ3 2 We can see that ψ transforms under rotations like an spin-1/2 object. For a rotation around a general direction nˆ, θ θ S = I cos + inˆ Σ sin . (112) 2 · 2

17 Example: Boost in xˆ direction (passive).

01 = 10 = η, (113) − i σ = [γ , γ ] = iγ γ , (114) 01 2 0 1 0 1 i i S = exp σ µν = exp ηiγ γ −4 µν −2 0 1 η η = exp γ γ = exp α1 2 0 1 −2 η η = I cosh α1 sinh . (115) 2 − 2 For a particle moving in the direction of nˆ in the new frame, we need to boost the frame in the nˆ direction, − η η S = I cosh + α nˆ sinh . (116) 2 · 2

6 Free Particle Solutions to the Dirac Equation

The solutions to the Dirac equation for a free particle at rest are

1 2m 0 ψ = e−imt, E = +m, 1 0 r V 0   0 2m 1 −imt ψ2 =   e , E = +m, 0 r V 0   0 2m 0 ψ = eimt, E = m, 3 1 r V − 0   0 2m 0 ψ = eimt, E = m, (117) 4 0 r V − 1     where we have set ~ = c = 1 and V is the total volume. Note that I have chosen a particular normalization d3xψ†ψ = 2m (118) Z

18 for a particle at rest. This is more convenient when we learn ﬁeld theory later, because ψ†ψ is not invariant under boosts. Instead, it’s the zeroth component of a 4-vector, similar to E. The solutions for a free particle moving at a constant velocity can be obtained by a Lorentz boost, η η S = I cosh + α nˆ sinh . (119) 2 · 2 Using the Pauli-Dirac representation, 0 σi αi = , σi 0 η η cosh 2 σ nˆ sinh 2 S = η · η , (120) σ nˆ sinh cosh · 2 2 and the following relations, E0 cosh η = γ0 = , sinh η = γ0β0, m η 1 + cosh η 1 + γ0 m + E0 cosh = = = , 2 r 2 r 2 r 2m η cosh η 1 E0 m sinh = − = − 2 r 2 r 2m p0 x0µ = p0 x0 E0t0 = p xµ = mt, (121) − µ + · − − µ − where p0 = p0 nˆ is the 3-momentum of the positive energy state, we obtain + | +|

η 1 cosh 2 0 0 2m 0 −imt ψ1(x ) = Sψ1(x) =   e V η 1 r σ nˆ sinh  · 2 0     1  1 0 0 0 0 0 = √m + E0   ei(p+·x −E t ) √V E0−m 1 0 σ nˆ  E +m · 0  q   1  1 0 0 0 0 0 √ 0 i(p+·x −E t ) = m + E  0  e (122) √V σ·p+ 1 0  E +m 0      where we have used

0 02 2 0 E m √E m p+ 0 − = 0 − = |0 | (123) rE + m E + m E + m in the last line.

19 1 0 ψ0 (x0) has the same form except that is replaced by . 2 0 1 0 0 2 2 0 0 For the negative energy solutions E− = E = p + m and p− = vnˆE− = vE0nˆ = p0 . So we have − − | | − − + p 0 σ·p− 1 E0+m 0 0 1 − 0 i(p0 ·x0+E0t0) ψ (x ) = √m + E0   e − , (124) 3 √V 1  0      1 0 and ψ0 (x0) is obtained by the replacement . 4 0 → 1 Now we can drop the primes and the subscripts, 1 χ+,− i(p·x−Et) 1 i(p·x−Et) ψ1,2 = √E + m σ·p e = u1,2e , √V E+m χ+,− √V σ·p 1 E+m χ+,− i(p·x+Et) 1 i(p·x+Et) ψ3,4 = √E + m − e = u3,4e , (125) √ χ − √ V +, V where 1 0 χ = , χ = (126) + 0 − 1 (V is the proper volume in the frame where the particle is at rest.) Properties of spinors u , u 1 · · · 4 u†u = 0 for r = s. (127) r s 6

† † † σ·p χ+ u u = (E + m) χ χ σ p 1 1 + + E+m · χ E+m + (σ p)(σ p) = (E + m)χ† 1 + · · χ . (128) + (E + m)2 + Using the following identity: (σ a)(σ b) = σ a σ b = (δ + i σ )a b = a b + iσ (a b), (129) · · i i j j ij ijk k i j · · × we have p 2 u†u = (E + m)χ† 1 + | | χ 1 1 + (E + m)2 + E2 + 2Em + m2 + p 2 = (E + m)χ† | | χ + (E + m)2 + 2E2 + 2Em = χ χ + E + m + † = 2Eχ+χ+ = 2E. (130)

20 † † Similarly for other ur we have urus = δrs2E, which reﬂects that ρ = ψ ψ is the zeroth component of a 4-vector. One can also check that u u = 2mδ (131) r s rs where + for r = 1, 2 and for r = 3, 4. − I 0 u u = u†γ0u γ0 = 1 1 1 1 0 I − p 2 = (E + m)χ† 1 | | χ + − (E + m)2 + E2 + 2Em + m2 p 2 = (E + m)χ† − | | χ + (E + m)2 + 2m2 + 2Em = χ χ + E + m + † = 2mχ+χ+ = 2m (132) is invariant under Lorentz transformation. Orbital angular momentum and spin Orbital angular momentum

L = r p or × Li = ijkrjpk. (133)

(We don’t distinguish upper and lower indices when dealing with space dimensions only.)

dL i = i[H, L ] dt i = i[cα p + βmc2, L ] · i = icαn[pn, ijkrjpk]

= icαnijk[pn, rj]pk = icα ( iδ )p n ijk − nj k = cijkαjpk = c(α p) = 0. (134) × i 6 We ﬁnd that the orbital angular momentum of a free particle is not a constant of the motion.

21 σ 0 Consider the spin 1 Σ = 1 i , 2 2 0 σ i dΣ i = i[H, Σ ] dt i 2 = i[cαjpj + βmc , Σi] σ 0 0 I 0 σ = ic[α , Σ ]p using Σ γ = i = i = α = γ Σ j i j i 5 0 σ I 0 σ 0 i 5 i i i = ic[γ5Σj, Σi]pj

= icγ5[Σj, Σi]pj = icγ ( 2i Σ )p 5 − ijk k j = 2cγ5ijkΣkpj

= 2cijkαkpj = 2c(α p) . (135) − × i Comparing it with Eq. (134), we ﬁnd d(L + 1 Σ ) i 2 i = 0, (136) dt 1 so the total angular momentum J = L + 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~ eφ, e < 0 for electron → − ∂t − e e p p A = i~∇ A. (137) → − c − − c In covariant form, ie ∂ ∂ + A ∂ + ieA ~ = c = 1. (138) µ → µ ~c µ → µ µ Dirac equation in external potential: iγµ(∂ + ieA )ψ mψ = 0. (139) µ µ − Two component reduction of Dirac equation in Pauli-Dirac basis: I 0 ψ 0 σ ψ ψ (E eφ) A (p eA) A m A = 0, 0 I − ψB − σ 0 − ψB − ψB − (E eφ)ψ σ (p eA)ψ mψ = 0 (140) ⇒ − A − · − B − A (E eφ)ψ + σ (p eA)ψ mψ = 0 (141) − − B · − A − B 22 where E and p represent the operators i∂t and i∇ respectively. Deﬁne W = E m, π = p eA, then we have − − − σ πψ = (W eφ)ψ (142) · B − A σ πψ = (2m + W eφ)ψ (143) · A − B From Eq. (143), ψ = (2m + W eφ)−1σ πψ . (144) B − · A Substitute it into Eq. (142),

(σ π)(σ π) · · ψ = (W eφ)ψ . (145) 2m + W eφ A − A − In non-relativistic limit, W eφ m, − 1 1 W eφ = 1 − + . (146) 2m + W eφ 2m − 2m · · · − 1 In the lowest order approximation we can keep only the leading term 2m , 1 (σ π)(σ π)ψ (W eφ)ψ . (147) 2m · · A ' − A Using Eq. (129),

(σ π)(σ π)ψ = [π π + iσ (π π)]ψ . (148) · · A · · × A

(π π)ψ = [(p eA) (p eA)]ψ × A − × − A = [ eA p ep A]ψ − × − × A = [+ieA ∇ + ie∇ A]ψ × × A = ieψ (∇ A) A × = ieBψA, (149) so 1 e (p eA)2ψ σ Bψ + eφψ = W ψ . (150) 2m − A − 2m · A A A Restoring ~, c, 1 e e~ (p A)2ψ σ Bψ + eφψ = W ψ . (151) 2m − c A − 2mc · A A A This is the “Pauli-Schr¨odinger equation” for a particle with the spin-magnetic moment, e~ e µ = σ = 2 S. (152) 2mc 2mc

23 In comparison, the relation between the angular momentum and the magnetic moment of a classical charged object is given by Iπr2 ω πr2 eωr2 e e µ = = e = = mωr2 = L. (153) c 2π c 2c 2mc 2mc

We can write e µ = g S (154) s 2mc in general. In Dirac theory, gs = 2. Experimentally,

g (e−) = 2 (1.0011596521859 38 10−13). (155) s × × α The deviation from 2 is due to radiative corrections in QED, (g 2)/2 = 2π + . The predicted value for g 2 using α from the quantum Hall eﬀect− is · · · s − (g 2) /2 = 0.0011596521564 229 10−13. (156) s − qH × They agree down to the 10−11 level. There are also spin-1/2 particles with anomalous magnetic moments, e.g., e e µproton = 2.79 | | , µneutron = 1.91 | | . (157) 2mpc − 2mnc This can be described by adding the Pauli moment term to the Dirac equation,

iγµ(∂ + iqA )ψ mψ + kσ F µνψ = 0. (158) µ µ − µν Recall i σ = (γ γ γ γ ), µν 2 µ ν − ν µ i i I 0 0 σ 0 σ i σ0i = iγ0γi = i − = i − = iα , 0 I σi 0 σi 0 − − − σk 0 σ = iγ γ = Σk = , ij i j ijk ijk 0 σk F 0i = Ei, − F ij = ijkBk. (159) − Then the Pauli moment term can be written as

iγµ(∂ + iqA )ψ mψ + 2ikα Eψ 2kΣ Bψ = 0. (160) µ µ − · − · The two component reduction gives

(E qφ)ψ σ πψ mψ + 2ikσ Eψ 2kσ Bψ = 0, (161) − A − · B − A · B − · A (E qφ)ψ + σ πψ mψ + 2ikσ Eψ 2kσ Bψ = 0. (162) − − B · A − B · A − · B 24 (σ π 2ikσ E)ψ = (W qφ 2kσ B)ψ , (163) · − · B − − · A (σ π + 2ikσ E)ψ = (2m + W qφ + 2kσ B)ψ . (164) · · A − · B Again taking the non-relativistic limit, 1 ψ (σ π + 2ikσ E)ψ , (165) B ' 2m · · A we obtain 1 (W qφ 2kσ B)ψ = (σ π 2ikσ E)(σ π + 2ikσ E)ψ . (166) − − · A 2m · − · · · A Let’s consider two special cases. (a) φ = 0, E = 0 1 (W 2kσ B)ψ = (σ π)2ψ − · A 2m · A 1 q W ψ = π2ψ σ Bψ + 2kσ Bψ ⇒ A 2m A − 2m · A · A q µ = 2k. (167) ⇒ 2m −

(b) B = 0, E = 0 for the neutron (q = 0) 6 1 W ψ = σ (p + iµ E) σ (p iµ E)ψ A 2m · n · − n A 1 = [(p + iµ E) (p + iµ E) + iσ (p + iµ E) (p iµ E)] ψ 2m n · n · n × − n A 1 = p2 + µ2 E2 + iµ E p iµ p E + iσ (iµ p E iµ E p) ψ 2m n n · − n · · n × − n × A 1 = p2 + µ2 E2 µ (∇ E) + 2µ σ (E p) + iµ σ (∇ E) ψ 2m n − n · n · × n · × A 1 = p2 + µ2 E2 µ ρ + 2µ σ (E p) ψ . (168) 2m n − n n · × A The last term is the spin-orbit interaction, 1 dφ 1 dφ σ (E p) = σ (r p) = σ L. (169) · × −r dr · × −r dr · The second to last term gives an eﬀective potential for a slow neutron moving in the electric ﬁeld of an electron, µ ρ µ V = n = n ( e)δ3(r). (170) − 2m 2m − It’s called “Foldy” potential and does exist experimentally.