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5 Lorentz Covariance of the Dirac Equation

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5 Lorentz Covariance of the Dirac Equation 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 (xµ) which obeys @ iγµ m (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 0(x0ν ) and 0(x0ν ) satisfies @ iγν m 0(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 to 0 under the Lorentz trans- formation? 0(Λx) = S (x): (100) Applying S to Eq. (97), @ iSγµS−1 S (xµ) mS (xµ) = 0 @xµ − @ iSγµS−1 0(x0ν) m 0(x0ν) = 0: (101) ) @xµ − Using @ @ @x0ν @ = = Λν ; (102) @xµ @x0ν @xµ µ @x0ν we obtain @ iSγµS−1Λν 0(x0ν ) m 0(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 transfor- mation 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 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). 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 transforms under rotations like an spin-1/2 object. For a rotation around a general direction n^, θ θ S = I cos + in^ Σ sin : (112) 2 · 2 17 Example: Boost in x^ 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 n^ in the new frame, we need to boost the frame in the n^ direction, − η η S = I cosh + α n^ 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 = e−imt; E = +m; 1 001 r V B0C B C @0A 2m 1 −imt 2 = 0 1 e ; E = +m; 0 r V B0C B C @0A 2m 0 = eimt; E = m; 3 011 r V − B0C B C @0A 2m 0 = eimt; E = m; (117) 4 001 r V − B1C B C @ A where we have set ~ = c = 1 and V is the total volume. Note that I have chosen a particular normalization d3x y = 2m (118) Z 18 for a particle at rest. This is more convenient when we learn field theory later, because y 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 n^ sinh : (119) 2 · 2 Using the Pauli-Dirac representation, 0 σi αi = ; σi 0 η η cosh 2 σ n^ sinh 2 S = η · η ; (120) σ n^ 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 n^ is the 3-momentum of the positive energy state, we obtain + j +j η 1 cosh 2 0 0 2m 0 −imt 1(x ) = S 1(x) = 0 1 e V η 1 r σ n^ sinh B · 2 0 C B C @ 1 A 1 0 0 0 0 0 = pm + E0 0 1 ei(p+·x −E t ) pV E0−m 1 0 σ n^ B E +m · 0 C Bq C @ 1 A 1 0 0 0 0 0 p 0 i(p+·x −E t ) = m + E 0 0 1 e (122) pV σ·p+ 1 0 B E +m 0 C B C @ A where we have used 0 02 2 0 E m pE m p+ 0 − = 0 − = j0 j (123) rE + m E + m E + m in the last line. 19 1 0 0 (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− = vn^E− = vE0n^ = p0 . So we have − − j j − − + p 0 σ·p− 1 E0+m 0 0 1 − 0 i(p0 ·x0+E0t0) (x ) = pm + E0 0 1 e − ; (124) 3 pV 1 B 0 C B C @ A 1 0 and 0 (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) 1;2 = pE + m σ·p e = u1;2e ; pV E+m χ+;− pV σ·p 1 E+m χ+;− i(p·x+Et) 1 i(p·x+Et) 3;4 = pE + m − e = u3;4e ; (125) p χ − p 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 uyu = 0 for r = s: (127) r s 6 y y y σ·p χ+ u u = (E + m) χ χ σ p 1 1 + + E+m · χ E+m + (σ p)(σ p) = (E + m)χy 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 uyu = (E + m)χy 1 + j j χ 1 1 + (E + m)2 + E2 + 2Em + m2 + p 2 = (E + m)χy j j χ + (E + m)2 + 2E2 + 2Em = χ χ + E + m + y = 2Eχ+χ+ = 2E: (130) 20 y y Similarly for other ur we have urus = δrs2E, which reflects that ρ = 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 = uyγ0u γ0 = 1 1 1 1 0 I − p 2 = (E + m)χy 1 j j χ + − (E + m)2 + E2 + 2Em + m2 p 2 = (E + m)χy − j j χ + (E + m)2 + 2m2 + 2Em = χ χ + E + m + y = 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~r 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 ) m = 0: (139) µ µ − Two component reduction of Dirac equation in Pauli-Dirac basis: I 0 0 σ (E eφ) A (p eA) A m A = 0; 0 I − B − σ 0 − B − B − (E eφ) σ (p eA) m = 0 (140) ) − A − · − B − A (E eφ) + σ (p eA) m = 0 (141) − − B · − A − B 22 where E and p represent the operators i@t and ir respectively.
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