Now, suppose the electron field Ψ(x) satisfies the covariant Dirac equation (i D m)ψ = 0.

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1 PHY 396 K. Solutions for homework set #7. Problem 1a: γ α γ α 1 {γα, γ β }g αβ g αβ g αβ 4; S.1 γ α γ ν γ α γ α γ ν g να γ ν γ α γ α γ ν γ ν γ α γ α 4 γ ν ; S. γ α γ µ γ ν γ α γ α γ µ g µα γ µ γ α γ ν γ α γ ν γ µ γ µ γ α γ ν γ α γ ν γ ν γ µ + γ µ γ ν 4g µν ; S.3 γ α γ λ γ µ γ ν γ α γ α γ λ g µα γ λ γ α γ µ γ ν γ α γ µ γ ν γ λ γ λ γ α γ µ γ ν γ α 4g µν γ µ γ ν 4g µν γ ν γ µ γ λ γ ν γ µ γ λ. S.4 Problem 1b: First, a Lemma: for any Lorentz vector a µ, the a def γ µ a µ matrix squares to a a γ µ a µ γ ν a ν a µ a ν γ µ γ ν g µν is µν a i[a µ, a ν ] S µν S.5 where the last equality comes from S µν S νµ. For vectors a µ whose components commute with each other, this formula simplifies to a a, in particular for the ordinary derivatives µ,. However, the covariant derivatives D µ do not commute with each other. Instead, [D µ, D ν ] iqf µν x where q is the electric charge of the field on which D µ act; for the electron field Ψx, q e and hence [D µ, D ν ]Ψx ief µν xψx. Hence, according to the lemma S.5, D Ψ D Ψ ef µν S µν Ψ. S.6 Now, suppose the electron field Ψx satisfies the covariant Dirac equation i D mψ. 1

2 Then for any differential operator D, D i D mψ, and in particular i D m i D mψ. S.7 The LHS of this formula amounts to i D m i D mψ D + m Ψ D ef µν S µν + m Ψ, S.8 which immediately leads to eq.. Problem 1c: The anti-commutation relations 1 imply γ µ γ ν ±γ ν γ µ where the sign is + for µ ν and otherwise. Hence for any product Γ of the γ matrices, γ µ Γ 1 n Γγ µ, where n is the number of γ ν µ factors of Γ. For the Γ γ 5 iγ γ 1 γ γ 3, n 3 for any µ, 1,, 3, hence γ µ γ 5 γ 5 γ µ. As to the spin matrices, γ 5 γ µ γ ν γ µ γ 5 γ ν +γ µ γ ν γ 5 and therefore γ 5 S µν +S µν γ 5. Problem 1d: First, the hermiticity: γ 5 iγ γ 1 γ γ 3 iγ 3 γ γ 1 γ +iγ 3 γ γ 1 γ +iγ 3 γ γ 1 γ 1 3 i γ γ 3 γ γ i γ γ 1 γ 3 γ i γ γ 1 γ γ 3 S.9 +iγ γ 1 γ γ 3 +γ 5. Second, the square: γ 5 γ 5 γ 5 iγ γ 1 γ γ 3 iγ 3 γ γ 1 γ γ γ 1 γ γ 3 γ 3 γ γ 1 γ +γ γ 1 γ γ γ 1 γ γ γ 1 γ 1 γ +γ γ +1. S.1

3 Problem 1e: Since the four Dirac matrices γ, γ 1, γ, γ 3 all anticommute with each other, we have ɛ κλµν γ κ γ λ γ µ γ ν γ [ γ 1 γ γ 3] 4γ γ 1 γ γ 3 4iγ 5. S.11 To prove the other identity, we note that a totally antisymmetric product γ [κ γ λ γ µ γ ν] vanishes unless the Lorentz indices κ, λ, µ, ν are all distinct which makes them, 1,, 3 in some order. For such indices, the anticommutativity of the Dirac matrices implies γ κ γ λ γ µ γ ν ɛ κλµν γ γ 1 γ γ 3 note that ɛ 13 1, and hence γ [κ γ λ γ µ γ ν] 4ɛ κλµν γ γ 1 γ γ 3 +4iɛ κλµν γ 5. S.1 Problem 1f: 6iɛ κλµν γ κ γ γ κ γ [κ γ λ γ µ γ ν] 4 1 γ κ γ κ γ [λ γ µ γ ν] γ [λ γ κ γ µ γ ν] + γ [λ γ µ γ κ γ ν] γ [λ γ µ γ ν] γ κ 4γ [λ γ µ γ ν] + γ [λ γ µ γ ν] + 4g [λµ γ ν] + γ [ν γ µ γ λ] 1 4 S γ[λ γ µ γ ν] γ [λ γ µ γ ν]. Problem 1g: Proof by inspection: In the Weyl basis and block form, the 16 matrices are 1 1 +σ i 1 4 4, γ, γ i 1 1 σ i, 1 σ k γ[i γ j] iɛ ijk 1 σ i σ k, γ [ γ i] +σ i, S.14 1 σ i 1 γ 5 γ, γ 5 γ i +1 σ i, γ 5, +1 and their linear independence is self-evident. Since there are only 16 independent 4 4 matrices altogether, any such matrix Γ is a linear combination of the matrices S.14. Q.E.D. 3

4 Algebraic Proof: Without making any assumption about the matrix form of the γ µ operators, let us consider the Clifford algebra 1. Using γ µ γ ν ±γ ν γ µ where the sign is + for µ ν and for µ ν, we may re-order any product of the γ matrices as ±γ γ γ 1 γ 1 γ γ γ 3 γ 3. Moreover, since each γ µ squares to +1 or 1, we may further simplify the product in question to to ±γ or 1 γ 1 or 1 γ or 1 γ 3 or 1. The net result is up to a sign or ±i factor one of the 16 matrices: 1, or a γ µ, or a γ µ γ ν for µ ν which equals to 1 γ[µ γ ν], or a γ λ γ µ γ ν for 3 different λ, µ, ν which equals to 1 6 γ[λ γ µ γ ν] iɛ λµνρ γ 5 γ ρ cf. part f, or γ γ 1 γ γ 3 iγ 5. Consequently, any operator Γ algebraically constructed of the γ s is a linear combination of these 16 matrices. Incidentally, this proof explains why the Dirac matrices are 4 4 in d 4 spacetime dimensions: the 16 linearly-independent products of Dirac matrices require matrix size to be Technically, we may also use matrices of size 4n 4n, but then we would have γ µ γ µ n n, and ditto for all their products. Physically, this means combining the Dirac spinor index with some other index i 1,..., n which has nothing to do with Lorentz symmetry. Nobody wants such an index confusion, so physicists always stick to 4 4 Dirac matrices in 4 spacetime dimensions. Problem 1 : As in problem 1c, for any product Γ of the Dirac matrices, γ µ Γ 1 n Γγ µ where n is the number of γ ν µ factors of Γ. For Γ γ γ 1 γ d 1 moduli the overall sign or ±i factor, each γ µ appear once, with the remaining d 1 factors being γ ν µ. Thus, n d 1 and γ µ Γ 1 d 1 Γγ µ : for even spacetime dimensions d, Γ anticommutes with all the γ µ matrices, but for odd d, Γ commutes with all the γ µ instead of anticommuting. Now consider the dimension of the Clifford algebra made out of Dirac matrix products and their linear combinations. As in problem 1g, any product of Dirac matrices can be re-arranged as ±γ or 1 γ 1 or 1 γ d 1 or 1, and the number of distinct products of this type is clearly d. Alternatively, we count a single unit matrix, d 1 d of γ µ matrices, d of independent 1 γ [µ γ ν] products, d 3 of 1 6 γ [λ γ µ γ ν] products, etc., etc., all the way up to d of γ [β γ ω] ɛ αβ ω γ α Γ, and d d d 1 d 1 of γ [α γ ω] ɛ αβ ω Γ; the net number of all 4

5 such matrices is d k d k d. For even dimensions d, all the d matrix products are distinct, and we need matrices of size d/ d/ to accommodate them. But for odd d, the Γ matrix commutes with the entire Clifford algebra since it commutes with all the γ µ, so to avoid redundancy we should set Γ 1 or Γ 1. Consequently, any product of k distinct Dirac matrices becomes identical up to a sign or ±i with the product of the other d k matrices, for example γ γ k 1 i?? γ k γ d 1. This halves the net dimension of the Clifford algebra from d to d 1 and calls for matrix size d 1/ d 1/. Thus, the Dirac matrices in or 3 space time dimensions are, in 4 or 5 dimensions they are 4 4, in 6 or 7 dimensions 8 8, in 8 or 9 dimensions 16 16, in 1 or 11 dimensions 3 3, etc., etc. Problem a: I have already written down the γ 5 and the 1γ[µ γ ν] is µν in eq. S.14, but here is the detailed calculation, in case you need it: σ σ 1 σ σ 3 γ 5 i σ σ 1 σ σ 3 σ σ 1 σ σ 3 +iσ 1 σ σ 3 i σ σ 1 σ σ 3 iσ 1 σ σ 3 1, S σ µ σ ν σ µ γ µ γ ν σ ν σµ σν σµ σ ν, S.16 and consequently S ij i σ [i σ j] 4 σ [i σ j] ɛijk σ k σ k S.17 while S k S k i γ γ k i σ k +σ k. S.18 5

6 Problem b: In 3-vector terms, eqs. S.17 and S.18 mean that the Lorentz generators Ĵ and ˆK are represented in the Dirac spinor multiplet by the S J 1 σ σ and S K 1 iσ. S.19 +iσ Consequently, for a rotationr of space through angle θ around axis n, M D R exp iθ n S J exp i θn σ exp i θn σ exp i θn σ i θn σ, S. which translates to the Weyl spinor language as M L R M R R exp i θn σ. 5. Similarly, a Lorentz boost B of rapidity r artanhβ in the direction n is represented in the Dirac spinor multiplet by the matrix M D B exp iθ n S K exp 1 θn σ exp 1 θn σ exp + 1 θn σ + 1 θn σ, S.1 which translates to the Weyl spinor language as M L B exp 1 θn σ, M R B exp + 1 θn σ. 6. 6

7 Problem c: For any unit 3 vector n, n σ 1, hence for any integer power n, n σ n { 1 for even n, n σ for odd n. S. Therefore, for any complex λ, exp λn σ n even n λ n n! n σ n λ n n! 1 + odd n λ n n! n σ coshλ 1 + sinhλ n σ. S.3 In particular, for λ ±rapidity r of some boost, exp rn σ γ βγn σ. S.4 Now, comparing this formula to eqs.6, we immediately see that M L B exp rn σ γ 1 βn σ, M R B exp +rn σ γ 1 + βn σ, S.5 and hence M L B γ 1 βn σ, M R B γ 1 + βn σ. 8 Problem d: First, let s write down the M L and the M R representation of a generic continuous Lorentz symmetry. Generally, we have L µ ν expθ µ ν for some antisymmetric matrix Θ µν Θ νµ, 7

8 and the Dirac spinor representation of this transform is M D L exp i Θ µνs µν. S.6 In 3D language, the Θ µν matrix is equivalent to the pair of 3 vectors a and b according to Θ ij ɛ ijk a k, Θ i Θ i b i. Consequently, 1 a ib σ 1 Θ µν S µν a S J + b S K 1 a + ib σ S.7 and therefore M D L exp 1 ia b σ exp 1 ia + b σ. S.8 Or in the Weyl spinor language, M L L exp 1 ia b σ, M R L exp 1 ia + b σ. S.9 Now let s show that the two matrices S.9 are always related to each other according to eq. 9. First, a Lemma: σ σ σ σ. S.3 Proof: The σ 1 and the σ 3 Pauli marices are real while the σ matrix is imaginary. On the other hand, the σ 1 and the σ 3 anticommute with the σ while the σ commutes with itself. Therefore, σ σ1 σ +σ 1 σ 1 σ σ 1 σ σ σ 1, σ σ σ σ σ σ σ σ σ σ, S.31 σ σ 3 σ +σ 1 σ 3 σ σ 3 σ σ σ 3, or in 3 vector notations, S.3. 8

9 Thanks to this lemma, for any complex 3 vector c, σ c σ σ c σ, S.3 σ c σ n σ c σ n, S.33 and therefore σ exp c σ σ exp c σ. S.34 In particular, for c 1 b ia and hence c 1 ±b ia, we have σ exp 1 b ia σ σ exp 1 ±b ia σ. S.35 Combining this formula with eqs. S.9 for the Weyl spinor representation of a generic continuous Lorentz transfrom, we immediately see that σ M L L σ σ exp 1 b ia σ σ exp 1 +b ia σ M R L S.36 and likewise σ M R L σ σ exp 1 +b ia σ σ exp 1 b ia σ M L L. S.37 Quod erat demonstrandum. Problem e: ψ L x M L ψ L x, ψ L x M L ψ Lx, S.38 σ ψ L x σ ML ψ L x σ ML σ σ ψl x M R σ ψl x, thus σ ψl x transforms under the continuous Lorentz transforms exactly like the ψ Rx. Likewise, ψ R x M R ψ R x, ψ R x M R ψ Rx, S.39 σ ψ R x σ M R ψ R x σ M R σ σ ψ R x M L σ ψ R x, thus σ ψ R x transforms under the continuous Lorentz transforms exactly like the ψ Lx. 9

10 Problem f: Given the γ µ matrices 3 and the decomposition 1 of the Dirac spinor field Ψx into Weyl spinor fields ψ L x and ψ R x, we have m i σ µ iγ µ µ ψl mψl + i σ µ µ ψ R µ mψ iσ µ µ m ψ R iσ µ S.4 µ ψ L mψ R while 1 Ψ Ψ γ ψ L ψ R 1 ψ R ψ L. S.41 Consequently, the Dirac Lagrangian becomes L Ψ iγ µ µ mψ ψ R mψ L + i σ µ µ ψ R + Ψ L iσ µ µ ψ L mψ R S.4 iψ L σµ µ ψ L + iψ R σµ µ ψ R m ψ L ψ R m ψ R ψ L. Problem g: For the massless fermions and only for the massless fermions, the last two terms in the Lagrangian L iψ L σµ µ ψ L + iψ R σµ µ ψ R m ψ L ψ R m ψ R ψ L S.4 go away, and we are left with L massless iψ L σµ µ ψ L + iψ R σµ µ ψ R S.43 where the two Weyl spinor fields ψ L x and ψ R x are completely independent from each other. In particular, they obey independent Weyl equations i σ µ µ ψ L and iσ µ µ ψ R. S.44 On the other hand, for m the two last terms in the Lagrangian S.4 connect the ψ L and ψ R to each other and we cannot have one without the other. In particular, their equations of 1

11 motion become mixed, i σ µ µ ψ L mψ R and iσ µ µ ψ R mψ L. S.45 Problem 3a: As we saw in class, the Dirac equation i mψ implies the Klein Gordon equation + m Ψ, hence any plane-wave solution pf the Dirac equation must have on-shell momentum p µ with p m. But there is more to the Dirac equation than the Klein Gordon equation, hence eqs. 1 for the spinor coefficients of the e ipx plane waves. Indeed, for Ψ α x e ipx u α the Dirac equation becomes i me ipx u e ipx p mu, S.46 so the spinor u α must obey the marix equation p mu. Likewise, for Ψ α x e +ipx v α the Dirac equation becomes i me +ipx v e +ipx p mv, S.47 so the spinor v α must obey the matrix equation p + mv. Conversely, for any on-shell momentum p and any spinors u α and v α obeying the matrix equations 1 the plane waves e ipx u α and e +ipx v α obey the Dirac equation. Problem 3b: For p, p +m and p m mγ 1. Hence, the up, s spinors satisfy γ 1u, or in the Weyl basis u u ζ S.48 ζ where ζ is an arbitrary two-component spinor. Its normalization follows from u u ζ ζ, so if we want u u E p m for p, then we need ζ ζ m. Equivalently, we want ζ m ξ and hence u as in eq. 14 for a conventionally normalized spinor ξ, ξ ξ. 11

12 Note that there are two independent choices of ξ, normalized to ξ sξ s δ s,s, so they give rise to two independent u α o, s spinors normalized to u, su, s mδ s,s E p δ s,s. They correspond to the two spin states of the p electron. In terms of the spin vector, S 1ξ sσξ s. Problem 3c: The Dirac equation is Lorentz-covariant, so we may obtain solutions for all p µ +E p, p by simply Lorentz-boosting the solutions 14 from the rest frame where p µ +m,. Thus, up, s M D B up, s ML M R m ξs m ξs m ML ξ s m MR ξ s S.49 where M D, M L, and M R are respectively Dirac-spinor, LH Weyl-spinor, and RH Weyl-spinor representations or the Lorentz boost B from p µ m, to p µ E, p. Specifically, this boost has γ E m, γβn p m S.5 so eqs. 8 for its Weyl-spinor representations M L and M R may be written as m ML E p σ, m MR E + p σ. S.51 Plugging in these formulae into eq. S.49 immediately gives us up, s E p σ ξ s E + p σ ξ s S.5 in accordance with eq

13 Problem 3d: The negative-frequency solutions e +ipx v α p, s have Dirac spinors v α satisfying p + mv. For a particle at rest, p µ +m,, this equation becomes mγ + 1v, or in the Weyl basis vp, s vp, s m +ηs η s S.53 for some two-component spinor η s. As in part b, the m factor translates the normalization v p, svp, s E p δ s,s mδ s,s for p to η sη s δ s,s. For p we proceed similarly to part c, namely Lorentz-boost the rest-frame solution S.53 to the frame where p µ +E p, p: ML + m ηs vp, s M D B vp, s M R m η s S.54 + m ML η s + E p σ η s m M R η s E + p σ η s in accordance with eq. 16 Problem 3e: Physically, a hole in the Fermi sea has opposite free energy, opposite momentum, opposite spin, etc., from the missing fermion see my notes for the explanation. A positron is a hole in the Dirac sea of electrons, so it should have opposite p µ E, p from the missing electron state that s why the vp, s spinors accompany the e +ipx e +iet ipx plane waves instead of the e ipx E iet+ipx factors of the up, s spinors. Likewise, the positron should have the opposite spin state from the missing electron, and that s why the η s should have the opposite spin from the ξ s. More accurately, the η s should carry the opposite spin vector η ssη s from the ξ ssξ s. The solution to this spin relation is η s σ ξs where denotes complex conjugation. Indeed, using the Pauli matrix relation σ σ σ σ from the problem d cf. eqs. S.31 13

14 for the proof, we see that η σ ξ η ξ σ η ση ξ σ σσ ξ ξ σ σ σ ξ ξ σξ ξ σξ η Sη 1 η ση ξ Sξ. S.55 Also, for η s σ ξ s we have + E p σ η s + E p σ vp, s σ ξs E + p σ η s E + p σ σ ξs +σ σ E p σ σ ξs σ σ E + p σ σ ξ s using σ σσ σ σ E p σ σ +σ E + p σ ξ s σ E p σ ξ s +σ + E p σ ξ s σ + E + p σ ξ s γ u p, s. E ± p σ S.56 Similarly, up, s γ v p, s. Indeed, γ v p, s γ γ u p, s γ γ up, s up, s S.57 since γ γ γ γ

15 Problem 3f: The 3D spinors ξ λ of definite helicity λ 1 satisfy p σξ p ξ. S.58 Plugging these ξ λ into the positive-energy Dirac spinors 15, we obtain E ± p ξ up, λ 1. S.59 E p ξ In the ultra-relativistic limit E p m, the square roots here simplify to E + p E and E p by comparison with the other root. Consequently, eq. S.59 simplifies to up, L E ξl, up, R E ξ R. S.6 In other words, the ultra-relativistic positive-energy Dirac spinors of definite helicity are chiral dominated by the LH Weyl components for the left helicity or by the RH Weyl components for the right helicity. Now consider the negative-energy Dirac spinors 16. Because the η s spinors have exactly opposite spins from the ξ s, their helicities are also opposite and hence p ση ± p η S.61 note the opposite sign from eq. S.58. Therefore, the negative-energy Dirac spinors v of definite helicity are + E p η vp, λ 1, S.6 E ± p η and in the ultra-relativistic limit they become vp, L E, vp, R + E η L ηr. S.63 Again, the ultra-relativistic negative-energy spinors are chiral, but this time the chirality is opposite from the helicity the left-helicity spinor has dominant RH Weyl components while the right-helicity spinor has dominant LH Weyl components. 15

16 Problem 4a: First, let s check the normalization conditions for the Dirac spinors 15 and 16 for the general momenta: u p, sup, s ξ s E p σ + E + p σ ξ s ξ seξ s Eδ s,s, v p, svp, s η s + E p σ + E + p σ η s η s+eη s Eδ s,s. S.64 Note that in the second equation η sη s ξ s σ σ ξ s ξ sξ s δs,s. Now consider the Lorentz invariant products ūu and vv. The ū and v are given by ūp, s u p, sγ E p σ ξ s 1 E + p σ ξ s 1 ξ s E + p σ, ξ s E p σ, + E p σ vp, s v p, sγ η s 1 E + p σ η s 1 S.65 η s E + p σ, +η s E p σ. Consequently, ūp, s up, s ξ s E + p σ E p σ ξ s + ξ s E p σ E + p σ ξ s S.66 because Likewise, m ξ sξ s mδ s,s E + p σ E p σ E p σ E + p σ E p σ E p m. vp, s vp, s η s E + p σ E p σ η s η s E p σ E + p σ η s m η sη s mδ s,s. S.67 S.68 16

17 Problem 4b: In matrix notations column row matrix, we have up, s up, s E pσ ξs ξ E + pσ s E + pσ, ξ s E pσ ξs vp, s vp, s E pσ ξs ξ s E + pσ E + pσ E pσ ξ s ξ s E + pσ ξs ξ s E + pσ ξ s ξ s + E pσ ηs η s E + pσ, +η s E pσ E + pσ η s E E pσ η s η s + pσ + E pσ η s η s E pσ E pσ, S.69 + E + pσ η s η s E + pσ E + pσ η s η s E pσ E pσ. S.7 Summing over two spin polarizations replaces ξ s ξ s with s ξ s ξ s 1 η s η s with s η s η s 1. Consequently, and likewise up, s up, s s [ E pσ s ξ s ξ s ] E + pσ E pσ [ E s ξ s ξ s] pσ ] E + pσ E + pσ [ E s ξ s ξ s] pσ [ E + pσ s ξ s ξ s E pσ E + pσ E pσ E pσ E + pσ E + pσ E + pσ E pσ m E pσ E + pσ m m E γ p γ p + m. S.71 17

18 vp, s vp, s s [ E pσ s η s η s ] E + pσ + E pσ [ E s η s η s] pσ + E + pσ [ ] E s η s η s + pσ E + pσ [ E s η s η s] pσ E pσ E + pσ + E pσ E pσ + E + pσ E + pσ E + pσ E pσ m E pσ E + pσ m m E γ p γ p m. S.7 Q.E.D. 18