PHY 396 K. Solutions for homework set #9.

Transcription

1 PHY 396 K. Solutions for homework set #9. Problem 2(a): The γ 0 matrix commutes with itself but anticommutes with the space-indexed γ 1,2,3. At the same time, the parity reflects the space coordinates but not the time coordinate, x x = x but t t = +t, hence the new space and time derivatives are related to the old derivatives as = but 0 = +. Together, these two facts give us γ 0 = ( γ 0 0 γ ) γ 0 = γ 0( γ γ ) = γ 0( +γ 0 0 γ ) = γ 0 (S.1) and hence (i m) γ 0 = γ 0 (i m). (S.2) Combining this formula with eq. (9.1) for the Dirac field, we find (i m) Ψ (x ) = (i m) ( ±γ 0 Ψ(x) ) = ±(i m)γ 0 Ψ(x) = ±γ 0 (i m)ψ(x) (S.3) the (i m)ψ(x) transforms under parity precisely like the Ψ(x) field itself. In other words, the Dirac equation transforms covariantly. Now consider the Dirac Lagrangian. Taking the Hermitian conjugate of eq. (9.1) we find Ψ ( x, t) = ±Ψ (x, t)γ 0 = ±Ψ (x, t)γ 0 (S.4) and hence Ψ ( x, t) = ±Ψ(x, t)γ 0. (S.5) Consequently, the Dirac Lagrangian L = Ψ(i m)ψ transforms into L(x ) = Ψ (x ) (i m) Ψ (x ) = ±Ψ(x)γ 0 ±γ 0 (i m)ψ(x) = +Ψ(x) (i m)ψ(x) (S.6) = L(x). In other words, the Dirac Lagrangian is invariant modulo x x = ( x, +t), and the Dirac action S = d 4 x L is invariant. 1

2 Problem 1(b): First, let s check how parity acts on the plane-wave solutions from the last homework, ( ) ( ) + E p σ ξ s + E p σ u(p, s) = + η s, v(p, s) = E + p σ ξ s. ( ) E + p σ η s Under parity, particles momenta reverse directions but the angular momenta including the spins remain unchanged, thus p p but s +s. Applying this rule to the Dirac spinors ( ), we see that indeed, ( ) + E + p σ ξ s u(p, s) u( p, +s) = + E p σ ξ s ( ) ( ) E p σ ξ s = = γ 0 u(p, s), E + p σ ξ s ( ) + E + p σ η s v(p, s) v( p, +s) = E p σ η s ( ) ( ) E p σ η s = 1 0 = γ 0 v(p, s). E + p σ η s (S.7) Now let s apply parity to the quantum Dirac field Ψ(x, t) = d 3 p 1 ( (2π) 3 e itep+ix p u(p, s) â 2E p,s + e +itep ix p v(p, s) ˆb ) p,s. p s (S.8) Everything besides the â p,s and ˆb p,s operators in this expansion is a c-number, hence the expression on the left hand side of eq. (9.2) expands to ( P Ψ(x, d 3 p 1 e ite p+ix p u(p, s) P ) â p,s P t) P = (2π) 3 2E p + e +itep ix p v(p, s) P ˆb P. (S.9) p,s s 2

3 At the same time, for the right hand side of eq. (9.2) we have ±γ 0 Ψ( x, t) = d 3 p (2π) 3 1 using eqs. (S.7) = d 3 p (2π) 3 1 2E p s 2E p s ± e itep ix p γ 0 u(p, s) â p,s ± e +itep+ix p γ 0 v(p, s) ˆb p,s ± e itep ix p u( p, s) â p,s e +itep+ix p v( p, s) ˆb p,s changing variable p p d 3 p 1 = ± e itep+ix p u(p, s) â p,s (2π) 3 2E p s e +itep ix p v(p, s) ˆb. p,s (S.10) By eq. (9.2), the right hand sides of eqs. (S.9) and (S.10) must be equal to each other. Since the Dirac plane waves e ipx u(p, s) and e +ipx v(p, s) are linearly independent from each other, this means P â p,s P = ±â p,s and P ˆb P p,s = ˆb p,s. (9.3.a) The rest of eq. (9.3) follows by hermitian conjugation: Since P = P 1 = P, P â p,s P = ) ( P âp,s P = ±â p,s, P ˆb p,s P = ( P ˆb p,s P) = ˆb p,s. (9.3.b) Finally, eqs. (9.4) follow from eqs. (9.3) for the creation operators and also from the parityinvariance of the vacuum state, P 0 = 0. Indeed, P F (p, s) = Pâ p,s 0 = Pâ P p,s P 0 = ±â p,+s 0 = ± F ( p, +s) (9.4.a) and likewise P F (p, s) = Pˆb p,s 0 = Pˆb p,s P P 0 = ˆb p,+s 0 = F ( p, +s). (9.4.b) As promised, the fermion and the antifermion end up with opposite intrinsic parities. 3

4 Problem 1(c): Despite anticommutativity of the fermionic fields, the Hermitian conjugation of an operator product reverses the order of operators without any extra sign factors, thus (Ψ αψ β ) = +Ψ β Ψ α. Consequently, for any 4 4 matrix Γ, (Ψ ΓΨ) = +Ψ Γ Ψ, and hence (ΨΓΨ) = ΨΓΨ where Γ = γ 0 Γ γ 0 is the Dirac conjugate of Γ. Now consider the 16 matrices which appear in the bilinears (9.5). Obviously 1 = +1, which gives us S = +S. Next, we saw in class that γ µ = +γ µ, and this gives us (V µ ) = +V µ. We also saw that 2 i γ[µ γ ν] = 2 i γ[ν γ µ] = + 2 i γ[µ γ ν], and this gives us (T µν ) = +T µν. As to the γ 5 matrix, we saw in the last homework (problem 1) that it s Hermitian and anticommutes with all the γ µ. Consequently, γ 5 = γ 0 (γ 5 ) γ 0 = +γ 0 γ 5 γ 0 = γ 5 = iγ 5 = +iγ 5, which gives us P = +P. Finally, γ 5 γ µ = γ µ γ 5 = γ µ γ 5 = +γ 5 γ µ, which gives us (A µ ) = +A µ. Thus, by inspection, all the bilinears (9.5) are Hermitian. Q.E.D. Problem 1(d): Under a continuous Lorentz symmetry x x = Lx, the Dirac spinor field and its conjugate transform according to Ψ (x ) = M(L)Ψ(x = L 1 x ), Ψ (x ) = Ψ(x = L 1 x )M 1 (L), (S.11) hence any bilinear ΨΓΨ transforms according to Ψ (x )ΓΨ(x ) = Ψ(x)Γ Ψ(x) (S.12) where Γ = M 1 (L)ΓM(L). (S.13) Now let s check the specific bilinears. Obviously for Γ = 1, Γ = M 1 M = 1, which makes S a Lorentz scalar. Also, for Γ = γ 5, Γ = M 1 γ 5 M = γ 5 because for continuous Lorentz symmetries L, the M(L) matrix commutes with the γ 5 cf. last homework, problem 2, which makes P another Lorentz scalar. 4

5 The transformation laws for the V µ and A µ bilinears follows from the formula M 1 (L)γ µ M(L) = L µ νγ ν (D.23) I had derived in class, see my notes on Dirac spinors, eq. (23). Consequently, V µ = L µ νv ν, which makes V µ a Lorentz vector. Also, since γ 5 commutes with the M(L), we have M 1 (L)γ µ γ 5 M(L) = L µ νγ ν γ 5, (S.14) thus A µ is also a Lorentz vector. Finally, M 1 γ µ γ ν M = M 1 γ µ M M 1 γ ν M = L µ κl ν λ γκ γ λ M 1( i 2 γ [µ γ ν]) M = L µ κl ν λ i 2 γ[κ γ λ], (S.15) which makes T µν a Lorentz tensor (with two antisymmetric indices). Q.E.D. Problem 1(e): Under parity, the Dirac fields transform as Ψ (x ) = ±γ 0 Ψ(x), Ψ (x ) = ±Ψ(x)γ 0, (S.16) cf. eqs. (9.1) and (S.5). Consequently, the Dirac bilinears transform as P : ΨΓΨ Ψ ΓΨ x = Ψ(γ 0 Γγ 0 )Ψ. x x (S.17) By inspection, out of 16 possible Γ matrices, 1, γ 0, 2 i γ[i γ j], and γ 5 γ i commute with the γ 0, while γ i, 2 i γ[0 γ i], γ 5 γ 0, and γ 5 anticommute with the γ 0. Therefore, the S, V 0, T ij, and A i remain invariant under parity, while the V i, T 0i, A 0, and P change their signs. From the 3D point of view, this means that S and V 0 are true scalars, P and A 0 are pseudoscalars, V is a true or polar vector, A is a pseudo-vector or axial vector, and the tensor T contains one true vector T 0i and one axial vector 2 1ɛijk T jk. In space-time terms, we call S a true (Lorentz) scalar, P a (Lorentz) pseudoscalar, V µ a true (Lorentz) vector, and A µ an axial (Lorentz) vector. Pedantically speaking, T µν is a true Lorentz tensor while T κλ 1 2 ɛκλµν T µν is a Lorentz pseudo-tensor, but few people bother with this distinction. 5

6 Problem 2(a): In the Weyl convention, the Dirac fields transform as in eq. (9.8), and we also have a convenient γ matrix relation γ µ γ 2 = γ 2 (γ µ ) µ. Consequently, (i m)γ 2 = γ 2 (i m) and hence Ĉ(i m)ψĉ = (i m) (ĈΨ Ĉ = ±γ 2 Ψ ) = ±γ 2 (i m) Ψ (S.18) which makes the Dirac equation covariant with respect to the charge conjugation. Now consider the Dirac Lagrangian L = Ψ(i m)ψ. Under charge conjugations it transforms into ( Ĉ Ψ(i m)ψ)ĉ = +Ψ γ 2 γ 2 (i m) Ψ = Ψ (i m) Ψ, which naively = L. (S.19) Since we expect the Dirac action and hence the Lagrangian L to be real, it seems to flip sign under charge conjugation instead of being invariant. However, the complex conjugation of the classical fermionic fields and their products is tricky because their values are anticommuting Grassmann numbers instead of the ordinary real or complex numbers. I have not discussed the mathematics of Grassmann numbers in class. All we need to know is that the odd GN anticommute with each other while the even GN commute with each other and with the odd GN. Also, the GN valued classical fermionic fields Ψ(x) must make sense as the classical limits of the operator-valued quantum fields Ψ(x). Thus, regardless of the precise definition of the complex conjugation for the Grassmann numbers, we know that it should work similarly to the hermitian conjugation of operators in the Fock space. particular, for any GN g, (g ) = g, (cg) = c g for any c-number c, and for any 2 GNs g 1 and g 2, (g 1 g 2 ) = g 2 g 1 under conjugation, the order of the product is reversed, similar to (Â1Â2) = Â 2Â 1 for any operators Â1 and Â2. For two odd GN such as classical fermionic fields F 1 and F 2, this means In (F 1 F 2 ) = F 2 F 1 = F 1 F 2. (S.20) Consequently, in eq. (S.19) Ψ (i m) Ψ = ( Ψ(i m)ψ ) (S.21) 6

7 and hence Ĉ LĈ = +L. (S.22) Since the Dirac Lagrangian is real up to a total derivative, this makes the Dirac action invariant under charge conjugation Ĉ. Alternative proof: In the solutions below for parts (b) and (c) of this problem we shall see that Ĉ ΨΨĈ = +ΨΨ while Ĉ Ψγµ ΨĈ = Ψγµ Ψ. (S.23) The first formula here provides for C invariance of the mass term mψψ in the Dirac Lagrangian, while the second formula can be generalized to 2 different Dirac spinors transforming in the same way. Ĉ Ψ 1 γ µ Ψ 2 Ĉ = Ψ 2 γ µ Ψ 1. (S.24) Now, let Ψ 1 = Ψ while Ψ 2 = i µ Ψ, then Ĉ Ψ(iγ µ µ )ΨĈ = i µψγ µ Ψ = +iψγ µ µ Ψ i µ ( Ψγ µ Ψ ) (S.25) where the second term is a total derivative. Altogether, Ĉ Ψ(i m)ψĉ = +Ψ(i m)ψ + a total derivative. (S.26) Problem 2(b): For the quantum Dirac fields, the complex conjugation is defined as hermitian conjugation in the Hilbert space of each component but without transposing the Dirac spinor from a column vector to a row vector or vice verse, Ψ = ˆΨ 1. = ( Ψ ), Ψ = ( Ψ ) γ 0 = Ψ γ 0 = ( ˆΨ1 ˆΨ4 ) γ 0. (S.27) ˆΨ 4 7

8 Thus, eqs. (9.8) for the charge conjugation of the quantum Dirac fields mean Ĉ Ψ(x)Ĉ = ±γ2 Ψ (x) = ±γ 2( Ψ (x) ), Ĉ Ψ(x)Ĉ = ± Ψ (x)γ 2 = ± Ψ (x)γ 0 γ 2 = Ψ (x)γ 2 γ 0. (S.28) Consequently, for the Dirac bilinears ΨΓ Ψ, the charge conjugations acts as Ĉ ΨΓ ΨĈ = Ψ (x)γ 2 γ 0 Γγ 2( Ψ (x) ) ( γ 2 γ 0 Γγ 2) ˆΨ α ˆΨ αβ β ( = + γ 2 γ 0 Γγ 2) ( ˆΨ ˆΨ αβ β α { ˆΨ β, ˆΨ ) α } (S.29) = + Ψ ( γ 2 γ 0 Γγ 2) Ψ the anticommutator term. In the classical limit the anticommutator term goes away, and we are left with C : ΨΓΨ Ψ ( γ 2 γ 0 Γγ 2) Ψ = Ψ γ 2 Γ γ 0 γ 2 Ψ = Ψγ 0 γ 2 Γ γ 0 γ 2 Ψ ΨΓ c Ψ. (S.30) Q.E.D. Problem 2(e): By inspection, 1 c γ 0 γ 2 γ 0 γ 2 = +1. The γ 5 matrix is symmetric and commutes with the γ 0 γ 2, hence γ5 c = +γ 5. Among the four γ µ matrices, the γ 1 and the γ 3 are anti-symmetric and commute with the γ 0 γ 2 while the γ 0 and the γ 2 are symmetric but anti-commute with the γ 0 γ 2 ; hence, for all four γ µ, γµ c = γ µ. Finally, because of the transposition involved, (γ µ γ ν ) c = γνγ c µ c = +γ ν γ µ, hence ( 2 i γ[µ γ ν] ) c = + 2 i γ[ν γ µ] = 2 i γ[µ γ ν]. Likewise, (γ 5 γ µ ) c = (γ µ ) c (γ 5 ) c = γ µ γ 5 = +γ 5 γ µ. Therefore, according to eq. (S.30), the scalar S, the pseudoscalar P, and the axial vector A µ are C even, while the vector V µ and the tensor T µν are C odd. 8

9 Problem 3(a): A bound state with a definite orbital angular momentum L is symmetric with respect to positions x 1 and x 2 of the two particles for even L and antisymmetric for odd L. After Fourier transforming from the relative position x 1 x 2 to the reduced momentum p red, this means ψ( p red, s 1, s 2 ) = ( 1) L ψ(+p red, s 1, s 2 ). (S.31) Likewise, a bound state with a definite net spin S is symmetric or antisymmetric with respect to the two particles spin states s 1 and s 2 depending on S = 1 or S = 0, ψ(p red, s 2, s 1 ) = ( 1) 1 S ψ(p red, s 1, s 2 ). (S.32) Eqs. (9.10) for the intrinsic C-parity and P-parity of the bound state follow from these two facts, plus the action of the Ĉ and P operators on the one-fermion+one-antifermions states â ˆb 0. Indeed, for the charge conjugation operator Ĉ we have Ĉ â (+p red, s 1 )ˆb ( p red, s 2 ) 0 = Ĉ â (+p red, s 1 )Ĉ Ĉ ˆb ( p red, s 2 )Ĉ 0 = (±1) 2 ˆb (+p red, s 1 ) â ( p red, s 2 ) 0 = +ˆb (+p red, s 1 )â ( p red, s 2 ) 0. (S.33) = â ( p red, s 2 )ˆb (+p red, s 1 ) 0. Consequently, applying Ĉ to the bound state (9.9) gives us Ĉ B = d 3 p red (2π) 3 = d 3 p red (2π) 3 ψ(p red, s 1, s 2 ) Ĉ â (+p red, s 1 )ˆb ( p red, s 2 ) 0 s 1,s 2 ψ(p red, s 1, s 2 ) â ( p red, s 2 )ˆb (+p red, s 1 ) 0 s 1,s 2 changing variable from p red to p red and also swapping s 1 s 2 d 3 p = red (2π) 3 ψ( p red, s 2, s 1 ) â (+p red, s 1 )ˆb ( p red, s 2 ) 0, s 1,s 2 (S.34) 9

10 where in light of eqs. (S.31) and (S.32) ψ( p red, s 2, s 1 ) = ( 1) L ψ(+p red, s 2, s 1 ) = ( 1) L ( 1) 1 S ψ(+p red, s 1, s 2 ) (S.35) = ( 1) L+S ψ(+p red, s 1, s 2 ). Therefore, Ĉ B = ( 1) L+S B (S.36) in accordance with the first eq. (10.10). Likewise, for the parity operator P we have P â (+p red, s 1 )ˆb ( p red, s 2 ) 0 = P â (+p red, s 1 ) P P ˆb ( p red, s 2 ) P 0 = (±1)â ( p red, s 1 ) ( )ˆb (+p red, s 2 ) 0 (S.37) = â ( p red, s 1 )ˆb (+p red, s 2 ) 0. where the overall sign comes from opposite intrinsic parities of the fermion and the antifermion. Consequently, P acting on the bound state (9.9) produces P B(p tot ) = d 3 p red (2π) 3 = d 3 p red (2π) 3 ψ(p red, s 1, s 2 ) P â (+p red, s 1 )ˆb ( p red, s 2 ) 0 s 1,s 2 ψ(p red, s 1, s 2 ) â ( p red, s 1 )ˆb (+p red, s 2 ) 0 s 1,s 2 changing variable p red p red, but no spin swap d 3 p = red (2π) 3 ψ( p red, s 1, s 2 ) â (+p red, s 1 )ˆb ( p red, s 2 ) 0, s 1,s 2 (S.38) where this time ψ( p red, s 1, s 2 ) = ( 1) L ψ(+p red, s 1, s 2 ). (S.39) Therefore, P B = (01) L+1 B (S.40) in accordance with the second eq. (9.10). 10

11 Problem 3(b) Since the EM interactions are symmetric under charge conjugations, the EM processes such as e +e + photons conserve C-parity. A photon of any momentum or polarization has C = 1, so the net C-parity of an n photon finite state is ( 1) n. Consequently, if the initial electron and positron are in a bound state with C = +1 they must annihilate into an even number of photons, e + e + 2γ, 4γ, 6γ,.... But if the bound state has C = 1, the electron and the positron must annihilate into an odd number of photons, e + e + 3γ, 5γ,.... (Annihilation into a single photon is forbidden by the energy-momentum conservation.) The ground state of a hydrogen-like positronium atom is 1S, meaning n rad = 1 and L = 0. Due to spins, there are actually 4 almost-degenerate 1S states; the hyperfine structure splits them into the 1S 3 triplet of S = 1 states called the ortho-positronium and the 1S 1 singlet S = 0 state called the para-positronium. According to eq. (S.36), the ortho-positronium states have C = ( 1) L ( 1) S = ( 1) 0 ( 1) 1 = 1 while the para-positronium state has C = ( 1) L ( 1) S = ( 1) 0 ( 1) 0 = +1. Consequently, the para-positronium decays into an even number of photons, (e + e + )@1S 1 2γ, 4γ,..., (S.41) while the ortho-positronium decays into odd numbers of photons, (e + e + )@1S 3 3γ, 5γ,.... (S.42) This difference affects the net decay rate of each state because QED (Quantum Electro Dynamics) has a rather small coupling constant α = (e 2 /4π) 1/137. For each photon in the final state, the decay amplitude carries a factor of e, so the decay rate of a positronium atom into n photons Γ(e e + nγ) is O(α n ). Consequently, the S = 0 para-positronium state usually decays into just 2 photons while decays into 4, 6, or more photons are allowed but much less common. Likewise, the S = 1 ortho-positronium states usually decays into 3 photons while decays into 5 or more photons are allowed but rare. Moreover, the decay rate into 3 photons is much slower than the decay rate into just 2 photons, Γ ( (e + e + )@1S 3 3γ ) Γ ( (e + e + )@1S 1 2γ ) = O(α3 ) O(α 2 ) = O(α), (S.43) hence the net decay rate of an S = 1 ortho-positronium state into anything it can decay to i.e., into any odd number of photons is much slower then the net decay rate of the S = 0 11

12 para-positronium state, Γ ( (e + e + )@1S 3 anything ) Γ ( ) (e + e + = O(α) 1. )@1S 1 anything (S.44) Experimentally, the ortho-positronium decays about 1100 times slower than the parapositronium, Γ(ortho) s 1 Γ(para) s 1. (S.45) Problem 4(a): In the last homework (problem 2) you saw that in terms of the Weyl fermions, the Dirac Lagrangian becomes L = Ψ(i m)ψ = iψ R σµ µ ψ R + iψ L σµ µ ψ L mψ L ψ R mψ R ψ L, (S.46) cf. eq. (S8.18) of the the solutions. In this problem we take ψ L = χ and ψ R = σ 2 χ, so the last three terms in the Lagrangian (S.46) become iψ L σµ µ ψ L iχ σ µ µ χ, mψ L ψ R +mχ σ 2 χ, (S.47) mψ R ψ L +m χ σ 2 χ. The remaining first term in the RHS of eq. (S.46) is slightly more tricky: iψ R σµ µ ψ R i ( χ ) σ 2 σ µ ( µ σ2 χ ) = i χ ( σ 2 σ µ σ 2 = ( σ µ ) ) µ χ i χ α ( σ µ ) βα µ χ β for classical fermionic fields χ α µ χ β = µ χ β χ α (S.48) = i µ χ β ( σµ ) βα χ α i µ χ σ µ χ = i µ ( χ σ µ χ ) + i χ σ µ µ χ. Assembling all these terms together, we obtain L = i µ ( χ σ µ χ ) + i χ σ µ µ χ + iχ σ µ µ χ + mχ σ 2 χ + m χ σ 2 χ, (S.49) where the first term on the RHS is a total derivative while the remaining terms are exactly as in eq. (9.12). Q.E.D. 12

13 Problem 4(b): Let s work through the matrices: Ψ1Ψ = Ψγ µ Ψ = Ψγ µ γ ν Ψ = Ψγ µ γ 5 Ψ = Ψγ 5 Ψ = thus ( χ ( ) ( ) σ 2 χ ) 1 0 χ 0 1 σ 2 χ = χ σ 2 χ χ σ 2 χ, ( χ ( ) ( ) σ 2 χ ) 0 σ µ χ σ µ 0 σ 2 χ = χ σ 2 σ µ σ 2 χ + χ σ µ χ, ( χ ( σ 2 χ ) σ µ σ ) ( ) ν 0 χ i 0 2 σ µ σ ν σ 2 χ = χ σ 2 σ µ σ ν χ χ σ µ σ ν χ ( χ ( ) ( ) σ 2 χ ) 0 σ µ χ σ µ 0 σ 2 χ = χ σ 2 σ µ σ 2 χ χ σ µ χ, ( χ ( ) ( ) σ 2 χ ) 1 0 χ 0 +1 σ 2 χ = + χ σ 2 χ χ σ 2 χ, (S.50) S Ψ1Ψ = χ σ 2 χ χ σ 2 χ, V µ Ψγ µ Ψ = χ σ 2 σ µ σ 2 χ + χ σ µ χ, T µν Ψ i 2 γ[µ γ ν] Ψ = i 2 χ σ 2 σ [µ σ ν] χ i 2 χ σ [µ σ ν] χ, (S.51) A µ Ψγ µ γ 5 Ψ = χ σ 2 σ µ σ 2 χ χ σ µ χ, P Ψiγ 5 Ψ = +i χ σ 2 χ iχ σ 2 χ. Note that the scalar S, the pseudoscalar P, and the tensor T all have form O + O where O is a complex (non-hermitian) bilinear of the two LH spinors χ and χ. On the other hand, the vector V and the axial vector A are sums of two independent real (hermitian) terms, one a bilinear of χ and its conjugate χ, and the other a bilinear of χ and χ. In fact, the two terms can be written in a symmetric way using χ σ 2 σ µ σ 2 χ = χ ( σ µ ) χ = χ σ µ χ, (S.52) cf. similar argument in part (a), eq. (S.48). Consequently, the vector and the axial vector 13

14 become V µ = χ σ µ χ + χ σ µ χ, A µ = χ σ µ χ χ σ µ χ. (S.53) In this form, it is clear that V µ and A µ are both currents of the respective vector / axial symmetries of the Dirac spinor fields. In terms of the LH Weyl fields χ and χ, these symmetries act as χ(x) exp( iθ V + iθ A ) χ(x), χ(x) exp(+iθ V + iθ A ) χ(x), (S.54) hence the currents (S.53). Problem 4(c): The parity P acts on the Dirac spinor Ψ(x) as in eq. (9.1). In terms of the LH Weyl spinors χ and χ, this means ( hence χ ( x, +t) σ 2 χ ( x, +t) ) = ± ( ) ( χ(x, t) σ 2 χ (x, t) ), (S.55) χ ( x, t) = σ 2 χ (x, t) (S.56) while σ 2 χ ( x, +t) = ±χ(x, t) = χ ( x, t) = ±σ 2 χ (x, t). (S.57) Altogether, χ ( x, t) = σ 2 χ (x, t), χ ( x, t) = ±σ 2 χ (x, t). (S.58) Similarly, the charge conjugation C acts on the Dirac spinor according to eq. (9.8). In terms of the LH Weyl spinors χ and χ, this means ( ) ( ) ( ) χ ( ) (x) 0 σ2 χ(x) χ(x) σ 2 χ = ± (x) σ 2 0 σ 2 χ = ± (x) σ 2 χ, (S.59) hence χ (x) = ± χ(x), χ (x) = ±χ(x). (S.60) In other words, the charge conjugation simply exchanges the two LH Weyl spinors, χ χ. 14

15 Finally, consider the combined symmetry CP. In light of eqs. (S.58) and (S.60), it acts as χ(x, t) P ±σ 2 χ ( x, +t) C ±σ 2 χ ( x, +t), χ(x, t) P σ 2 χ ( x, +t) C σ 2 χ ( x, +t). (S.61) Note that unlike C and P that mix χ and χ, the combined CP symmetry acts separately on each Weyl spinor. Consequently, for theories with multiple Weyl spinor fields χ i (x), we may figure the action of CP by simply generalizing eqs. (S.61) or (9.15) without knowing which pairs of Weyl spinors combine into Dirac spinors as in eq. (9.11). On the other hand, the C and P symmetries naturally combine Weyl fermions into pairs, and we do need to who is whose partner to find how C and P act assuming the theory even has those symmetries. Problem 4(d): First, let s derive the Weyl equations from the Lagrangian (9.13). Taking the derivative of L WRT χ i, etc., and minding the order of the fermionic fields, we obtain L ( µ χ i ) = iχi σ µ, L ( µ χ i ) = 0, and hence Euler Lagrange equations L χ i = M ji χ j σ 2, L χ i = +i σµ µ χ i + M ijσ 2 χ j, (S.62) σ µ µ χ i = im ijσ 2 χ j (S.63) and µ χ i σ µ = im ij χ j σ 2 = ( σ µ ) µ χ i = +im ij σ 2 χ j = σ µ σ 2 µ χ i = +im ij χ j (where the last version follows from σ 2 ( σ µ ) σ 2 = σ µ σ 2 ( σ µ ) = σ µ σ 2 ). (S.64) Next, let s derive the Klein Gordon equations. immediately obtain Combining eqs. (S.63) and (S.64), we σ ν ν σ µ µ χ i = im ijσ ν ν ( σ2 χ j ) = M ijm jk χ k. (S.65) Now let s simplify the differential operator on the LHS. The derivatives µ commute with each 15

16 other, hence σ ν ν σ µ µ = 1 2 { µ, ν } σ µ σ ν = 1 2 µ ν ( σ µ σ ν + σ ν σ µ) = 1 2 µ ν 2g µν = 2. (S.66) Consequently, eqs. (S.65) become Klein Gordon equations for multiple χ i fields with the mass 2 matrix M M, 2 χ i = M ijm jk χ k. (S.67) Since the fermionic mass matrix M jk is symmetric, the mass 2 matrix here M M = M M is hermitian and 0, so its eigenvalues are real and non-negative. Physically, these eigenvalues are the physical masses 2 of the N species of fermions. Problem 4(e): First, let s check that the kinetic terms in the Lagrangian are invariant (up to a total derivative) under the CP symmetry (9.15). The CP includes space reflection, (t, x) (+t, x), so the derivatives acting on fields before and after CP transform are related as 0 = + 0 but i = i. In light of the definitions of the σµ and σ µ matrices, this gives us σ µ µ = σ µ µ and σ µ µ = σ µ µ. (S.68) Consequently, for each Weyl spinor χ, CP : iχ (x) σ µ µ χ(x) iχ (x ) σ µ µχ (x ) = iχ (x)σ 2 σ µ µ σ 2 χ (x) = iχ (x)σ 2 σ µ σ 2 µ χ (x) = iχ (x) ( σ µ ) µ χ (x) = i µ χ (x) σ µ χ(x) (S.69) = +iχ (x) σ µ µ χ(x) + a total derivative. Thus, each kinetic term in the Lagrangian (9.13) is separately CP invariant. Now consider the mass terms. For any two spinors χ j and χ k, CP : χ j (x)σ 2 χ k (x) χ j (x )σ 2 χ k (x ) = χ j (x)σ 2σ 2 σ 2 χ k (x) = + χ j (x)σ 2 χ k (x), CP : χ j (x)σ 2 χ k (x) χ j (x )σ 2 χ k (x ) = χ j (x)σ 2 σ 2σ 2χ k (x) = +χ j (x)σ 2 χ k (x). (S.70) 16

17 Consequently, when we put all the mass terms together, we obtain 1 CP : 2 M jk χ j σ 2 χ k 2 1 M jk χ j σ 2 χ k, j,k 1 CP : 2 Mjk χj σ 2 χ k 1 2 Mjk χ j σ 2 χ k, j,k j,k j,k (S.71) and hence CP : 1 2 M jk χ j σ 2 χ k Mjk χj σ 2 χ k j,k 1 2 j,k j,k Mjk χ j σ 2 χ k M jk χ j σ 2 χ k. j,k (S.72) Thus, the CP symmetry transforms the mass terms into similar mass terms but with complexconjugated mass matrices, CP : M jk M jk and M jk M jk. (S.73) Therefore, the whole Lagrangian (9.13) is invariant under the CP symmetry acting as in eqs. (9.15) if and only if the mass matrix M jk is real. Q.E.D. Problem 4(f): The modified CP action (9.16) is a combination of the original CP action (9.15) with a unitary transform U : χ j (x) ic k j χ k (x), CP mod = U CP orig. (S.74) In part (e) we saw that the kinetic terms in the Lagrangian are invariant under CP orig. Also, the sum of the kinetic term for all N Weyl spinor fields is clearly invariant under for any unitary matrix Cj k. Consequently, the kinetic part of the Lagrangian is invariant under any modified CP transform (9.16), so all we need to worry about are the mass terms. In part (e) we also saw that the effect of the CP orig on the sum of all the mass terms is equivalent to complex-conjugating the mass matrix as in eq. (S.73). As to the unitary 17

18 transform U, its effect also amounts to changing the mass matrix: U : M jk χ j σ 2 χ k M jk j,k j,k l,m M jk χj σ 2 χ k Mjk j,k j,k l,m ic l j ic m k χ l σ 2χ m, (ic l j ) (ic m k ) χl σ 2χ m, (S.75) which is equivalent to U : M lm j,k M jk ic l j ic m k, M lm j,k M jk (ic l j ) (ic m k ), (S.76) or in matrix notations, U : M CMC, M C M C. (S.77) Combining the effects of the CP orig and the U transforms, we see that the modified CP transform (9.16) has the net effect of changing the mass matrix, CP mod : M M CM C, M M C MC. (S.78) Therefore, the modified CP transform (9.16) is a symmetry of the Lagrangian (9.13) if and only if CM C = M. Q.E.D. Problem 4( ): Any complex matrix M can be written as a product M = UH where H = (M M) 1/2 is a hermitian matrix and U is a unitary matrix. Since H is hermitian, it can be diagonalized; that is, H = V DU where D is real and diagonal while V is another unitary matrix. Altogether, we have M = UV DV. (S.79) When we apply such a decomposition to a symmetric matrix M as is the fermion mass 18

19 matrix we end up with UV DV = ( UV DV ) = V D(UV ) = V = (UV ), UV = V (S.80) and hence M = V DV (S.81) for a real diagonal D and a unitary V = (V ) 1. Next, complex conjugating eq. (S.81), we obtain M = V DV, (S.82) hence V M V = V V D V V = D, (S.83) and therefore M = V ( V M V ) V = (V V ) M (V V ). (S.84) Now let C = iv V. This matrix is unitary since V = V 1 and also V are both unitary. Also, C is symmetric, C = ( iv V ) = i(v ) (V ) = iv V = C. (S.85) Consequently, in light of eq. (S.84), CM C = V V M V V = M, (S.86) which is exactly what we need. 19