Physics 513, Quantum Field Theory Examination 1

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1 Physics 513, Quantum Field Theory Examination 1 Due Tuesday, 28 th October 2003 Jacob Lewis Bourjaily University of Michigan, Department of Physics, Ann Arbor, MI

2 2 JACOB LEWIS BOURJAILY 1. a We are to verify that in the Schödinger picture we may write the total momentum operator, P d 3 x πx φx, in terms of ladder operators as P 2π 3 p a pa p. Recall that in the Schrödinger picture, we have the following expansions for the fields φ and π in terms of the bosonic ladder operators 1 φx 2π 3 e ipx a p + a p ; 1.1 πx 2π 3 i Ep 2 eipx a p a p. 1.2 To begin our derivation, let us compute φx. 1 φx ap 2π 3 e ipx + a pe ipx, 1 ipap 2π 3 e ipx ipa pe ipx, 1 2π 3 ipe ipx a p a p. Using this and 1.2 we may write the expression for P directly. P d 3 x πx φx, d 3 k 2π 6 d 3 x d3 k 1 2π E k pe ip+kx a k a k a p a p, E p E k p2π 3 δ 3 p + k a k a k a p a p, E p 2π p a p a p a p a p Using symmetry we may show that a p a p a p a p. With this, our total momentum becomes, 1 P 2π 3 2 p a pa p + a p a p. By adding and then subtracting a pa p inside the parenthesis, one sees that 1 P 2π 3 2 p 2a pa p + [a p, a p], 2π 3 p a pa p + ½ [a p, a p]. Unfortunately, we have precisely the same problem that we had with the Hamiltonian: there is an infinite baseline momentum. Of course, our justification here will be identical to the one offered in that case and so P. 2π 3 p a pa p. 1.3 óπɛρ ɛδɛι δɛ ιξαι

3 PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 3 b We are to verify that the Dirac charge operator, Q d 3 x ψ xψx, may be written in terms of ladder operators as Q a s 2π 3 p a s p b s p b s p. s Recall that we can expand our Dirac ψ s in terms of fermionic ladder operators. 1 ψ a x 2π 3 e ipx a s pu s ap + b s pva p s ; 1.4 s ψ b x 1 2π 3 e ipx r Therefore, we can compute Q by writing out its terms explicitly. Q d 3 x ψ xψx, d 3 x d3 k 1 2π 6 2 e ip kx E k E p r,s d 3 k 1 2π 6 2 2π 3 δ 3 p k E k E p r,s 1 2π 3 2E p r,s p u r b p + br pv r b p. 1.5 [ ] k ur b k + br kv r b k a s pu s ap + b s pva p s, [ ] k ur b k + br kv r b k a s pu s ap + b s pva p s, [ p u r b p + br pv r b p a s pu s ap + b s pv s a p 1 2π 3 p a s 2E pu r b pus ap + p b s pu r b pvs a p p r,s 2π 3 1 2E p r,s 2π 3 1 2E p 2E p δ rs r,s 2π 3 s ], +b r pa s pv r b pus ap + b r pb s pv r p a s pu r b pus ap + b r pb s pv r b pvs a p a s p a s p + b s pb s p, p a s p + b r pb s p,, b pvs a p, We note that by symmetry b s pb s p b s pb s p. By using its anticommutation relation to rewrite b s pb s p and then dropping the infinite baseline energy as we did in part a, we see that Q a s 2π 3 p a s p b s p b s p. 1.6 s óπɛρ ɛδɛι δɛ ιξαι

4 4 JACOB LEWIS BOURJAILY 2. a We are to show that the matrices generate the Lorentz algebra, J µν αβ i δ µ αδ ν β δ µ β δν α, [J µν, J ρσ ] i g νρ J µσ g µρ J νσ g νσ J µρ + g µσ J νρ. We are reminded that matrix multiplication is given by AB αγ A αβ B β γ. Recall that in homework 5.1, we showed that J µν α β i g µα δ ν β g να δ µ β. Let us proceed directly to demonstrate the Lorentz algebra. g [J µν, J ρσ ] δ µ αδ ν β δ µ β α δν ρβ δ σ γ g σβ δ ρ g γ + δ ρ αδ σ β δ ρ β α δσ µβ δ ν γ g νβ δ µ γ, δ µ αδ ν βg ρβ δ σ γ + δ µ αδ ν βg σβ δ ρ γ + δ µ β δν αg ρβ δ σ γ δ µ β δν αg σβ δ ρ γ δ ρ αδ σ βg µβ δ ν γ δ ρ αδ σ βg νβ δ µ γ δ ρ β δσ αg µβ δ ν γ + δ ρ β δσ αg νβ δ µ γ, δ µ αδ ν βg ρβ δ σ γ δ ρ β δσ αg νβ δ µ γg νρ + δ µ αδ ν βg σβ δ ρ γ δ ρ αδ σ βg νβ δ µ γ 1&8 2&6 g νσ + δ µ β δν αg ρβ δ σ γ δ ρ β δσ αg µβ δ ν γg µρ δ µ β δν αg σβ δ ρ γ δ ρ αδ σ βg µβ δ ν γg µσ, 3&7 4&5 [J µν, J ρσ ] i g νρ J µσ g µρ J νσ g νσ J µρ + g µσ J νρ. 2.1 óπɛρ ɛδɛι δɛ ιξαι b Like part a above, we are to show that the matrices generate the Lorentz algebra, S µν i 4 [γµ, γ ν ], [S µν, S ρσ ] i g νρ S µσ g µρ S νσ g νσ S µρ + g µσ S νρ. As Pascal wrote, I apologize for the length of this [proof], for I did not have time to make it short. Before we proceed directly, let s outline the derivation so that the algebra is clear. First, we will fully expand the commutator of S µν with S ρσ. We will have 8 terms. For each of those terms, we will use the anticommutation identity γ µ γ ν 2g µν γ ν γ µ to rewrite the middle of each term. By repeated use of the anticommutation relations, it can be shown that γ µ γ ρ γ ν γ σ γ σ γ ν γ ρ γ µ + 2g νσ γ µ γ ρ g ρσ γ µ γ ν + g µσ γ ρ γ ν g ρν γ σ γ µ + g µν γ σ γ ρ g µρ γ σ γ ν, 2.2 This will be used to cancel many terms and multiply the whole expression by 2 before we contract back to terms involving S µν s. Let us begin. [S µν, S ρσ ] 1 16 [γµ γ ν γ ν γ µ, γ ρ γ σ γ σ γ ρ ], 1 16 [γµ γ ν, γ ρ γ σ ] [γ µ γ ν, γ σ γ ρ ] [γ ν γ µ, γ ρ γ σ ] + [γ ν γ µ, γ σ γ ρ ], 1 16 γµ γ ν γ ρ γ σ γ ρ γ σ γ µ γ ν γ µ γ ν γ σ γ ρ + γ σ γ ρ γ µ γ ν γ ν γ µ γ ρ γ σ + γ ρ γ σ γ ν γ µ + γ ν γ µ γ σ γ ρ γ σ γ ρ γ ν γ µ, 1 2g νρ γ µ γ σ γ µ γ ρ γ ν γ σ 2g νρ γ σ γ µ + γ σ γ ν γ ρ γ µ 16 2g σµ γ ρ γ ν + γ ρ γ µ γ σ γ ν + 2g σµ γ ν γ ρ γ ν γ σ γ µ γ ρ 2g νσ γ µ γ ρ + γ µ γ σ γ ν γ ρ + 2g νσ γ ρ γ µ γ ρ γ ν γ σ γ µ + 2g µρ γ σ γ ν γ σ γ µ γ ρ γ ν 2g µρ γ ν γ σ + γ ν γ ρ γ µ γ σ.

5 PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 5 Now, the rest of the derivation is a consequence of 2.2. Because each γ µ γ ν γ ρ γ σ term is equal to its complete antisymmetrization γ σ γ ρ γ ν γ µ together with six g νσ -like terms, all terms not involving the metric tensor will cancel each other. When we add all of the contributions from all of the cancellings, sixteen of the added twenty-four terms will cancel each other and the eight remaining will have the effect of multiplying each of the g νσ -like terms by two. So after this is done in a couple of pages of algebra that I am not courageous enough to type, the commutator is reduced to [S µν, S ρσ ] 1 4 gνρ γ µ γ σ γ σ γ µ g µσ γ ν γ ρ γ ρ γ ν + g νσ γ µ γ ρ γ ρ γ µ + g µρ γ ν γ σ γ σ γ ν. Λη [S µν, S ρσ ] i g νρ S µσ g µρ S νσ g νσ S µρ + g µσ S νρ. 2.3 óπɛρ ɛδɛι δɛ ιξαι c We are to show the explicit formulations of the Lorentz boost matrices Λη along the x 3 direction in both vector and spinor representations. These are generically given by Λω e i 2 ωµνj µν, where J µν are the representation matrices of the algebra and ω µν parameterize the transformation group element. In the vector representation, this matrix is, So, Λη In the spinor representation, this matrix is coshη 0 0 sinhη sinhη 0 0 coshη. 2.4 coshη/2 sinhη/ coshη/2 + sinhη/ coshη/2 + sinhη/ coshη/2 sinhη/2 Λη e η/ e η/ e η/ e η/2 d No components of the Dirac spinor are invariant under a nontrivial boost e Like part c above, we are to explicitly write out the rotation matrices Λθ corresponding to a rotation about the x 3 axis. In the vector representation, this matrix is given by Λθ cos θ sin θ 0 0 sin θ cos θ In the spinor representation, this matrix is given by Λθ e iθ/ e iθ/ e iθ/ e iθ/ f The vectors are symmetric under 2π rotations and so are unchanged under a complete rotation. Spinors, however, are symmetric under 4π rotations are therefore only half-way back under a 2π rotation.

6 6 JACOB LEWIS BOURJAILY 3. a Let us define the chiral transformation to be given by ψ e iαγ5 ψ. How does the conjugate spinor ψ transform? We may begin to compute this transformation directly. ψ ψ ψ γ 0, e iαγ5 ψ γ 0, ψ e iαγ5 γ 0. When we expand e iαγ5 in its Taylor series, we see that because γ 0 anticommutes with each of the γ 5 terms, we may bring the γ 0 to the left of the exponential with the cost of a change in the sign of the exponent. Therefore ψ ψe iαγ b We are to show the transformation properties of the vector V µ ψγ µ ψ. We can compute this transformation directly. Note that γ 5 anticommutes with all γ µ. Therefore, V µ ψγ µ ψ ψe iαγ5 γ µ e iαγ5 ψ, ψγ µ e iαγ5 e iαγ5 ψ, ψγ µ ψ V µ. V µ V µ. 3.2 c We must show that the Dirac Lagrangian L ψiγ µ µ mψ is invariant under chiral transformations in the the massless case but is not so when m 0. Note that because the vectors are invariant, µ µ. Therefore, we may directly compute the transformation in each case. Let us say that m 0. L ψiγ µ µ ψ L ψie iαγ5 γ µ e iαγ5 µ, ψiγ µ e iαγ5 e iαγ5 µ ψ, ψiγ µ µ ψ L. Therefore the Lagrangian is invariant if m 0. On the other hand, if m 0, L ψiγ µ µ ψ ψmψ L ψiγ µ µ ψ ψe iαγ5 me iαγ5 ψ, ψiγ µ µ ψ ψme 2iαγ5 ψ L. It is clear that the Lagrangian is not invariant under the chiral transformation generally. d The most general Noether current is j µ µ φ δφx µ φ νφx Lδ µ ν δx ν, where δφ is the total variation of the field and δx ν is the coordinate variation. In the chiral transformation, δx ν 0 and φ is the Dirac spinor field. So the Noether current in our case is given by, Now, first we note that j µ 5 µ ψ δψ + µ ψ ψiγ µ and µ ψ δ ψ. µ ψ 0. To compute the conserved current, we must find δψ. We know ψ ψ e iαγ5 1 + iαγ 5 ψ, so δψ iγ 5 ψ. Therefore, our conserved current is j µ 5 ψγ µ γ 5 ψ. 3.3 Note that Peskin and Schroeder write the conserved current as j µ 5 ψγ µ γ 5 ψ. This is essentially equivalent to the current above and is likewise conserved.

7 PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 7 e We are to compute the divergence of the Noether current generally i.e. when there is a possibly non-zero mass. We note that the Dirac equation implies that γ µ µ ψ imψ and µ ψγ µ im ψ. Therefore, we may compute the divergence directly. µ j µ 5 µ ψγ µ γ 5 ψ ψγ µ γ 5 µ ψ, µ ψγ µ γ 5 ψ + ψγ 5 γ µ µ ψ, im ψγ 5 ψ im ψγ 5 ψ, µ j µ 5 i2m ψγ 5 ψ. 3.4 Again, this is consistent with the sign convention we derived for j µ 5 Schroeder. but differs from Peskin and 4. a We are to find unitary operators C and P and an anti-unitary operator T that give the standard transformations of the complex Klein-Gordon field. Recall that the complex Klein-Gordon field may be written φx φ x 1 ap 2π 3 e ipx + b pe ipx ; 1 a 2π 3 p e ipx + b p e ipx. We will proceed by ansatz and propose each operator s transformation on the ladder operators and then verify the transformation properties of the field itself. Parity We must to define an operator P such that Pφt, xp φt, x. Let the parity transformations of the ladder operators to be given by Pa p P η a a p and Pb p P η b b p. We claim that the desired transformation will occur with a condition on η. transformations imply that Pφt, xp 1 η 2π 3 a a p e ipx + ηb b pe ipx φt, x. Clearly, these If we want Pφt, xp φt, x up to a phase η a, then it is clear that η a must equal ηb in general. More so, however, if we want true equality we demand that η a ηb 1. Charge Conjugation We must to define an operator C such that Cφt, xc φ t, x. Let the charge conjugation transformations of the ladder operators be given by Ca p C b p and Cb p C a p. These transformations clearly show that Cφt, xc 1 bp 2π 3 e ipx + a pe ipx φ t, x. Time Reversal We must to define an operator T such that T φt, xt φ t, x. Let the anti-unitary time reversal transformations of the ladder operators be given by T a p T a p and T b p T b p. Note that when we act with T on the field φ, because it is anti-unitary, we must take the complex conjugate of each of the exponential terms as we bring T in. This yields the transformation, T φt, xt 1 a 2π 3 p e ipx + b pe ipx φ t, x.

8 8 JACOB LEWIS BOURJAILY b We are to check the transformation properties of the current under C, P, and T. Let us do each in turn. Parity Note that under parity, µ µ. Charge Conjugation J µ i[φ µ φ µ φ φ], PJ µ P Pi[φ µ φ µ φ φ]p, i[pφ P P µ φp P µ φ P PφP ], i[φ t, x µ φt, x µ φ t, xφt, x], PJ µ P J µ. 4.1 CJ µ C Ci[φ µ φ µ φ φ]c, i[cφ C C µ φc C µ φ C CφC ], i[φ µ φ µ φφ ], CJ µ C J µ. 4.2 Time Reversal Note that under time reversal, µ µ and that T is anti-unitary. T J µ T T i[φ µ φ µ φ φ]t, i[t φ T T µ φt T µ φ T T φt ], i[ T φ T µ T φt + µ T φ T T φt ], i[φ t, x µ φ t, x µ φ t, xφ t, x], T J µ T J µ. 4.3