Quantum Field Theory Exam 1 Solutions - Advanced Quantum Mechanic | PHYSICS 513, Exams of Quantum Mechanics

Download Quantum Field Theory Exam 1 Solutions - Advanced Quantum Mechanic | PHYSICS 513 and more Exams Quantum Mechanics in PDF only on Docsity! Physics 513, Quantum Field Theory Examination 1 Due Tuesday, 28th October 2003 Jacob Lewis Bourjaily University of Michigan, Department of Physics, Ann Arbor, MI 48109-1120 1 2 JACOB LEWIS BOURJAILY 1. a) We are to verify that in the Schödinger picture we may write the total momentum operator, P = − ∫ d3x π(x)∇φ(x), in terms of ladder operators as P = ∫ d3p (2π)3 p a†pap. Recall that in the Schrödinger picture, we have the following expansions for the fields φ and π in terms of the bosonic ladder operators φ(x) = ∫ d3p (2π)3 1√ 2Ep eipx ( ap + a † −p ) ; (1.1) π(x) = ∫ d3p (2π)3 (−i) √ Ep 2 eipx ( ap − a†−p ) . (1.2) To begin our derivation, let us compute ~∇φ(x). ∇φ(x) = ∇ ∫ d3p (2π)3 1√ 2Ep ( ape ipx + a†pe −ipx), = ∫ d3p (2π)3 1√ 2Ep ( ipapeipx − ipa†pe−ipx ) , = ∫ d3p (2π)3 1√ 2Ep ipeipx ( ap − a†−p ) . Using this and (1.2) we may write the expression for P directly. P = − ∫ d3x π(x)∇φ(x), = − ∫ d3x d3kd3p (2π)6 1 2 √ Ek Ep pei(p+k)x ( ak − a†−k )( ap − a†−p ) , = ∫ d3kd3p (2π)6 −1 2 √ Ek Ep p(2π)3δ(3)(p + k) ( ak − a†−k )( ap − a†−p ) , = ∫ d3p (2π)3 1 2 p ( a†p − a−p ) ( ap − a†−p ) . Using symmetry we may show that a−pa † −p = apa † p. With this, our total momentum becomes, P = ∫ d3p (2π)3 1 2 p ( a†pap + apa † p ) . By adding and then subtracting a†pap inside the parenthesis, one sees that P = ∫ d3p (2π)3 1 2 p ( 2a†pap + [ap, a † p] ) , = ∫ d3p (2π)3 p ( a†pap + ½ [ap, 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 = ∫ d3p (2π)3 p a†pap. (1.3) ‘óπ²ρ ’²́δ²ι δ²ιξαι PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 5 Now, the rest of the derivation is a consequence of (2.2). Because each γµγνγργσ term is equal to its complete antisymmetrization γσγργνγµ together with six gνσ-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 gνσ-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 [Sµν , Sρσ] = 1 4 (−gνρ(γµγσ − γσγµ)− gµσ(γνγρ − γργν) + gνσ(γµγρ − γργµ) + gµρ(γνγσ − γσγν)) . ∴ [Sµν , Sρσ] = i (gνρSµσ − gµρSνσ − gνσSµρ + gµσSνρ) . (2.3) ‘óπ²ρ ’²́δ²ι δ²ιξαι c) We are to show the explicit formulations of the Lorentz boost matrices Λ(η) along the x3 direction in both vector and spinor representations. These are generically given by Λ(ω) = e− i 2 ωµνJ µν , where Jµν are the representation matrices of the algebra and ωµν parameterize the transforma- tion group element. In the vector representation, this matrix is, Λ(η) =   cosh(η) 0 0 sinh(η) 0 1 0 0 0 0 1 0 sinh(η) 0 0 cosh(η)   . (2.4) In the spinor representation, this matrix is Λ(η) =   cosh(η/2)− sinh(η/2) 0 0 0 0 cosh(η/2) + sinh(η/2) 0 0 0 0 cosh(η/2) + sinh(η/2) 0 0 0 0 cosh(η/2)− sinh(η/2)   So, Λ(η) =   e−η/2 0 0 0 0 eη/2 0 0 0 0 eη/2 0 0 0 0 e−η/2   . (2.5) 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 Λ(θ) corresponding to a rotation about the x3 axis. In the vector representation, this matrix is given by Λ(θ) =   1 0 0 0 0 cos θ − sin θ 0 0 sin θ cos θ 0 0 0 0 1   . (2.6) In the spinor representation, this matrix is given by Λ(θ) =   e−iθ/2 0 0 0 0 eiθ/2 0 0 0 0 e−iθ/2 0 0 0 0 eiθ/2   . (2.7) 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 JACOB LEWIS BOURJAILY 3. a) Let us define the chiral transformation to be given by ψ → eiαγ5ψ. How does the conjugate spinor ψ̄ transform? We may begin to compute this transformation directly. ψ̄ → ψ̄′ = ψ′†γ0, = (eiαγ 5 ψ)†γ0, = ψ†e−iαγ 5 γ0. When we expand e−iαγ 5 in its Taylor series, we see that because γ0 anticommutes with each of the γ5 terms, we may bring the γ0 to the left of the exponential with the cost of a change in the sign of the exponent. Therefore ψ̄ → ψ̄eiαγ5 . (3.1) b) We are to show the transformation properties of the vector V µ = ψ̄γµψ. We can compute this transformation directly. Note that γ5 anticommutes with all γµ. V µ = ψ̄γµψ → ψ̄eiαγ5γµeiαγ5ψ, = ψ̄γµe−iαγ 5 eiαγ 5 ψ, = ψ̄γµψ = V µ. Therefore, V µ → V µ. (3.2) c) We must show that the Dirac Lagrangian L = ψ̄(iγµ∂µ −m)ψ is invariant under chiral trans- formations in the the massless case but is not so when m 6= 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 = ψ̄iγµ∂µψ → L′ = ψ̄ieiαγ 5 γµeiαγ 5 ∂µ, = ψ̄iγµe−iαγ 5 eiαγ 5 ∂µψ, = ψ̄iγµ∂µψ = L. Therefore the Lagrangian is invariant if m = 0. On the other hand, if m 6= 0, L = ψ̄iγµ∂µψ − ψ̄mψ → L′ = ψ̄iγµ∂µψ − ψ̄eiαγ 5 meiαγ 5 ψ, = ψ̄iγµ∂µψ − ψ̄me2iαγ 5 ψ 6= L. It is clear that the Lagrangian is not invariant under the chiral transformation generally. d) The most general Noether current is jµ = ∂L ∂(∂µφ) δφ(x)− ( ∂L ∂(∂µφ) ∂νφ(x)− Lδµν ) δxν , where δφ is the total variation of the field and δxν is the coordinate variation. In the chiral transformation, δxν = 0 and φ is the Dirac spinor field. So the Noether current in our case is given by, jµ5 = ∂L ∂(∂µψ) δψ + ∂L ∂(∂µψ̄) δψ̄. Now, first we note that ∂L ∂(∂µψ) = ψ̄iγµ and ∂L ∂(∂µψ̄) = 0. To compute the conserved current, we must find δψ. We know ψ → ψ′ = eiαγ5 ∼ (1 + iαγ5)ψ, so δψ ∼ iγ5ψ. Therefore, our conserved current is jµ5 = −ψ̄γµγ5ψ. (3.3) Note that Peskin and Schroeder write the conserved current as jµ5 = ψ̄γ µγ5ψ. This is essen- tially equivalent to the current above and is likewise conserved. 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 γµ∂µψ = −imψ and ∂µψ̄γµ = imψ̄. Therefore, we may compute the divergence directly. ∂µj µ 5 = −(∂µψ̄)γµγ5ψ − ψ̄γµγ5∂µψ, = −(∂µψ̄)γµγ5ψ + ψ̄γ5γµ∂µψ, = −imψ̄γ5ψ − imψ̄γ5ψ, ∴ ∂µjµ5 = −i2mψ̄γ5ψ. (3.4) Again, this is consistent with the sign convention we derived for jµ5 but differs from Peskin and Schroeder. 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 φ(x) = ∫ d3p (2π)3 1√ 2Ep ( ape −ipx + b†pe ipx ) ; φ∗(x)− ∫ d3p (2π)3 1√ 2Ep ( a†pe ipx + bpe−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 Pφ(t,x)P† = φ(t,−x). Let the parity transfor- mations of the ladder operators to be given by PapP† = ηaa−p and PbpP† = ηbb−p. We claim that the desired transformation will occur (with a condition on η). Clearly, these transformations imply that Pφ(t,x)P† = ∫ d3p (2π)3 1√ 2Ep ( ηaa−pe−ipx + η∗b b † −pe ipx ) ∼ φ(t,−x). If we want Pφ(t,x)P† = φ(t,−x) up to a phase ηa, then it is clear that ηa must equal η∗b in general. More so, however, if we want true equality we demand that ηa = η∗b = 1. Charge Conjugation We must to define an operator C such that Cφ(t,x)C† = φ∗(t,x). Let the charge conjugation transformations of the ladder operators be given by CapC† = bp and CbpC† = ap. These transformations clearly show that Cφ(t,x)C† = ∫ d3p (2π)3 1√ 2Ep ( bpe −ipx + a†pe ipx ) = φ∗(t,x). Time Reversal We must to define an operator T such that T φ(t,x)T † = φ(−t,x). Let the anti-unitary time reversal transformations of the ladder operators be given by T apT † = a−p and T bpT † = 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 φ(t,x)T † = ∫ d3p (2π)3 1√ 2Ep ( a−peipx + b † −pe −ipx ) = φ(−t,x).