Quantum Field Theory 2011 – Solutions

Yichen Shi Easter 2014

Note that we use the metric convention (− + ++).

State and prove Noethers theorem in the context of a classical Lagrangian ﬁeld theory deﬁned in Minkowski space.

See Q4 2013.

Yukawa theory has a Lagrangian density

1 1 L = − ∂ φ ∂aφ − m2φ2 − iψ¯(γa∂ + m)ψ + g ψ¯ φ ψ. 2 a 2 a Find the Noether current and conserved charge associated with the phase rotation ψ → e−iαψ. Find the Noether current and conserved charge associated with time translation symmetry.

The question implies that ψ → e−iαψ ' (1 − iα)ψ is a symmetry of the theory, so we do not need to verify them. Then by Noether’s theorem, the conserved current

a ∂L ∂L ¯ αj = δψ + ¯ δψ ∂(∂aψ) ∂(∂aψ) = −iψγ¯ a(−iαψ) = −αψγ¯ aψ.

And the conserved charge Z Z Q = d3x j0 = − d3x ψγ¯ aψ.

For a space-time translational symmetry, we have xa → xa + αa. Then,

a a φ(x) → φ(x) − α ∂aφ(x); ψ(x) → ψ(x) − α ∂aψ(x).

By Noether’s theorem, the associated current is the energy-momentum tensor ∂L ∂L T ab = − ∂bψ − ∂bφ + ηabL ∂(∂aψ) ∂(∂aφ) 1 1 = iψγ¯ a∂bψ + ∂aφ∂bφ − ηab ∂ φ ∂cφ + m2φ2 + iψ¯(γc∂ + m)ψ − g ψ¯ φ ψ 2 c 2 c 1 1 = iψγ¯ a∂bψ + ∂aφ∂bφ − ηab ∂ φ ∂cφ + m2φ2 , 2 c 2

1 by using the equations of motion for ψ¯. We can do so since a current is conserved only when the equations of motion are obeyed. So for a time translational symmetry, 1 1 ⇒ T 00 = iψγ¯ 0ψ˙ + φ˙2 + ∂ φ ∂cφ + m2φ2 2 c 2 1 1 1 = ψ¯(iγi∂ − m)ψ + φ˙2 + (∇φ)2 + m2φ2, i 2 2 2 where we again used the equations of motion for ψ¯. And the conserved charge is the total energy, Z E = d3x T 00.

Brieﬂy describe any additional Noether currents and charges that this theory has.

There are conserved currents and charges associated with spatial translational invariances, i.e., total momentum. Note that there is no Noether current from the phase rotation of φ since gψφψ¯ leads to a term in δL that is not a total derivative.

Free scalar field theory in momentum space is defined by a set of Schrodinger picture operators a~p and † a~p satisfying † 3 3 [a~p, a~q] = (2π) δ (~p − ~q) and a Hamiltonian Z d3~p H = E a† a . (2π)3 ~p ~p ~p H Define what is meant by a Heisenberg picture operator, and find the Heisenberg picture operators a~p H† † and a~p obtained from a~p and a~p.

In the Heisenberg picture, operators evolve with time whereas states are time-independent, in contrast with the Schrodinger picture. For any operator,

iHt −iHt OH = e OS e .

Given the Schrodinger picture ﬁelds

Z d3~p 1 φ(~x) = a ei~p·~x + a† e−i~p·~x ; 3 p ~p ~p (2π) 2E~p r Z d3~p E π(~x) = (−i) ~p a ei~p·~x − a† e−i~p·~x , (2π)3 2 ~p ~p ﬁnd the corresponding Heisenberg picture ﬁelds. Show that φH obeys the Klein-Gordon equation.

Since † † [H, a~p] = −E~pa~p;[H, a~p] = E~pa~p we have H iHt −iHt −iE~pt a~p = e a~p e = e a~p;

†H iHt † −iHt iE~pt † a~p = e a~p e = e a~p.

2 H Hence if we let a~p → a~p in the Schodinger picture mode expansions, then

Z d3~p 1 φH (x) = a eip·x + a† e−ip·x ; 3 p ~p ~p (2π) 2E~p r Z d3~p E πH (x) = (−i) ~p a eip·x − a† e−ip·x . (2π)3 2 ~p ~p Now,

φ˙H = iH eiHtφS e−iHt + eiHtφS e−iHt(−iH) = i[H, φH ] i Z = d3~x π2(~y) + (∇φ(~y))2 + m2φ2(~y), φ(~x) 2 i Z = d3~x π2(~y), φ(~x) 2 = π(~x), and

π˙ H = iH eiHtπS e−iHt + eiHtπS e−iHt(−iH) = i[H, πH ] i Z = d3~x (∇φ(~y))2 + m2φ2(~y), φ(~x) 2 Z = − d3~x ∇φ(~y)∇δ3(~x − ~y) + m2φ(~y)δ3(~x − ~y) Z = d3~x ∇2φ(~y)δ3(~x − ~y) − m2φ(~x)

= ∇2φ(~x) − m2φ(~x).

Combining the two calculations, we have

φ¨H − ∇2φH + m2φH = 0, which is the KG equation.

Deﬁne the vacuum state |0i and ﬁnd an expression for h0|φH (x)φH (y)|0i as a 3-momentum integral.

The vacuum state |0i is deﬁned such that it is annihilated by an annihilation operator, i.e.,

a~p|0i = 0.

Z d3~p Z d3~q 1 0|φH (x)φH (y)|0 = (eip·x−iq·y) 0|[a , a†]|0 3 3 p p q (2π) (2π) 2 E~pE~q Z d3~p Z d3~q 1 = (eip·x−iq·y)(2π¨¨)3δ3(~p − ~q) 3 ¨¨3 p ¨ (2π) ¨(2π) 2 E~pE~q Z 3 d ~p 1 ip·(x−y) = 3 e . (2π) 2E~p

3 Let u and v be arbitrary Dirac spinors. Show that (¯uγav)∗ =vγ ¯ au. Using the Dirac algebra, determine Tr(γaγb) and Tr(γaγbγcγd).

Since γa∗ = γ0γaγ0, γ0∗ = −γ0, and

(¯uγav)∗ = v∗γa∗γ0∗u†∗ = v†γ0γa(−γ0)γ0u =vγ ¯ au.

By the cyclic permutation of elements in Tr(), and noting that we are tracing over spinor (not Lorentz) indices, 1 1 Tr(γaγb) = Tr(γaγb + γbγa) = Tr(2ηabI ) = 4ηab. 2 2 4

Tr(γaγbγcγd) = Tr(γaγb(2ηcd − γdγc)) = 2ηcdTr(γaγb) − Tr(γaγbγdγc) = 2ηcdTr(γaγb) − Tr(γa(2ηbd − γdγb)γc) = 2ηcdTr(γaγb) − 2ηbdTr(γaγc) + Tr(γaγdγbγc) = 2ηcdTr(γaγb) − 2ηbdTr(γaγc) + Tr((2ηad − γdγa)γbγc) = 2ηcdTr(γaγb) − 2ηbdTr(γaγc) + 2ηadTr(γbγc) − Tr(γdγaγbγc) = 8ηcdηab − 8ηbdηac + 8ηadηbc − Tr(γaγbγcγd).

Hence Tr(γaγbγcγd) = 4(ηcdηab − ηbdηac + ηadηbc)

Now let {us(p), vs(p)}, s = ±1/2 be an orthonormalised basis set of spinors satisfying

a a (iγ pa + m)us(p) = 0, (−iγ pa + m)vs(p) = 0;

u¯sur = −2imδsr, v¯svr = 2imδsr. Why is it necessary that p2 + m2 = 0 here? Establish the spin sums

X a X a us(p)¯us(p) = −γ pa − im, vs(p)¯vs(p) = −γ pa + im. s s

We are given a (iγ pa + m)us(p) = 0. b a ⇒ (−iγ pb + m)(iγ pa + m)us(p) = 0. b a 2 ⇒ (γ γ pbpa + m )us(p) = 0. But we similarly have a b 2 (γ γ papb + m )us(p) = 0. Assuming the solution is nontrivial, adding the expressions and using the Cliﬀord algebra, we obtain

ab 2 2η papb + 2m = 0.

⇒ p2 + m2 = 0.

Now, since {us(p), vs(p)} is a basis of the four dimensional vector space of Dirac spinors for every ﬁxed value of p, linear operators X X A(p) := us(p)¯us(p),B(p) := vs(p)¯vs(p) s s

4 are determined by their action on the basis spinors ur(p) and vr(p). X A(p)ur(p) = us(p)¯us(p)ur(p) s X = −2im us(p)δsr s

= −2imur(p) a = −imur(p) − i(−iγ pa)ur(p) a = (−γ pa − im)ur(p).

Hence X a us(p)¯us(p) = −γ pa − im. s Similarly, a B(p)vr(p) = (−γ pa + im)vr(p). Hence X a vs(p)¯vs(p) = −γ pa + im. s Use the results above to simplify the expression

X a ∗ b (¯us(p)γ vr(q)) (¯us(p)γ vr(q)), s,r

where p2 = q2 = −m2. Describe brieﬂy how this quantity could arise in a QED calculation.

Note that an 1x1 matrix is equal to its trace. Also recall the identities

Tr(γaγbγc) = 0 due to the antisymmetry of γa and the cyclic permutation of trace, and

Tr(γaγb) = 4ηab, Tr(γaγbγcγd) = 4(ηabηcd − ηacηbd + ηadηbc).

X a ∗ b X a b ⇒ (¯us(p)γ vr(q)) (¯us(p)γ vr(q)) = v¯r(q)γ us(p)¯us(p)γ vr(q) s,r s,r ! X a b = Tr v¯r(q)γ us(p)¯us(p)γ vr(q) s,r ! X a b = Tr vr(q)¯vr(q)γ us(p)¯us(p)γ s,r c a d b = Tr (−γ qc + im)γ (−γ pd − im)γ c a d b 2 a b = Tr(γ qcγ γ pdγ ) + m Tr(γ γ ) = 4(qapb − p · qηab + qbpa + m2ηab).

This quantity can arise in, for example, muon-electron scattering, when ﬁnding |A|2 from

0 a 0 A ∼ u¯(p1)γ u(p1)¯u(p2)γau(p2).

5 Write down the Feynman rules for a QED scattering amplitude involving only photons on external lines.

1. Draw the Feynman diagrams and label with momentum consistent with conservation. 2. For each internal fermion propagator, give a factor of −i(−p/ + m) . p2 + m2 − i 3. For each interaction vertex, add −ieγa, impose vertex momentum conservation and overall momentum conservation with 4 4 X X (2π) δ pin − pout .

d4k 4. Integrate over momenta associated with loops with measure (2π)4 . ∗ 5. Add polarisation vectors for external photons: a incoming; a outgoing. 6. Take into account minus signs due to the commutation of fermions. 7. Divide by symmetry factors.

a) Explain the constraints on the photon polarization vectors and how they arise.

Using the mode expansion of A~, Z d3~p 1 X A~ = (a eip·x + a† e−ip·x), (2π)3 p r r,~p r,~p 2E~p r the gauge condition ∇ · A~ = 0 gives · ~p = 0, hence the polarisation vectors are transverse to spatial momentum.

b) Explain the key steps in the derivation of the photon propagator.

We can simply read oﬀ the photon propagator from Maxwell’s Lagrangian in the Coulomb gauge: 1 1 L = − F F ab = − A Ai. Coulomb 4 ab 2 i 2 Fourier transform of the propagator, being the inverse of δijk , is, δ D (k) = ij . F ij k2 − i

c) Draw the lowest order Feynman diagram for two photon to two photon scattering, and give the integral expression for the scattering amplitude to this order. (Simpliﬁcation of the expression is not required.)

Z 4 −i(−p + m) −i(−p + m) −i(−p + m) −i(−p + m) 4 d k a /1 b /2 c /3 d /4 −(−ie) 4 γ 2 2 γ 2 2 γ 2 2 γ 4 2 (2π) p1 + m − i p2 + m − i p3 + m − i p1 + m − i d) Draw the diagrams that contribute at next to lowest order to this scattering amplitude.