Solutions to Problems in Peskin and Schroeder, An Introduction To Quantum Field Theory. Chapter 9

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1 Solutions to Problems in Peskin and Schroeder, An Introduction To Quantum Field Theory Homer Reid June 3, 6 Chapter 9 Problem 9.1 Part a. Part 1: Complex scalar propagator The action for the scalars alone is S[φ, φ, J, J d 4 x { µ φ µ φ m φ φ + J φ + Jφ d 4 x { φ [ µ µ + m φ + J φ + Jφ d 4 x { φ O KG φ + J φ + Jφ where the differential operator O KG is O KG µ µ + m. Introduce shifted fields: φ(x) φ (x) + i d 4 yd F (x y)j(y) φ (x) φ (x) i d 4 ydf (x y)j (y) φ (x) + i d 4 yd F (x y)j (y) where id F (x y) is the solution to O KG id F (x y) δ (4) (x y). 1

2 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 Then the action for the scalars becomes { [ [ S d 4 x φ + i d 4 yd F (x y)j (y) O KG φ + i { d 4 x + J [φ + i φ O KG φ i + i + i d 4 xφ O KG φ + i [ d 4 yd F (x y)j(y) + J φ + i d 4 yj(x)d F (x y)j (y) d 4 yj (x)d F (x y)j(y) d 4 yj(x)d F (x y)j (y) d 4 xd 4 yj (x)d F (x y)j(y). The generating functional is then Z[J, J DφDφ e is[φ,φ,j,j { Z exp d 4 xd 4 yj (x)d F (x y)j(y), d 4 yd F (x y)j(y) d 4 yd F (x y)j (y) where Z DφDφ e iφ O KG φ, and the two-point function is φ (x)φ(y) 1 [ [ i i Z J(x) J Z[J, J (y) J [ { i i d 4 xj(x)d F (x y)z[j, J J(x) D F (x y). Part : Photon propagator J The action for the photon field is { S[A µ, j µ d 4 x 1 4 F µν(x)f µν (x) + j µ (x)a µ (x) (1) Integrate by parts as described pedantically in Appendix 1: { 1 d 4 x Aµ (x) [ g µν µ ν A ν (x) + j µ (x)a µ (x)

3 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 3 Go over to momentum space: d 4 { k 1 (π) 4 Aµ (k) [ k g µν + k µ k ν A ν ( k) + j µ ( k)a µ (k) Go through mysterious Faddeev Popov procedure to come up with magical factor of 1/ξ to desingularize the kernel of the kinetic term, which is otherwise singular as illustrated by brute force in Appendix : d 4 { k 1 (π) 4 d 4 k (π) 4 where the differential operator O EM µν is or in momentum space [ ( k g µν ) k µ k ν A ν ( k) + j µ ( k)a µ (k) ξ Aµ (k) { 1 Aµ (k)o EM µν A ν ( k) + j µ ( k)a µ (k) O EM µν g µν ( 1 1 ) µ ν ξ ( Oµν EM k g µν ) k µ k ν ξ () and has inverse id µν 1 k ( g µν (1 ξ) kµ k ν defined such that the 4x4 matrix equation k ) (3) O EM id 1 is satisfied. Now introduce shifted fields: A ν (x) A ν (x) d 4 y id νρ (x y)j ρ (y) The action becomes { S d 4 x 1 1 A µ (x)o EM µν A ν (x) j µ (x)a µ (x) j µ (x) [ A µ (x) d 4 xa µ (x)o EM µνa ν (x) 1 d 4 y id µρ (x y)j ρ (y) d 4 y j µ (x)id µρ (x y)j ρ (y) d 4 xd 4 y j µ (x)id µρ (x y)j ρ (y). (4)

4 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 4 Aside. Effective fermion-fermion interaction in QED. I think it is interesting to see how this procedure may be used to integrate the photon out of the QED lagrangian to give an effective electron-electron interaction. If we wanted to calculate the expectation value of a product of electron field operators, say for example the electron density at a point x, we would be led to consider an expression of the form ψ(x)γ ψ(x) DψDψDA ψ(x)γ ψ(x)e il[ψ,ψ,a DψDψDA e il[ψ,ψ,a (5) where the QED lagrangian is the spacetime integral of the QED lagrangian density, L[ψ, ψ, A d 4 x L[ψ, ψ, A d 4 x { 14 F µνf µν + ψ(i/ m)ψ eψγ µ ψa µ. Then the photon part of the path integral is { [ DAexp i d 4 x 14 F µνf µν ej µ A µ where j µ eψγ µ ψ is the electron current. The argument of the exponential here is just the action (1). Transforming it into the form (4), the photon path integral is [ { DA exp i [ d 4 x A µ (x)oµν EM A ν (x) e R ie d 4 xd 4 y j µ(x) id µν (x y)j ν(y) The first term is independent of the fermions, and since there are no photon fields in the payload of the numerator in (6), this term cancels, leaving DψDψ ψ(x)γ ψ(x)e il eff [ψ,ψ ψ(x)γ ψ(x) (6) DψDψ e il eff [ψ,ψ where the effective Lagrangian now contains a nonlocal term and hence cannot be written as the space integral of a local density function: { L eff [ψ, ψ d 4 x ψ(x)(i/ m)ψ(x) { e d 4 x d 4 y ψ(x)γ µ ψ(x)id µν (x y)ψ(y)γ ν ψ(y).

5 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 5 To look at the interaction term in k-space, we put d 4 p 1 ψ(x) (π) 4 e ip1x ψ(p 1 ) d 4 p ψ(x) (π) 4 e+ipx ψ(p ) id µν d 4 q e iq(x y) (x y) (g µν (π) 4 q (1 ξ) qµ q ν ) + iɛ q d 4 k 1 ψ(y) (π) 4 e ik1y ψ(k 1 ). d 4 k ψ(y) (π) 4 e+iky ψ(k ) The fermion-fermion interaction term becomes { e d 4 x d 4 y ψ(x)γ µ ψ(x)id µν (x y)ψ(y)γ ν ψ(y) e d 4 p 1 d 4 k 1 d 4 q ψ(p 1 q)γ µ ψ(p 1 ) [g µν (1 ξ) qµ q ν q ψ(k 1 + q)γ ν ψ(k 1 ) (π) 1 q + iɛ. The longitudinal part of id µν actually doesn t contribute here, because e.g. [ ψ(k 1 +q)/qψ(k 1 ) ψ(k 1 +q) [(/k 1 + /q) /k 1 ψ(k 1 ) ψ(k 1 +q) m m ψ(k 1 ) so the interaction term reduces to e d 4 p 1 d 4 k 1 d 4 { q ψ(p1 q)γ µ ψ(p 1 )ψ(k 1 + q)γ µ ψ(k 1 ) (π) 1 q. + iɛ This looks exactly like the interaction term you get when you write down the hamiltonian for the interacting homogenous electron gas, although in that case there are no γ µ matrices and the k integrals are only over 3-dimensional space. Part b. The lowest-order diagram is p k p + p µ ν p k [ ( ie) (p p ) µ igµν (p p ) u s (k)γ ν v s (k )

6 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 6 The squared, averaged matrix element is (ignoring the electron mass) M 1 4 e4 (p p ) µ (p p ) ν (p p ) 4 Tr [ /kγ µ /k γ ν e 4(p p ) µ (p p ) ν (p p ) 4 [ kµ k ν + k νk µ (k k )g µν e 4 { (p p ) 4 We work in a frame such that [ k (p p ) [ k (p p ) (k k )(p p ) (7) k (E,,, E), k (E,,, E) and p (E, p sin θ,, p cosθ), p (E, p sin θ,, p cosθ). Then (7) reads M e4 E p [ cos θ + 1 e4 E p sin θ. Inserting the kinematic factors from P&S equation 4.84, The total cross section is Part c. dσ dω e 4 56π p E sin θ σ π e 4 56π E E m sin θ. π The two first-order diagrams are dσ sin θdθ dω e 4 18πE E m e 4 96πE E m. π sin 3 θdθ

7 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 7 M 1 ( ie) k + q µ ρ σ ν q q d 4 k (π) 4 k [ [ igµρ (k + q) ρ q i (k + q) m (k + q) σ [ [ i igσν k m q k M µ ρ σ ν q q (ie ) d 4 k (π) 4 [ [ [ igµρ ig ρσ igσν q k m q The sum of the diagrams is [ [ igµρ M 1 + M Π ρσ igσν (q) q q where Π ρσ (q) e [ d 4 k (k + q) ρ (k + q) σ g ρσ (k + q) m (π) 4 [(k + q) m [k m (8) Rewrite the denominator: 1 [(k + q) m [k m 1 1 dx {x[(k + q) m + (1 x)[k m dx [k + xk q + xq m Shift variables to l k + xq in (8): [ Π ρσ (q) e d 4 l 4l ρ l σ + (1 x) q ρ q σ g ρσ l + (1 x) q m dx (π) 4 [l, where we ignored numerator terms of odd order in l, and where x(1 x)q + m. (9)

8 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 8 Using l ρσ 1 d l g ρσ under the integral sign, we rewrite this as { [ Π ρσ (q) e dx g ρσ I 1 ( ) + (1 x) q ρ q σ [(1 x) q m g ρσ I ( ) where ( ) 4 I 1 ( ) d i (4π) i (4π) Γ d d l l (π) d (l ) ( 4 d ) d Γ ( ) 1 d ( d d 1 d (here we have used the fact that ( ( ) ( 1 ) d Γ 1 d Γ d ) ) and d d l 1 I ( ) (π) d (l ) i ( ) Γ d (4π). ) d (1) Then (1) is Π ρσ (q) ie (4π) ie (4π) ie (4π) dx Γ ( ) d { [ (1 x) q + m g ρσ + (1 x) q ρ q σ d dx Γ ( ) d { (1 x)(1 x)q g ρσ + (1 x) q ρ q σ d dx Γ ( ) d (1 [ x) q g ρσ q ρ q σ (1 x)q g ρσ {{ d where the last term in the curly brackets integrates to, being odd under x (1 x) while the denominator, which involves, is even. Then the final result is iπ ρσ (q) (q g µν q ρσ )iπ(q ) where iπ(q ) e (4π) e (4π) 1 1 (1 x) Γ( d ) dx d [ (1 x) const. log dx const. γ + log 4π ɛ

9 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter ( m m x(1 x)q )dx 1 (1 x) log q /m Figure 1: Integral in (11) as evaluated by numerical quadrature. - is infinite but q-independent at d 4. Isolating the q-dependence of Π, we have 1 ( iπ(q ) iπ() e (4π) (1 x) m ) log m x(1 x)q dx. (11) Values of this integral are plotted in Figure 1. The divergence at q 4m is clear. Appendix: Integration-by-parts of kinetic term in photon Lagrangian Someday I hope to be able to perform in my head the kind of manipulations that are needed to go from the first to the second lines of P&S equation However, at the moment I need to work it all out in detail. The Lagrangian

10 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 1 density is L 1 4 F µνf µν 1 4 ( µa ν ν A µ )( µ A ν ν A µ ) 1 ( A 1 1 A )( A 1 1 A ) 1 ( A A )( A A ) 1 ( A 3 3 A )( A 3 3 A ) 1 ( 1A A 1 )( 1 A A 1 ) 1 ( 1A 3 3 A 1 )( 1 A 3 3 A 1 ) 1 ( A 3 3 A )( A 3 3 A ). I get scared whenever I see something like A µ or µ because I know there are a bunch of hidden minus signs in there and they confuse me. To make sure that all minus signs are explicit, let s rewrite F µν F µν with all minus signs displayed explicitly. Basically we just go through and replace, i i, A A,

11 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 11 A i A i. This gives L 1 ( A 1 1 A )( A A ) 1 ( A A )( A + A ) 1 ( A 3 3 A )( A A ) 1 ( 1A + 1 A )( 1 A + 1 A ) 1 ( 1A A 3 )( 1 A A 3 ) 1 ( A 3 + A 3 )( A 3 + A 3 ) + 1 [ ( A 1 ) + ( 1 A ) + ( A 1 )( 1 A ) + 1 [ ( A ) + ( A ) + ( A )( A ) + 1 [ ( A 3 ) + ( 3 A ) + ( A 3 )( 3 A ) 1 [ ( 1 A ) + ( A 1 ) ( 1 A )( A 1 ) 1 [ ( 1 A 3 ) + ( 3 A 1 ) ( 1 A 3 )( 3 A 1 ) 1 [ ( A 3 ) + ( 3 A ) ( A 3 )( 3 A ). The next step is to integrate by parts, which entails making replacements like ( 1 A 3 )( 3 A 1 ) A A 1. Also, we break up the terms with prefactors of into two separate terms, each of which we integrate by parts in a different way. This gives a different Lagrangian that integrates to the same thing as the old Lagrangian did, so we will just call the new Lagrangian L as well: L 1 [A 1 A1 + A 1 A + A 1 1 A + A 1 A 1 1 [A A + A A + A A + A A 1 [A 3 A 3 + A 3A + A 3 3 A + A 3 A [A 1A + A 1 A 1 A 1 A 1 A 1 1 A [A 3 1 A3 + A 1 3 A1 A A 1 A A [ A 3 A3 + A 3 A A 3 3 A A 3 A 3.

12 Homer Reid s Solutions to Peskin and Schroeder Problems: Chapter 9 1 I will think of this as a kind of bilinear form between 4-dimensional vectors: ( L 1 1 A A 1 A A ) 1 3 B ) ( ) C B ( ) A A 1 A A 3 1 C A Appendix : Convincing the skeptics among us that k g µν + k µ k ν is singular, but k g µν + (1 1 ξ )k µk ν is invertible with inverse given by (3) These should all work in both octave or matlab. octave:1> g[1 ; -1 ; -1 ; -1; octave:> kurand(4,1) % k with raised index ku octave:3> klg*ku; % k with lowered index octave:4> kkl * ku k octave:5> kmat-k*g + kl * kl kmat octave:6> rank(kmat) ans 3 Sure enough, the matrix has less than full rank! The problem is the existence of an eigenvector with eigenvalue, namely k µ itself. octave:7> kmat*ku ans e e-17.e e-17 On the other hand, the matrix modified to contain the magical Fadeev-Popov factor is nonsingular:

13 octave:8> xirand xi octave:9> kmat-k*g + (1-1/xi)*kl*kl kmat octave:1> rank(kmat) ans 4 Moreover, its inverse is just the matrix (3): octave:11> id-(g - (1-xi)*ku*ku /k) / k id octave:1> kmat*id ans