# Evaluation of QED amplitude with 1 external photon

I'm trying to compute the exact QED amplitude with one external photon. Suppose that the photon has 4-momentum $q$ and polarization $\varepsilon^\mu$.

Peskin and Schroeder (p318) claim that ignoring the contribution from the external photon line we get the amplitude

$$\langle\Omega |S|q\rangle = -ie\int d^4xe^{-iq.x}\langle\Omega|T\bar\psi\gamma_\mu\psi|\Omega\rangle$$

I have no idea how they derive this however. I've tried two ideas and would appreciate if somebody would review both.

Approach 1

The result would seem to follow directly from the definition of the $S$-matrix if we replaced the $\Omega$ on the RHS by the vacuum of the free theory, namely $0$. Have P&S just made a typo?

Approach 2

Using the LSZ formula I get something completely different, namely

$$\frac{\sqrt{Z}}{q^2 + i\epsilon}\langle\Omega |S|q\rangle\epsilon^\mu = \int d^4xe^{-iq.x}\langle\Omega|TA^\mu|\Omega\rangle$$

It looks like I've got the VEV of completely the wrong field now! Have I applied LSZ wrongly? And if so what's my conceptual mistake?

• – Jia Yiyang Aug 25 '13 at 1:30
• To be brief, the first matrix element is the matrix element of a current $j_\mu$ while the latter has $A_\mu$. Roughly speaking, in some gauge and with some normalization coefficients, $\Box A_\mu = j_\mu$ so the two expressions only differ by the box which is essentially equal to the simple factor $q^2$ you see in the LSZ formula. – Luboš Motl Aug 25 '13 at 17:57

I've figured out the answer myself, so I thought I would post it here in case anyone is interested. I'll treat each of the approaches in turn.

Approach 1

What I wrote above is patently wrong. We can derive the correct result using the definition of the $S$-matrix, and it's relation to free theory amplitudes. This is given by

$$\langle\Omega|S|q\rangle_\Omega = \left(\langle 0|T \exp\left\{-i\int d^4 x \mathcal{H}_I\right\}|q\rangle_0\right)_{\textrm{connected, amputated}}$$

where $\mathcal{H}_I$ is the Hamiltonian density in the interaction picture (i.e. involves the Heisenberg fields of the free theories only).

Now we can write out the terms of this using Wick's theorem and observing (as usual) that only fully contracted terms count. The first nontrivial terms are schematically

$$\int \langle0|T-ie\bar{\psi}\gamma^\mu\psi A^\mu|q\rangle + \int\int\int\langle0|(-ie)^3 \langle0|T\bar{\psi}\gamma^\mu\psi A^\mu\bar{\psi}\gamma^\mu\psi A^\mu\bar{\psi}\gamma^\mu\psi A^\mu|q\rangle + \dots$$

where the integrals are over spacetime. Now the contraction of $A^\mu$ with $q$ gives an $e^{-iq.x}\epsilon^\mu$. If you ignore the polarization vector and check that the factors work out correctly you verify the result in P&S.

Note that the amputated prescription is not a problem, since this only applies to the external photon line, which we ignore.

Approach 2

The LSZ formula as I wrote it above was largely correct. To be honest, I've realised that the formula itself is not a good thing to memorise. Rather, one should note that the $S$-matrix element can be computed by

• Fourier transforming the relevant correlation function
• computing the residue of the multiparticle pole
• dividing by the appropriate field strength renormalization factors

The relevant correlation function in this case is $\langle\Omega|TA^\mu|\Omega\rangle$ as I wrote above. We'd like to relate this to the correlation function $\langle \Omega|Tj^\mu|\Omega\rangle$. The key idea (as pointed out by Lubos) is to use the classical equation of motion which has form

$$\Box A^\mu = -j^\mu$$

where $\Box$ is some differential operator, with inverse given by the Feynman propagator for photons (under an integral). The quantum generalisation of this is the Schwinger-Dyson equation

$$\Box\langle\Omega|TA^\mu|\Omega\rangle + \langle\Omega|Tj^\mu|\Omega\rangle = \textrm{ contact terms }$$

which is straightforward to derive in the path-integral formalism. The contact terms don't contribute to the $S$-matrix because they have the wrong singularity structure. Thus inverting $\Box$ we find up to contact terms

$$\langle \Omega | TA^\mu(x) |\Omega\rangle = \int d^4y D_F^{\mu\nu}(x-y)\langle\Omega|Tj_\nu(y)|\Omega\rangle$$

whence the Fourier transform is

$$\int d^4x e^{-iq.x}\langle\Omega|TA^\mu|\Omega\rangle = \int d^4y e^{-iq.y}\frac{(-i)}{q^2}\langle\Omega|Tj^\mu|\Omega\rangle$$

by effecting the Fourier transform of the Green's function. Now taking the residue as the photon goes "on-shell" and ignoring the polarization and field strength renormalization we regain the result from P&S again.