# Why is $\mathcal{M}(k)$ given by this? (Ward Identity derivation in Peskin & Schroeder)

In page 160 of Peskin & Schroeder we are considering an amplitude $$\mathcal{M}(k)$$ with an external photon as given in equation (5.77) $$\sum_{\epsilon}|\epsilon_\mu^*(k)\mathcal{M}^\mu(k)|^2=|\mathcal{M}^1(k)|^2+|\mathcal{M}^2(k)|^2.\tag{5.77}$$ Then we recall that external photons are given by the term $$\int{d^4x}ej^\mu A_\mu$$ and hence $$\mathcal{M}^\mu(k)$$ is given by $$\mathcal{M}^\mu(k)=\int{d^4x e^{ikx}\langle f|\bar{\psi}(x)\gamma^\mu\psi(x)|i\rangle}.\tag{5.78}$$ I don't understand how we come to this conclusion.

It is due to the simplest application of relativistic perturbation theory. The S-matrix is defined with $$H_I$$ as interaction Hamiltonian

$$S= T\exp\left(-i \int_{-\infty}^\infty \hat{H_I} dt \right)$$

where the symbol T stands for time ordering to be applied to the development of the exponential in a power series. But as we will only deal with 1. order of perturbation theory here, the time ordering does not come into effect. So the S-matrix is up to first order:

$$S = id - i\int_{-\infty}^\infty \hat{H_I} dt$$

Now in QED the interaction operator is just the volume integral of the electron current $$j^\mu$$ and the electromagnetic 4-potential operator $$\hat{A}_\mu$$

$$\hat{H_I} = \int d^3x e \hat{j}^\mu \hat{A}_\mu$$

so plugging this into the precedent expression gives:

$$S = id - i \int d^4x e \hat{j}^\mu \hat{A}_\mu$$

Now we only need to plug in the electromagnetic potential operator into the last expression ($$c$$ and $$c^\dagger$$ are the annihilation and creation operators of photons):

$$\hat{A}_\mu = \sum_{\epsilon=1,2}\sum_p [c_{pe} \epsilon_\mu e^{-ipx} + c^\dagger_{pe} \epsilon^\ast_\mu e^{ipx} ]$$

$$\epsilon_\mu$$ are components of the polarisation vector of the photon.

Considering an emission of a photon means that in the final state the is an additional photon with 4-momentum k, we take the matrix element of the S-matrix $$\langle 1_k f| S| 0i\rangle$$. Here $$1_k$$ means a final state with 1 photon of 4-momentum k and a transition of the participating electron from initial state $$i$$ to final state $$f$$ (note that the corresponding matrix element $$\langle 1|c_{pe}|0\rangle =0$$):

$$\langle 1_k f|S| 0i\rangle = \langle 1_k f| 0i\rangle - i \int d^4x e \langle f|j^\mu|i\rangle \sum_{\epsilon=1,2}\sum_p \langle 1_k| c^\dagger_{pe}| 0\rangle \epsilon^\ast_\mu e^{ipx} = - i\int d^4x \sum_{\epsilon=1,2} e \langle f|\bar{\psi}\gamma^\mu \psi|i\rangle e^{ikx} \epsilon^\ast_\mu$$

by using $$\langle 1_k| c^\ast_{pe}| 0\rangle =\delta_{pk}$$ and the orthogonality of Fock states ($$\langle 1| 0\rangle =0$$)

By defining $$\cal{M}^\mu(k) = \int d^4x \sum_{\epsilon=1,2}\langle f|\bar{\psi}\gamma^\mu \psi|i\rangle e^{ikx}$$

we get:

$$\langle 1_k f|S| 0i\rangle = -i e\sum_{\epsilon=1,2}\epsilon^\ast_\mu \cal{M}^\mu(k)$$

so the S-matrix element between a transition of an electron state $$i$$ to and electron state $$f$$ and an emitted photon of the 4-momentum k and polarisation state $$\epsilon$$ is given as above indicated.

Finally if we consider the transition probability we would take the modulus of the $$S$$-matrix element:

$$| \langle 1_k f|S| 0i\rangle |^2 =e^2 \sum_{\epsilon=1,2}|\epsilon^\ast_\mu \cal{M}^\mu(k)|^2$$

because the mixed term cancels out due the orthogonality of the polarisation states. Well, we get an additional $$e^2$$, but actually for the derivation flow, I guess, it does not matter.

Important remark: the emission of a single real photon is actually not possible for free electrons (only possible for bound electrons), so we have to imagine the states $$i$$ and $$f$$ of the participating electron as bound.

• Precise and to the point. Thank you! That is exactly what I was looking for. And for anyone who reads this in the future. The definition of $\hat{A}_\mu$ is the Fourier Transform of equation 4.131 of Peskin & Schroeder. Commented May 20, 2022 at 12:54
• @twistedmanifold Happy to hear this. I actually used another source (Landau/Lifschitz Vol.4) as a guideline. I will have look later into P&S on eq. 4.131. Commented May 20, 2022 at 12:57
• Actually, it's the same relation, just in the continuum case of the momenta. Commented May 20, 2022 at 13:21
• Sorry, I'd like to make a step clear from the derivation. How do you get the matrix elements $\langle f|j^\mu|i\rangle$ seperateley from the rest of the amplitude? I'm not sure how you compute that. I understand the annihilation/creation operators but not how you take that seperately. Commented May 30, 2022 at 14:00
• Well, $\langle 1_p| c_{pe} \epsilon_\mu e^{-ipx} + c^\dagger_{pe} \epsilon^\ast_\mu e^{ipx}|0\rangle = \langle 1_p| c_{pe}| 0\rangle \epsilon_\mu e^{-ipx} + \langle 1_p| c^\dagger_{pe}|0\rangle \epsilon^\ast_\mu e^{ipx} = 0 + \epsilon^\ast_\mu e^{ipx}$ due to the linearity of the Hilbert space. An annihilation operator acting on$|0 \rangle$ is zero. $\langle 1_k| c^\dagger_{pe}| 0\rangle =0$ too if $p\neq k$. Commented May 30, 2022 at 14:50