# Why can we use the equation of motion to calculate the amplitude in “Quantum Field Theory”?

I am reading the chapter on electron-proton scattering from "Quantum Field Theory in a Nutshell". The author calculates the amplitude of the electron-proton scattering (up to the second order). The Feynman diagram used in this calculation is

We start by using the massive spin 1 boson as an exchange, and hope to get rid of the mass later. Let the mass of the boson be $$\mu$$

Using the Feynman rules, we obtained the amplitude to be

$$\mathcal{M}(P,P_N) = (-ie)(ie)\frac{i}{(P-p)^2-\mu^2}\left(\frac{k_\mu k_\nu}{\mu^2}-\eta_{\mu\nu}\right)\bar{u}(P)\gamma^\mu u(p)\bar{u}(P_N)\gamma^\nu u(p_N).$$

He claims that the term $$k_\mu k_\nu$$ will just disappear because

$$k_\mu \bar{u}(P)\gamma^\mu u(p) = (P-p)_\mu \bar{u}(P)\gamma^\mu u(p) = \bar{u}(P)(\not P - \not p) u(p)=\bar{u}(P)(m - m) u(p)=0.$$

However, the last equation follows the Dirac equation, which is the equation of motion of the electron. The point that I do not understand is, how can we use the equation of motion in the quantum field theory? Clearly, the equation of motion comes from minimising (extremising) the action, which leads to the classical field rather than quantum field.

• How well have you studied the quantization of the Dirac free field before diving into the interacting theory? – OON Aug 9 '19 at 11:01
• I am not sure how well, but I have been through it once. I have studied both canonical quantisation and path integral quantisation. – TangBear Aug 9 '19 at 11:10
• well, probably not too well. Look for the decomposition of the Dirac field operator in terms of the creation-annihilation operators. There will be those functions $u(p)$, note that those are simply c-numbers, not operators – OON Aug 9 '19 at 11:52
• I still don't quite understand what you are trying to guide me. Would you clarify it a little bit, please? – TangBear Aug 9 '19 at 12:05
• It seems you may be trying to learn QFT from Zee. This is a great book, but imo not the best to learn this topic. Peskin & Schroeder or Schwartz are much more pedagogical and give more details. – Oбжорoв Aug 9 '19 at 12:40

The spinor $$u(p)$$ satisfies the equation $$\not{\!p} u(p) = m\, u(p)$$ by construction from the expansion of the Dirac spinor $$\psi(x)$$ in solutions of the Dirac equation. The quantisation (in the canonical approach) happens by making the coefficients of these spinors operators. These two things should not be confused with one another.
The whole point of canonical quantisation is to expand the solution of the classical equation of motion in plane waves and make the coefficients operators. For scalar fields that is straightforward and it is basically the harmonic oscillator. For spin one field you need an extra $$\epsilon^\mu$$ to describe the polarisation of the field (the degrees of freedom of $$A^\mu$$). These polarisations don't become operators, it is the coefficients $$a$$ and $$a^\dagger$$. For spin 1/2 you need to attach a spinor to the plane wave (because a fermion is described by a spinor). But these spinors need to satisfy the equations $$\not{\!p} u(p) = m\, u(p)$$ for the expansion to satisfy the Dirac equation. Once more these are not promoted to operators after quantisation. It is the coefficients $$a, a^\dagger, b$$ and $$b^\dagger$$ that become operators.
• I brought up the Ehrenfest theorem as an example of why the equation of motion can be written in quantum system. I got confused because when I look at the Feynman prescription, equation of motion does not have any meaning (since we can get the equation of motion only when the action is extremised). When I make an analogy to Ehrenfest, things become more transparent. In classical system, you can have equation of motion $\frac{dx}{dt}=\frac{p}{m}$, which is meaningless in quantum mechanics. It can be correct in quantum only if you promote them to the operator level (obtained using Ehrenfest) – TangBear Aug 10 '19 at 10:24