# Deriving the comparator to second order in Peskin and Schroeder

In section 15.1 of Peskin and Schroeder, expression (15.9) is given for the comparator $$U(y,x)$$ in an infinitesimal expansion to second order:

$$U(x+\epsilon n, x)=\exp \left[-i e \epsilon n^{\mu} A_{\mu}\left(x+\frac{\epsilon}{2} n\right)+\mathcal{O}\left(\epsilon^{3}\right)\right]$$

This uses the assumption that $$U(y,x)$$ is pure phase and the restriction that $$(U(x, y))^{\dagger}=U(y, x)$$. I understood how expression (15.5) was achieved on the previous page, as it's a simple taylor expansion:

$$U(x+\epsilon n, x)=1-i e \epsilon n^{\mu} A_{\mu}(x)+\mathcal{O}\left(\epsilon^{2}\right)$$

But (15.9) is a mystery to me. I'm not sure how it is derived.

I had trouble understanding this as well. The assumption that $$U(y,x)$$ is a pure phase means that we can write $$U(y,x)=\exp[f(y,x)].$$ We can Taylor expand $$f(y,x)$$ as a function of $$y$$ around the point $$y=x$$ to get $$f(x+\epsilon n,x)=f(x,x)+\epsilon n^\mu \frac{\partial f(x+\epsilon n,x)}{\partial n^\mu}\Big|_{x+\frac{\epsilon}{2}n}+\mathcal{O}(\epsilon^3).$$ Note that the error is $$\mathcal{O}(\epsilon^3)$$ instead of $$\mathcal{O}(\epsilon^2)$$ because we evaluate the derivatives at the midpoint. We must have $$f(x,x)=0$$ so that $$U(x,x)=1$$. Furthermore, the requirement that $$\big(U(x,y)\big)^\dagger=U(y,x)$$ translates to the same requirement on $$f$$. This means the order $$\epsilon$$ term has to be purely anti-Hermitian (or an imaginary number times a Hermitian operator). Then it is just a matter of convention to define $$-ieA_\mu(x+\frac{\epsilon}{2}n)\equiv \frac{\partial f(x+\epsilon n,x)}{\partial n^\mu}\Big|_{x+\frac{\epsilon}{2}n}$$, where $$A_\mu$$ is a Hermitian operator.
• Because that gives a more accurate approximation. If we evaluated derivatives at the starting point, we would have to worry about $\mathcal{O}(\epsilon^2)$ errors in $f$ (although I believe you could show that these must vanish anyway). – JoshuaTS Aug 18 at 23:30