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.


1 Answer 1


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.

  • $\begingroup$ Why do we evaluate this derivatives at the midpoint? $\endgroup$
    – user244484
    Commented Aug 18, 2020 at 0:25
  • $\begingroup$ 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). $\endgroup$ Commented Aug 18, 2020 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.