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In Peskin's book (an introduction to QFT), Page 655, the axial vector current is defined as follows, \begin{eqnarray*} j^{\mu5} & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)\psi(x-\frac{\epsilon}{2})\bigg\}. \tag{19.22} \end{eqnarray*}

Then he obtained $\partial_{\mu}j^{\mu5}$ as follows, \begin{eqnarray*} \partial_{\mu}j^{\mu5} & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{[\partial_{\mu}\bar{\psi}(x+\frac{\epsilon}{2})]\gamma^{\mu}\gamma^{5}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)\psi(x-\frac{\epsilon}{2})\\ & & +\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)[\partial_{\mu}\psi(x-\frac{\epsilon}{2})]\\ & & +\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}[-ie\epsilon^{\nu}\partial_{\color{Red}{\mu}}A_{\color{Red}{\nu}}(x)]\psi(x-\frac{\epsilon}{2})\bigg\}. \tag{19.24} \end{eqnarray*}

I know the last line of above formula (19.24) comes from \begin{eqnarray*} \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}\bigg[\partial_{\mu}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)\bigg]\psi(x-\frac{\epsilon}{2})\bigg\}. \end{eqnarray*}

But when I carefully calculate it, I find \begin{eqnarray*} & & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}\bigg[\partial_{\mu}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)\bigg]\psi(x-\frac{\epsilon}{2})\bigg\}\\ & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}\exp\bigg(-ie\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg)\\ & & \times(-ie)\partial_{\mu}\bigg[\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg]\psi(x-\frac{\epsilon}{2})\bigg\}\\ & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}(-ie)\partial_{\mu}\bigg[\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg]\psi(x-\frac{\epsilon}{2})\bigg\}\\ & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}(-ie)[A_{\mu}(x+\frac{\epsilon}{2})-A_{\mu}(x-\frac{\epsilon}{2})]\psi(x-\frac{\epsilon}{2})\bigg\}\\ & = & \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}[-ie\epsilon^{\nu}\partial_{\color{Red}{\nu}}A_{\color{Red}{\mu}}(x)]\psi(x-\frac{\epsilon}{2})\bigg\}. \end{eqnarray*}

The factor $\epsilon^{\nu}\partial_{\color{Red}{\nu}}A_{\color{Red}{\mu}}(x)$ in the last line of my calculation is different from the factor $\epsilon^{\nu}\partial_{\color{Red}{\mu}}A_{\color{Red}{\nu}}(x)$ in the last line of Peskin's calculation. So my question is: How this difference comes from?

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2 Answers 2

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The pertinent contour integral reads $$I(x)~:=~\int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)~=~\epsilon^{\nu} A_{\nu}(x)+{\cal O}(\epsilon^2)$$ so differentiation yields $$ \partial_{\mu}I(x)~=~ \epsilon^{\nu} \partial_{\color{Red}{\mu}}A_{\color{Red}{\nu}}(x)+{\cal O}(\epsilon^2),$$ as Peskin & Schroeder write in eq. (19.24). It is not a misprint.

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The point is that in general cases we cannot ensure the existence of $\phi$ such that $$ d\phi(z) =A(z) \cdot dz \quad \Leftrightarrow \quad \frac{\partial \phi(z)}{\partial z^\mu} =A_\mu(z) \quad \text{for } \forall z \quad . $$ For $\phi$ to exist, the condition $ \frac{\partial A_\nu}{\partial z_\mu} = \frac{\partial A_\mu}{\partial z_\nu} $ is needed.
Setting $ z=x-\epsilon/2 +t\epsilon $, The correct calculation goes as follows. $$ \frac{\partial}{\partial x_\mu} \int_{x-\epsilon/2}^{x+\epsilon/2} dz \cdot A(z) = \frac{\partial}{\partial x_\mu} \int_0^1A_\nu(x -\epsilon/2+t \epsilon) \, d(t \epsilon)^\nu = \epsilon^\nu \int_0^1 \frac{\partial}{\partial x_\mu} A_{\nu}(x -\epsilon/2+t \epsilon) dt \\ = \epsilon^{\nu} \int_0^1 \left( \frac{\partial}{\partial x_\mu} A_\nu(x) +{\cal O}(\epsilon) \right) dt = \epsilon^\nu \frac{\partial}{\partial x_\mu} A_\nu(x) +{\cal O}(\epsilon^2) $$
I think the last line of (19.24) should be replaced by $$ + \bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}[-ie\epsilon^{\nu}\partial_{\mu}A_{\nu}(x) + O(\epsilon^2) ]\text{exp}\bigg[-ie \int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg] \psi(x-\frac{\epsilon}{2})\bigg\} . $$ Because the authors doesn't restrict terms within order $\epsilon$ at this stage, and the Wilson line exists in the first and the second line of (19.24). It is a misprnt. ($ \epsilon^{\nu}\partial_{\mu}A_{\nu}(x) $ is correct there.)

In this case considering $$ \text{symm }\lim_{\epsilon\rightarrow0}=\left\{\frac{\epsilon^{\mu}}{\epsilon^2}\right\} =\infty ,\quad \text{symm }\lim_{\epsilon\rightarrow0} \Bigl\{\frac{\epsilon^{\mu}\epsilon^{\nu}}{\epsilon^2}\Bigr\} = \frac{g^{\mu \nu}}{d} , \tag{19.23} $$ (the first of these is misprinted in the textbook. And since the real identity of $\epsilon$ is a mystery, we may take absolute value like $\text{symm }\lim_{\epsilon\rightarrow0}=\lvert\frac{\epsilon^{\mu}}{\epsilon^2}\rvert =\infty$. Reference:here), the path of the line integral is fixed. Hence we cannot ensure the existence of $\phi$ for the reason that we cannot claim the line integral gives the same value for any path. Even if we define $\varphi(x)= \int_{x_i}^{x} dz \cdot A(z) $ for a fixed path(at least numerically calculable), $\partial_{\mu} \varphi(x)$ doesn't exist. When we could ensure the existence of $\phi$, using $ \frac{\partial A_\nu}{\partial z_\mu} = \frac{\partial A_\mu}{\partial z_\nu} \Leftrightarrow F_{\mu\nu}=0 $ the result of your calculation would be equal to $$ \text{symm }\lim_{\epsilon\rightarrow0}\bigg\{\bar{\psi}(x+\frac{\epsilon}{2})\gamma^{\mu}\gamma^{5}[-ie\epsilon^{\nu}\partial_{\color{Red}{\mu}}A_{\color{Red}{\nu}}(x)]\psi(x-\frac{\epsilon}{2})\bigg\}. $$ It's the same as that of the texbook. And then the last line of (19.25) and (19.29) would be $$ \partial_\mu j^{\,\mu 5} =\text{symm }\lim_{\epsilon\rightarrow0}\bigg\{ \bar{\psi}(x+\frac{\epsilon}{2}) [(O(\epsilon^2))_{\mu}] \gamma^{5}\psi(x-\frac{\epsilon}{2})\bigg\} . ,\tag{19.25} $$ $$ \partial_\mu j^{\,\mu 5}= \text{symm }\lim_{\epsilon\rightarrow0} \frac{\epsilon_{\mu}\epsilon^{\nu}}{\epsilon^2} \epsilon^{\mu\alpha} ({\cal O}(\epsilon))_{\nu\alpha}=0 , \tag{19.29}$$ since $\displaystyle \text{symm }\lim_{\epsilon\rightarrow0} \Bigl\{\frac{\epsilon^{\mu}\epsilon^{\nu}\epsilon^{\lambda}}{\epsilon^2}\Bigr\}=0 . $

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  • $\begingroup$ And in the last sentence of page 656. "If we had defined the axial vector current by reversing the sign of the Wilson line in (19.22), a prescription that would have done violence to local gauge invariance, we would have found the various contributions canceling on the right-hand side of (19.29)." $\endgroup$
    – GotchaP
    Dec 1, 2016 at 1:52
  • $\begingroup$ If we replaced the Wilson line into $\text{exp}\bigg[+ie \int_{x-\epsilon/2}^{x+\epsilon/2}dz\cdot A(z)\bigg]$ , the effect would be in order ${\cal O}(\epsilon^2)$ on the last line of (19.25) and in order ${\cal O}(\epsilon^3)$ on (19.29). So I think the wording "vanishing" is better rather than "canceling" since $\displaystyle \text{symm }\lim_{\epsilon\rightarrow0} \Bigl\{\frac{\epsilon^{\mu}\epsilon^{\nu}\epsilon^{\lambda}}{\epsilon^2}\Bigr\}=0 .$. $\endgroup$
    – GotchaP
    Dec 1, 2016 at 1:53
  • $\begingroup$ The question I posted at math.SE about this has got resolved. $\endgroup$
    – GotchaP
    Dec 17, 2016 at 3:14

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