4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 . $

$\endgroup$
  • $\begingroup$ 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 '16 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 '16 at 1:53
  • $\begingroup$ The question I posted at math.SE about this has got resolved. $\endgroup$ – GotchaP Dec 17 '16 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.