I am following Schwarz Quantum Field Theory textbook. In particular, I am looking at triangle diagrams with massless fermions. On pg. 623 - 624 Schwarz attempts to calculate $q_\mu^1 M_{5}^{\alpha\mu\nu}$ which should be zero by the Ward Identity of QED. This gives (Eq. 30.29),

\begin{multline} q_\mu^1 M_{5}^{\alpha\mu\nu} = \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{\text{Tr}[ \gamma^\nu(\not\! k + \not\!q_2)\gamma^\alpha\gamma^5(\not\! k - \not\!q_1) ]}{(k - q_1)^2 (k + q_2)^2} - \frac{\text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} \right.\\[0.25cm] \left. + \frac{ \text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] }{k^2(k - q_2)^2} - \frac{\text{Tr}[ \gamma^\nu(\not\! k + \not\! q_1)\gamma^\alpha\gamma^5(\not\! k - \not\! q_2) ]}{(k + q_1)^2(k - q_2)^2} \right]. \end{multline}

Schwarz says that after completing the traces, the result is (Eq. 30.30) \begin{equation} q_\mu^1 M_{5}^{\alpha\mu\nu} = - 4i\epsilon^{\alpha\nu\rho\sigma} \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{(k - q_1)^\rho(k + q_2)^\sigma}{(k - q_1)^2(k + q_2)^2} - \frac{(k - q_2)^\rho(k + q_1)^\sigma}{(k - q_2)^2(k + q_1)^2} \right]. \end{equation}

This leads me to believe that $$ I = \int \frac{d^4 k}{(2\pi)^2} \left[ - \frac{\text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} + \frac{ \text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] }{k^2(k - q_2)^2} \right] = 0 $$ which I am having difficulty proving. So far, I have evaluated the traces as follows \begin{align} \text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ] &= \text{Tr}[ \gamma^\alpha\gamma^5 \not\!k \gamma^\nu(\not\! k + \not\! q_2) ]\\[0.25cm] &= -\text{Tr}[\gamma^5\gamma^\alpha\gamma^\rho\gamma^\nu\gamma^\sigma ]k_\rho(k + q_2)_\sigma\\[0.25cm] &= -4i\epsilon^{\alpha\rho\nu\sigma} k_\rho(k + q_2)_\sigma\\[0.25cm] &= 4i\epsilon^{\alpha\nu\rho\sigma} k_\rho(q_2)_\sigma. \end{align}

Similarly, we can show that \begin{equation} \text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] = -4i\epsilon^{\alpha\nu\rho\sigma}(q_2)_\rho k_\rho . \end{equation}

From here, the integral $I$ becomes, \begin{align} I &= -4i\epsilon^{\alpha\nu\rho\sigma}\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2} \left[\frac{k_\rho(q_2)_\sigma}{(k + q_2)^2} + \frac{k_\sigma(q_2)_\rho}{(k - q_2)^2}\right]\\[0.25cm] &= -4i\epsilon^{\alpha\nu\rho\sigma}\int\frac{d^4k}{(2\pi)^4} \frac{k_\rho(q_2)_\sigma}{k^2}\left[\frac{1}{(k + q_2)^2} - \frac{1}{(k - q_2)^2}\right] \overset{?}{=} 0. \end{align}

Here is where I am stuck. Simplifying the term in square brackets doesn't seem to help. My only other thought is that the integrand is odd as there is a $k_\rho$ term and may vanish.

Any help would be appreciated!

  1. It is potentially inconsistent to shift the integration variable $k$ of divergent integrals, cf. e.g. subsections 30.2.2-3. This is particular sensitive in a discussion of quantum anomalies from triangle diagrams!

  2. Pull $(q_2)_\sigma$ outside of the integral $I$ in OP's last expression.

  3. The integral now has a lower external Lorentz index $\rho$. Since the integral only depends on $q_2$, any Lorentz-covariant regularization of the integral must produce a factor $(q_2)_{\rho}$.

  4. Contraction of $\epsilon^{\alpha\nu\rho\sigma} (q_2)_\sigma$ with $(q_2)_{\rho}$ then yields that $I=0$ as OP wanted to show.

| cite | improve this answer | |
  • $\begingroup$ Didn't knew point 1. Thanks a lot for pointing that out! $\endgroup$ – vin92 May 24 at 21:04

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.