While studying notes on anomaly by Adel Bilal (http://arxiv.org/abs/0802.0634), I stuck in a calculation. Here it goes as follows:

The three-current correlator in perturbation theory as a one-loop triangle diagram with the three currents $J_{5}^{\mu}=\bar{\psi}\gamma^{\mu}\gamma_{5}\psi(x)t$ and $J^{\mu}=\bar{\psi}\gamma^{\mu}t\psi$ given by \begin{equation} \langle T(J^{\mu}_{5}(x)J^{\nu}(y)J^{\rho}(z))\rangle=-\Gamma_{5}^{\mu\nu\rho}(x,y,z) \end{equation} The abelian anomaly is given by (following Fujukawa method) \begin{equation} \partial_{\mu}\langle J_{5}^{\mu}(x)\rangle_{A}= \cal{A}(x)=\frac{1}{16\pi^{2}}\epsilon^{\mu\nu\rho\sigma}\mathrm {tr}_{\cal{R}}tF_{\mu\nu}(x)F_{\rho\sigma}(x) \end{equation} where $F_{\mu\nu}=F^{a}_{\mu\nu}t_{a}$, $t's$ are the generators of the underlying gauge group in some representation $\cal{R}$ and $\langle\dots\rangle_{A}$ indicates the vacuum expectation value computed in a fixed $A_{\mu}$ background.

In a theory with only a single $U(1)$ gauge field the generators $t_{a}$ can be replaced by single $t$ whose eigen values are $U(1)$ charges of the fields. Lets call it $q_{j}$. Then $ \text{Tr}(tt_{a}t_{b})=\sum_{j}q_{j}^{3}$. Then we have for the abelian anomaly given in last equation \begin{equation} \frac{\delta}{\delta A_{\nu}(y)}\frac{\delta}{\delta A_{\rho}(z)}\cal{A}(x)=\frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}\left(\frac{\partial}{\partial y^{\lambda}}\delta^{(4)}(y-x)\right)\left(\frac{\partial}{\partial z^{\sigma}}\delta^{(4)}(z-x)\right) = \frac{{\partial}}{\partial x^{\mu}}\left(\langle T(J^{\mu}_{5}(x)J^{\nu}(y)J^{\rho}(z))\rangle\right)=-\frac{{\partial}}{\partial x^{\mu}}\Gamma_{5}^{\mu\nu\rho}(x,y,z) \end{equation} Leads the following equality \begin{equation} -\frac{{\partial}}{\partial x^{\mu}}\Gamma_{5}^{\mu\nu\rho}(x,y,z) = \frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}\left(\frac{\partial}{\partial y^{\lambda}}\delta^{(4)}(y-x)\right)\left(\frac{\partial}{\partial z^{\sigma}}\delta^{(4)}(z-x)\right) \end{equation} My question is how can i take the Fourier transform of the last equation to show the Ward identity \begin{equation} -i(p+q)_{\mu}\Gamma_{5}^{\mu\nu\rho}(-p-q,p,q) = \frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}p_{\lambda}q_{\sigma} \end{equation} Any suggestion?


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If the Fourier transform of $f(x)$ is $\tilde{f}(k)$, then the Fourier transform of $df/dx$ is $ik\tilde{f}(k)$. Proof: $$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$ This explains the momentum factors, so we've reduced the task to showing $$\Gamma(x, y, z) \sim \delta(y-x) \delta(z-x) \quad \to \quad \tilde{\Gamma}(-p-q, p, q) \sim 1$$ where irrelevant constants/indices have been dropped. Fourier transform the first equation for $$\tilde{\Gamma}(k, p, q) \sim \int dx dy dz\, e^{-ikx}e^{-ipy}e^{-iqz} \delta(y-x) \delta(z-x) = \int dx\, e^{-ix(k+p+q)} = \delta(k+p+q).$$ Integrating over $k$ gives the result.

  • $\begingroup$ Following your steps, we would find that $\tilde{\Gamma}(-p-q,p,q)$ is infinite, not one. The resolution is that the text has extracted a delta function out of the fourier transform of $\Gamma$, so that $\mathcal{F}(\Gamma) = (2\pi)^4\delta(k +p+q) \tilde{\Gamma}$ according to their convention. With this convention, your steps give the equation $\delta(k+p+q) \tilde{\Gamma} = \delta(k+p+q)$, so $\Gamma(-p-q,p,q)=1$. See their equation 4.29 $\endgroup$ – Brian Moths May 21 '16 at 3:23
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Ah, that makes sense. I figured they were playing loose with notation, but I didn't know exactly where. $\endgroup$ – knzhou May 21 '16 at 3:28
  • $\begingroup$ What is not clear to me is how $\Gamma(x,y,z)$ same as $\delta(y-x)\delta(z-x)$ up to some numerical coefficients. $\endgroup$ – AMS May 22 '16 at 8:27
  • $\begingroup$ @mas Sorry, I should have explained this better. That equation isn't strictly true. What I did was drop constants, and also drop the derivatives, because they just turn into "constants" in momentum space. That is, we already know exactly what they're going to do (i.e. produce the desired momentum factors), and they don't affect the rest of the computation, so they can be dropped to save space. $\endgroup$ – knzhou May 22 '16 at 8:40

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