# Anomaly, Ward identity [closed]

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 $$\langle T(J^{\mu}_{5}(x)J^{\nu}(y)J^{\rho}(z))\rangle=-\Gamma_{5}^{\mu\nu\rho}(x,y,z)$$ The abelian anomaly is given by (following Fujukawa method) $$\partial_{\mu}\langle J_{5}^{\mu}(x)\rangle_{A}= \cal{A}(x)=\frac{1}{16\pi^{2}}\epsilon^{\mu\nu\rho\sigma}\mathrm {tr}_{\cal{R}}tF_{\mu\nu}(x)F_{\rho\sigma}(x)$$ 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 $$\frac{\delta}{\delta A_{\nu}(y)}\frac{\delta}{\delta A_{\rho}(z)}\cal{A}(x)=\frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}\left(\frac{\partial}{\partial y^{\lambda}}\delta^{(4)}(y-x)\right)\left(\frac{\partial}{\partial z^{\sigma}}\delta^{(4)}(z-x)\right) = \frac{{\partial}}{\partial x^{\mu}}\left(\langle T(J^{\mu}_{5}(x)J^{\nu}(y)J^{\rho}(z))\rangle\right)=-\frac{{\partial}}{\partial x^{\mu}}\Gamma_{5}^{\mu\nu\rho}(x,y,z)$$ Leads the following equality $$-\frac{{\partial}}{\partial x^{\mu}}\Gamma_{5}^{\mu\nu\rho}(x,y,z) = \frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}\left(\frac{\partial}{\partial y^{\lambda}}\delta^{(4)}(y-x)\right)\left(\frac{\partial}{\partial z^{\sigma}}\delta^{(4)}(z-x)\right)$$ My question is how can i take the Fourier transform of the last equation to show the Ward identity $$-i(p+q)_{\mu}\Gamma_{5}^{\mu\nu\rho}(-p-q,p,q) = \frac{1}{2\pi^{2}}\left(\sum_{j}q_{j}^{3}\right)\epsilon^{\nu\rho\lambda\sigma}p_{\lambda}q_{\sigma}$$ Any suggestion?

## closed as off-topic by ACuriousMind♦, user10851, DilithiumMatrix, Gert, honeste_vivereMay 23 '16 at 20:48

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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.
• 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 – Brian Moths May 21 '16 at 3:23
• 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. – AMS May 22 '16 at 8:27