Can we derive Ward identity from Noether current of a Lagrangian density, under the assumption that path integral measure is invariant?

Suppose that $\delta\psi(x) = \frac{d\hat{\psi}(x, \epsilon)}{d\epsilon}\bigr|_{\epsilon = 0}$ is a smooth variation such that $\psi(x) = \hat{\psi}(x, 0)$. In fact,

$$\hat{\psi}(x, \epsilon) = \psi(x)+\epsilon\delta \psi(x)\tag{1}$$


$$\begin{align} \delta\mathcal{L} &= \left(\frac{\partial \mathcal L}{\partial\psi} - \partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\right)\delta\psi + \partial_\mu J^\mu,\quad J^\mu = \frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi.\tag{2} \end{align}$$

For the action $\displaystyle S[\psi] = \int \mathcal{L}(\psi, \partial_{\mu}\psi)d^nx$, the variation of the path integral $$Z[\psi] = \int D[\psi]e^{\frac{i}{\hbar}S[\psi]}\tag{3}$$


$$\delta Z[\psi] = \frac{d}{d\epsilon}\biggr|_{e = 0}\biggr(\int D[\hat{\psi}]e^{\frac{i}{\hbar}S[\hat{\psi}]}\biggr) = \int \frac{d\mathcal{D}[\hat{\psi}]}{de}\biggr|_{\epsilon = 0}e^{\frac{i}{\hbar}S[\hat{\psi}]} + \frac{i}{\hbar}\int \mathcal{D}[\hat{\psi}]\frac{dS[\hat{\psi}]}{de}\biggr|_{\epsilon = 0}e^{\frac{i}{\hbar}S[\psi]}\tag{4}$$


$$\frac{dS[\hat{\psi}]}{d\epsilon}\biggr|_{\epsilon = 0} = \int \left(\frac{\partial \mathcal L}{\partial\psi} - \partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\right)\delta\psi + \partial_{\mu}\biggr(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi\biggr)\biggr)d^nx\tag{5}$$

Under the assumption that Lagrangian is on-shell, this simplifies to

$$\frac{dS[\hat{\psi}]}{d\epsilon}\biggr|_{\epsilon = 0} = \int\partial_{\mu}\biggr(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi\biggr)d^nx\tag{6}$$

From which we conclude that

$$\begin{align}\displaystyle\frac{i}{\hbar}\int \mathcal{D}[\hat{\psi}]\frac{dS[\hat{\psi}]}{d\epsilon}\biggr|_{\epsilon = 0}e^{\frac{i}{\hbar}S[\psi]} &= \frac{i}{\hbar}\int \mathcal{D}[\hat{\psi}]\biggr(\int\partial_{\mu}\bigr(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi\bigr)d^nx\biggr)e^{\frac{i}{\hbar}S[\psi]}\\ &= \frac{i}{\hbar}\int \mathcal{D}[\hat{\psi}]\biggr(\int\partial_{\mu}J^{\mu}d^nx\biggr)e^{\frac{i}{\hbar}S[\psi]}\end{align}\tag{7}$$


$$\delta Z[\psi] = \int \delta\mathcal{D}[\psi]e^{\frac{i}{\hbar}S[\psi]} + \frac{i}{\hbar}\int \mathcal{D}[\psi]\biggr(\int\partial_{\mu}J^{\mu}d^nx\biggr)e^{\frac{i}{\hbar}S[\psi]}\tag{8}$$

However, I'm not sure whether this is the right way to derive Ward identity, and whether I have a clear interpretation of my derivations.


2 Answers 2


For what it's worth the usual argument is as follows:

  1. If the path integral measure $\delta_{\epsilon} \left({\cal D}\phi\right)=0$ has a local(=$x$-dependent) symmetry, one may derive the following Schwinger-Dyson (SD) equation $$\langle \delta_{\epsilon}F[\phi]\rangle + \frac{i}{\hbar} \langle F[\phi]\delta_{\epsilon}S[\phi]\rangle~=~0,\tag{A}$$ cf. this related Phys.SE post.

  2. If moreover the action $S[\phi]$ has the corresponding global quasisymmetry $\delta_{\epsilon} S[\phi] = \int \mathrm{d}^n x ~ J^{\mu}~ d_{\mu}\epsilon$, one gets the following Ward-Takahashi identity $$\langle \frac{\delta \delta_{\epsilon}F[\phi]}{\delta \epsilon(x)}\rangle ~=~ \frac{i}{\hbar} \langle F[\phi]d_{\mu}J^{\mu}(x)\rangle.\tag{B}$$


I think what you want to do but got confused along the way is the following:

Regardless of whether you have a symmetry or not the following equation is always true:$\newcommand{\D}{\mathrm{D}}\newcommand{\ex}[1]{\mathrm{e}^{#1}}$ $$ Z = \int \D{\phi}\ \ex{-S[\phi]} = \int \D{\phi'}\ \ex{-S[\phi']}.\tag{1}\label{1}$$ So far I've just relabeled my integration dummy variables.

Now take $\phi'(x) = \phi(x) + \epsilon(x)\, \delta\phi(x)$ and assume that it is a non-anomalous symmetry, i.e. both the action and the measure are invariant. Then what you get is (see e.g. page 19 of David Tong's QFT notes): $$ Z = \int \D{\phi}\ \ex{-S[\phi]}\left(1 - \int \mathrm{d}^d x\ J^\mu(x) \partial_\mu \epsilon(x)\right),$$ which using \eqref{1} gives $$ \int \D{\phi}\ \ex{-S[\phi]}\left(\int \mathrm{d}^d x\ J^\mu(x) \partial_\mu \epsilon(x)\right) = 0,$$ which, since it's true for arbitrary $\epsilon(x)$, is the same as $$\left<\partial_\mu J^\mu(x) \right> = 0.$$

You can repeat the same for correlation functions, i.e. change the dummy variable in the path integral expression for $\left<\mathcal{O}_1(x_1)\cdots\mathcal{O}_n(x_n)\right>$ and play the same game. Now you have to include the variations of the operators: $$\mathcal{O}_i(x_i) \mapsto \mathcal{O}_i(x_i) + \epsilon(x_i)\, \delta\mathcal{O}_i(x_i),$$ finding, after the same steps: $$ \left<\partial_\mu J^\mu(x)\ \mathcal{O}_1(x_1)\cdots\mathcal{O}_n(x_n)\right> = \sum_{i=1}^n \delta(x-x_i) \left<\mathcal{O}_1(x_1)\cdots\delta\mathcal{O}_i(x_i)\cdots\mathcal{O}_n(x_n)\right>.$$ This is a Ward identity.

  • $\begingroup$ This makes perfect sense, although I'm still curious (assuming that my math is flawless along the computations) whether what I derived in my question. $\endgroup$
    – user375448
    Oct 8, 2023 at 9:32
  • $\begingroup$ @Supergravity your math is not correct. For example all expressions containing a $\int\mathrm{D}\psi$ hitting nothing are big red flags. $\endgroup$ Oct 8, 2023 at 9:36
  • $\begingroup$ Yes, that term was what I was particularly concerned about. However, are you suggesting that we cannot simply take $\frac{dS[\hat{\psi}]}{ds}\biggr|_{s = 0}$ out of the path integral? $\endgroup$
    – user375448
    Oct 8, 2023 at 9:38
  • $\begingroup$ @Supergravity yes, you cannot do that $\endgroup$ Oct 8, 2023 at 9:46
  • $\begingroup$ @Supergravity you have defined $\psi(x)$ as $\hat\psi(x,0)$ so they're not independent. Anyways, all of these are comments to your computations and not suitable to be discussed under this answer. $\endgroup$ Oct 8, 2023 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.