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.

  • $\begingroup$ I think formula (6) is not valid in this context, since the term $\frac{dS[\psi]}{d \epsilon}_{\epsilon =0}$ is in the path integral in (4) and the path integral integrates over all field configurations, also the ones that are not solution of the lagrange equations, so it's not necessarily true that $\frac{\partial L}{\partial \psi}-\partial _\mu \frac{\partial L}{\partial (\partial _\mu \psi)}=0$. $\endgroup$
    – dallla
    Mar 26 at 22:35

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

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