I'm trying to bring together and understand the concepts of anomaly, quantum symmetries, and Ward (or Ward-Takahashi, or Slavnov-Taylor) identity in QFT. I think I know what the ideas mean, but I'm not sure if my unified understanding of the subject is correct. To avoid a mess, I will first lay out what I think I know, highlighting the focal points, and then explicitly ask the questions.

From what I understand, a symmetry (global or gauge) is said to be anomalous if it doesn't hold after renormalization, and a current $J$ that was conserved classically isn't conserved after quantization. The anomaly $\mathcal{A}$ is how much this current isn't conserved: $\partial_\mu J^\mu=\mathcal{A}$. If the said symmetry is a gauge one, then one requires its anomaly to be zero. Also, the anomaly is absent at the 0-loop level but is exact at 1-loop (meaning that 2+ loop calculations will yield the same result).

The Ward (or W-T or S-T) identity is an identity between correlation functions that holds iff a certain symmetry (global or gauge) holds. The identity is there because even after gauge fixing, the observables can't be gauge dependent, but the correlation functions (that can be gauge dependant) are linked to the observables and therefore can't be arbitrary (1). They remain valid after renormalization, meaning that if they hold in the classical case they also hold in the quantum one (2). The Ward identity in QED can be obtained directly from $\partial_\mu J_{EM}^\mu=0$, and more generally a Ward identity can be obtained directly from Noether's theorem (3, 4).


(1): this definition of the Ward identity obviously relies on the fact that the symmetry is gauge. What is the definition of the Ward identity for global symmetry (provided that there's one)?
(2): do they also remain valid at every perturbation order? It seems to me that they should because if the anomaly is an exact result and the anomaly measures how much the current is not conserved, the Ward identity should hold at every order too.
(3): is the "Noether's theorem" here meant to be the first one? So, global symmetry, and physical conserved current?
(4): what is the relationship between Ward identity and anomaly? It seems to me they are somehow linked since they are both related to the quantum version of the equation of the conservation of current $\partial_\mu J^\mu=...$ but I can't grasp how they are connected.

  • 1
    $\begingroup$ I think AccidentalFourierTransform's answer here answers a couple of these, at least in the context of QED $\endgroup$ – Nihar Karve Apr 9 at 13:20


Here we work in the Euclidean theory throughout. I also preface this with a disclaimer that I have been a bit lax with indices, but hopefully the message remains clear.

The Ward identities in their broadest form possible are actually a statement about any infinitesimal shift in the fields, not necessarily symmetries of the action or otherwise:

$$ \delta\phi=\epsilon\cdot f\qquad[\delta\phi_i=\epsilon^r(x^\mu) f_r^i(\phi, \partial\phi)] \\Z=\int\mathcal D\phi'\exp\left(-S'+\int J\cdot\phi'\right) \\=\int\mathcal D\phi\left(1+\epsilon^r\mathrm{Tr}\frac{\delta f_r}{\delta\phi_j}\right)\exp\left(-S-\delta_\epsilon S+\int J\cdot(\phi+\epsilon\cdot f)\right) $$ $$ \\\Rightarrow\left\langle-\delta_\epsilon S+\int \epsilon J\cdot f\right\rangle=-\left\langle\epsilon^r\mathrm{Tr}\frac{\delta f_r}{\delta\phi_j}\right\rangle\tag{1} $$ Some people call this the Ward identity, but it's far too general to be of any immediate use. Note that $J$ is not the analogue of the classical Noether current: it's simply a background source. Let me stress again: this holds for all transformations $\delta_\epsilon$.

As you have probably seen in the derivation of Noether's first theorem, any symmetry of the action must have $$ \delta_\epsilon S=\int\delta_\epsilon\mathcal L=-\int\epsilon\cdot\ \partial_\mu\left(f\cdot\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\right)=-\int\epsilon\cdot\ \partial_\mu j^\mu $$ with the appropriate boundary conditions on $\epsilon(x)$. Here $j^\mu(x^\mu)$ is the classically conserved Noether current.

Now for a symmetry in the quantum theory, this means that $$ Z=\int\mathcal D\phi'\exp\left(-S'+\int J\cdot\phi'\right) \\=\int\mathcal D\phi\left(1+\epsilon^r\mathrm{Tr}\frac{\delta f_r}{\delta\phi_j}\right)\exp\left(-S+\int J\cdot\phi\right)\left(1-\int\epsilon\cdot(\partial_\mu j^\mu-J\cdot f)\right) $$ $$ \Rightarrow\partial_\mu\langle j^\mu\rangle-\left\langle J\cdot f\right\rangle=\left\langle\mathrm{Tr}\frac{\delta f^i}{\delta\phi_j}\right\rangle\tag{2} $$

This is what is usually called the Ward identity. The right-hand side of the equation denotes the anomaly from the path integral measure, and is zero in case of a non-anomalous transformation (it is also a very formal expression, and actually requires quite a large toolset to evaluate in most cases). Also recall that the source $J$ is literally there for us to play around with, so we can take derivatives with respect to it before setting it to zero in order to derive the Ward identities for correlators: taking a non-anomalous transformation for simplicity, $$ 0=\int\mathcal D\phi\exp\left(-S+\int J\cdot\phi\right)\left(\partial_\mu j^\mu-J\cdot f\right) \\ J=0\Rightarrow\partial_\mu\langle j^\mu\rangle=0 \\-i\frac{\delta}{\delta J(x_1)}Z\bigg|_{J=0}=0\Rightarrow\partial_\mu \langle j^\mu(x)\phi(x_1)\rangle=\delta(x-x_1)\langle f(\phi)\rangle $$

and so on, to derive the relations between the correlators.

Another important observation is that for linear global symmetries, equation $(2)$ is equivalent to $$ \epsilon^r\partial_\mu\langle j^\mu\rangle=\delta_\epsilon\Gamma[\varphi] $$ where $\Gamma[\varphi]$ is the 1PI effective action. This also formalises the idea of "classical symmetries carrying over to the quantum theory": the 1PI effective action will normally have the same symmetries as the classical theory - this violation is measured by the anomaly. This, however, does not work for gauge theories, since we need to gauge-fix the action under the path integral. You could of course of course take the BRST route, but a simpler method is to consider a different sort of effective action - one where the gauge fields are integrated out into the background, say $W[A']$, which I will discuss qualitatively. The variation of this object under $\delta_\epsilon$ correctly gives the anomaly. Finally, one writes the conserved current in the presence of the background fields as $$ \frac{\delta W[A]}{\delta A^a_\mu}=\langle J_a^\mu(x)\rangle $$ whereupon $$ D_\mu\langle J_a^\mu(x)\rangle=\left\langle\mathrm{Tr}\frac{\delta f^i}{\delta\phi_j}\right\rangle $$ is the Slavnov-Taylor identity for non-abelian gauge symmetries with an anomaly.


  1. As mentioned, the Ward identities hold for all symmetries of the action. So the Ward-Takahashi identities hold for global symmetries too, although the trick of "promoting" a global symmetry to a local one (as is done for using Noether's theorem in a classical field theory) and dropping spacetime dependence right at the end is a quick and easy way to derive the identity for a given theory - this does not qualify as a gauge transformation because we do not take the gauge field itself to transform. Also note that in the case when the operators in the theory only change via a replacement $\phi\to\phi'$, then the global Ward identities are particularly simple:$\langle\mathcal O_1(\phi(x_1))\dots\mathcal O_n(\phi(x_n))\rangle=\langle\mathcal O_1(\phi'(x_1))\dots\mathcal O_n(\phi'(x_n))\rangle$

  2. Since the Ward identities are in terms of exact correlators, they hold to all orders in perturbation theory: you can also write out the identities in terms of the effective action $W[J,\{\Phi\}]$ with background gauge fields $\{\Phi\}$, which generates relations between all of the connected Green's functions.

  3. Yes, this uses Noether's first theorem, as shown. Of course, $\partial_\mu j^\mu=0$ does not imply $\partial_\mu\langle j^\mu\rangle=0$ unless the path integral measure is invariant. However, recall that even local transformations of the fields that are symmetries of the action generate an on-shell conserved current when the corresponding infinitesimal parameters $\epsilon^r(x^\mu)$ are made constant.

  4. OP has summarised this well themselves: the anomaly measures the extent of the violation of the naïvely expected Ward identity, as also shown above. For example, we would naïvely expect $\partial_\mu \langle j^\mu_A\rangle=0$ for the axial current, but the path integral measure transforms as $$ \int\mathcal D\psi\mathcal D\bar\psi\rightarrow\int\mathcal D\psi\mathcal D\bar\psi\exp\left(-\frac{ie^2}{16\pi^2}\int\epsilon \ F_{\mu\nu}\tilde F^{\mu\nu}\right) $$ and so $$ \partial_\mu \langle j^\mu_A\rangle=\frac{e^2}{16\pi^2} F_{\mu\nu}\tilde F^{\mu\nu} $$ which is the chiral anomaly. This is an example of a global anomaly, which is entirely harmless, but not entirely useless: it helps in e.g. the predicting $\pi^0\to\gamma\gamma$ decay rate. On the other hand, gauge anomalies ought to be cancelled otherwise, roughly, the lack of gauge invariance will prevent us from removing the negative norm states while trying to restrict to a well-defined Hilbert space for our states. These cancellation conditions lead to very important consistency requirements on the theory - a wild example is determining the gauge groups for the type I and heterotic string theories.
  • 1
    $\begingroup$ Your answers are clear, now I have a better picture of what the WI is and how is related to anomaly and Noether's theorem, but I think I'm missing some points on the localization of these objects (which is what I'm mostly interested in). From what I understand, the real anomaly is the one associated with gauge transformation, while the global case can be easily taken care of. How does this relate to what you wrote? Could you bring this out in points (1) and (3)? $\endgroup$ – Mauro Giliberti Apr 21 at 15:10
  • 1
    $\begingroup$ @MauroGiliberti updated. $\endgroup$ – Nihar Karve Apr 22 at 10:46

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.