0
$\begingroup$

For local observables $\{O_i(x_i)\}^n_{i = 1}$, one defines the Ward identity as

$$\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O(x_i)\rangle = \sum^n_{i = 1}\delta(x-x_i)\langle O_1(x_1)\cdots\delta O_i(x_i)\cdots O(x_n)\rangle\tag{1}$$

Under the local variation of the field $$\Psi(x)\mapsto \Psi(x)+\epsilon\delta\Psi(x)\tag{2}$$

Whenever $\delta_{\epsilon}\mathcal{D}[\Psi(x)] = 0$, $\delta_{\epsilon} Z[\Psi(x)] = 0$, one obtains the following variation

$$\begin{align}\delta_{\epsilon}\langle \prod^n_{i = 1}O_i(x_i)\rangle &= \delta_{\epsilon}\biggr(\frac{\int \mathcal{D}[\Psi(x)]\prod^n_{i = 1}O_i(x_i)e^{\frac{i}{\hbar}S[\Psi(x)]}}{Z[\Psi(x)]}\biggr)\\&= \frac{\delta_{\epsilon}\biggr(\int \mathcal{D}[\Psi(x)]\prod^n_{i = 1}O_i(x_i)e^{\frac{i}{\hbar}S[\Psi(x)]}\biggr)Z[\Psi(x)] - \delta_{\epsilon} Z[\Psi(x)]\biggr(\int \mathcal{D}[\Psi(x)]\prod^n_{i = 1}O_i(x_i)e^{\frac{i}{\hbar}S[\Psi(x)]}\biggr)}{Z^2[\Psi(x)]}\\&= \frac{1}{Z[\Psi(x)]}\delta_{\epsilon}\biggr(\int \mathcal{D}[\Psi(x)]\prod^n_{i = 1}O_i(x_i)e^{\frac{i}{\hbar}S[\Psi(x)]}\biggr)-\underbrace{\frac{1}{Z[\Psi(x)]}\delta_{\epsilon} Z[\Psi(x)]\langle \prod^n_{i = 1}O_i(x_i)\rangle}_{ = 0} \\ &= \sum^n_{i = 1}\langle O_1(x_1)\cdots\delta O_i(x_i)\cdots O_n(x_n)\rangle+\frac{i}{\hbar}\langle\delta_{\epsilon} S[\Psi(x)]\prod^n_{i = 1}O_i(x_i)\rangle\\&= \sum^n_{i = 1}\langle O_1(x_1)\cdots\delta O_i(x_i)\cdots O_n(x_n)\rangle+\frac{i}{\hbar}\int_{M} d^n x\epsilon(x)\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O_i(x_i)\rangle\end{align}\tag{3}$$

Where we assumed that $$Z[\Psi(x)] = \displaystyle\int D[\Psi(x)]e^{\frac{i}{\hbar}S[\Psi(x)]}\tag{4}$$

Further recalling that $$\int_{V}O_i(x)\delta(x-x_i)d^nx = \begin{cases}O_i(x_i), x_i\in V\\ 0, x_i\notin V\end{cases}\tag{5}$$

We observe that under arbitrary spacetime-dependent parameter $\epsilon : M\rightarrow \mathbb{R}$,

$$\begin{align}\delta_{\epsilon}\langle \prod^n_{i = 1}O_i(x_i)\rangle &= \int_{M}d^nx\sum^n_{i = 1}\delta(x-x_i)\langle O_1(x_1)\cdots\delta O_i(x)\cdots O_n(x_n)\rangle+\frac{i}{\hbar}\int_{M} d^nx\epsilon(x) \partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O_i(x_i)\rangle\end{align}\tag{6}$$

From which we conclude that whenever $\delta_{\epsilon}\langle \prod^n_{i = 1}O_i(x_i)\rangle = 0$,

$$-i\hbar\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O(x_i)\rangle = \sum^n_{i = 1}\delta(x-x_i)\langle O_1(x_1)\cdots\delta O_i(x)\cdots O(x_n)\rangle\tag{7}$$

However, I am not sure how the Dirac delta function $\delta(x-x_i)$ arises in the Ward identity in $(1)$. Can you provide clarification?

$\endgroup$
2
  • $\begingroup$ @Qmechanic Hello. If you like, you can remove the link you put in the comment so I can move it inside the question as a reference. $\endgroup$
    – user361492
    Dec 30, 2023 at 15:15
  • $\begingroup$ $\uparrow$ Done. $\endgroup$
    – Qmechanic
    Dec 30, 2023 at 15:18

1 Answer 1

0
$\begingroup$

It might be completely wrong, if it sounds fishy, please tell me.

You have to keep in mind the difference between $\delta\langle O\rangle$ and $\langle \delta O\rangle$. The former is 0 by definition of the invariance of the system, while the latter is not (somehow it accounts for the variation of the observable that compensate the variation of the probability density so that the total variation is 0). This implies that your equation (2) (which corresponds to the former case) is equal to 0. Moreover, the variation here: $$O_i(x_i)\mapsto O_i(x_i)+\epsilon\delta O_i(x_\color{red}i\color{black})$$ is with respect to the space time variable you're considering (and not an arbitrary $x$). These two things lead us to: $$\sum_{i = 1}\langle O_1(x_1)\cdots\delta O_i(x_i)\cdots O_n(x_n)\rangle=-i\int_{M} d^n x\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O_i(x_i)\rangle$$

This gives us, using the delta identity: $$\int_M d^nx\sum_{i = 1}\delta(x_i - x)\langle O_1(x_1)\cdots\delta O_i(x)\cdots O_n(x_n)\rangle=-i\int_{M} d^n x\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O_i(x_i)\rangle$$.

Since this integral holds for every variation, the equality holds for the objects inside the integrals:

$$\sum_{i = 1}\delta(x_i - x)\langle O_1(x_1)\cdots\delta O_i(x)\cdots O_n(x_n)\rangle=-i\partial_{\mu}\langle j^{\mu}(x)\prod^n_{i = 1}O_i(x_i)\rangle$$

You can make the last argument clearer by clearly using a $x$ dependent infinitesimal: $\epsilon(x)$ (which we omitted here). The argument is the same for the derivation of the Euler-Lagrange equations.


Edit: Concerning your question in the comments. I don't know a damn thing about QFT :) ! So my answer would be to ask directly with another question (I'm also interested by this). My (classical mechanics) feeling would be that, your infinitesimal spacetime variations leaving unchanged your average values belong to a Lie algebra and thus, through the exponential map can form the full Lie group associated to this algebra. This means that, even if you only consider infinitesimal variation, you can reconstruct, FROM THESE INFINITESIMAL variations, any finite element of the Lie group (through exponentiation). What this means, to me, is that there is as much information in the $\epsilon$ Ward identity that there would be in an $\epsilon^2$ (or even finite variation) Ward identity.

Here are a PSE link about these ideas of Lie algebras. Here is also a PSE post, with an answer by ACuriousMind who says at the end:

This is "exact" since we consider an infinitesimal gauge transformation, anyways. There are rigorous foundations in Lie theory for this "throwing away" of higher order terms. Nevertheless, it is quite important to carry these tricks the physicist likes so much oneself, since the answer coming out is correct.

Here is an Article saying that my argument does not always hold (albeit, in a rather specific context):

Indeed, due to the group structure of the transformation under consideration, it is expected that a first-order development is sufficient because in principle any finite transformation can be deduced from the composition of infinitesimal transformations of the first-order. Thus, it was always assumed that the identities obtained in the second order or beyond should be redundant with respect to the identities in the first order because of the internal composition law of the symmetry group. In the present work, we show that this is not the case. Second-order transformations cannot be deduced from the composition of first-order transformations. While one might think that the origin of this phenomenon lies in the specific non-locality of tensor interactions, a detailed analysis shows that this is not the case. Moreover, the redundancy of Ward identities has been demonstrated for a tensor theory including a closure constraint in [66]. It seems that the origin of this phenomenon lies in the kinetic term, and in the presence or absence of constraints on the field. Indeed, it seems that this anomaly is a specificity of TGFTs without closure constraints, and we provide more details on the perspectives of this observation in the conclusion.

$\endgroup$
5
  • $\begingroup$ Thanks, I just realized that I had mistakenly written $O_i(x_i)\mapsto O_i(x_i)+\epsilon\delta O_i(x)$. However, in the third part, what is it that you're doing to get $\delta(x-x_i)$? If I am not mistaken, $\frac{\delta O_i(x)}{\delta O_i(x_i)} = \delta(x-x_j)$, and so, integrating both sides leads us to the third part. $\endgroup$
    – user361492
    Nov 8, 2023 at 21:31
  • $\begingroup$ I'm just using the fact that $\int dx \delta(x - x_i)\delta O(x) = \delta O(x_i)$ and then taking the integral outstide the sum and the $\delta(x - x_i)$ outside the average. $\endgroup$
    – Syrocco
    Nov 8, 2023 at 21:36
  • $\begingroup$ If you wish to see a cleaner derivation: Conformal field theory by Francesco Eq. 2.153 to 2.157 $\endgroup$
    – Syrocco
    Nov 8, 2023 at 21:39
  • $\begingroup$ That makes perfect sense! $\endgroup$
    – user361492
    Nov 8, 2023 at 21:45
  • $\begingroup$ @Supersymmetry I updated my answer $\endgroup$
    – Syrocco
    Nov 20, 2023 at 14:22

Your Answer

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