# Meaning of $\delta A$ in Ward’s identity in Polchinski

In eq $$(2.3.7)$$ the symbol $$\delta A(\sigma_0)$$ is introduced in Polchinski: $$\delta A(\sigma_0)+\frac\epsilon{2\pi i}\int_R d^d\sigma g^{1/2}\nabla_a j^a(\sigma)A(\sigma_0)=0\tag{2.3.7}$$ but it is not clear to me what it stands for. Following the calculation line mentioned in paragraph above the mentioned equation I get following: $$0=\int\mathcal{D \phi’}e^{-S[\phi’]}A(\sigma_0)-\int\mathcal{D\phi}e^{-S[\phi]}A(\sigma_0)$$ I have taken LHS to be zero because of reparameterization invariance of path integral and since the operator $$A(\sigma_0)$$ doesn’t depend on $$\phi$$ therefore reparameterization invariance should hold. Doing the expansion part of first term of RHS is major headache if the instruction of paragraph above 2.3.7 is followed $$\rho(\sigma)$$ is $$1$$ inside $$R$$ (region of integration) $$0$$ outside. This calculation then leads to trivial result $$0=0$$ since order $$1$$ term in $$\epsilon$$ in $$(2.3.4)$$ will be zero.

I think I ‘m making two mistakes

1. Assuming LHS to be $$0$$.

2. Neglecting the idea $$\rho(\sigma)$$ being some sort of distribution function the boundary of $$R$$.

• $A(\sigma)$ is an operator and most definitely depends on $\phi$. Jun 23, 2021 at 7:53
• @PraharMitra Thanks! 5 page later there was an explicit example for it. Jun 24, 2021 at 16:56

Under the infinitesimal global symmetry $$\delta_\epsilon$$, both the action and the measure are invariant. For an infinitesimal local shift $$\delta_{\epsilon(\sigma)}$$, neither are invariant in general, but the change in the path integral will be proportional to $$\partial_\mu\epsilon$$ since setting $$\epsilon(\sigma)=\text{const}$$ recovers the case for the global symmetry.

Consider a local operator (or "field") $$A$$, which is a local expression formed from the various $$\phi_i$$, their derivatives $$\partial\phi_i, \partial^2\phi_i$$, etc. Consequently it will also shift under $$\delta_{\epsilon(\sigma)}$$: $$A\overset{\delta_{\epsilon(\sigma)}}\rightarrow A+\epsilon(\sigma)\delta A$$. This shift is non-zero provided that $$\epsilon(\sigma)$$ is non-zero at the point $$\sigma_1$$ where the operator is inserted, otherwise it remains unaffected.

The expectation value of this operator, possibly sandwiched with other operators is:

$$\langle A(\sigma_1)...\rangle\sim\int\mathcal D\phi\exp(-S[\phi]) A(\sigma_1)...\tag{1}$$

Assume that $$\epsilon(\sigma)$$ only has support in a region $$R$$ containing $$\sigma_1$$, but not any of the other operator insertion points. Under the local shift, $$\overset{\delta_{\epsilon(\sigma)}}\rightarrow\int\mathcal D\phi'\exp(-S[\phi'])\left(1-\frac{i}{2\pi}\int\mathrm d^2\sigma\sqrt g\ j^\alpha(\sigma)\partial_\alpha\epsilon(\sigma)\right)(A(\sigma_1)+\epsilon(\sigma)\delta A)...\tag{2}$$

Now, by reparameterization invariance (or actually, just a relabelling of a dummy variable in $$(1)$$), this must be equal to $$\int\mathcal D\phi'\exp(-S[\phi'])A(\sigma_1)...$$, so all of the additional baggage is $$(2)$$ is just zero. Dropping higher-order terms in $$\epsilon$$, integrating by parts ($$\epsilon(\sigma)$$ clearly has compact support) and finally setting $$\epsilon(\sigma)=\text{const}$$, we have: $$-\frac{i}{2\pi}\int_R\mathrm d^2\sigma\sqrt g\ \nabla_\alpha\langle j^\alpha A(\sigma_1)...\rangle=\langle\delta A(\sigma_1)...\rangle\tag{3}$$

up to conventions, which is of course the Ward identity.