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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 $R$.

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

1 Answer 1

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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.

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