Stringy Things

# Notes on String Theory: Conformal Field Theory – Ward Identities and Noether’s Theorem

Introduction

We now turn our attention to an introduction to Ward identities, which extends the ideas of Noether’s theorem in quantum field theory. Polchinski notes (p.41), A continuous symmetry in field theory implies the existence of a conserved current (Noether’s theorem) and also Ward identities, which constrain the operator products of the current’.

In this post we want to derive a particular form of the Ward identity, coinciding with Section 2.3 in Polchinski’s textbook. And we shall proceed with the following discussion by emphasising again the perspective we have been building for some time, which goes all the way back to the definition of local operators. Moreover, Ward identities are in fact operator equations generally satisfied by the correlation functions, which, of course, are tied to the symmetry of the theory. So we take this as a starting point. As Polchinski comments, symmetries of the string worldsheet play a very important role in string theory. A large part of our study here is to consider some general field theory and derive similarly general consequences of symmetry in that field theory, extracting what we may learn as a result. It turns out that what we learn is how, among other things, we may derive Ward identities through the functional integral of the correlation functions, utilising the method of a change of variables.

1. Ward Identities and Noether Currents

We start by taking the path integral. Now, suppose we have a general field theory. For an arbitrary infinitesimal transformation of the form ${\phi_{\alpha}^{\prime}(\sigma) = \phi_{\alpha}(\sigma) + \epsilon \cdot \delta\phi_{a}(\sigma)}$, where ${\epsilon}$ is the infinitesimal parameter, $\displaystyle \int [d\phi^{\prime}]e^{-S[\phi^{\prime}]} = \int [d\phi]e^{-S[\phi]} \ \ (1)$

What we have done is considered the symmetry ${\phi_{\alpha}^{\prime}(\sigma) = \phi_{\alpha}(\sigma) + \epsilon \cdot \delta\phi_{a}(\sigma)}$ of our general field theory and found that both the measure and the action are left invariant (1). They are invariant because what we have is in fact an exact or continuous symmetry of our field theory. A continuous symmetry implies the existence of a conserved current, which, of course, infers Noether’s Theorem and also Ward identities. So, from this basic premise, we want to consider another transformation of the form ${\phi_{a} \rightarrow \phi_{a}^{\prime}(\sigma) = \phi_{a}(\sigma) +\rho(\sigma)\delta\phi_{a}(\sigma)}$, where ${\rho(\sigma)}$ is an arbitrary function. Consider the following comments for clarity: in this change of variables what we are doing is basically promoting ${\epsilon}$ to be ${\epsilon(\sigma)}$. In that this transformation is not a symmetry of the theory, because one will notice that the action and the measure are no longer invariant, what we find is that to leading order of ${\epsilon}$ the variation of the path integral actually becomes proportional to the gradient ${\partial_{a} \rho}$. Notice, $\displaystyle \int [d\phi^{\prime}]e^{-S[\phi^{\prime}]} = \int [d\phi]e^{-S[\phi]}[1 + \frac{i\epsilon}{2\pi} \int d^{d}\sigma \sqrt{-g} J^{a}(\sigma) \partial_{a}\rho(\sigma) + \mathcal{O}(\epsilon^2)] \ \ (2)$

Where ${J^{a}(\sigma)}$ is a local function that comes from the variation of the measure and the action. Indeed, it should be emphasised, both the measure and the action are local (p.41). The picture one should have in their mind is the same we have been building for some time: namely, we are working in some localised region within which all the operators we’re considering reside. This is one of the big ideas at this point in our study of CFTs.

Now, the idea from (2) is that, whilst we have technically changed the integrand, the partition function has actually remained the same. Why? In the change of variables, we have simply redefined the dummy integration variable ${\phi}$. This invariance of the path integral under change of variables gives the quantum version of Noether’s theorem ${\frac{\epsilon}{2\pi i} \int d^{d}\sigma\sqrt{g} \rho(\sigma) \langle \nabla_{a}J^{a}(\sigma)... \rangle = 0}$, where …’ are arbitrary additional insertions outside of the small local region in which ${\rho}$ is taken to be zero. This is precisely why Polchinski comments that, when we take the function ${\rho}$ to be nonzero only in a small region’, it follows we may consider a path integral with general insertions `…’ outside this region’ (p.41). In other words, as ${\rho}$ is taken to be nonzero in a small region, insertions outside this region are invariant under the change of variables.

From this clever logic, where we have ${\nabla_{a}J^{a} = 0}$ as an operator statement (p.42), we want to proceed to derive the Ward identity. It follows that, as motivated by Polchinski, given (2) we now want to insert new operators into the path integral, noting ${\rho(\sigma)}$ has finite support. Therefore, we may write, $\displaystyle \int [d\phi^{\prime}] e^{-S[\phi^{\prime}]} A^{\prime}(\sigma_{0}) = \int [d\phi]e^{-S}[A(\sigma_{0}) + \delta A + \frac{i\epsilon}{2\pi} \int d^2\sigma\sqrt{-g} J^{a}(\sigma)A(\sigma_{0})\partial_{a}\rho + \mathcal{O}(\epsilon)^2] \ \ (3)$

Where, again, $\displaystyle \phi_{a} \rightarrow \phi^{\prime}_{a} = \phi_{a} + \epsilon\cdot \rho(\sigma) \cdot \delta \phi_{a}(\sigma)$

And, now, $\displaystyle A(\sigma) \rightarrow A^{\prime}(\sigma) = A(\sigma) + \delta(A) \ \ (4)$

Then, we may use ${\int d\phi^{\prime}e^{-S^{\prime}}A^{\prime} = \int d\phi e^{-S}A}$ to show, $\displaystyle 0 = -\delta A(\sigma_{0}) - \frac{i\epsilon}{2\pi} \int d^2 \sqrt{-g} J^{a}\partial_{a}\rho$ $\displaystyle 0 = - \delta A(\sigma_{0}) + \frac{i\epsilon}{2\pi} \int d^2 \sqrt{-g} \nabla_{a}J^{a}\rho \ \ (5)$

Notice, at this point, that while we now have an integral equation, we can write it without the integral. This implies the following, $\displaystyle \nabla_{a}J^{a}A(\sigma_{0}) = \frac{1}{\sqrt{-g}}\delta^{d}(\sigma - \sigma_{0}) \frac{2\pi}{i\epsilon} \delta A(\sigma_{0}) + \text{total derivative} \ \ (6)$

Where we have a total ${\sigma}$-derivative. But this statement is equivalent to, more generally, $\displaystyle \delta A(\sigma_{0}) + \frac{\epsilon}{2\pi i} \int_{R} d^{d}\sigma \sqrt{-g}\nabla_{a}J^{a}(\sigma)A(\sigma_{0}) = 0 \ \ (7)$

Which is precisely the operator relation Polchinski gives in eqn. (2.3.7). In (7) above, what we have done is let ${\rho(\sigma) = 1}$ in some region R and ${0}$ outside that region. In the context of our present theory, the divergence theorem may then be invoked to give, $\displaystyle \int_{R} d^2 \sigma \sqrt{-g} \nabla_{a}[J^{a}A(\sigma_{0})] = \int_{\partial R}dA n_{a}J^{a} A(\sigma_{0}) = \frac{2\pi}{i \epsilon} \delta A(\sigma_{0}) \ \ (8)$

Where the area element is ${dA}$ and ${n^{a}}$ the outward normal. As Polchinski explains, what we have is a relation between the integral of the current around the operator and the variation of that same operator (p.42). We can see this in the structure of the above equation.

If the current is divergenceless, then the surface interior should give zero – i.e., it should vanish. One might say, more simply, there should therefore be a conservation current. But that would be prior to the insertion of the operator. In other words, we are assuming the symmetry transformation acts on the operator.

The next thing we want to do is convert to holomorphic and antiholomorphic coordinates, instead of ${(\sigma)}$ coordinates. To do this we may rewrite (8) in flat 2-dimensions as, $\displaystyle \oint_{\partial R} (J_{z}dz - \bar{J}_{z}d\bar{z})A(z_{0}, \bar{z}_{0}) = \frac{2\pi}{\epsilon}\delta A(z_{0}, \bar{z}_{0}) \ \ (9)$

In general, it is difficult to evaluate this integral exactly. We can evaluate it in cases, for example, where the LHS simplifies. It simplifies when, ${J_z}$ is holomorphic, meaning ${\partial J_{z} = 0}$. Therefore, also, ${J_{\bar{z}}}$ is antiholomorphic, meaning ${\partial J_{\bar{z}} = 0}$. In these cases we use residue theorem, $\displaystyle 2\pi i [Res J_{z}A(z_{0}, \bar{z}_{0}) + Res J_{\bar{z}}A(z_{0}, \bar{z}_{0})] = \frac{2\pi}{\epsilon}\delta A(z_{0}, \bar{z}_{0}) \ \ (10)$

Another way to put it is that the integral (9) selects and gathers the residues in the OPE. And what we find is the Ward identity, $\displaystyle Res_{z \rightarrow z_{0}} J_{z} A(z_{0}, \bar{z}_{0}) + \bar{Res}_{\bar{z} \rightarrow \bar{z}_{0}} J_{\bar{z}}A(z_{0}, \bar{z}_{0}) = \frac{1}{i\epsilon}\delta A(z_{0}, \bar{z}_{0}) \ \ (11)$

Where ${\text{Res}}$ and ${\bar{\text{Res}}}$ are the coefficients of ${(z - z_{0})^{-1}}$ and ${(\bar{z} - \bar{z}_{0})^{-1}}$.

Now, it should be stated that this Ward identity is extremely powerful. It tells us the variation of any operator in terms of currents. One will see it in action quite a bit in bosonic string theory. Moving forward, we will also be using all the tools that we so far defined or studied. For example, we will eventually look at the OPEs to extract ${\frac{1}{z}}$ like dependence and ${\frac{1}{z} - z_{0}}$ like dependence. And in this way we will learn how operators transform.

All of this is to say, once we find out what is our conformal symmetry group, we will see there is a very close relation between OPEs in the CFT and the singular path of the transformations of the operators. And this will lead us to some rather deep insights.

It should be mentioned, again, that from these introductory notes we will go on to compute numerous detailed examples. For now, the focus is very much on introducing key concepts and familiarising ourselves with some of the deeper ideas in relation to stringy CFTS.

References

Joseph Polchinski. (2005). ‘String Theory: An Introduction to the Bosonic String’, Vol. 1.

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