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  Operator product expansion energy momentum tensor

+ 2 like - 0 dislike

We have the following equation from Polchinski (2.4.6) $$ T(z)X^{\mu}(0) \sim \frac{1}{z}\partial X^{\mu}(0) , \tag{2.4.6} $$ where $T(z)$ is defined as $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ and : : is the normal ordering defined by $$:X^{\mu}(z,\bar{z}): = X^{\mu}(z,\bar{z})\tag{2.1.21a}$$ and $$:X^{\mu}(z_{1},\bar{z_{1}})X^{\nu}(z_{2},\bar{z_{2}}): = X^{\mu}(z_{1},\bar{z_{1}})X^{\nu}(z_{2},\bar{z_{2}}) + \frac{\alpha'}{2}\eta^{\mu \nu}\ln|z_{12}|^{2}.\qquad \tag{2.1.21b}$$

How exactly do we arrive at equation 2.4.6 from these definitions? I understand the previous assertions in the chapter where they simply taylor expanded in the normal ordering, but I can't see how the above is derived.

In particular, from http://arxiv.org/abs/0812.4408 (exercise 2.7), how is it concluded that $$T(z)\partial X^{\mu}(0) \sim \frac{1}{z^{2}}\partial X^{\mu}(z) \tag{18}$$

Edit: Just one more question: We have the expansion $$:F::G: = exp(-\frac{\alpha'}{2}\int d^{2}z_{1}d^{2}z_{2}ln|z_{12}|^{2}\frac{\delta}{\delta X^{\mu}_{F}(z_{1},\bar{z_{1}}}\frac{\delta}{\delta X_{G\mu}(z_{2},\bar{z_{2}})}):FG:, \tag{2.2.10}$$

given in Polchinski.

Is there any relation between this and the Ward Identity given by Polchinski (2.3.11) $$Res_{z \to z_{0}}j(z)A(z_{0},\bar{z_{0}}) + \bar{Res}_{\bar{z}\to \bar{z_{0}}}\tilde{j}(\bar{z})A(z_{0},\bar{z_{0}}) =\frac{1}{i\epsilon}\delta A(z_{0},\bar{z_{0}}), \tag{2.3.11}$$

Do these give two different ways to compute the weight of a given operator?

The reason for this question is when I attempt to compute the above following the answer given here Identity of Operator Product Expansion (OPE) , I can't seem to find how equation (18) would follow. It seems as if the solutions manual somehow concludes rhs of equation (18) and then taylor expands which gives $\frac{1}{z^{2}} \partial X^{\mu}(0) + \frac{1}{z} \partial^{2}X^{\mu}(0)$. If one were to follow the computation given in the above link, wouldn't one automatically arrive at this?


This post imported from StackExchange Physics at 2015-06-29 21:00 (UTC), posted by SE-user combustion1925
asked Jun 27, 2015 in Theoretical Physics by combustion1925 (10 points) [ no revision ]
retagged Jun 29, 2015

1 Answer

+ 1 like - 0 dislike

Equation (2.4.6): $T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0)$ means that the RHS is the most singular term of the LHS. $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial X^{\nu}(z):}{\alpha'}\left\langle \partial X_{\nu}(z)X^{\mu}(0)\right\rangle \\ & \sim-\frac{2\partial X^{\nu}(z)}{\alpha'}\partial\left(-\eta_{\nu}^{\ \mu}\frac{\alpha'}{2}ln\left|z\right|^{2}\right)\\ & \sim\partial X^{\mu}(z)\partial\left(lnz+ln\bar{z}\right)\\ & \sim\frac{\partial X^{\mu}\left(z\right)}{z}\\ & \sim\frac{1}{z}\partial X^{\mu}(0) \end{align*}

where the second line is from Wick's Theorem (Equation 2.2.9 of Polchinski's book), and the factor 2 is because you have two ways of contraction. And the last line is from Taylor expansion.

I'm learning this chapter now, too, so some places may not be clear in my calculation.

This post imported from StackExchange Physics at 2015-06-29 21:00 (UTC), posted by SE-user HChen
answered Jun 28, 2015 by HBChen (35 points) [ no revision ]

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