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  Factor of two differences for free field Green's functions in conformal field theory

+ 2 like - 0 dislike

I have a question about the expressions for free field Green's functions in conformal field theory. It comes from three origins

1) In Polchinski's string theory volume I p36, it is given

$$ \frac{1}{\pi \alpha'} \partial_z \partial_{\bar{z}} X^{\mu}(z,\bar{z}) X^{\nu} (z',\bar{z}') = -\eta^{\mu \nu} \delta^2 (z-\bar{z}', \bar{z}-\bar{z}') \tag{2.1.20} $$

2) In David Tong's string theory lecture note, p 186 (in the number at the bottom) or p 193 (counting by acrobat reader),

$$\langle Y^a(z,\bar{z}) Y^b(\omega,\bar{\omega}) \rangle = G^{ab} (z,\bar{z}; \omega,\bar{\omega} )$$ $$ \partial \bar{\partial } G^{ab} (z,\bar{z}) = - 2 \pi \delta^{ab} \delta (z,\bar{z}) \tag{7.33} $$

here $\alpha'$ has been scaled by $X^a = \bar{x}^a + \sqrt{\alpha'} Y^a$ in p 185/192. Why there is a factor of 2 difference in the RHS of (2.1.20) (Polchinski) and (7.33) (Tong)?

In page 77, it is given

$$ \langle \partial^2 X(\sigma) X(\sigma') \rangle = - 2\pi \alpha' \delta(\sigma-\sigma') \tag{4.20}$$

I am not sure what is the definition of $\partial^2$. Since $\partial \bar{\partial}=\frac{1}{4} (\partial_1^2 + \partial_2^2) $ and $\delta^2(z,\bar{z})=\frac{1}{2} \delta(\sigma^1) \delta(\sigma^2)$ (Polchinski p33),the factor at RHS of Eq. (4.20) in David Tong's lecture note seems to agree with Polchinski (2.1.20), but not Tong's (7.33)

In addition, David Tong's lecture is related to

We'll be fairly explicit here, but if you want to see more details then the best place to look is the original paper by Abouelsaood, Callan, Nappi and Yost, \Open Strings in Background Gauge Fields", Nucl. Phys. B280 (1987) 599

3) In the original paper, it is given

$$\frac{1}{2\pi\alpha'} \square G(z,z')= - \delta(z-z') \tag{2.7}$$

where $\square=\partial_{\tau}^2 + \partial_{\sigma}^2$.

Again, a factor of two difference between (2.7) and (2.1.20) of Polchinski. In Eq. (2.1) of that paper states

$$ S= \frac{1}{2\pi \alpha'} \left[ \frac{1}{2} \int_{M^2} d^2 z \partial^a X_{\mu} \partial_a X^{\mu} + i \int_{\partial M} d \tau A_{\mu} \partial_{\tau} X^{\nu} \right] $$

In Polchinski's string theory volume I, p32, it is given

$$ S = \frac{1}{4\pi \alpha'} \int d^2 \sigma \left( \partial_1 X^{\mu} \partial_1 X_{\mu} + \partial_2 X^{\mu} \partial_2 X_{\mu} \right) \tag{2.1.1}$$

Since $d^2 z = 2 d \sigma^1 d \sigma^2 \tag{2.1.7}$ , is that the origin of factor of 2 difference? Because combininig Eqs.(2.1.1) and (2.1.7) leads to $$ S = \frac{1}{2\pi \alpha'} \int d^2 z \left( \partial_1 X^{\mu} \partial_1 X_{\mu} + \partial_2 X^{\mu} \partial_2 X_{\mu} \right) $$

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user user26143
asked Aug 29, 2013 in Theoretical Physics by user26143 (360 points) [ no revision ]
Pol. $2.1.20$ is an equality between operators, but in $2.1.19$, if I take $... = 1$, I have an expectation value, so your remarks are relevant. I deleted my previous answer ,well, I think there was an interesting part concerning chapter 6.2 of Polchinksi, but I think that the correct answer is more subtle (or more simple). i am trying to investigate that.

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user Trimok
As you notice in your question, Tong.$4.20$ and Polchinski $2.1.20$ are coherent, and you gave the right explanation. The problem is Tong $7.33$

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user Trimok
I gave a new answer.

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user Trimok

2 Answers

+ 2 like - 0 dislike
Yes. that's correct.

Instead of say, $\mbox d^2\sigma=\mbox d\sigma_0 \mbox{ }d\sigma_1 $, if we havce, $\mbox d^2 z =2\mbox{ d}\sigma_0 \mbox{ d} \sigma_1 $ , then clearly, say,

$$\int F \mbox{ d}^2z = 2\int F\mbox{ d}{} ^2 \sigma $$
answered Aug 29, 2013 by dimension10 (1,950 points) [ revision history ]
Thanks a lot ^_^

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user user26143
+ 2 like - 0 dislike

The difference is due in fact to a topological difference between the sphere and the disk, and this corresponds to tree-level amplitudes for closed string and open string.

For the sphere, we have :

$$G'(z_1, z_2) = -\frac{\alpha'}{2} \ln |z_1-z_2|^2+...\tag{Pol 6.2.9}$$

So, the Green function corresponds to the expectation value, is coherent with the one that we may obtain with Pol. $2.1.19$. The topology of the sphere is used when we calculate tree-level closed string amplitudes (Pol. Chapter $6.6$)

On another way, with open strings, we have manifolds with boundary like the disk. to respect Neumann conditions - $\partial_\sigma G(z,w)_{|\sigma=0}=0$ (Tong $4.52$ p.$106$) , we have to use the method of images, so that :

$$G(z_1, z_2) = -\frac{\alpha'}{2} \ln |z_1-z_2|^2 -\frac{\alpha'}{2} \ln |z_1-\bar z_2|^2\tag{Pol 6.2.32}$$

For $2$ points on the boundary ($Im (z_1)= Im(z_2)=0$), the $2$ terms in Pol $6.2.32$ are equals, and, if we consider boudary operators, we have to consider boundary normal ordering (see Pol $6.2.34$), so you have a factor $2$ (if you compare to Pol $2.1.21b$):

$$*^*X^\mu(y_1)X^\nu(y_2)*^* = X^\mu(y_1)X^\nu(y_2) + \alpha' \ln|y_1-y_2|^2 \tag{Pol $6.2.34$}$$ where $y_1,y_2$ are real axis coordinates.

When we calculate open string tree-level, we use the disk topology, and the vertex operators (in and out) are lying on the boundary of the disk, we have to use the Green function $6.2.32$, and we have to use boundary normal ordering $6.2.34$.

So, if you take the Green Fuction $6.2.32$,at $z_2=0$, you have :

$$\partial_{z_1} \partial_{\bar z_1}G(z_1, \bar z_1,0,0) = -2\pi \alpha' \delta(z_1, \bar z_1)$$

And this is exactly Tong $7.33$. So you have the supplementary factor 2 in this case. If you read the all chapter Tong $7.5.1$, you will see that one is talking about open strings (and D-branes), so all is coherent here.

This post imported from StackExchange Physics at 2014-03-07 16:36 (UCT), posted by SE-user Trimok
answered Aug 30, 2013 by Trimok (950 points) [ no revision ]

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