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