Stringy Things

# Notes on String Theory: Conformal Field Theory – Massless Scalars in Flat 2-dimensions

In past entries we familiarised ourselves very briefly with conformal transformations and the 2-dimensional conformal algebra. To progress with our study of Chapter 2 in Polchinski, we need to equip ourselves with a number of other essential tools which will assist in building toward computing operator product expansions (OPEs). In this post, we will focus on notational conventions, transforming to complex coordinates, and utilising holomorphic and antiholomorphic functions. Then, in the next post, we will focus on the path integral and operator insertions, before turning attention to the general formula for OPEs.

To start, it will be beneficial if we define a toy theory of free massless scalars in 2-dimensions. For consistency, we will use the same toy theory that Polchinski describes on p.32. (Please also note, this theory is very similar to the theory on the string WS and this is why it will be useful for us, as it will support an introductory study within an applied setting).

In the context of our toy theory, the Polyakov action takes the form, $\displaystyle S_{P} = \frac{1}{4\pi \alpha^{\prime}} \int d^{2}\sigma [\partial_{1}X^{\mu} \partial_{1}X^{\nu} + \partial_{2}X^{\mu}\partial_{2}X^{\nu}] \ \ (1)$

This is the Polyakov action with ${\gamma_{ab}}$ being replaced by a flat Euclidean metric ${\delta_{ab}}$ and with Wick rotation. What is the benefit of the Euclidean metric and what is meant by Wick rotation?

In general, a lot of calculation in string theory is performed on a Euclidean WS, in which case, for flat metrics, standard analytic continuation may be used to relate Euclidean and Minkowksi amplitudes. The benefit is that the Euclidean metric enables us to study ordinary geometry and to use conformal field theory on the string. But one will note that in previous constructions of the Polyakov action we used a Minkowski metric, and in past discussions we have also been using light-cone coordinates. So let’s consider transforming from a Minkowski measure to a Euclidean one. To achieve a flat Euclidean metric, the idea is simple: we use a Wick rotation to rewrite the Minkowski metric. Moreover, recall that in Euclidean space, namely the x-y plane, the infinitesimal measure is given by ${ds^{2} = dx^{2} + dy^{2}}$. Compare this with the Minkowski measure, which we may write in WS coordinates as ${ds^{2} = -d\tau^{2} + d\sigma^{2}}$. Notice that in the Euclidean picture all quantities are positive (or at least share the same sign). Now, by Wick rotation, we make a transformation on the time coordinate in the Minkowski measure such that ${\tau \rightarrow -i\tau}$. This means ${d\tau \rightarrow -id\tau}$ and from this it follows ${ds^{2} = - (-id\tau)^{2} + d\sigma^{2} = d\tau^{2} + d\sigma^{2}}$. This is a Euclidean metric.

Hence, by Wick rotation, we are working in imaginary time signature such that the new metric in Euclidean coordinates may be written as, $\displaystyle \delta = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \ \ (2)$

As suggested, the goal is to end up with a Euclidean theory of massless scalars in flat 2-dimensions. Note, also, that as a result of Wick rotation, $\displaystyle (\sigma_{0}, \sigma_{1}) \rightarrow (\sigma_{2}, \sigma_{1}) \ \ (3)$

Where we define ${\tau \equiv \sigma_{0}}$ and ${\sigma \equiv \sigma_{1}}$, and where ${\sigma_{2} \equiv i\sigma_{0}}$.

Moving forward, one should note that after Wick rotation from LC coordinates ${(+, -)}$, we enter into the use of complex coordinates ${(z, \bar{z})}$. We first observed these coordinates in the last section on the 2-dimensions conformal algebra. Further clarification may be offered. Most notably, the description of the WS is now performed using complex variables by defining these complex coordinates ${(z, \bar{z}}$ that are, in fact, a function of the variables ${(\tau, \sigma)}$ with which we have already grown accustomed. Hence, ${z = \tau + i\sigma}$ and ${\bar{z} = \tau - i\sigma}$. The benefit of setting up complex coordinates is that it enables us to employ holomorphic (left-moving) and antiholomorphic (right-moving) indices, where holomorphic = ${z}$ and antiholomorphic = ${\bar{z}}$ as also observed in our discussion on the conformal generators.

Now that our field theory has been sketched, and complex coordinates have been formally established, to understand how to transform these coordinates we must understand how to compute the derivatives. The first step is to invert the coordinates and then we will differentiate, $\displaystyle \tau = \frac{z + \bar{z}}{2}, \ \ \sigma = \frac{z - \bar{z}}{2i} \ \ (4)$

Differentiating with respect to ${z}$ and ${\bar{z}}$ coordinates we obtain the following, $\displaystyle \frac{\partial \tau}{\partial z} = \frac{\partial \tau}{\partial \bar{z}} = \frac{1}{2} \ \ (5)$

And, $\displaystyle \frac{\partial \sigma}{\partial z} = \frac{1}{2i}, \ \ \ \frac{\partial \sigma}{\partial \bar{z}} = -\frac{1}{2i} \ \ (6)$

With these results we can then compute for the holomorphic coordinates, $\displaystyle \frac{\partial}{\partial z} = \frac{\partial \tau}{\partial z}\frac{\partial}{\partial \tau} + \frac{\partial \sigma}{\partial}\frac{\partial}{\partial \sigma} = \frac{1}{2}\frac{\partial}{\partial \tau} + \frac{1}{2i}\frac{\partial}{\partial \sigma} = \frac{1}{2}(\frac{\partial}{\partial \tau} - i\frac{\partial}{\partial \sigma}) \ \ (7)$

One can also repeat the same steps for the antiholomorphic case ${\bar{z}}$, $\displaystyle \frac{\partial}{\partial \bar{z}} = \frac{\partial \tau}{\partial \bar{z}}\frac{\partial}{\partial\tau} + \frac{\partial \sigma}{\partial \bar{z}}\frac{\partial}{\partial \sigma} = \frac{1}{2}\frac{\partial}{\partial \tau} - \frac{1}{2i}\frac{\partial}{\partial \sigma} = \frac{1}{2}(\partial_{\tau} + i\partial_{\sigma}) \ \ (8)$

Hence, the shorthand notation as read in Polchinksi and which we will use from this point forward, $\displaystyle \partial \equiv \partial_{z} = \frac{1}{2}(\partial_1 - i \partial_2) \ \ (9)$ $\displaystyle \bar{\partial} \equiv \partial_{\bar{z}} = \frac{1}{2}(\partial_1 + i\partial_2) \ \ (10)$

Where ${\partial_zz = 1}$ and ${\partial_{\bar{z}}z = 0}$.

To continue setting things up, we must now also register that we may set ${\sigma = (\sigma^{1},\sigma^{2})}$ and ${\sigma^{z} = \sigma^{1} + i\sigma^{2}}$ and ${\sigma^{\bar{z}} = \sigma^{1} - i\sigma^{2}}$. The reason for this will become clear in a moment. For the metric, given the above ${\gamma_{ab} \rightarrow \delta_{ab} \rightarrow g_{ab}}$, $\displaystyle g_{ab} = \begin{bmatrix} g_{zz} & g_{z\bar{z}} \\ g_{\bar{z}z} & g_{\bar{z}\bar{z}} \\ \end{bmatrix} (11)$

From this we can also say that ${\det g = \sqrt{g} = \frac{1}{2}}$, which is true for Minkowski and indeed ${\delta_{ab}}$ since we have Wick rotated. When we raise indices a factor of ${2}$ is returned: ${g^{z\bar{z}} = g^{\bar{z}z} = 2}$. Lastly, the area element transforms as ${d\sigma^{1}d\sigma^{2} \equiv d^{2}\sigma \equiv 2 d\sigma^{1}d\sigma^{2}}$. So, we see, ${d^{2}z \sqrt{g} \equiv d^{2}\sigma}$.

We now need to study how the delta function transforms. Given ${\int d^{2}\sigma \delta^{2}(\sigma_{1},\sigma_{2}) = \int d^{2}\sigma \delta(\sigma_{1})\delta(\sigma_{2}) = 1}$, we find that in our new coordinates: $\displaystyle \int d^{2}z \delta^{2}(z,\bar{z}) = 1 \implies \delta^{2}(z,\bar{z}) = \frac{1}{2} \delta^{2}(\sigma_{1},\sigma_{2}) \ \ (12)$

We can continue establishing relevant notation by focusing on how we may rewrite the Polyakov action. In the notation that we’ve constructed we find, $\displaystyle S_{P} = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu} \bar{\partial}X_{\mu} \ \ (13)$

Where ${d^{2}z = dzd\bar{z}}$. Using identities constructed throughout this post, the tools are available to see how we arrive at this simpler form of the action. The task now is to see what returns when we vary (13).

Proposition: We vary the action (13) and find the EoM to be ${\partial\bar{\partial}X^{\mu} = 0}$.

Proof: The string coordinate field, if not obvious, is now ${X(z, \bar{z}) = X(z) + \bar{X}(\bar{z})}$. It will become clear in the following discussion that we want to compute a quantity without linear dependence on ${\tau}$. To that end we use the derivative of the coordinate field ${\partial X(z)}$ and ${\bar{\partial}\bar{X}(\bar{z})}$.

Now, when we vary the simplified action we find, $\displaystyle 0 = \frac{\delta S}{\delta X^{\mu}} = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu}\bar{\partial}(X_{\mu} + \delta X_{\mu})$ $\displaystyle = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu} (\bar{\partial}X_{\mu} + \bar{\partial}\delta X_{\mu})$ $\displaystyle = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu}\bar{\partial}X_{\mu} + \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu}(\bar{\partial}\delta X_{\mu}) \ \ (14)$

Continuing with the conventional procedure, where we now integrate by parts (and for convenience discard the boundary terms), we find the EoM to be $\displaystyle \delta S = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial X^{\mu}(\bar{\partial}\delta X_{\mu})$ $\displaystyle = - \frac{1}{2\pi \alpha^{\prime}} \int d^{2}z \partial\bar{\partial}X^{\mu} (\delta X_{\mu}) = 0$ $\displaystyle \implies \partial\bar{\partial}X^{\mu} (z, \bar{z}) = 0 \ \ (15)$ $\Box$

Using the fact that partial derivatives commute. This completes the proof. The classical solution can be solved by, or in other words it decomposes as, ${X(z) + \bar{X}(z)}$. And we should also note, for pedagogical purposes, that we may write the EoM as $\partial (\bar{\partial} X^{\mu}) = \bar{\partial} (\partial X^{\mu}) = 0$ such that $\displaystyle \partial X^{\mu} = \partial X^{\mu}(z) \ \ \ \text{holomorphic function}$ $\displaystyle \bar{\partial}X^{\mu} = \bar{\partial}X^{\mu} (\bar{z}) \ \ \ \text{antiholomorphic function}$

References

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

David Tong. (2009). ‘String Theory’ [lecture notes].

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