Vertex insertions as functionals

+ 4 like - 0 dislike
44 views

In lecture notes on string theory by David Tong (available for free here http://www.damtp.cam.ac.uk/user/tong/string.html) a brief explanation of the vertex operator for tachyon is given. The main points are:

1. Because it has to be diffeomorphism-invariant, it is a result of an integration of some quantity over the world-sheet.

2. Because it is conformal-invariant, the quantity has to correspond to the primary operator with weight $(+1, +1)$ in order to compensate the $d^2z$ measure

3. Because tachyon is the vacuum of the string, it has to correspond to the highest-weight state of the Verma module. This is fulfilled if

$$L_n \left| \text{tachyon} \right> = \tilde{L}_n \left| \text{tachyon} \right> = 0, \quad n > 0;$$ $$L_0 \left| \text{tachyon} \right> = \tilde{L}_0 \left| \text{tachyon} \right> = a \left| \text{tachyon} \right>.$$

Based on these assumptions he identifies the vertex operator with

$$V = \int d^2 \sigma \: \cdot \, : e^{i k \cdot X } :,$$

where $:...:$ denotes normal-ordering, radial-ordering is implicit and $k$ is on-shell with negative (tachyonic) $m^2$.

$V$ is conformal-invariant. As far as I understand, this can be showed on the quantum level by evaluating the $VT$ OPE (or, equivalently, by evaluating the residues of $VL_n$, where $L_n$ are the Virasoro generators).

I am trying to understand why this is a valid choice. Specifically, it would be great to make sense of this in terms of path integrals of functionals.

It is well known that observables of a quantum theory are expectation values (path integrals weighted by the classical action) of gauge-invariant functionals. These expectations can be evaluated using the Faddeev-Popov procedure which fixes the gauge and ensures that expectations weighted by the new action are correct, but only if the functional of interest is gauge-invariant.

This point of view makes a lot of sense in QFT.

Different ways to fix the gauge exist. These give rise to different actions and (in general) different expectation values. Since the choice of gauge-fixing is arbitrary, only gauge-invariant expectations (which coincide) make sense as physical observables.

Now back to the tachyon vertex operator. I would expect it to correspond to some (classical) gauge-invariant functional over $X^{\mu}$ and the world-sheet metric $g_{\alpha \beta}$.

Such a functional has to be an integral with a diffeomorphism-invariant measure:

$$\sim \int d^2 \sigma \sqrt{g}.$$

Then there are Weyl transformation. In order to be Weyl-invariant, this integral has to have an inverse world-sheet metric which cancels the square root of the determinant in two dimensions:

$$\sim \int d^2 \sigma \sqrt{g} g^{\alpha \beta}$$

The last formula contains loose indices which (again, diffeomorphism-invariance) have to be contracted with something. For example,

$$\sim \int d^2 \sigma \sqrt{g} g^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} e^{i p \cdot X },$$

which looks much like the graviton insertion operator.

On the contrary, it can be shown that the tachyon vertex operator is conformal-invariant in some weird quantum sense, by taking its residue with the $L_n$ conformal generator (or just with $T$).

But what about diffeomorphism- and Weyl- invariance, separately? $L_n$s don't generate them. One could expect that, for example, products of $V$ (which correspond to tachyon scattering amplitudes) have different expectations when evaluated on different world-sheet backgrounds related by diffeomorphisms and/or Weyl transformations. Because, again, Faddeev-Popov method gives different expectations of a non-gauge-invariant functional when different gauge-fixings are used.

My question is: what happened to $\sqrt{g}$ and $g^{\alpha \beta}$ in the tachyon vertex operator; why is it considered diffeomorphism- and Weyl- invariant? And is it even possible to represent the Virasoro-Shapiro amplitude, for example, in terms of the expectation of some classical diffeo- and Weyl- invariant functional?

This post imported from StackExchange Physics at 2015-02-03 13:00 (UTC), posted by SE-user Hindsight
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.