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:

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

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

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