# Questions on string theory vacuum and quantizing gravity

Actually I have few questions. The answer to a question related to string field theory helped me, but let me still ask it as its probably a bit different.

When e.g we do bosonic/super string theory we see that the vacuum is $|o,p>$ that is not a true space time vacuua (I mean a unique |0>) (depends on CM momenta of strings). Its okay I thought, as our action to start with, describes a string or even the interaction between them. So, our states can never probably be free of strings.

1. So, let me just verify that, is it possible (in first quantize string theory) to get a space time vacuum state as in QFT? Is it because a string never annihilates? Or does it annihilate and can give a unique translationally invariant vacuum state?

I was thinking about this. In QFT (rather in 1 particle case first), we impose space time and momentum commutation relation. In QFT we just apply commutation relation between $\Phi$ and $\Pi$. In string theory we apply between $X^\mu$ and its corresponding $\Pi^\mu$. So, if we don't get a vacuum (like in QFT) in string theory, is it because $X^\mu$ are really not space time coordinate, but only describing strings in Space time and we're quantizing them? Though after learning that a single string action can actually describe the splitting, joining of the string, it seems to me that we can also get a space time vacuum state like in field theory (a state with no strings like a state with no particle). Please verify/clarify?

Related to this quantization, let me ask another question. I think when we get a massless spin two particle in string spectrum, we immediately recognize it as graviton. We also consider this gravitons in the background and try to quantize the action (That's the SUGRA action.. right(as in low energy, effective) which we fail to quantize)?

1. So, is it that, in string theory we get the graviton (which will appear also if we can quantize space time metric fluctuation $h_{\mu\nu}$) and that's it from "quantizing gravity" point of view? I mean we actually don't quantize the metric fluctuations in string theory to see the graviton arising. In fact what will I mean by the term Quantizing Gravity at all if not this. Please verify/clarify! Of course I am not blaming string theory for this but just want to know our goal if we say "we will try to quantize gravity" and how much string theory does for us. Thanks..
• It is not clear to me what you are asking precisely, maybe you can try to separate this to a few questions. One thing that is clear: when we first quantize we determine the excitations of a single object (string or particle), the "vacuum" is the state of an unexcited one string. QFT is a much larger system, and the vacuum there is the lowest energy state of any number of objects, usually (and imprecisely) interpreted as the state containing no particles. – user566 Feb 23 '11 at 18:19

The closed string contains two separate sets of modes, left and right (or clockwise and counter-clockwise) modes. The ground state is a scalar tachyon with $\alpha’M^2~=~-4$. The $SO(24)$ multiplet of physical states for the open string are doubled, into $SO(24)\times SO(24)$ in the closed string. The gravitational massless state is the symmetric part of $|\Omega^{\mu\nu}\rangle~=~\alpha^\mu_{-1}{\tilde\alpha}^\nu_{-1}|0\rangle$ in the closed string. The state ${1\over 2}|\Omega^{\{\mu\nu\}}\rangle$ is a spin $2$ field, which is the graviton. The antisymmetric portion is a second rank tensor.
Various combinations or tensor products of left and right movers on the open string can be used to describe the closed string modes. The for $\alpha’M^2~=~4(N~-~1)$, which from the two modes $N~=~2$ or Young’s Tableaux $\square\square\times\square\square$ you can construct positive states.