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  Curved spacetime as a coherent state in string theory

+ 6 like - 0 dislike

I have a question about Polchinski's string theory book, volume I, p 108. When we write the Polyakov action in curved spacetime, it is said

$$ S_{\sigma} = \frac{1}{4\pi\alpha'} \int_M d^2 \sigma g^{1/2} g^{ab} G_{\mu\nu}(X) \partial_a X^{\mu} \partial_b X^{\nu} \tag{3.7.2} $$ [...] A curved spacetime is, roughly speaking, a coherent background of gravitons, and therefore in string theory it is a coherent state of strings.

My question is, how to see a coherent state here? Is coherent state defined as
$$ \hat{a} |\alpha \rangle = \alpha | \alpha \rangle $$ as in wiki?

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user user26143
asked Sep 8, 2013 in Theoretical Physics by user26143 (405 points) [ no revision ]

2 Answers

+ 6 like - 0 dislike

Yes. In order to create a curved background in string theory you incorporate an expansion of the metric around a flat background. It turns out that the terms in the expansion can be realized by taking the vertex operator of a graviton state and inserting it on the string worldsheet. If you want large curvature, you do not just take one graviton state, but a coherent state of many gravitons.


In string theory, the action of creation operators on the vacuum, and as such the existence of excited states, can be realized by and is equivalent to the "insertion of vertex operators". This basically means adding terms of a certain form to the action. To make this more clear, we will take a look at the partition function.

Suppose that your curved background is given by a split between a flat background and a fluctuation:


Then one can write the partition function as


where $S_p$ is the flat Polyakov action and V is given by

$$V=\frac{1}{4\pi\alpha'}\int d^2\sigma\sqrt{g}g^{\alpha\beta}\partial_\alpha, X^\mu\partial_\beta X^\nu h_{\mu\nu}$$

which is the vertex operator for a graviton and $S_\sigma=S_P+V$. A single vertex operator corresponds to a single graviton which is created from the vacuum, but an exponential of this vertex operator, as we have inserted it into the partition function, corresponds to the exponential of a combination of ladder operators, corresponding to a coherent state of gravitons.

For more information, see these lecture notes.

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Frederic Brünner
answered Sep 9, 2013 by Frederic Brünner (1,130 points) [ no revision ]
Excuse me, would you explain a bit more about how the coherent state appears? Thx

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user user26143
Sorry for the delay, I have edited the answer and added some detail.

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Frederic Brünner
+ 3 like - 0 dislike

The answer given above is somewhat incomplete, in that it doesn't answer the question. Independently of the fact that expanding around a flat background is equivalent to inserting an exponential of a graviton vertex operator (which was nicely reviewed by @Frederic Brünner above), the question is why is this a coherent state and what is the definition of a coherent state in this context.

The word coherent state (in string theory and beyond) has a very precise meaning: any coherent state must:

(1) depend on continuous quantum numbers (position and momentum do not qualify as continuous because there does not exist a smooth limit for the corresponding 2-pt amplitudes*);

(2) there should exist a resolution of unity with respect to these quantum numbers.

Since the above state, $\exp(-V)$, does not satisfy these conditions (as it stands) it is not a coherent state and we have to work harder. Secondly, my understanding is that eigenstates of annihilation operators do not exist in closed string theory** (there is a global obstruction that is the Euler number), for the same reason that charged coherent states are not eigenstates of annihilation operators (in standard quantum mechanics), the relevant charge in the string case being $L_0-\tilde{L}_0$. The above definition (associated to the continuous quantum numbers) nevertheless applies for all coherent states in string theory and beyond. So for all these reasons, the honest answer is that the connection between string coherent states and curved backgrounds is not well understood (and the answer is certainly not in any popular qualitative lecture notes). In fact the question is a very good one.

Nevertheless, one can make progress. Firstly there are two types of coherent states in string theory: coherent states of string fields and coherent states of modes on a single string, both of which have vertex operator realisations. The former is more general than the latter (the former is non-local on the worldsheet and in spacetime whereas the latter is only non-local in spacetime), but there is a corner in moduli space where there are equivalent. The coherent states of modes on a single string have been constructed and are only now starting to become understood (see https://arxiv.org/abs/1304.1155 and refs therein). Coherent states of string fields are much much less well understood and i do not know of any reference. The relation of coherent states to curved backgrounds is in principle understood but in practice remains elusive. Needless to say, there are groups around the world working on this very interesting question because it provides one of the few quantum gravity handles associated to non-trivial backgrounds (and clearly does not assume validity of low energy effective field theory as the approach is entirely stringy).

*For example, in flat spacetime and for momentum eigenstates the overlap is proportional to a delta function in momenta.

** This is true in standard gauges, such as covariant or lightcone gauge, and is in fact an offshell statement (and hence very difficult to circumvent). It is simply the statement that $L_0-\tilde{L}_0$ doesn't commute with the annihilation operators, and therefore there cannot exist simultaneous eigenstates of annihilation operators and of $L_0-\tilde{L}_0$. Given that invariance under the latter is more fundamental (for the reason mentioned above), the conclusion follows (https://arxiv.org/abs/1006.2559). I should mention there are gauges where this is not the case (such as static gauge), but here one does not know how to quantise the string beyond an ''effective theory type'' of quantisation. Another possible way out is to compactify a lightlike direction of spacetime, known as DLCQ quantisation in the M-theory literature but again there are fundamental difficulties in trying to make sense of this at the quantum level (https://arxiv.org/abs/hep-th/9711037). So this is why I say eigenstates of annihilation operators do not exist in closed string theory.

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Wakabaloola
answered Feb 1, 2018 by Wakabaloola (30 points) [ no revision ]
Great stuff. You should think about what the new gravitational explanation of muon g-2 would look like when expressed in these terms...

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Mitchell Porter
Minor comment to the post (v3): In the future please link to abstract pages rather than pdf files.

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Qmechanic
@MitchellPorter: it would be fun indeed if the new gravitational explanation of g-2 was correct (arxiv.org/abs/1802.00651)

This post imported from StackExchange Physics at 2018-03-02 22:48 (UTC), posted by SE-user Wakabaloola

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