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How did Nima Arkani-Hamed come up with the following nonzero space-like commutation relation in string theory?

$$\left\langle \left| \left[ \hat{\mathcal{O}}(\vec{x}), \hat{\mathcal{O}}(\vec{y}) \right] \right|^2 \right\rangle \sim \exp \left( -\frac{k\left| \vec{x} - \vec{y} \right|^{D-2}}{8\pi G} \right)$$

His explanation isn't too clear.

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Are you getting this from arXiv:0704.1814 by N A-H, Dubovsky, Nicolis, Trincherini, and Villadoro? If not, you might want to read the discussion there. – Matt Reece Mar 17 '11 at 15:52

This is a cool estimate of the inherent non-locality in quantum gravity but one shouldn't over-interpret it. It is just an estimate.

The vacuum expectation value is inserted only to turn the operator into a number. The squaring is only inserted to make it positive and to avoid cancellations; if you omitted the squaring, the order-of-magnitude estimate would only differ by having the constant $k$ from the right hand side divided by two.

Also, the formula applies to a hardcore quantum gravity regime - because it depends on $G$, i.e. Newton's constant, rather than $L_{string}$ or something similar that is more specific for perturbative string theory or any other "more well-defined" stringy vacuum.

Now, why it's at least morally correct? Well, it could be justified by a quantum-foam-like calculation. Perturbatively, the commutator vanishes for spacelike separations, by causality. Non-perturbatively, there are instanton-like contributions. One may imagine that some energy sources are inserted into points $\vec x,\vec y$ that allow one to write down a solution - a configuration of general relativity - with things similar to deficit angles at these two points.

The typical size of this solution is $|\vec x - \vec y|$, and the Einstein-Hilbert action is equal, according to dimensional analysis, to $$ \frac{1}{16 \pi G} |\vec x - \vec y|^D / |\vec x - \vec y|^2 $$ So the commutator may get a contribution from this gravitational instanton whose action is given by something like the expression above - I wrote $1/16\pi G$ to emphasize that it comes from the gravitational action. The $D$-th power comes from the volume integral over the "body" of the instanton and the second power is from the derivatives in the Ricci scalar.

You see that my displayed equation is identical to the exponent in your Ansatz. My expression has to be taken with the minus sign and exponentiated, because it's an action - $\exp(-S)$ appears in non-perturbative contributions in the Euclidean path integral calculations.

I am not 100% certain whether his result or my justification of it is correct in any calculable sense in string theory. If Nima has a more rigorous calculation, then I apologize, but my guess is that it won't differ too much.

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Sorry, Can you please provide the reference to this calculation of Nima? – user1349 Mar 17 '11 at 12:38

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