Take two identical closed strings, both tracing out exactly the same path in space. These two strings are coincident everywhere. Call this state I.

Take a single closed string following exactly the same closed path as in the first case, but not closing just yet. The string goes around exactly the same path once again before closing in on itself. Two cycles around the same closed path. Call this state II.

String field theory tells us unambiguously states I and II are distinct.

Stretch this closed path to make it much larger than the string scale. Supposedly, stringy nonlocality only happens at the string scale. States I and II still differ.

Partition target space into local regions the size of the string scale. The path cuts across a chain of such local regions. If string theory were local, we can reconstruct the state of the entire universe from the restricted states of each subregion if we allow for quantum entanglement between regions. Locally, states I and II ought to be indistinguishable over each local region. For each local region, we always see two string segments passing through it. Thus, states I and II have to be identical?

This can't be. Either string theory is inherently nonlocal over scales much larger than the string scale, or it obeys Maxwell-Boltzmann statistics and not Bose-Einstein statistics.

This isn't some Aharonov-Bohm effect. Even if we include all the local regions in the "interior" of the closed loop, this doesn't change matters the least bit.

PS. Please reread my question more carefully. What you call configuration III is actually my configuration I.

PPS: Let me try to understand your explanation. If we have N coincident strings, or a string which winds round the same loop N times or any other combination in between, this can be described by an $S_N$ discrete gauge symmetry. The conjugacy class of the holonomy of this discrete gauge symmetry around the loop distinguishes between the various combinations. Feel free to correct me if I am wrong. This has the flavor of parastatistics, does it not?


1 Answer 1


Here, state I is just one closed string going along the path. State III will be used for two strings on the same path - which is called I in the question. Sorry for the permutations: I also discussed a more trivial issue and I don't want to delete it. ;-)

The states I and II are always distinct and there is no non-locality implied by this fact.

In particular, strings usually carry a charge with respect to the B-field - a two-form field in the NS-NS sector; among the 5 ten-dimensional superstring theories, type I theory is the only exception because its string are unoriented. If the path in I,II goes around a non-contractible circle in spacetime, the charge will manifest itself as a "winding number" $w$ - the configurations I,II will have winding numbers $w=1,2$, respectively, and those values of $w$ will behave as different values of electric charges.

The configuration II simply carries twice as high a charge than the configuration I and is completely different - just like two point-like W-bosons on top of each other are different from one W-boson. In fact, the string example of yours only differs by an added dimension to the strings' shape.

It is not hard to agree that two W-bosons on top of one another are different from one W-boson. Would you agree that one doesn't need any non-locality to deduce this fact? The configurations of strings I,II differ at each point along the string, not just globally.

Also, your more general point is totally valid: the interactions of strings are totally local in space as long as we look at strings as extended objects, and the right world sheet theory describing these interactions is also completely local on the world sheet. This condition requires that two interactions (processes) where strings are rearranged locally in some region must always have the same amplitude regardless of what the strings in both cases are doing away from the interaction point.

What you wanted to distinguish in the question was your configuration II and another configuration III (you called I) which has two coincident strings closing around the same path. The configurations II and III may be very hard to distinguish, indeed - in some sense, II is a bound-state version of III.

In the context of matrix string theory, II is a "long string" while III is a pair of strings - and these two configurations may be continuously connected on the configuration space of a Yang-Mills description of perturbative string theory which is what matrix string theory really is:


The configurations II, III only differ by a global monodromy of the matrix theory's gauge field along the long path in space that the strings are wrapped around: the configuration III involves a permutation of the two strands while the configuration II doesn't. The permutation group $S_N$ is embedded into the gauge group $U(N)$ with the same value of $N$ in the context of any matrix-theory model.

In matrix string theory, all closed-string-like (crossing-over) interactions in string theory are represented by the addition of a transposition to the permutation that remembers the monodromy. Such a transposition may occur whenever the points of several strands are close to each other in the physical spacetime and may be described as the DVV interaction - see the 2nd paper above for details. Once we study it rationally, it is counterproductive to draw the strings in a singular configuration. The same mechanism is really taking place whether or not the remainders of the strings coincide.

Recall that the $U(N)$ gauge group arises from D-branes - the fundamental strings as modeled by matrix string theory are really D1-branes which are dual to them, and which appear in a particular extreme kinematic regime: the duality is the pillar of a derivation of matrix string theory (by Seiberg). At any rate, the comments about the differences between II and III really apply to D1-branes, too (or especially to them).

Matrix string theory is discussed here on physics stack exchange:

Good introductory text for matrix string theory


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.