# D-branes conserving only half of the supersymmetries

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Concerning D-branes as BPS states from T-duality. In D-Branes by Johnson (and other sources) I find the following statement (p.196):

"Only the total [supersymmetric] charge of $Q_a+\tilde{Q}_a$ of the left- and right-movers is conserved. Under T-duality this becomes $Q_a+(\prod_m P_m)\tilde{Q}_a$ where the product of reflections $P_m$ runs over all the dualised dimensions."

This I can see, T-dualizing changes the parity of the right-movers and not the left-movers so a supersymmetry is broken, what I fail to understand is

"Closed strings couple to open, so the general amplitude has only one linearly realised supersymmetry. That is, the vacuum without D-branes is invariant under N=2 supersymmetry, but the state containing the D-brane is invariant under only N=1: it is a BPS state."

My question is: Why are half of the supersymmetries broken and not just one per T-dualised dimension? And what do closed coupling to open strings have to do with this picture?

This post imported from StackExchange Physics at 2016-05-14 17:16 (UTC), posted by SE-user MattiV

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It's a bit strange because the OP basically repeats the same statement twice. First, he says that he understands it and second, he says that he doesn't.

But let's try to avoid these detailed surprising aspects of the question and try to explain what's going on, anyway or again.

Type I/II strings have spacetime supercharges $Q_a$ that are constructed purely out of left-moving fields on the world sheet, and $\bar Q_a$ that are constructed out of the right-moving ones. The second-quantized spacetime vacuum of a type II string theory (with no D-branes) is invariant under both.

However, when we add a D-brane to the vacuum, the supersymmetry parameters are no longer arbitrary. Only if the left-moving parameters $\epsilon_a$ i.e. coefficients in front of $Q_a$ are equal to the right-moving ones $\bar\epsilon_a$ up to the action of the product of gamma matrices, one gets a symmetry that the D-brane is invariant under. It's because the D-brane is imposing (Neumann or Dirichlet for bosons; and some conditions for fermions) boundary conditions relating the left-moving fields and the right-moving at the open strings' end points. That's why the left-movers and right-movers can no longer be independent for the symmetry to be preserved by the D-brane.

You may view the open strings' endpoints as connections between two strings, one that carried the open-string degrees of freedom and the other that carried the right-moving ones. The reflection of a wave from the boundary turns one to the other (left to right or vice versa). So it's similarly as if you had two strings which had some $SU(3)$ group each, to mention a random group, but the $SU(3)\times SU(3)$ would be broken to the diagonal $SU(3)$ because the connection between the two strings would say that they're no longer independent – the parameters must be identified or equal due to the connection between the strings. The endpoint is playing the same role except that the identified parameters are fermionic spinors living on the left-moving and right-moving sector of the same string.

Instead of the $Q_a$ and $\bar Q_a$ basis, one may pick another basis on the space of supercharges, those composed of $Q_a \pm (\prod \Gamma^\mu)\bar Q_a$. Those generators with the relative plus sign are preserved by the D-brane (because the SUSY parameters obey the right boundary conditions at the endpoints of open string); those with the minus sign are broken (they maximally violate the boundary conditions, by a minus sign).

The number of broken generators is exactly one-half because the number of objects in the list $Q_a + (\prod \Gamma^\mu)\bar Q_a$ is the same as the number of objects in the list $Q_a - (\prod \Gamma^\mu)\bar Q_a$. This halving is how Dirac matrices work in general. Every Dirac matrix or their product has the degenerate eigenvalues $1$ (or $i$) for one-half of the spectrum, and $-1$ (or $-i$) for the other half.

The supersymmetries transform as spinors (the dimension of a spinor representation is a power of two) so the doubling or halving is omnirepresent. The idea that one spacetime direction would break one supercharge could be reasonable if supersymmetries were transforming as vectors which they can't. But it would also be incompatible with the T-duality because the two branes of different dimensions must have the same number of unbroken supercharges if there is a T-duality relating them: the number of unbroken supercharges is the same for all dual descriptions (a necessary condition for the equivalence). And indeed, in the correct theory, that's the case. Every type II D-brane breaks 1/2 of the supercharges.

The coupling of closed strings to open strings is mentioned to emphasize that even the general amplitude for closed strings will only be invariant under one-half of the supercharges. Even though the theory of closed strings only could have the doubled amount of SUSY, the open strings are allowed to appear as virtual particles, and the "Lagrangian terms" describing those virtual particles do not respect the whole doubled SUSY but only one-half of it, the SUSY preserved by the D-branes. So these D-branes (or open strings attached to them) are breaking one-half of the supersymmetry even for amplitudes between purely closed strings because the SUSY breaking is felt in the loops.

This post imported from StackExchange Physics at 2016-05-14 17:16 (UTC), posted by SE-user Luboš Motl
answered May 13, 2016 by (10,248 points)

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