# BPS states and SUSY

1. Jan 17, 2014

### nrqed

A quick and simple question: one always talks about BPS states annihilating half the supercharges. What does that mean exactly?

For example, in a pedagogical article by Alvarez-Gaumé and Hassan they give the anticommutator of one set of supercharges to be

$\{b_\alpha, b_\beta^\dagger \} = \delta_{\alpha \beta} (M-\sqrt{2} Z)$

They also give anti commutation relations for other modes which are not annihilated by BPS states.
They then say that from this it is obvious that BPS states (with $M = \sqrt{2} Z$) annihilate half the suzy generators.

What does this mean, exactly? The above relation implies that on a BPS state, $b_\alpha b_\beta^\dagger - b_\beta^\dagger b_\alpha$ gives zero, but how does that lead to the state annihilating half the generators? I could imagine that maybe given that a BPS state is a state of lowest energy in a given sector we possibly impose that the annihilation operators $b_\alpha$ may be annihilating the states. But that still would not imply that a creation operator applied on the BPS state will give zero.

I know this is a trivial question and probably just a question of terminology but I would appreciate it if someone could clear that up for me.

Thanks!

2. Jan 20, 2014

### Haelfix

I haven't read the paper and I don't know what notations they are using and how they are defining their variables, but in general to count the amount of supersymmetries a state possesses is by definition to analyze how many zero eigenvalues there are in the full matrix that defines the supersymmetry algebra (so it is a symmetric matrix given by the antibracket of all of the supercharges).

In the relevant case here where there are central extensions, the matrix is positive definite and the energy of the system will be bounded from below by the largest eigenvalue of the Z matrix (the matrix of central charges). A bps state is one that saturates this energy bound, and will lead to an nfold degeneracy in these values and therefore will preserve n of the supersymmetries.

I should add that in general it is not creation operators perse, but rather certain linear combinations that annihilate the state.

Last edited: Jan 20, 2014
3. Feb 1, 2014

### karlzr

For BPS states, $b b^\dagger + b^\dagger b =0$. Impose this relation on some arbitrary state $|X>$, one can find $$| b^\dagger|X>|^2+|b|X>|^2=0$$. Since $|X>$ can be any state, this implies that $$| b^\dagger|X>|=0=|b|X>|$$. In other words, $b$ or $b^\dagger$ generates states of zero norm.

4. Feb 8, 2014

### nrqed

Thank you very much Haelfix and Karlzr for your help (and my apologies for the suzy instead of susy, it turns out that the change was automatically made by my spell checker and I did not notice it! But I did make a mistake by typing a minus sign in the anticommutator.)

This is helpful but it brings up new questions in my mind. I thought that the M appearing in the algebra was to be thought of as an operator with eigenvalue equal to the mass of the state on which it is acting. So I thought that only when applied to a BPS state could we set $M= \sqrt{2} Z$. Is that correct? If that is the case, then in Karlzr's post the state $|X \rangle$ is not an arbitrary state but a BPS state, is that correct?

In Haelfix's post, the matrix is built by sandwiching anticommutators between states, right? And then, in the case of BPS state one obtains entries that are zero, so this does correspond to my idea that M is an operator which gives an actual mass only after being applied to a state, right?

A last question: when they say that a state "preserves" a certain number of supersymmetries, this refers to operators whose anticommutators do NOT vanish when sandwiched between those states, correct?

Thank you again for your help,

Patrick

5. Feb 9, 2014

I agree.