Explaining BPS Saturation in Superalgebras

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SUMMARY

BPS saturation refers to the condition where the mass of a particle satisfies the equality M = \frac{c_{1}}{\lambda}|W|, indicating that the BPS bound is fully met. This concept is integral to supersymmetry, which generalizes the Poincare group by introducing supercharges Q alongside traditional charges. BPS states are defined as quantum states that maintain some unbroken supersymmetry, meaning not all supercharges are conserved. The discussion highlights the importance of understanding the relationship between invariant mass and supercharges in the context of superalgebras.

PREREQUISITES
  • Understanding of supersymmetry and its role in quantum field theory.
  • Familiarity with the Poincare group and its representations.
  • Knowledge of superalgebras and their mathematical structure.
  • Basic grasp of string theory and the concept of string coupling constants.
NEXT STEPS
  • Study the mathematical framework of superalgebras and their applications in physics.
  • Explore the implications of BPS states in string theory and quantum field theory.
  • Learn about the significance of supercharges in the context of supersymmetry.
  • Investigate the relationship between invariant mass and supercharges in various physical systems.
USEFUL FOR

The discussion is beneficial for theoretical physicists, particularly those specializing in supersymmetry, string theory, and quantum field theory, as well as graduate students seeking to deepen their understanding of BPS states and superalgebras.

AlphaNumeric
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I've tried searching, but without much luck. Can someone please explain what it means for a state to be BPS saturated? It's in relation to a superalgebra \{ Q_{a},Q_{\dot{a}} \} = W \delta_{a\dot{a}} and then the mass of a particle satisfying M \geq \frac{c_{1}}{\lambda}|W| where lambda is the string coupling constant \lambda = e^{\phi}.

Any search I do for 'BPS states' comes out with some way over my head stuff.

I think it just means equality in the \geq sign, but I want to be sure :)
 
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"BPS saturated" is indeed when the equality holds. "BPS states" refer to (quantum) states that have such a mass. The concept originated in quantum field theory and was later extended to supersymmetric and string theories.
 
Javier said:
The concept originated in quantum field theory and was later extended to supersymmetric and string theories.

Well, no. Though you're both right about saturation meaning that the inequality is an equality, BPS states are all about supersymmetry, the latter in fact having been discovered in the context of string theory (by Pierre Ramond).

So what are BPS states and what does the inequality mean? As you guys may know, supersymmetry is a kind of generalization of the usual Poincare group. In the context of the original question, one may view this generalization as involving the addition of a new kind of charge - supercharge Q - to the other charges of the Poincare group, these including the momentum , which generates spacetime translations, together with the generators of the lorentz group.

The addition of supersymmetry means that not only must states transform in the usual representations of the inhomogeneous lorentz (i.e. Poincare) group, but they also must be organized into representations labelled by the supercharges. The dimensions of these representations, which tell us how much unbroken supersymmetry there is, are determined by the relation between the invariant mass and the supercharges. In general, the Q are always less than or equal to the invariant mass m of a system. This upper bound on the supercharges is called the BPS bound.

The representations of lowest dimension - called ultrashort representations - occur when there is no unbroken supersymmetry (i.e., when all the supercharges are conserved). This is the case when all the Q equal m, i.e., when the BPS bound is completely saturated.

Proper BPS states are defined to be those that have some unbroken supersymmetry, by which is meant that some but not all of the supercharges are conserved.
 

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