This question stems from page 14 of John Terning's Modern Supersymmetry.

Suppose we consider $\mathcal{N} = 2$ supersymmetry, and denote the spin-0 Clifford vaccuum by $\Omega_0$. Now, the non-trivial anticommutators of the SUSY algebra for a massive supermultiplet of mass $m$ are

$\{Q_\alpha^a, {Q^\dagger}_{\dot{\alpha}b}\} = 2 m \delta_{\alpha\dot{\alpha}}\delta^a_b$

where $a, b = 1, \ldots, \mathcal{N}$. Let's build the states.

$|\Omega_0\rangle$ spin = 0

${Q^\dagger}_{\dot{\alpha}a} |\Omega_0\rangle$ spin = 0 + 1/2 = 1/2

${Q^\dagger}_{\dot{\alpha}a} {Q^\dagger}_{\dot{\beta}b} |\Omega_0\rangle$ naively, spin = 0 + 1/2 + 1/2 = 1, except this doesn't seem to be true, because you can get both spin 0 and spin 1 in this case.

In the book, the author first provides a schematic construction of the supermultiplet starting from a Clifford vacuum of spin s. Then, he considers a specific example for $\mathcal{N} = 2$.

But in the general case of spin s, he doesn't seem to distinguish between states of spin $s + 1/2$ and $s - 1/2$. Why not?

I am guessing this is more to do with notation than anything else, but doesn't $s$ denote the TOTAL SPIN? A massive state should be labeled then by $|m, s, s_3\rangle$ and not just the total spin.

To be perfectly clear, what happens when $Q^\dagger$ acts on a general state of spin $s$? Does it produce a rung of states with total spin ranging from $-|s-1/2|$ to $|s+1/2|$?

  • $\begingroup$ Try p.172 of Muller-Kirsten/Wiedemann (Introduction to Supersimmetry). $\endgroup$
    – Rexcirus
    Jan 1 '15 at 11:41
  • 2
    $\begingroup$ There is an answer at physicsoverflow.org/25692 $\endgroup$ Mar 7 '15 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.