Timeline for Supersymmetry charge $Q$ as anti-linear and anti-unitary operator
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2021 at 23:57 | vote | accept | wonderich | ||
| Sep 24, 2020 at 11:34 | comment | added | Theo Johnson-Freyd | @wonderich Yes. | |
| Sep 24, 2020 at 1:33 | comment | added | wonderich | Thanks so much. Is your Lie product exactly the same or some special case o the Lie superalgebra's super bracket? en.wikipedia.org/wiki/Lie_superalgebra#Definition | |
| Sep 22, 2020 at 15:39 | comment | added | Theo Johnson-Freyd | Again this is no fundamental difference: the two conventions are intertwined by multiplying $\langle v, w\rangle$ by $i^{|v|}$. (But be a bit careful with the Hilbert pairing on a tensor product of super Hilbert spaces!) These decisions propagate along. For example, the "spectrum" of an odd operator is a funny thing, because the eigenvectors are inhomogeneous, and so not something a category theorist can talk about directly. The category theorist chose her sign conventions so that some things would be clean, and the cost is that the spectrum of $Q$ is dirty: it is in $\sqrt[4]{-1}\mathbb{R}$. | |
| Sep 22, 2020 at 15:33 | comment | added | Theo Johnson-Freyd | ... the meaning of "odd self-adjoint operator". The conventions will be intertwined by multiplying odd operators by an 8th root of unity. Actually, you can see the convention question really early. What is the symmetry law for the Hilbert pairing? I.e. if $v,w \in \mathcal{H}$ are homogeneous odd elements, do we have $\langle w, v\rangle = (\pm1) \overline{\langle v, w\rangle}$? A category theorist would say $(-1)$, but a physicist might say $+1$. In particular, a category theorist would commit to deciding that for $v$ a homogeneous odd element, $\|v\| = \langle v, v\rangle$ is pure-imaginary! | |
| Sep 22, 2020 at 15:30 | comment | added | Theo Johnson-Freyd | @wonderich I meant the following. If you are a category theorist, then the thing that would seem most natural to you is to think of $Q^2 = \frac12[Q,Q]$ as the Lie product. The Lie product of self-adjoint operators is skew-adjoint, and $Q$ was supposed to be self-adjoint, so we should have an equation like $Q^2 = \sqrt{-1} \hat{H}$. But a physicist might prefer to work with underlying vector spaces of a super vector space. In terms of underlying vector spaces, $Q^2$ is the Jordan product. The result is that the category theorist and the physicist will adopt different conventions about ... | |
| Sep 22, 2020 at 4:28 | comment | added | wonderich | Thanks, I had voted up -- can you clarify the opening question remark you made: "But we also expect that the Lie product of self-adjoint operators is skew-adjoint. Well, is $Q^2$ the Jordan product of $Q$ with itself (as it would be in the even case) or the Lie product (since $[Q,Q] = QQ - (-1)^{|Q||Q|} QQ$)? " What follows next? | |
| Sep 22, 2020 at 1:50 | history | answered | Theo Johnson-Freyd | CC BY-SA 4.0 |