5
$\begingroup$

I'm watching this lecture on introductory Supersymmetry (Clay Cordova, 2019 TASI lecture 2 on Supersymmetry). My question relates to the first 20minutes or so. The lecturer is introducing Superfields in the context of Quantum Mechanics.

He supposes we have $\mathcal{N} = 2$ supersymmetry, and we have a single complex supercharge (and its conjugate) and so $\{Q, \bar{Q}\} = 2H $. He introduces a complex Grassmann variable, and its complex conjugate $\theta$ and $\bar{\theta}$. He then defines a superfield as a function of time, and the complex grassmann variables $\theta$ and $\bar{\theta}$,

\begin{align*} X(t, \theta, \bar{\theta}) &= x(t) + \theta \psi(t) + \bar{\theta}\chi(t) + \theta \bar{\theta}D(t), \end{align*}

where $x(t), D(t)$ are classical bosonic functions of time, and $\psi(t), \chi(t)$ are Grassmann functions of time.

We define

\begin{align*} \{\theta, \partial_{\theta}\} &= 1 \\ \{\bar{\theta}, \partial_{\bar{\theta}}\} &= 1 \\ Q &= \partial_{\theta} + \bar{\theta}\partial_t \\ \bar{Q} &= \partial_{\bar{\theta}} + i \theta\partial_t. \end{align*}

It is then confirmed that $\{Q, \bar{Q}\} = 2i\partial_t (=2H)$.

He then says how these charges act as operators once we quantise,

\begin{align*} Q_{op}\left(X(t, \theta, \bar{\theta}) \right) &= [Q_{op}, X(t, \theta, \bar{\theta})] \\ &= [Q_{op}, x] + \theta\{ Q_{op}, \psi\} + \bar{\theta}\{Q_{op}, \chi\} + \theta \bar{\theta} [Q_{op}, D]. \end{align*}

And then says that it is simple algebra to check that

\begin{align*} [Q, x] &= \psi \\ \{Q, \psi\} &= 0 \\ \{ \bar{Q}, \psi \} &= i \dot{x} - D. \end{align*}

I have two issues:

  1. I can't figure out how he gets those three results.

As far as I can tell it should be

\begin{align*} [Q, x] &= [\partial_{\theta} + i\bar{\theta}\partial_t, x(t)] \\ &= i \bar{\theta}(\dot(x) + x\partial_t - x\partial_t)\\ &= i \bar{\theta}\dot{x} \\ \{Q, \psi\} &= \{\partial_{\theta} + i\bar{\theta}\partial_t, \psi(t)\} \\ &= \psi \partial_{\theta} + i\bar{\theta} \dot{\psi} + i\bar{\theta}\psi \partial_t + \psi \partial_{\theta} + i\psi \bar{\theta}\partial_t \\ &= i \bar{\theta} \dot{\psi} \\ \{ \bar{Q}, \psi \} &= i \theta \dot{\psi}. \end{align*}

Could someone explain how this is supposed to work please?

  1. In what sense is $\bar{Q}$ the adjoint of $Q$? I can see that the only change was in changing the $\theta$ to its complex conjugate $\bar{\theta}$, but given an operator acting on some Hilbert space, the adjoint $\bar{Q}$ is the operator such that $\langle \phi, Q \eta \rangle = \langle \bar{Q}\phi, \eta \rangle$ for every $\phi, \eta$ in the space. Shouldn't we know the Hilbert space, and in particular the inner product, and be able to deduce that the given $\bar{Q}$ is in fact the adjoint of $Q$? How would one do that?
$\endgroup$
1
  • $\begingroup$ I thought there was an answer provided here at some point. Where did it disappear to? Can it be recovered? $\endgroup$
    – Gleeson
    Apr 7 at 14:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.