In this paper on supersymmetry, the Hamiltonian for the supersymmetric oscillator is given: $$H = \frac12 p^2 + \frac12 \omega^2 x^2 + \omega\bar\psi\psi.$$ Furthermore, its factorisation is given as $$H = \omega(a_B^{\dagger}a_B + a_F^{\dagger}a_F)$$ where the creation/annihilation operators have been defined as follows: $$a_B^{\dagger}=\frac{1}{\sqrt{2\omega}}(-ip+\omega x),\quad a_B=\frac{1}{\sqrt{2\omega}}(ip+\omega x),\quad a_F^{\dagger}=\bar\psi,\quad a_F=\psi.$$ Questions:

  1. In order for this factorisation to work, surely $x$ and $p$ must commute? So I assume $x$, $p$ are just commuting bosonic variables (not operators) and $\psi$, $\bar\psi$ are just anti-commuting fermionic (Grassmann) variables?
  2. Later in the paper it states that $\{Q,\bar Q\}=H/\omega$, this being the anti-commutator between the operators $Q=a_B^{\dagger}a_F$ and $\bar Q=a_F^{\dagger}a_B$. But this can't be true if $x$, $p$ (or $\psi$, $\bar\psi$) commute (or anti-commute). So have they been promoted to operators at this point?
  3. If so and they obey the relations $[x,p]=i$ and $\{\psi,\bar\psi\}=1$, then I find that $\{Q,\bar Q\}\ne H/\omega$. Am I missing something?
  • $\begingroup$ I do, however, find that $\{Q,\bar Q\}=H'/\omega$, where $H' = H - \omega/2$. $\endgroup$
    – Feyn
    Oct 22, 2019 at 19:25

1 Answer 1

  1. No. It's like just rewriting of classical Hamiltonan.
  2. Yes. At this point you need consider fields as operators.
  3. Be more accurate. You formulas for commutators are true. I done this calculation and obtained true result.

Details (about operators ambiguities in QM see Operator Ordering Ambiguities, I choosen hermitean hamiltonian):

\begin{equation} \hat{H}_B = \frac{\omega}{2}(\hat{a}_B^\dagger \hat{a}_B + \hat{a}_B \hat{a}_B^\dagger) = \omega (\hat{a}_B^\dagger \hat{a}_B +\frac{1}{2}) \end{equation} \begin{equation} \hat{H}_F = \frac{\omega}{2} (\hat{a}^\dagger_F \hat{a}_F - \hat{a}_F \hat{a}^\dagger_F ) = \omega(\hat{a}^\dagger_F \hat{a}_F - \frac{1}{2}) \end{equation} \begin{equation} \hat{H} = \hat{H}_B+\hat{H}_F = \omega (\hat{a}_B^\dagger \hat{a}_B + \hat{a}^\dagger_F \hat{a}_F) \end{equation}

\begin{equation} \{\hat{Q},\hat{Q}^\dagger \} = \{\hat{a}^\dagger_B\hat{a}_F\;,\hat{a}^\dagger_F\hat{a}_B \} = \hat{a}^\dagger_B\underbrace{\hat{a}_F \hat{a}^\dagger_F}_{-\hat{a}^\dagger_F \hat{a}_F - 1}\hat{a}_B + \hat{a}^\dagger_F\underbrace{\hat{a}_B\hat{a}^\dagger_B}_{\hat{a}^\dagger_B\hat{a}_B + 1}\hat{a}_F = (\hat{a}_B^\dagger \hat{a}_B + \hat{a}^\dagger_F \hat{a}_F) = \frac{\hat{H}}{\omega} \end{equation}

  • $\begingroup$ Makes sense. However, in point 3, did you do your calculations using the factorised or unfactorised Hamiltonian? I find that the factorised Hamiltonian doesn't yield the correct result. $\endgroup$
    – J-J
    Dec 28, 2019 at 16:34
  • 1
    $\begingroup$ I add details, to clarify calculation. $\endgroup$
    – Nikita
    Dec 28, 2019 at 20:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.