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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?
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  • $\begingroup$ 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

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  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}

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  • $\begingroup$ 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
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    $\begingroup$ I add details, to clarify calculation. $\endgroup$
    – Nikita
    Dec 28, 2019 at 20:37

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