Factorising the supersymmetric oscillator Hamiltonian: what (anti)commutes? 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:


*

*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?

*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?

*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?

 A: *

*No. It's like just rewriting of classical Hamiltonan.

*Yes. At this point you need consider fields as operators.

*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}
