# 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:

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?
• I do, however, find that $\{Q,\bar Q\}=H'/\omega$, where $H' = H - \omega/2$.
– Feyn
Oct 22, 2019 at 19:25

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):

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

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

• 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.
– J-J
Dec 28, 2019 at 16:34
• I add details, to clarify calculation. Dec 28, 2019 at 20:37