# Supersymmetric Bogoliubov transformation

Given the simplest system containing one bosonic and one fermionic degree of freedom with the Hilbert space spanned by \begin{align} |n,m\rangle\quad\text{with}\quad n\in\mathbb{N}\quad\text{and}\quad m\in\{0,1\}\,, \end{align} such that $\hat{a}$ and $\hat{a}^\dagger$ are bosonic creation and annihilation operators for the occupation number $n$ and $\hat{b}$ and $\hat{b}^\dagger$ are fermionic creation and annihilation operators for the occupation number $m$. This means we have \begin{align} [\hat{a},\hat{a}^\dagger]=1\quad\text{and}\quad[\hat{b},\hat{b}^\dagger]_+=1\,. \end{align} I was wondering if there is a Bogoliubov-like transformation that mixes fermionic and bosonic operators to find a new operators \begin{align} \tilde{a}&=A\,a+B\,a^\dagger+C\,b+D\,b^\dagger\,\\ \tilde{b}&=E\,a+F\,a^\dagger+G\,b+H\,b^\dagger\,\\ \end{align} such that we can build a new Fock space on it with new vectors \begin{align} |\tilde{n},\tilde{m}\rangle\,. \end{align} Can you recommend a reference on this? In particular, I read a lot about complex structures'' that one can use to parametrize the different choices of raising & lowering operators and would be interested if one can also apply this framework in the supersymmetric case.

• Comment to the post (v2): Are $C$, $D$, $E$, $F$ Grassmann-odd constants? What are they physically supposed to mean? – Qmechanic Aug 11 '16 at 18:50
• Well, I played around with it and I believe that C, D, E and F need to be Grassmann odd, but I'm mostly looking for some references where somebody has already worked out these things. I just don't believe that I'm the first person thinking about these structures - I even thought that a supersymmetry transformation should do something like that: interchange the notion of boson and fermion... – LFH Aug 11 '16 at 19:06
• Without actually doing the calculation my initial possibly wrong guess is that the Bogoliubov transformation is governed by the supergroup $U(1,1|2)$. Why don't you just work it out? – Qmechanic Aug 11 '16 at 21:21
• Well, I essentially already did: Without loss of generality, we can set $A=1$, $B=0$, $G=1$ and $H=0$. From that, one can check that the commutation relations are satisfied. I agree that it should be some supergroup, I would have assumed it should be the orthosymplectic supergroup. However, I would still like to get a more detailed reference: in particular, I would be interested to see if the ground states can still be characterized by complex structures. – LFH Aug 12 '16 at 2:13