# Supersymmetric Bogoliubov transformation

+ 1 like - 0 dislike
169 views

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.

This post imported from StackExchange Physics at 2016-08-13 17:01 (UTC), posted by SE-user LFH
Comment to the post (v2): Are $C$, $D$, $E$, $F$ Grassmann-odd constants? What are they physically supposed to mean?

This post imported from StackExchange Physics at 2016-08-13 17:01 (UTC), posted by SE-user Qmechanic
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...

This post imported from StackExchange Physics at 2016-08-13 17:01 (UTC), posted by SE-user LFH
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?

This post imported from StackExchange Physics at 2016-08-13 17:01 (UTC), posted by SE-user Qmechanic
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.

This post imported from StackExchange Physics at 2016-08-13 17:01 (UTC), posted by SE-user LFH

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.