Fermionic subspace in a quantum harmonic oscillator

Question

I have a bosonic harmonic oscillator with annihilation and creation operators $a$ and $a^\dagger$. These operators are defined with the position and momentum operators $\hat{X}$ and $\hat{P}$ and verify the usual commutation rules $$ a = \hat{X} + i\hat{P}\text{ ,} \quad a^\dagger = \hat{X} - i\hat{P}$$ $$ [a,a^\dagger] = 1$$ In this bosonic Hilbert space, is there an operator $A$ and a state $|\psi\rangle$ that verify the following relations $$ A|\psi\rangle = 0$$ $$ A^\dagger |\psi\rangle \neq 0 $$ $$ (A^\dagger)^2 |\psi\rangle = 0$$ In the subspace defined by $\text{Span}\left\{|\psi\rangle, A^\dagger |\psi\rangle\right\}$, the operator $A$ would then somehow act as a fermionic annihilation operator, with Fock states defined by $|0\rangle = |\psi\rangle$ and $|1\rangle = A^\dagger|\psi\rangle$.

I am able to find operators and states that verify the first two relations, but not the third. For instance using coherent states, we can have $|\psi\rangle = |\alpha\rangle$ and $A = a - \alpha$, but they do not verify $(A^\dagger)^2 |\psi\rangle = 0$.

Any tips, references or ideas to show that such objects exist (or not) and how to find them would be very much appreciated.

Answer 1

Indeed, "chopping up the Fock ladder" is a game that goes back to the mists of time: Cigler 1979, and Heisenberg's students, apocryphally to him!

Here is a paper that reviews some of them . The basic idea is to terminate the rise of representations, in your case after two steps (but could be n ...).

Recall $$ N=a^\dagger a , \qquad \leadsto ~~~ [N,a^\dagger]= a^\dagger $$ whence, given the projection operator $$ P_N ={1-(-)^N \over 2}, \qquad \leadsto ~~~ P_N^2= 1, $$ take $$ A^\dagger = P_N a^\dagger , \qquad \leadsto ~~~ A= a P_N,\\ P_N P_{N+1}=0, \qquad P_N + P_{N+1}=1, $$ so that $$ A^{\dagger ~~2}=P_N a^{\dagger} P_N a^{\dagger}= a^{\dagger} P_{N-1} P_N a^{\dagger} =0, $$ when acting on the conventional integer ladder of states.

Further see here.