# Fermionic subspace in a quantum harmonic oscillator

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.

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.