Bosonic operator eigenvalues in second quantization

[RicardoMP]

Homework Statement

Following from $$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j<br /> \mid \Psi \rangle<br /> )=(|B_-^j|^2-1)\hat{b}_j<br /> \mid \Psi \rangle$$, I want to prove that if I keep applying $\hat{b}_j$, $n_j$times, I'll get: $$(|B_-^j|^2-n_j)\hat{b}_j\hat{b}_j\hat{b}_j ...<br /> \mid \Psi \rangle$$.

Homework Equations

Commutation relations: $$[\hat{b}^\dagger_j\hat{b}_j,\hat{b}_j]=-\hat{b}_j$$
Also: $$\langle \psi \mid<br /> \hat{b}^\dagger_j\hat{b}_j<br /> \mid \Psi \rangle<br /> =||\hat{b}_j<br /> \mid \Psi \rangle<br /> ||^2$$

The Attempt at a Solution

After proving for n=1, I went for n=2 and from there on the proof would be trivial. However, for n=2 I get:
$$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle<br /> )=(|B_-^j|^2-1)\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle$$ after using the commutation relation I referred above (the same for n>2). How do I get $$(|B_-^j|^2-2)\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle$$?

 

[nrqed]

Homework Statement

Following from $$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j<br /> \mid \Psi \rangle<br /> )=(|B_-^j|^2-1)\hat{b}_j<br /> \mid \Psi \rangle$$, I want to prove that if I keep applying $\hat{b}_j$, $n_j$times, I'll get: $$(|B_-^j|^2-n_j)\hat{b}_j\hat{b}_j\hat{b}_j ...<br /> \mid \Psi \rangle$$.

Homework Equations

Commutation relations: $$[\hat{b}^\dagger_j\hat{b}_j,\hat{b}_j]=-\hat{b}_j$$
Also: $$\langle \psi \mid<br /> \hat{b}^\dagger_j\hat{b}_j<br /> \mid \Psi \rangle<br /> =||\hat{b}_j<br /> \mid \Psi \rangle<br /> ||^2$$

The Attempt at a Solution

After proving for n=1, I went for n=2 and from there on the proof would be trivial. However, for n=2 I get:
$$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle<br /> )=(|B_-^j|^2-1)\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle$$ after using the commutation relation I referred above (the same for n>2). How do I get $$(|B_-^j|^2-2)\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle$$?

If the question is really that you must keep applying $\hat{b}_j$, then it means that it must be applied from the left. For n=2 you should be doing
$$\hat{b}_j \hat{b}^\dagger_j(\hat{b}_j\hat{b}_j<br /> \mid \Psi \rangle<br /> )$$