# Bosonic operator eigenvalues in second quantization

## Homework Statement

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

## Homework Equations

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

## 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:
$$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j\hat{b}_j \mid \Psi \rangle )=(|B_-^j|^2-1)\hat{b}_j\hat{b}_j \mid \Psi \rangle$$ after using the commutation relation I referred above (the same for n>2). How do I get $$(|B_-^j|^2-2)\hat{b}_j\hat{b}_j \mid \Psi \rangle$$?

nrqed
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## Homework Statement

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

## Homework Equations

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

## 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:
$$\hat{b}^\dagger_j\hat{b}_j(\hat{b}_j\hat{b}_j \mid \Psi \rangle )=(|B_-^j|^2-1)\hat{b}_j\hat{b}_j \mid \Psi \rangle$$ after using the commutation relation I referred above (the same for n>2). How do I get $$(|B_-^j|^2-2)\hat{b}_j\hat{b}_j \mid \Psi \rangle$$?
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
$$\hat{b}_j \hat{b}^\dagger_j(\hat{b}_j\hat{b}_j \mid \Psi \rangle )$$