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<br />, 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<br />.

Homework Equations

Commutation relations: [\hat{b}^\dagger_j\hat{b}_j,\hat{b}_j]=-\hat{b}_j
Also: <br /> \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<br /> 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<br />?

 

[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<br />, 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<br />.

Homework Equations

Commutation relations: [\hat{b}^\dagger_j\hat{b}_j,\hat{b}_j]=-\hat{b}_j
Also: <br /> \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<br /> 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<br />?

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 /> )