- #1

- 49

- 2

## Homework Statement

Following from [tex] \hat{b}^\dagger_j\hat{b}_j(\hat{b}_j

\mid \Psi \rangle

)=(|B_-^j|^2-1)\hat{b}_j

\mid \Psi \rangle

[/tex], I want to prove that if I keep applying ##\hat{b}_j##, ## n_j##times, I'll get: [tex] (|B_-^j|^2-n_j)\hat{b}_j\hat{b}_j\hat{b}_j ...

\mid \Psi \rangle

[/tex].

## Homework Equations

Commutation relations: [tex] [\hat{b}^\dagger_j\hat{b}_j,\hat{b}_j]=-\hat{b}_j[/tex]

Also: [tex]

\langle \psi \mid

\hat{b}^\dagger_j\hat{b}_j

\mid \Psi \rangle

=||\hat{b}_j

\mid \Psi \rangle

||^2[/tex]

## 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:

[tex] \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

[/tex] after using the commutation relation I referred above (the same for n>2). How do I get [tex](|B_-^j|^2-2)\hat{b}_j\hat{b}_j

\mid \Psi \rangle

[/tex]?