- #1
RicardoMP
- 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]?