Bosonic operator eigenvalues in second quantization

  • #1
47
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]?
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,737
279

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]?
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
[tex]\hat{b}_j \hat{b}^\dagger_j(\hat{b}_j\hat{b}_j
\mid \Psi \rangle
) [/tex]
 

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