- #1

patric44

- 296

- 39

- Homework Statement
- derive the following expression involving boson operator

- Relevant Equations
- B=\sum_{i}\alpha_{i}b_{i}

Hi all

I found this expression in a paper that concerns the derivation of some relations about boson operators but it is not very clear to me how the results were obtained. The derivation starts as, let B be an operator as a linear combination of different boson operators:

$$

B=\sum_{i}\alpha_{i}b_{i}

$$

then the expectation value of the identity operator in the n-boson state is :

$$

\bra{B^{n}}\hat{1}\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}\sum_{i}\alpha_{i}\frac{\partial}{\partial b^{\dagger}_{i}}\ket{(B^{\dagger})^{n}}=n\alpa^{2}N_{n-1}

$$

where the partial derivative came from? and what is big N,the paper doesn't mention that, shouldn't the expression be :

$$

\bra{B^{n}}\hat{1}\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}B\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}\sum_{i}\alpha_{i}b_{i}\ket{(B^{\dagger})^{n}}

$$

can any one clarify, I will appreciate any help.

Thanks in advance

I found this expression in a paper that concerns the derivation of some relations about boson operators but it is not very clear to me how the results were obtained. The derivation starts as, let B be an operator as a linear combination of different boson operators:

$$

B=\sum_{i}\alpha_{i}b_{i}

$$

then the expectation value of the identity operator in the n-boson state is :

$$

\bra{B^{n}}\hat{1}\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}\sum_{i}\alpha_{i}\frac{\partial}{\partial b^{\dagger}_{i}}\ket{(B^{\dagger})^{n}}=n\alpa^{2}N_{n-1}

$$

where the partial derivative came from? and what is big N,the paper doesn't mention that, shouldn't the expression be :

$$

\bra{B^{n}}\hat{1}\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}B\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}\sum_{i}\alpha_{i}b_{i}\ket{(B^{\dagger})^{n}}

$$

can any one clarify, I will appreciate any help.

Thanks in advance

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