Derivation of an expression involving boson operators

[patric44]

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

 

[strangerep]

Homework Statement: derive the following expression involving boson operator
Relevant Equations: B=\sum_{i}\alpha_{i}b_{i}

Did you forget some hash hash symbols?

I found this expression in a paper

It's usually a good idea to include a link to your source, just in case you've mistyped or misunderstood something.

where the partial derivative came from?

It's possible to prove a general formula like $$[a, f(a^\dagger)] ~=~ i\hbar \, \partial_{a^\dagger} f(a^\dagger) ~.$$The constant ##i\hbar## factor might be different depending on what conventions you're using for the canonical commutation relations. (Exercise: use induction to prove this formula for simple functions like ##f(x) = x^n##, then use linearity of the commutator to generalize the formula to polynomials.)

and what is big N,

I'm guessing it's the number operator, something involving ##\sum_i b_i^\dagger b_i##.

the paper doesn't mention that, shouldn't the expression be :
$$
\bra{B^{n}}\hat{1}\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}B\ket{(B^{\dagger})^{n}}=\bra{B^{n-1}}\sum_{i}\alpha_{i}b_{i}\ket{(B^{\dagger})^{n}}
$$

Without seeing the paper, it's impossible to say for sure. But my guess is "no". Write out the expression properly and apply the rule I described above.

 

[patric44]

Did you forget some hash hash symbols?It's usually a good idea to include a link to your source, just in case you've mistyped or misunderstood something.It's possible to prove a general formula like $$[a, f(a^\dagger)] ~=~ i\hbar \, \partial_{a^\dagger} f(a^\dagger) ~.$$The constant ##i\hbar## factor might be different depending on what conventions you're using for the canonical commutation relations. (Exercise: use induction to prove this formula for simple functions like ##f(x) = x^n##, then use linearity of the commutator to generalize the formula to polynomials.)

I'm guessing it's the number operator, something involving ##\sum_i b_i^\dagger b_i##.

Without seeing the paper, it's impossible to say for sure. But my guess is "no". Write out the expression properly and apply the rule I described above.

the paper isn't open access so I thought I would write the question separably, here is the link of the paper:
the paper, the commutation relation is included in the paper but i am not interested in proving them, rather my concern is equations 4a,4b,4c