# Derivation of an expression involving boson operators

• patric44
In summary, the conversation discusses an expression found in a paper regarding the derivation of relations involving boson operators. There is confusion about the derivation process and the use of a partial derivative, as well as the meaning of the symbol "N" in the expression. The source of the paper is provided and the conversation delves into the details of the equations involved.
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

Last edited:
patric44 said:
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?

patric44 said:
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.

patric44 said:
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.)
patric44 said:
and what is big N,
I'm guessing it's the number operator, something involving ##\sum_i b_i^\dagger b_i##.
patric44 said:
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.

vanhees71
strangerep said:
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

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