# Annihilation and Creation operator for bosons properties in second quantization

In second quantization the commutation relation of annihilation and creation operators of bosons is

$$$$[b,b^\dagger]=bb^\dagger-b^\dagger b=1$$$$

I am wondering what the general commutation relation is:

$$$$[(b)^n,(b^\dagger)^m]=?$$$$

I am able to find that $$[(b)^n,b^\dagger]=n(b)^{n-1}+b^\dagger b^n$$ and $$[b,(b^\dagger)^m]=m(b^\dagger)^{m-1}+(b^\dagger)^mb$$, but am having difficulties with the more general form

• Eqn. 35 here ? Nov 16, 2019 at 23:57

## 1 Answer

First, consider the expression $$[f^n,g]$$. The product rule for commutators is $$[fg,h]=[f,h]g+f[g,h] \\$$ Applying this to some powers of f, we get: $$[f^2,h] = [f,h]f+f[f,h] \\ [f^3,h] = [f^2,h]f+f[f^2,h] \\ = [f,h]f^2+2f[f,h]f+f^2[f,h]$$ This looks a lot like a binomial expansion, so we can infer that $$[f^n,h] = \sum_{i=0}^{n-1} {n-1\choose i} f^{i}[f,h]f^{(n-1)-i}$$ Now, let us upgrade to the case where both operators are raised to some power: $$[f^n,h^m] = \sum_{i=0}^{n-1} {n-1\choose i} f^{i}[f,h^m]f^{(n-1)-i} \\ [f^n,h^m] = \sum_{i=0}^{n-1} {n-1\choose i} f^{i}\left(\sum_{j=0}^{m-1} {m-1\choose j} h^{(m-1)-i}[f,h]h^{i}\right)f^{(n-1)-i}$$ Where in the second line the $$[f,h^m]$$ was expanded with the same rule as $$[f^n,h]$$, since $$[f,h^2] = [f,h]h+h[f,h]$$)

Letting $$f:=b$$ and $$h:=b^\dagger$$ and noting that $$[b,b^\dagger] = 1$$, we obtain:

$$[b^{(n)},b^{\dagger (m)}] = \sum_{i=0}^{n-1} \sum_{j=0}^{m-1} {n-1\choose i}{m-1\choose j} b^{(i)}\left( b^{\dagger (m-1)}\right)b^{(n-1)-i}$$