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

\begin{equation} [b,b^\dagger]=bb^\dagger-b^\dagger b=1 \end{equation}

I am wondering what the general commutation relation is:

\begin{equation} [(b)^n,(b^\dagger)^m]=? \end{equation}

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

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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:

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

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