Binomial expansion for noncommutative operator

Question

Is it possible to find a closed formula for $(A^\dagger -kA)^n$ with $[A,A^\dagger]=1$ ?

I am looking for the normal ordinate form: $\sum (A)^{n-j}(A^\dagger)^j$— possibly something to do with the commutator, I don't know.

Answer 1

As I said in the above comment, you will not get an answer in the exact form you specify; the sum of the powers will not necessarily be $n$, as the corresponding algebra is filtered, not graded. Even for $n = 2$, you get:

$(A^\dagger - kA)^2 = A^{\dagger 2} - k A^\dagger A - k A A^\dagger + k^2 A^2 = A^{\dagger 2} - 2 k A^\dagger A - k + k^2 A^2$

In order to get the answer:

First, we need to find $[A^\dagger, (A^\dagger - kA)^n]$; induction (or Leibniz rule) shows that it will equal $n(A^\dagger - kA)^{n - 1}$.

Then: $(A^\dagger - kA)^{n+1} = (A^\dagger - kA)^n(A^\dagger - kA)$

$ = A^\dagger (A^\dagger - kA)^n - k (A^\dagger - kA)^n A - [A^\dagger, (A^\dagger - kA)^n]$

$ = A^\dagger (A^\dagger - kA)^n - k (A^\dagger - kA)^n A - n(A^\dagger - kA)^{n - 1}$

If both the $(A^\dagger - kA)^i$ are well-ordered (with $A^\dagger$ in front and $A$ in back), then this expression for $(A^\dagger - kA)^{n + 1}$ is too, so we have our recursion relation.

Some induction then shows that:

$(A^\dagger - kA)^n = \sum_{i = 0}^n (-k)^i a_{n, i} \sum_{j = 0}^{n - i} {{n - i}\choose{j}} A^{\dagger j} (-k)^{n - i - j} A^{n - i - j}$

where $a_{n, i} = 0$ if $i$ is odd, and $\frac{n!}{(n - i)! i!!}$ if $i$ is even, where $i!!$ denotes the double factorial $i!! = 2*4*6*...*i = 2^{\frac{i}{2}} (\frac{i}{2})!$

Answer 2

I'm not certain what you would like to find. See if this paper and its references

Tewodros Amdeberhan, Valerio De Angelis, Atul Dixit, Victor H. Moll and Christophe Vignat, From sequences to polynomials and back, via operator orderings Journal of Mathematical Physics 54 (2013) 123502, doi:10.1063/1.4836778, arXiv:1303.6587 (pdf)

would help you in any way.