What could be a reference about binomial expansions for non-commutative elements?
Specifically, where can I find a closed formula for the expansion of $(A+B)^n$ where $[A,B]=C$ and $[C,A]=[C,B]=0$?
I've found some ideas about that and also a proof using PDE's in the following website: link. But I haven't found a such formula in a published scientific paper or book.
I don't know if you prefer a particular presentation of the formula, but this is essentially covered by the Baker-Campbell-Hausdorff formula, or actually it's dual, Zassenhaus formula, which in your case reduces to $$e^{(A+B)t}=e^{At}e^{Bt}e^{-[A,B]t^2/2},$$ where one side is the generating function for $(A+B)^n$ while the other has terms of the form $f(n,m,p)A^nB^mC^p$. The binomial theorem here is given by equating the coefficients of $t^n$ on both sides. $$(A+B)^n=\sum_{n\equiv k\pmod{2}} \left(\sum_{r=0}^k \binom{k}{r}A^rB^{k-r}\right)\left(-\frac{C}{2}\right)^{\frac{n-k}{2}}\frac{n!}{k!(\frac{n-k}{2})!}$$
Here are the calculations steps, in case it can help anyone:
Left hand side can be written $$e^{t(a+b)} = \sum_{k} \frac{(a+b)^k t^k}{k!}$$ where one recognizes the term we want to compute.
Right hand side is given by $$e^{ta} e^{tb} e^{- c t^2 / 2} = \sum_{i, j, k} t^{i + j + 2k} \frac{a^i b^j (-c / 2)^k}{i! \, j! \, k!}$$
Identifying both sides, one has: $$\frac{(a+b)^n}{n!} = \sum_{i + j + 2k = n} \frac{(-c / 2)^k}{i! \, j! \, k!} a^i b^j$$
$$\frac{(a+b)^n}{n!} = \sum_{\substack{i + j \leq n \\ i+j = n [2]}} \frac{(-c / 2)^{(n - i - j) / 2}}{i! \, j! \, \left(\frac{n - i - j}{2}\right)!} a^i b^j$$
Let us note $k = m+n$ and $r = m$, $$\frac{(a+b)^n}{n!} = \sum_{\substack{0 \leq r \leq k \leq n \\ k = n [2]}} \frac{(-c / 2)^{(n - k) / 2}}{\left(\frac{n - k}{2}\right)! \, r! \, (k-r)!} a^r b^{k-r}$$
$$\frac{(a+b)^n}{n!} = \sum_{\substack{k = 0 \\ k = n [2]}}^n \frac{(-c / 2)^{(n - k) / 2}}{k! \left(\frac{n - k}{2}\right)!} \sum_{r = 0}^{k} \binom{k}{r} a^r b^{k-r}$$
Which is the expected formula.
See a paper entitled "Binomial expansion for non-commutative operators" written by Martin Pépin and Lucas Verney (February 3, 2016).
NB: I do not have sufficient reputation to add a comment, so that this post is in a separate 'answer'.
The question is twofold.
First, for clarity, the clarifying comment "sum for $k = 0$ to $n$, having the same parity as $n$" is not clarifying to me. The answer reads $n\equiv k\:\text{mod}(2)$. Together, they seem to combine that $k$ runs over all numbers from $0$ to $n$ that are even (odd) if $n$ is even (odd).
Second, this question originally asked for a reference and an answer; the latter has been satisfied but a reference has not been given. I am rather sure my target audience does not know this formula, making it almost imperative to add a reference.
My sincere thanks, Josko
