# Binomial Expansion for non-commutative setting

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.

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A simple google search led me to this : voofie.com/content/110/… I think it pretty much answers your question. As for an actual reference, I'll have to dig a little more! – Somnath Basu Oct 22 '11 at 2:51
@Somnath Basu: This is the same link that I posted with my question! The issue is that there is no actual reference in there! – Chris Oct 22 '11 at 3:36

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. $$(A+B)^n=\sum_{n\equiv k\pmod{2}} \left(\sum_{r=0}^k \binom{k}{r}A^rB^{k-r}\right)\left(-\frac{C}{2}\right)^{\frac{n-k}{2}}\frac{n!}{k!(\frac{n-k}{2})!}$$

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@Gjergji Zaimi: Thank you so much! That is all I really needed. – Chris Oct 22 '11 at 3:41
@Gjergji Zaimi I am a bit confused by the outmost sum, is k allowed to be grater than n and less than 0? – Prastt Apr 23 '14 at 22:38
@Prastt, I think the outmost sum should be read as "sum for k = 0 to n, having the same parity as n". Sorry for answering rather than commenting, seems I do not have enough reputation for it. If someone has more details about the calculation, it would be cool however. Especially since the original link is dead. – Phyks Jan 27 at 10:10

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.

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Since this question was asked so long ago you should say more about how you are specifically answering this question. – Chris Ramsey Jan 27 at 14:28
I am providing extra calculation steps from @GjergjiZaimi answer. – Phyks Jan 28 at 16:28