# 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.

• 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! Oct 22, 2011 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! Oct 22, 2011 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. $$(A+B)^n=\sum_{n\equiv k\pmod{2}} \left(\sum_{r=0}^k \binom{k}{r}A^rB^{k-r}\right)\left(-\frac{C}{2}\right)^{\frac{n-k}{2}}\frac{n!}{k!(\frac{n-k}{2})!}$$

• @Gjergji Zaimi: Thank you so much! That is all I really needed. Oct 22, 2011 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? Apr 23, 2014 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. Jan 27, 2016 at 10:10
• Can this be made to work in characteristic 2? Feb 15, 2018 at 2:22

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.

• Since this question was asked so long ago you should say more about how you are specifically answering this question. Jan 27, 2016 at 14:28
• I am providing extra calculation steps from @GjergjiZaimi answer. Jan 28, 2016 at 16:28

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

• 1) Yes, same parity means even if $n$ is even, odd if $n$ is odd. The clarified part is that $0 \le k \le n$. Of course you'd have trouble with the $k!$ and $((n-k)/2)!$ if that were not the case. 2) Did you not notice the link to the Zassenhaus formula in Wikipedia? The Wikipedia page has extensive references. Mar 18, 2016 at 17:57
• @RobertIsrael, if C is positive, can I be sure that the result of this binomial expansion is also positive? Mar 19, 2016 at 12:01
• I don't see any reason why it would be positive. Mar 20, 2016 at 5:23
• I concur. The reason I asked is because the form the result takes is usually positive, but I just realised that the important part is going to use the norm anyway. Thanks for your time, @RobertIsrael. Mar 20, 2016 at 9:46