I would like to determine the general expansion of


where $[\hat{A},\hat{B}]\neq 0$, i.e. $\hat{A}$ and $\hat{B}$ are two generally non-commutative operators. How could I express this in terms of summations of the products of $\hat{A}$ and $\hat{B}$ operators?

  • 1
    $\begingroup$ You just want it as a summation of product of operators, or are you wondering whether there is a nice expression using the commutators? $\endgroup$
    – Ali Moh
    Mar 26, 2015 at 19:54
  • $\begingroup$ @Ali, I wanted the full expansion of $(A+B)^n$. Since it is a form of the Binomial expansion (although A and B are non-commutative), I would expect the final result to be in terms of a sum of operator products. $\endgroup$
    – Sid
    Mar 26, 2015 at 21:41
  • $\begingroup$ Someone has posted a nice formula in another thread: mathoverflow.net/q/78813. However, I did not find any other reference to this formula on the web. $\endgroup$
    – Phyks
    Jan 27, 2016 at 10:12
  • 1
    $\begingroup$ @Phyks: Note that the MO.SE thread and e.g. this Phys.SE post additionally assumes that $[A,C]=0=[B,C]$. $\endgroup$
    – Qmechanic
    Jan 27, 2016 at 10:30

4 Answers 4


I would like to determine the general expansion of


where [A,B]≠0

The expansion of $(A+B)^n$ for non-commuting A and B is the sum of $2^n$ different terms. Each term has the form $$ X_1X_2...X_n\;, $$ where $X_i=A$ or $X_i=B$, for all the different possible cases (there are 2^n possible cases). For example: $$ (A+B)^3=AAA+AAB+ABA+ABB+BAA+BAB+BBA+BBB $$

You can understand how these terms are always generated as described above by considering binary numbers. Let "A" represent "0" and "B" represent "1". Then each term corresponds to a number in binary from 0 to 2^n-1. E.g., in the n=3 case, 000,001,010,011,100,101,110,111.

You can prove by induction that the statements above are true for arbitrary n by considering what the application of another factor of $(A+B)$ does to $(A+B)^{n-1}$. The new terms from distributing the "A" in (A+B) just make copies of the previous "binary numbers" (from 0 to $2^{n-1}-1$) but with a different "bit-length". The new "B" terms generate the rest of the "binary numbers from $2^{(n-1)}$ to $2^{n}-1$ because they correspond to $2^{n-1}$ plus the previously generated terms.

  • 1
    $\begingroup$ This is great since it is easy to see how the expansion works. However, is there a nice and compact form of writing this since I have a term where the power index is arbitrarily large. I chose $n=3$ only as an example. Namely, I have $(A+B)^n$, and so I would like to write the resulting expression in a compact form in terms of $n$. $\endgroup$
    – Sid
    Mar 27, 2015 at 14:12
  • 2
    $\begingroup$ @Sid the compact form is $(A+B)^n$. $\endgroup$
    – user121330
    Sep 5, 2017 at 19:11

A non-commutative binomial formula is not a unique notion. Here we consider the formula$^1$ $$\begin{align} (\hat{A}+\hat{B})^n ~=~&\sum_{k=0}^n \begin{pmatrix}n \\k \end{pmatrix}(\hat{C}^k\hat{\bf 1})\hat{B}^{n-k}, \cr \hat{C}~\equiv~&\hat{A}+ [\hat{B}, \cdot]~\equiv~\hat{A}+ L_\hat{B}- R_\hat{B}, \end{align}\tag{1}$$ from Ref. 1, which in turn is equivalent to $$ e^{\hat{A}+\hat{B}}~=~(e^{\hat{C}}1)e^{\hat{B}}, \tag{2}$$ $$ (e^{\hat{C}}\hat{\bf 1})~=~e^{\hat{A}+\hat{B}}e^{-\hat{B}}, \tag{3}$$ or $$ (e^{t\hat{C}}\hat{\bf 1})~=~e^{t\hat{A}+t\hat{B}}e^{-t\hat{B}}. \tag{4}$$ where $t\in\mathbb{R}$ is a parameter.

Proof of eq. (4): First notice that it is trivially true for $t=0$. Next differentiate its left- & right-hand sides wrt. $t$ in order to show that the left- & right-hand sides (collectively called $\hat{S}$) satisfy the same ODE:

$$ \hat{S}^{\prime}(t)~=~\hat{C}\hat{S}(t)~\equiv~\hat{A}\hat{S}(t)+ [\hat{B}, \hat{S}(t)].\tag{5} $$ Hence the left- & right-hand sides of eq. (4) must be equal. $\Box$


  1. W. Wyss, arXiv:1707.03861 (Hat tip: Dan & BMRodriguez-Lara.)


$^1$ Notation: Note that the operator $\hat{C}$ takes operators to operators, e.g. $$\begin{align} (\hat{C}\hat{\bf 1})~=~&\hat{A}, \cr (\hat{C}^2\hat{\bf 1})~=~&\hat{A}^2 + [\hat{B},\hat{A}],\cr (\hat{C}^3\hat{\bf 1})~=~&\hat{A}^3 + [\hat{B},\hat{A}^2] + [\hat{B},[\hat{B},\hat{A}]],\end{align}\tag{6}$$ and so forth.

  • $\begingroup$ What is the $1$ in $(e^{\hat{C}}1)$ meant to represent? If it's just an identity operator, why is it needed there? $\endgroup$
    – Ruslan
    Apr 18, 2021 at 15:15
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 18, 2021 at 17:36
  • $\begingroup$ Check your C^2(1) (I think it should be A^2 at the beginning) and C^3(1) (it should be A^3 at the beginning and there's also a [B,A^2] term that must be added). $\endgroup$
    – JuanC97
    Mar 9, 2022 at 4:23
  • 1
    $\begingroup$ Ups. Thanks. Corrected. $\endgroup$
    – Qmechanic
    Mar 9, 2022 at 7:02

if $[A,B]=0$ then as you know you get the usual $$ (A+B)^n = \sum_{p=0}^n C^n_p A^{n-p}B^p $$ Now if $[A,B]\neq 0$ each term in the sum (for each $p$) splits into a sum of $C^n_p$ terms of all possible permutations of $(n-p)$ $A$s and $p$ $B$s, without regard to the order of $A$s and $B$s. Equivalently to the sum of all possible permutations of $(n-p)$ $A$s and $p$ $B$s divided by $p!(n-p)!$ \begin{align*} (A+B)^n &= \sum_{p=0}^n \left(\frac{1}{p!(n-p)!}\sum_{\text{perm}} \left\{A^{n-p}B^p\right\}\right)\\ &= \sum_{p=0}^n \left(\sum_{\text{perm no order}} \left\{A^{n-p}B^p\right\}\right) \end{align*}

I don't know if there is a nice formula that looks like $$ (A+B)^n = \sum_{p=0}^n C^n_p A^{n-p}B^p + \text{commutators} $$ Of course you can always rearrange the terms in each of the permutations, but I doubt that it will give something nice and concise in terms of commutators alone.


There is a nice formula that provides the result in terms of the binomial expansion plus terms related to the noncommutative algebra


  • 7
    $\begingroup$ Hi BMRodriguez-Lara it is more acceptable on PSE if you provide a summary of what the link says such that other people viewing this question don't have to click though it - rather then just providing an essentially link-only answer. $\endgroup$ Sep 5, 2017 at 17:47

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