Binomial expansion of non-commutative operators

Question

I would like to determine the general expansion of

$$(\hat{A}+\hat{B})^n,$$

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?

Answer 1

I would like to determine the general expansion of

$(A+B)^n$,

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.

Answer 2

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. $\Box$

References:

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

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$^1$ 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.

Answer 3

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.

Answer 4

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

https://arxiv.org/abs/1707.03861