Tag: work

It is a known result that, given generically noncommuting operators $ A,B$ , we have $ $ A^n B=\sum_{k=0}^n \binom{n}{k} \operatorname{ad}^k(A)(B) A^{n-k}.\tag A $ $ This can be proved for example via induction with not too much work.

However, while trying to get a better understanding of this formula, I realised that there is a much easier way to derive it, at least on a formal, intuitive level.

The trick

Let $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ (standing for “shift” and “commute”, respectively) denote operators that act on expressions of the form $ A^k D^j A^\ell$ (denoting for simplicity $ D^j\equiv\operatorname{ad}^j(A)(B)$ ) as follows:

\begin{align} \hat{\mathcal S} (A^k D^j A^\ell) &= A^{k-1} D^j A^{\ell+1}, \ \hat{\mathcal C} (A^{k-1} D^{j+1} A^\ell) &= A^{k-1} D^j A^{\ell+1}. \end{align} In other words, $ \hat{\mathcal S}$ “moves” the central $ D$ block on the left, while $ \hat{\mathcal C}$ makes it “eat” the neighboring $ A$ factor.

It is not hard to see that $ \hat{\mathcal S}+\hat{\mathcal C}=\mathbb 1$ , which is but another way to state the identity $ $ A[A,B]=[A,B]A+[A,[A,B]].$ $ Moreover, crucially, $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ commute. Because of this, I can write

$ $ A^n B=(\hat{\mathcal S}+\hat{\mathcal C})^n (A^n B)=\sum_{k=0}^n\binom{n}{k} \hat{\mathcal S}^{n-k} \hat{\mathcal C}^{k}(A^n B),$ $ which immediately gives me (A) without any need for recursion or other tricks.

The question

Now, this is all fine and dandy, but it leaves me wondering as to why does this kind of thing work? It looks like I am somehow bypassing the nuisance of having to deal with non-commuting operations by switching to a space of “superoperators”, in which the same operation can be expressed in terms of commuting “superoperators”.

I am not even sure how one could go in formalising this “superoperators” $ \hat{\mathcal S},\hat{\mathcal C}$ , as they seem to be objects acting on “strings of operators” more than on the elements of the operator algebra themselves.

Is there a way to formalise this way of handling the expressions? Is this a well-known method in this context (I had never seen it but I am not well-versed in this kinds of manipulations)?