I'm curious if there's a chain rule for the commutator (I'll explain what I mean) just like there's a product rule ([AB,C]).(adsbygoogle = window.adsbygoogle || []).push({});

So, say you have an operator, which can be expressed in terms of another operator, and we know the commutation relationship between x and another operator, y. I'll call this operator F(x) where you can express F in terms of x. I'm curious about [itex][F(x), y][/itex].

I know the answer for a specific case in quantum mechanics, but I'm curious about the general case.

Here's my specific case: [itex][V(x), p][/itex] where we know [itex][x, p] = i \hbar[/itex].

I'll have this commutator operate on [itex]\psi[/itex] so that it's very clear how I found this.

[itex][V(x), p]\psi = V(x)p\psi - pV(x)\psi = V(x)p\psi + i\hbar(V(x)\frac{\partial\psi}{\partial x} + \frac{\partial V(x)}{\partial x}\psi) = V(x)p\psi - V(x)p\psi + i\hbar\frac{\partial V(x)}{\partial x}\psi[/itex]

canceling some terms you get: [itex][V(x),p] = i\hbar \frac{\partial V(x)}{\partial x}[/itex]

Now the answer seems to be: [itex][V(x), p] = -(pV(x))[/itex].

The parenthesis are needed to differentiate it from [itex]pV(x)[/itex], which would be very different from [itex](pV(x))[/itex].

I can't be very general with this example. I expanded the commutator and used the p operator in the position representation, and I used the product rule for derivatives. That's not general at all. I am curious of the case of general operators F(x), x, y, and we know [x, y].

Does anybody have any insights into this? Is it even possible to make such a relationship without knowing more information? I've never taken an abstract algebra course (that'll be next semester probably), so I've turned to the internet for help. Thanks!

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# Chain rule for commutator (Lie derivative)?

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