Chain rule for commutator (Lie derivative)?

[WraithM]

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]).

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 $[F(x), y]$.

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

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

I'll have this commutator operate on $\psi$ so that it's very clear how I found this.
$[V(x), p]\psi = V(x)p\psi - pV(x)\psi = V(x)p\psi + i\hbar(V(x)\frac{\partial\psi}{\partial x} + \frac{\partial V(x)}{\partial x}\psi) = V(x)p\psi - V(x)p\psi + i\hbar\frac{\partial V(x)}{\partial x}\psi$

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

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

The parenthesis are needed to differentiate it from $pV(x)$, which would be very different from $(pV(x))$.

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!

 

[Fredrik]

First find the commutator $[x^n,p]$ for arbitrary n. (Try to guess the correct result, and then prove it by induction). Once you have that, it's easy to evaluate [F(x),p] whenever F(x) can be expressed as a series.

This approach should still work when you replace p with something else.

 

[WraithM]

First find the commutator $[x^n,p]$ for arbitrary n. (Try to guess the correct result, and then prove it by induction). Once you have that, it's easy to evaluate [F(x),p] whenever F(x) can be expressed as a series.

This approach should still work when you replace p with something else.

Of course! I have to add the condition that the operators are linear! Silly me :)

Thanks!

 

[strangerep]

Arnold Neumaier recently taught some extensions (beyond series) to what's already been described above.

I'll use Arnold's Lie product notation (since that's what the original document is written with). It's defined as:
$$\def\lp{\angle\,}<br /> \def\eps{\varepsilon}<br /> \def\implies{~~~\Rightarrow~~~}<br /> A \lp B ~:=~ \frac{i}{\hbar} [A,B] ~.$$
In the following, $X,f,g$ are (functions of) elements of a Lie algebra (actually a Poisson algebra since the Leibniz product rule holds). The first result below is easy. The second less so. (I can post the proofs if anyone wants them.)

Proposition 1:
If $f$ is invertible, then
$$X \lp f^{-1} ~=~ -f^{-1} (X\lp f) f^{-1}$$

Proposition 2:
For any smooth expression $F(f,g)$,
$$f \lp F(f,g)<br /> ~=~ \lim_{\eps\to 0} \frac{F(f, g + \eps f\lp g) - F(f,g)}{\eps}.$$

Corollary (of Prop 2):
If $f\lp g$ commutes with $g$ then for any smooth expression $F(f,g)$,
$$f\lp F(f,g) = F_g(f,g)(f\lp g) = (f\lp g) F_g(f,g) ,~~~~<br /> \left(\mbox{where}~~ F_g := \partial_g F \right).$$

In particular,
$$p \lp q = 1 \implies<br /> p \lp f(q) = f'(q) ,~~~<br /> f(p) \lp q = f'(p) .$$

 

[sweet springs]

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

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

The parenthesis are needed to differentiate it from $pV(x)$, which would be very different from $(pV(x))$.

Short comment.

It would be more clear to rewrite your -(pV(x)) to - hbar/i V'(x).

Regards.