Issues with operator associativity in quantum mechanics

Question

Textbook authors often point put that operators aren't in general commutative which is a source of confusion. Another point of confusion, which I haven't seen mentioned anywhere is the issue with associativity.

One might naively perform a calculation like $$\hat{p}\hat{x}\psi(x)=-i\hbar\frac{d}{dx}x\psi(x)=-i\hbar\frac{dx}{dx}\psi(x)=-i\hbar\psi(x).$$ Which is what I did when I was first trying to evaluate this. Because I didn't know operators don't associate like this. No textbook that I know of mentions it and I am probably not the only one who fell into this pitfall.

I am looking for more information about issues concerning operator associativity in QM.

EDIT: Consider a case like this: $\hat{A}\hat{p}\hat{x}$.

Answer 1

QM operators are associative, but you do have to be careful about how you calculate things. $\hat p$ and $\hat x$ are linear maps from functions to functions. $\hat p$ maps the function $f(x)$ to $-i\hbar f'(x)$ while $\hat x$ maps a function $f(x)$ to $xf(x)$. To multiply them is equivalent to composing the linear maps. Here's how to compute $\hat p \hat x$:

To exactly know what the final operator is, we need to know it's behaviour when applied to all functions $f$. So let $f$ be an arbitrary function. Then we have:

$$ \hat p\hat x f(x) = \hat p (\hat x f(x)) = \hat p (xf(x)) = -i\hbar\frac{\partial}{\partial x}(xf(x)) = -i\hbar(f(x) + xf'(x)) $$

Now that we know the behaviour when applied to all functions, we can remove the unknown function $f$ and just write the operator as:

$$ \hat p \hat x = -i\hbar(1 + x\frac{\partial}{\partial x}) $$

I think Griffiths has some discussion of this "dummy function" technique for multiplying operators.