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For example, given two operators:
$$A = \frac{\partial}{\partial x}+\frac{\partial}{\partial y},$$
$$B =\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} + 1.$$ Deriving commutator $[A,B]=0$ involves mixed partial derivative of higher order with opposite signs, and it is very temptative to zero them to make commutator equals zero. But is this allowed, i.e., does any valid quantum wave function must have continues (mixed) partial derivatives of higher order so one can use Schwartz theorem?

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2 Answers 2

Yes because $p_{i} \propto \displaystyle\frac{\partial}{\partial x_{i}}$ and the canonical commutation relations for momentum are $[p_{i},p_{j}]=0$

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But is this allowed, i.e., does any valid quantum wave function must have continues (mixed) partial derivatives of higher order so one can use Schwartz theorem?

Not as simply as that. The problem is that momentum is not defined on the full Hilbert Space, and so, you must be careful how to write this things. Imagine a function that is $\partial f/\partial x = N_0H(a^2-x^2) e^{-y^2-z^2}$ where H is the heaviside step function.

There is no classical (in opposition to weak, not to quantum) second x derivate of the function above, i.e., you can`t define as that traditional limit $\partial^2 f/\partial x^2$.

Its possible (I must check, but I'm pretty sure that's true) that you can suitably define momentum in such a way that, in the common domain, of $p_ip_jp_k$ work as Nivalth says, and you also can have this definition on a large enough domain, but this is a non-trivial result, even though its true. Thats the point I wanted to emphasize.

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