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This may be a silly question, but given that the momentum operator (say in the $x$-direction) can be written as $$p_x = -i \hbar \frac{\partial}{\partial x},$$ would it be correct to say that $$p_x^2 x = - i\hbar p_x,$$ where $x$ is the position operator in the $x$-direction?

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    $\begingroup$ Can you explain why do you think the equality should hold? $\endgroup$ Oct 31, 2023 at 8:50

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No, it wouldn't, because those operators are meant to be applied to some wavefunctions, so that (in the position basis) : $$ \begin{align} \hat{p}^2\hat{x}\psi(x) &= -\hbar^2\frac{\partial^2}{\partial x^2}(x\psi(x)) \\ &= -\hbar^2\frac{\partial}{\partial x}(\psi(x) + x\psi'(x)) \\ &= -\hbar^2(2\psi'(x) + x\psi''(x)) \\ &= (-2i\hbar\hat{p} + \hat{x}\hat{p}^2)\psi(x) \end{align} $$ hence $\hat{p}^2\hat{x} = \hat{x}\hat{p}^2 - 2i\hbar\hat{p}$. Actually, this result is a consequence of the canonical commutator $[\hat{x},\hat{p}] = i\hbar$.

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The equation $$p_x = -i \hbar \frac{\partial}{\partial x}$$ is an operator equation, i.e. its left and right side are meant to operate on a wave function. So this actually means $$p_x\psi(x) = -i \hbar \frac{\partial}{\partial x}\psi(x) \tag{1}$$ for every function $\psi(x)$.

Therefore your second equation actually means $$p_x^2 x \psi(x)= - i\hbar p_x \psi(x) \tag{2}$$ or using (1) $$-\hbar^2 \frac{\partial^2}{\partial x^2}x \psi(x)= -\hbar^2 \frac{\partial}{\partial x} \psi(x) $$ which is obviously wrong.

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