# Momentum operator and Space operator

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?

• Can you explain why do you think the equality should hold? Oct 31, 2023 at 8:50

## 2 Answers

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 $$[\hat{x},\hat{p}] = i\hbar$$.

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.