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The position and momentum operators do not commute because :

  • A.\(\hat{x}\hat{P}_{x}\Psi = x * \frac{\hbar}{i}\frac{d\Psi}{dx}\) and \(\hat{P}_{x}\hat{x}\Psi = \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right)\)
  • B.\(\hat{x}\hat{P}_{x}\Psi = \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right)\) and \(\hat{P}_{x}\hat{x}\Psi = \frac{x\hbar}{i}\frac{d\Psi}{dx}\)
  • C.\(\hat{x}\hat{P}_{x}\Psi = \hat{P}_{x}x\Psi = \frac{x\hbar}{i}\frac{d\Psi}{dx}\)
  • D.\(\hat{x}\hat{P}_{x}\Psi = \frac{\hbar}{i}\frac{d\Psi}{dx}\)

Answer: A

The correct pair is the first one: x̂P̂xΨ gives only the derivative term, while P̂xx̂Ψ also gives Ψ.

With \(\hat{P}_{x} = \frac{\hbar}{i}\frac{d}{dx}\):

\[ \begin{aligned} \hat{x}\hat{P}_{x}\Psi &= x\,\frac{\hbar}{i}\frac{d\Psi}{dx} \\ \hat{P}_{x}\hat{x}\Psi &= \frac{\hbar}{i}\frac{d(x\Psi)}{dx} \\ &= \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right) \end{aligned} \]

The product rule produces the extra \(\Psi\), so the two orders differ and \([\hat{x}, \hat{P}_{x}] = i\hbar\). The second option assigns the two results to the wrong orders, the third claims they are equal, and the fourth drops the factor x.

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