The momentum operator for one spatial dimension is $-i \hbar d/dx$ (which isn't a vector operator) but for 3 spatial dimensions is $-i\hbar\nabla$ which is a vector operator. So is it a vector or a scalar operator?


Momentum is a vector operator. Period.

When restricted to one-dimensional problems, momentum becomes a one-dimensional vector, which coincides with scalars in that space.

  • $\begingroup$ Shouldn't then be $-i\hbar\frac{d}{dx}\hat{i}$ ? Thanks in advantage. $\endgroup$ – ado sar Jul 22 at 8:38
  • $\begingroup$ The difference doesn't matter. $\endgroup$ – Emilio Pisanty Jul 22 at 9:08

$-i\hbar \frac{d}{dx}$, a scalar, is the position space representation of $\hat{p}_x$, the $x$ component of the momentum operator, a scalar. The momentum operator itself, $\hat{\textbf{p}}$, is a vector operator. The position space representation of $\hat{\textbf{p}}$ would be $-i\hbar \nabla$, a vector.

Again, the momentum operator is a vector operator. The components of the momentum operator are scalars operators.

  • $\begingroup$ Thanks for the answer .Shouldn't then be $-i\hbar \frac{d}{dx}\hat{i}$ ? $\endgroup$ – ado sar Jul 22 at 8:36
  • $\begingroup$ Don’t have time for proper response. My answer is actually a little messed up. In differential geometry $\frac{d}{dx}$ is actually thought of as a vector not a scalar. Regardless, you don’t usually ever see $\frac{d}{dx}\hat{i}$ so I wouldn’t write that down. Maybe later I can give a more thorough correction $\endgroup$ – jgerber Jul 22 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.