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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?

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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.

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  • $\begingroup$ 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
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$-i\hbar \frac{d}{dx}$, 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.

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  • $\begingroup$ 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

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