The momentum operator for one spatial dimension is $-i \hbar\frac{\mathrm d}{\mathrm dx}$ (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.
Emilio Pisanty
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Shouldn't then be $-i\hbar\frac{d}{dx}\hat{i}$ ? Thanks in advantage. – Antonios Sarikas Jul 22 '19 at 08:38
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5The difference doesn't matter. – Emilio Pisanty Jul 22 '19 at 09:08
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2Actually there is a remnant of orthogonal group in $R^1$ which is the parity inversion. $P$ is a vector operator with respect to it. – Valter Moretti Aug 02 '22 at 16:53
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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.
Jagerber48
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Thanks for the answer .Shouldn't then be $-i\hbar \frac{d}{dx}\hat{i}$ ? – Antonios Sarikas Jul 22 '19 at 08:36
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1Don’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 – Jagerber48 Jul 22 '19 at 14:13
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1@AntoniosSarikas there is only one dimension so the direction is implicit. You don't need an explicit $\hat i$ since there is only one possible axis along which to travel. – hft Aug 02 '22 at 16:44