# Momentum operator in QM - scalar or vector?

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.

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

• Thanks for the answer .Shouldn't then be $-i\hbar \frac{d}{dx}\hat{i}$ ? – ado sar Jul 22 at 8:36
• 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 – jgerber Jul 22 at 14:13