Direction of momentum given by the de Broglie relation

Question

The momentum of an electron can be computed by the well-known classical mechanics equation:

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p = h / \lambda \\ p = \hbar k$

if the Planck constant $h$ is scalar and the wavelength $\lambda$ is also scalar. Similarly the reduced Planck constant $\hbar$ is scalar and the wavenumber $k=2\pi/\lambda$ is also scalar.

Answer 1

In general the wavenumber is a vector. That is, $e^{i(\vec{k}\cdot\vec{x}-\omega t)}$ is a solution to the wave equation in 3 (or any number) dimensions. We say this solution is a plane wave propagating in the $\hat{k}$ direction with wavenumber $|\vec{k}|$ or wavelength $\lambda = 2\pi/|\vec{k}|$.

So properly the de Broglie relation is $\vec{p} = \hbar \vec{k}$. The momentum of a plane wave is in the same direction as the propagation of the wave.