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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.

plasmacel
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    the de Broglie relation just equates the magnitude of the momentum as equal to $h/\lambda$. The direction of the momentum is the same as the direction of the velocity – philip_0008 Jun 08 '16 at 20:00
  • p in the de Broglie relation is just the magnitude. – philip_0008 Jun 08 '16 at 20:04
  • and how can you determine the velocity, from a matter wave, if that has only wavelength, amplitude and phase? – plasmacel Jun 08 '16 at 20:04
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    In quantum mechanics, we have expectation values for position $\langle x \rangle$, (which works like an average position). the rate of change of the expectation value $\langle v \rangle = d\langle x \rangle/dt$ determines the expectation value for velocity. – philip_0008 Jun 08 '16 at 20:11
  • http://physics.stackexchange.com/questions/257826/about-de-broglie-relations-what-exactly-is-e-its-energy-of-what – Frobenius Jun 08 '16 at 21:14

1 Answers1

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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.

Luke Pritchett
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