Given that one usually defines two different velocities for a wave, these being the phase velocity and the group velocity, I was asking their meaning for the associated particle in quantum mechanics.

And is one of them more representative for a particle?

  • $\begingroup$ The whole wave function represents the particle in a sense that many measurements will give $|\psi (x)|^2$ as a probability density. Don't think that you will obtain the same result in each measurement. $\endgroup$ Oct 23, 2011 at 9:09
  • $\begingroup$ Same for particle velocity/momentum. $\endgroup$ Oct 23, 2011 at 13:31
  • $\begingroup$ The particle's velocity through space is its group velocity. $\endgroup$ Oct 23, 2011 at 13:48

2 Answers 2


Much like the position itself, the velocity in quantum mechanics isn't just a single number; it is an operator with different probabilities of different outcomes that may result from the measurement of the velocity.

The operator of velocity in the simplest quantum mechanical model is $$ v = p/m = -\frac{i\hbar}{m} \frac{\partial}{\partial x} $$ You may Fourier-transform your wave function to the momentum representation and then you see different values of the momentum, and therefore velocity, and the probability densities of different values are given by $|\tilde \psi (p)|^2$.

If you consider a simple plane wave, $$ \psi (x,t) = \exp( ipx/\hbar - iEt /\hbar ) $$ then the operator $v$ above has an eigenstate in the vector above and the eigenvalue is $p/m$. On the other hand, the phase velocity is given by $$v_p = \omega / k = \frac{E}{p} = \frac{pv}{2p} = \frac{v}2 $$ so the velocity of the particle is equal to twice the phase velocity, assuming that your energy (determine the change of phase in time) is only given by the non-relativistic piece, without any $mc^2$. One may also calculate the group velocity of the wave $$ v_g = \frac{\partial \omega}{\partial k } = \frac{\partial E}{\partial p} = \frac{p}m = v$$ which is exactly the velocity of the particle. The advantage of this relationship is that it holds even in relativity. If $E=\sqrt{p^2+m^2}$, then the derivative of $E$ with respect to $p$ is $1/2E\cdot 2p =p/E = v$ which is exactly the right velocity, too. It's not too surprising because if a wave packet is localized, the group velocity measures how the "center of mass" of this packet is moving but the packet's position coincides with the particle's position, so the two velocities must be equal.

  • $\begingroup$ You forgot to divide by the mass when defining the velocity operator. $\endgroup$
    – mmc
    Oct 23, 2011 at 17:20
  • $\begingroup$ How did you get from $\frac{\partial E}{\partial p}$ to here $\frac{p}{m}$? $\endgroup$
    – 71GA
    Jan 27, 2013 at 12:33
  • 1
    $\begingroup$ 71GA: For a free particle, i.e. a particle with no potential energy, $E=p^2/2m$. If you differentiate that with respect to momentum, you get $p/m$. $\endgroup$
    – jabirali
    Mar 8, 2013 at 18:46
  • $\begingroup$ What about photons in photonic crystals or other dispersive mediums where the group velocity can be in a different direction than the wavevector (and therefore phase velocity?)? Does a single photon particle move in the direction of the group velocity or the wave vector? Or is the "group velocity" a weird result of the frequency/wavevector uncertainty and superposition of corresponding waves/particles travelling at phase velocities $\omega/k$ and in directions $\vec{k}$? $\endgroup$
    – KIAaze
    Sep 4, 2015 at 16:58
  • $\begingroup$ Dear @KIAaze - the photon as a wave packet, something localized, always moves in the direction and by the speed given by the group velocity, even if its magnitude or direction differs from the phase velocity. $\endgroup$ Sep 5, 2015 at 14:55

To the first approximation, you may consider that phase velocity corresponds to $\frac{energy}{momentum}$ and group velocity to a "normal" velocity. However, there is not much sense in this analogy.

  • $\begingroup$ I am just trying to make it intuitive, but being aware of its defaults. $\endgroup$
    – Isaac
    Oct 23, 2011 at 10:01
  • $\begingroup$ Sorry, not momentum. Energy to momentum ratio, of course. $\endgroup$
    – Misha
    Oct 23, 2011 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.