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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?

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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. –  Vladimir Kalitvianski Oct 23 '11 at 9:09
Same for particle velocity/momentum. –  Vladimir Kalitvianski Oct 23 '11 at 13:31
The particle's velocity through space is its group velocity. –  Mike Dunlavey Oct 23 '11 at 13:48

2 Answers 2

up vote 6 down vote accepted

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.

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You forgot to divide by the mass when defining the velocity operator. –  mmc Oct 23 '11 at 17:20
How did you get from $\frac{\partial E}{\partial p}$ to here $\frac{p}{m}$? –  71GA Jan 27 '13 at 12:33
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 $p/m$. –  jabirali Mar 8 '13 at 18:46

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.

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I am just trying to make it intuitive, but being aware of its defaults. –  Isaac Oct 23 '11 at 10:01
Sorry, not momentum. Energy to momentum ratio, of course. –  Misha Oct 23 '11 at 10:34

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