I Phase and group velocity for a free particle

[hokhani]

Why for the free particle, the group velocity and phase velocity are not the same while we have only one wave? What is the envelope here?

 

[jfizzix]

The phase velocity is the velocity that the peaks of each frequency component move in space. It will be different for different frequency components.

The group velocity is (approximately) the rate of change of the mean position of the particle. You don't have to think of it as a separate envelope.

However, if you break up the wave components in terms of frequency, and the range of frequencies is narrow enough that any frequency in it is linearly related to momentum (i.e., so that \omega\approx \omega_{0} + \frac{d\omega}{dk} \Delta k, then you can express the wave as though it were a wave of the average frequency moving at speed \omega/k, times an envelope traveling at speed d\omega/dk .

 

[hokhani]

However, if you break up the wave components in terms of frequency, and the range of frequencies is narrow enough that any frequency in it is linearly related to momentum (i.e., so that \omega\approx \omega_{0} + \frac{d\omega}{dk} \Delta k, then you can express the wave as though it were a wave of the average frequency moving at speed \omega/k, times an envelope traveling at speed d\omega/dk .

Thanks. In the case of the free particle, we only have one wave component with one frequency. Why its phase and group velocities are different?

 

[Nugatory]

In the case of the free particle, we only have one wave component with one frequency.

Which free particle state do you mean? The eigensolutions to Schrödinger's equation for a free particle are infinite plane waves with different frequencies according to the energy eigenvalues (here I'm glossing over some mathematical subtleties about continuous spectra and normalizability).

The general states of a free particle are superpositions of these plane waves, which can be chosen to give us a physically reasonable wave packet for which the notion of group velocity makes sense. But when you specify that you're considering the case in which only one frequency is present... well, what is the "mean position of the particle" to which @jfizzix refers?

 

[hokhani]

The general states of a free particle are superpositions of these plane waves, which can be chosen to give us a physically reasonable wave packet for which the notion of group velocity makes sense. But when you specify that you're considering the case in which only one frequency is present... well, what is the "mean position of the particle" to which @jfizzix refers?

Please consider one dimensional free particle at an energy E which has two components with k and -k wave vectors but with the same frequency. Therefore, there are two phase velocities in opposite directions. Could you please say what is the group velocity of this particle?

 

[jfizzix]

Thanks. In the case of the free particle, we only have one wave component with one frequency. Why its phase and group velocities are different?

In the case of a free particle, for it to have a well-defined location, it has to be made up of many different frequency and momentum components, since a single frequency wave is a sine wave that extends over all space.

For a free particle (not counting relativistic effects):
\hbar\omega=E=\frac{p^{2}}{2m}=\frac{\hbar^{2}k^{2}}{2m}
Solving for frequency, we get the relation:
\omega = \frac{\hbar k^{2}}{2m} \approx \omega_{0} + \frac{\hbar k}{m} \Delta\omega

So the phase velocity function of a free particle is: \frac{\omega}{k}=\frac{\hbar k}{2m}
while the group velocity function of a free particle is: \frac{d\omega}{dk}=\frac{\hbar k}{m}
In this case, they are proportional, but always off by a factor of 1/2. The relation when including relativity can be more complicated, but one beautiful relation comes out (I'll spare you the derivation):
v_{g}v_{ph}=c^{2}

 

[BPHH85]

Please consider one dimensional free particle at an energy E which has two components with k and -k wave vectors but with the same frequency. Therefore, there are two phase velocities in opposite directions. Could you please say what is the group velocity of this particle?

Obviously you are mixing some models. Let's take a look at the finite quantum well example. Here the bounded eigen states from the time-independent Schrödinger equation are standing waves. A standing wave results from two waves of same frequency and amplitude but opposite direction of moving that interfere. If the quantum well has a depth of $E_0$, than eigen states with energy $E > E_0$ are unbound and referred to as "free" states. But I guess here is the point where you are mixing it up. This free states are not the states that can describe free particles moving in space, because these free states are plane waves that lead to location probability of an electron everywhere the same in the whole univers. Its momentum is sharply defined but its location is absolutely uncertain.

To define something particle like with a wave, you need to make it more locatable. As the uncertainty principle states, the more certain the location is the more uncertain is the momentum. So to have a particle with zero propability to find it in nearly most of the free space but about one at a point r(x,y,z,t), you need a certain bandwidth of the momentum as jfizzix stated in #6. This leads to the model of a wave package. Here it might be helpful to take a look a the concept of the Fourier transformation.

 

[hokhani]

In the case of a free particle, for it to have a well-defined location, it has to be made up of many different frequency and momentum components, since a single frequency wave is a sine wave that extends over all space.

For a free particle (not counting relativistic effects):
\hbar\omega=E=\frac{p^{2}}{2m}=\frac{\hbar^{2}k^{2}}{2m}
Solving for frequency, we get the relation:
\omega = \frac{\hbar k^{2}}{2m} \approx \omega_{0} + \frac{\hbar k}{m} \Delta\omega

So the phase velocity function of a free particle is: \frac{\omega}{k}=\frac{\hbar k}{2m}
while the group velocity function of a free particle is: \frac{d\omega}{dk}=\frac{\hbar k}{m}
In this case, they are proportional, but always off by a factor of 1/2. The relation when including relativity can be more complicated, but one beautiful relation comes out (I'll spare you the derivation):
v_{g}v_{ph}=c^{2}

Thank you very much for your thorough explanation. I can understand the phase velocity concept, as the velocity by which the phase is moving, but what is the concept of group velocity in this case (free particle)?

 

[jfizzix]

Thank you very much for your thorough explanation. I can understand the phase velocity concept, as the velocity by which the phase is moving, but what is the concept of group velocity in this case (free particle)?

It's a bit underwhelming, but from a technical standpoint phase and group velocities are functions, with different values at different frequencies. Phase velocity by definition is \omega/k and group velocity is by definition the function d\omega/dk

Since you already understand the phase velocity concept, the group velocity of a pulse is well-defined only if d\omega/dk is more or less constant over the range of frequencies the pulse is made up of. If this is condition is satisfied, the pulse travels at a speed equal to the average value of d\omega/dk over all frequencies. If this condition is not satisfied, the pulse can distort and spread out as it propagates.

 

[vanhees71]

Thanks. In the case of the free particle, we only have one wave component with one frequency. Why its phase and group velocities are different?

The reason is that the dispersion relation is not linear for non-relativistic free ("Schrödinger") particles. We have
$$E=\hbar \omega=\vec{p}^2/(2m)=\hbar^2 \vec{k}^2/(2m).$$
The group velocity is
$$\vec{v}_{\text{g}}=\partial_{\vec{p}} E=\frac{\vec{p}}{m},$$
which is expected from classical mechanics, and that's the velocity a particle is moving on average, if prepared in a true state represented by a wave packet. A plane wave is a generalized momentum eigenfunction but doesn't represent a true state, because it's not square integrable.

The phase velocity is the velocity for which the phase of the plane wave is stationary, i.e.,
$$\vec{v}_{\text{p}}=\vec{k} \omega/|\vec{k}|^2=\hbar \vec{k}/(2m).$$