- #1
hokhani
- 483
- 8
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?
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?jfizzix said: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 [itex]\omega\approx \omega_{0} + \frac{d\omega}{dk} \Delta k[/itex], then you can express the wave as though it were a wave of the average frequency moving at speed [itex]\omega/k[/itex], times an envelope traveling at speed [itex]d\omega/dk[/itex] .
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).hokhani said:In the case of the free particle, we only have one wave component with one frequency.
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?Nugatory said: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 said: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?
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.hokhani said: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?
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 said: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):
[itex]\hbar\omega=E=\frac{p^{2}}{2m}=\frac{\hbar^{2}k^{2}}{2m}[/itex]
Solving for frequency, we get the relation:
[itex]\omega = \frac{\hbar k^{2}}{2m} \approx \omega_{0} + \frac{\hbar k}{m} \Delta\omega[/itex]
So the phase velocity function of a free particle is: [itex]\frac{\omega}{k}=\frac{\hbar k}{2m}[/itex]
while the group velocity function of a free particle is: [itex]\frac{d\omega}{dk}=\frac{\hbar k}{m}[/itex]
In this case, they are proportional, but always off by a factor of [itex]1/2[/itex]. The relation when including relativity can be more complicated, but one beautiful relation comes out (I'll spare you the derivation):
[itex]v_{g}v_{ph}=c^{2}[/itex]
hokhani said: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)?
The reason is that the dispersion relation is not linear for non-relativistic free ("Schrödinger") particles. We havehokhani said: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 phase velocity of a free particle refers to the speed at which the phase of its wave function propagates, while the group velocity refers to the speed at which a wave packet (a group of waves) propagates. In other words, the phase velocity describes the motion of individual waves, while the group velocity describes the overall motion of a wave packet.
The phase velocity and group velocity are related through the dispersion relation, which is a mathematical relationship between a wave's frequency and wavelength. The group velocity is equal to the derivative of the phase velocity with respect to the wave number (the inverse of wavelength).
No, according to the theory of relativity, the phase velocity of a free particle cannot exceed the speed of light. However, the group velocity can exceed the speed of light in certain cases, such as in the phenomenon of superluminal propagation.
The phase velocity of a free particle is directly proportional to its energy. This means that as the energy of a particle increases, its phase velocity also increases.
The phase and group velocity for a free particle have important physical implications, such as determining the behavior of particles in quantum mechanics and understanding the properties of waves in classical mechanics. They also play a crucial role in various technological applications, such as in the development of quantum computers and telecommunications systems.