# Wave group and phase velocities

When talking about phase velocity and group velocity they tend to be both expressed in terms of another $$v$$. From my understanding this $$v$$ comes from the wave equation, but what does it represent? And regarding phase velocity my understanding was that it is the velocity of a specific frequency component, however within a guided wave it is possible for this velocity to tend to infinity, which does not agree so well with SR. Any help for where I'm getting this wrong are appreciated.

• I think there are two questions here, one is what the meaning of a phase vs. group velocity is in a dispersive context; this is probably a duplicate of other questions on the site. The other more interesting question is what this mystery $v$ is. Can you give us more information by illustrating some contexts in which you've encountered it? – CR Drost Nov 12 '18 at 21:46
• In most derivations of the dispersion relationship, or in fact relations between phase velocity and group velocity. In the most recent case the phase velocity multiplied by the group velocity gave this other v? Unfortunately I can't send screenshots of my lecture notes due to some restrictions by the university. I would quote some specific examples but most of them require a dozen other equations to be made sense of. Have I helped illustrate what Im talking about ? – Jake Rose Nov 12 '18 at 22:17
• – Qmechanic Nov 12 '18 at 22:30
• @Qmechanic I was considering more classical waves as opposed to QM. Will the ideas be the same? – Jake Rose Nov 12 '18 at 22:32
• @JakeRose what is "the dispersion relationship" you are talking about derivations of, exactly? I have certainly seen occasional cases where there was some sort of wave speed $c$ such that $v_p~v_g=c^2$ but it would be dimensionally not a velocity if that exponent weren't there... and solving for constant $c$ would imply $\omega~d\omega = c^2~k~dk$ so $\omega = \sqrt{c^2~k^2 + C}$ for some $C$, which is not a universal relationship (it is tantalizingly satisfied in relativity though, with $k\propto p$, if $\omega\propto E$). – CR Drost Nov 12 '18 at 22:41

## 2 Answers

Phase velocity and group velocity are determined by the dispersion relation, which is typically the angular frequency as a function of wavenumber: $$\omega(k)$$. The dispersion relation can be measured and used to infer properties of the wave and its medium; this is a very common thing to do in Condensed Matter physics. A wonderfully complex example of this is explored in this question: What information can we extract from the electronic band structure?

Because they don't teach this in chemistry or basic physics courses usually, it's likely that you didn't even know that band structure IS a dispersion relation. It is the dispersion relation for electrons in a crystal of that material. Except the dispersion relation is then usually written equivalently as $$E(p)$$. ($$E=\hbar\omega$$, $$p=\hbar k$$ for planes wave solutions of the Schrodinger Equation).

Based on your question, it seems to me that you don't know where the group and phase velocity come from, so I will illustrate it here. Phase velocity and group velocity are defined as $$v_{phase}=\frac{\omega}{k}$$ and $$v_{group}=\frac{\partial \omega}{\partial k}$$. To calculate these, we need $$\omega(k)$$, the dispersion relation. We can either measure this through experiment as I already mentioned, or if we have a wave equation, we can calculate it by plugging in a plane wave $$e^{i(kx-\omega t)}$$. This is one way a band structure diagram, like the first link in my post, can be read. The slope of the curve tells you have fast wave packets move that have different mean wavenumbers (since wavepackets have a range of wavenumbers, and they aren't plane waves).

I'll do a couple examples to show you how this goes. First, "the" wave equation: $$(\partial_{tt} - c^2 \partial_{xx})\psi = 0$$. If we plug in the plane wave, we get $$(\partial_{tt} - c^2 \partial_{xx})e^{i(kx-\omega t)} = (-\omega^2 + c^2k^2)e^{i(kx-\omega t)} = 0 \implies \omega^2 = c^2k^2$$. This is our dispersion relation. Now we can get the phase and group velocities: $$v_{phase} = \frac{\omega}{k} = \pm c$$ and $$v_{group} = \frac{\partial \omega}{\partial k} = \pm c$$. So in this particular case, the phase and group velocity are the same, and they are $$c$$. Really, $$c$$ is written in the original equation the way it is because of this fact. In general, these velocities depend on the wavenumber.

With a waveguide, the dispersion relation given in Quantum Physics (Gasiorowicz, 3rd edition) Problem 2.2 is $$\lambda = \frac{c}{\sqrt{\nu^2 - \nu_0^2}}$$. Substituting $$\lambda = \frac{2\pi}{k}, \nu = \frac{\omega}{2\pi}$$, we get $$v_p = \pm \sqrt{c^2+\frac{\omega_0^2}{k^2}}$$. As you said, the phase velocity has a singularity. The first reason this doesn't disagree with special relativity is that special relativity puts a limit on how fast wave packets can travel (the group velocity), not the phase velocity, which is $$v_g = \frac{c^2}{\sqrt{c^2+\frac{\omega_0^2}{k^2}}}$$, and doesn't ever go above $$c$$.

The second reason why it can be ok for a wave to travel faster than light is that the equation describing it is not relativistic, so of course it can. The Schroedinger equation is one such wave equation, so I will work out the velocities for it as an example. I will do the free particle in 1D for simplicity's sake, but the result is still generally relevant: $$i\hbar \partial_t e^{i(kx-\omega t)} = -\frac{\hbar^2}{2m}\partial_{xx}e^{i(kx-\omega t)} \implies \hbar\omega = \frac{\hbar^2}{2m}k^2 \implies v_p = \frac{\hbar k}{2m}, v_g = \frac{\hbar k}{m}$$. Note that k can be anything, so the velocity is unlimited, which is NOT ok in SR! Also note that the plane wave (a momentum eigenstate) has a definite momentum $$p=\hbar k$$. So $$v_g = \frac{p}{m}$$ which is the velocity that a wave packet will move at, and the velocity we expect. This is what a free particle wave packet looks like: https://giphy.com/gifs/dimension-coil-packet-ddYwgktdZxTA4 Note that the phase velocity is half the group velocity.

Ok, so what if we do want to be relativistic? For particles with spin, this gets very complicated, but for spinless particles, we can use the Klein-Gordon equation (from $$E^2 - (pc)^2 = (mc^2)^2$$): $$(-\hbar^2\partial_{tt} + \hbar^2c^2\partial_{xx} - (mc^2)^2)e^{i(kx-\omega t)} = 0 \implies \hbar^2\omega^2 - \hbar^2c^2k^2 - (mc^2)^2 = 0 \implies \omega^2 = k^2c^2 + \frac{m^2c^4}{\hbar^2} \implies v_p = \pm c\sqrt{1 + (\frac{mc}{\hbar k})^2}, v_g = \frac{kc^2}{\sqrt{k^2c^2 + \frac{m^2c^4}{\hbar^2}}} = \frac{c}{\sqrt{1+(\frac{mc}{\hbar k})^2}}$$. Clearly, the phase velocity is bonkers. It is ALWAYS greater than $$c$$. Also, it's nothing at all like the phase velocity in the Schroedinger equation when our momentum is small. However, the group velocity is always less than c, and has the same behavior as the waveguide from earlier! The reason for this correspondence is that the waveguide has symmetry that makes the problem one dimensional, and $$\omega_0 = \frac{mc^2}{\hbar}$$ depends on the mass of the particle. We can also see that $$\beta = \frac{v_g}{c} \implies \hbar k= \frac{v_g m }{\sqrt{1-\beta^2}} = \gamma v_g m$$, which is the relativistic momentum, and thus exactly what we expect. This is what a wavepacket solution to this would look like (color is phase, brightness is amplitude): https://en.wikipedia.org/wiki/Dispersion_relation#/media/File:DeBroglie3.gif What is most intriguing to me here is that the phase velocity is higher when the momentum is lower.

Another point about phase velocity: in quantum mechanics, the waves really are complex valued. A plane wave has uniform amplitude everywhere, which means it is equally likely to find the particle anywhere. The phase doesn't matter. So we should reasonably expect that the phase shouldn't affect observables, but clearly in the Schrodinger equation this is not true, since $$\hat{p}=-i\hbar \nabla$$ obviously changes if the phase is changed locally. To fix this, we instead use the Pauli-Schrodinger equation: https://www.youtube.com/watch?v=V5kgruUjVBs&t=625s So when phase is treated correctly, it doesn't affect any observable. That's why SR doesn't care about phase velocity.

I hope this helps, and I'd be happy to make any edits to further clarify things or straighten out any mistakes I may have made. Cheers!

Edit in response to comment: Phase velocity tells you how fast a particular phase travels. In a plane wave the phase is $$e^{i(kx - \omega t)}$$. So if we set the exponent to $$0$$, which gives a phase of $$1$$, then $$x = \frac{\omega}{k}t$$. That is what defines phase velocity. Deriving group velocity is a bit more slippery though, because it's how fast the packet or group moves, not how fast each peak moves (the phase velocity). Honestly, beyond this the Wikipedia page is pretty good: https://en.wikipedia.org/wiki/Phase_velocity.

As far as the group velocity indicating "how fast energy travels", look at wave packets like these: https://gfycat.com/gifs/detail/happyleafybighornsheep and http://www.math.uwaterloo.ca/~kglamb/course_animations/sech_packet_small.gif It should be evident that "where" the wave is and how fast the packet is moving is determined by the group velocity. Say you're considering a particle bouncing off a wall, it doesn't matter if the phase velocity is much higher than the group velocity, it won't bounce off the wall (i.e. transfer momentum to the wall) until it gets there, and that's determined by the group velocity. In addition, for quantum wavefunctions we can only ever directly observe local phase shifts through the Aharanov-Bohm effect, but the whole wavefunction's phase could change by the same amount and you'd have no way to measure this. You can't measure phase in other words, the only way it matters is with interference effects, which only depend on relative phase anyway. Please let me know if this helps clear it up.

• I love this answer, very well thought out thank you! I’m still confused about what the notion of phase velocity is though. If you had to say in a couple of sentences how would you describe it? And one more question, why does energy only travel at the group velocity and not the phase velocity? – Jake Rose Nov 22 '18 at 13:04
• I had another thought on how to explain this. A nice way to think about this is to consider a wave packet, like: desmos.com/calculator/r6apvpnvg8 . The wave has a slowly varying "envelope", and a quickly varying "carrier". If you pick a spot where the wave is rising, the phase velocity tells you how fast and which way that rising part will move. The group velocity tells you how fast and which way the envelope moves (roughly). – Liam Clink Jan 25 '20 at 14:45

https://g.redditmedia.com/D3s3fxYWZ1289gdavyhk0yu_ZIhkNTBExtNapKvzDYM.gif?fm=mp4&mp4-fragmented=false&s=af9c9fd7b6665056bc0c13c494142395 this gif above will give you a lot of intuition about what phase and group velocities are.

• This is a helpful comment, but not an answer, no? – Cosmas Zachos Nov 12 '18 at 23:12