# Calculating Relativistic Phase and Group Velocity

• I
In Quantum Mechanics Concepts and Applications by Zettili the following formulas are used
for phase and group velocities.

${\rm{ }}{v_{ph}} = \frac{w}{k} = \frac{{E\left( p \right)}}{{p}}{\rm{ }}\\ {\rm{ }}{v_g} = \frac{{dw}}{{dk}}{\rm{ = }}\frac{{dE\left( p \right)}}{{dp}}{\rm{ }}$

In the low velocity classical case for a free particle, E means Kinetic Energy. In the
relativistic case, E means Kinetic plus Rest Energy, according to the book. I suspect
that we should use Kinetic Energy on both cases. You seem to get the same results when
it matters whether you use Kinetic or Kinetic plus Rest in the relativistic case, so it
makes more sense to be consistent and use the Kinetic in both cases.

Is there a real reason why you have to use Kinetic Energy classically but Kinetic plus Rest
relativisticly?

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vanhees71
Gold Member
2019 Award
Yes, a constant contribution to the energy is unobservable. To include the mass contribution to relativistic energy is a good choice, because then the energy is the time component of the energy-momentum four-vector.

The problem I have with this answer is that it violates the general principle that any relativistic formula has to approach the classical formula as velocity becomes small.
If under relativistic QM you have to use Kinetic plus Rest Energy, then as velocity approaches zero you are left with Rest Energy. If that is true then the classical formula for the Phase Velocity of a particle with velocity=0 should be Rest Energy/Momentum, which equals infinity.

My point is this: There is a lot of weird crazy stuff in QM that doesnt make any sense and you just have to accept that that is the way the world works. I am trying to figure out whether this is one of those crazy things that doesnt make any sense and I have to deal with it, or did the author of the book got a little careless, not realizing
how confusing it would be to a person learning this for the first time. Perhaps it would help if I restated the question.

Is there some principle in classical QM that states that you have to use Kinetic Energy to calculate Group and Phase Velocities (if you use Kinetic Energy plus Rest Energy you sometimes get wrong answers), and another principle in relativistic QM that states you have to use Kinetic plus Rest Energy (if you use just Kinetic Energy you sometimes get wrong answers)? That is just the way the world works and I have to deal with it, and relativistic QM does not approach classical QM at low velocities in this case.

PeterDonis
Mentor
2019 Award
the general principle that any relativistic formula has to approach the classical formula as velocity becomes small
Where are you getting this "general principle" from?

If under relativistic QM you have to use Kinetic plus Rest Energy, then as velocity approaches zero you are left with Rest Energy. If that is true then the classical formula for the Phase Velocity of a particle with velocity=0 should be Rest Energy/Momentum, which equals infinity.
No, it's undefined, because you can't divide by zero. But that's also true if you use kinetic energy, since then you have 0/0, which is also undefined. So you have to fix the issue (one way would be to just say that the concept of "phase velocity" is undefined for the case of zero momentum; another would be to just define phase velocity to equal zero for zero momentum) either way.

1. I agree that I should have said undefined rather than infinite.
2. In the relativistic case, at first glance you seem to get 0/0 if you just use Kinetic Energy. But if you take the limit as momentum approaches 0 and apply L'Hopital's Theorem, you get 0.
3. None of this addresses my question.
4. If you are disputing that a relativistic theory has to approach the classical theory in the low velocity limit, then I dont think this exchange is going to be productive.

PeterDonis
Mentor
2019 Award
None of this addresses my question.
Note, first, that the issue you describe only arises for phase velocity. For group velocity, since you are taking the derivative of energy, any constant term, such as the rest energy, drops out. So you get the same answer in the relativistic case as in the classical case for group velocity, even if you use total energy instead of just kinetic energy.

Second, as @vanhees71 pointed out, a constant offset to energy is unobservable. That means that rest energy itself is unobservable as long as it's constant--which means, as long as the particle we're talking about isn't undergoing any reactions that create or destroy particles and change rest energies around. So in this restricted case, where we just have a free particle moving through space, rest energy can be ignored even in relativity. Another way of putting the restriction would be that we are assuming that the rest energies of all particles are so much larger than the kinetic energies that there is no significant probability of any interactions that would create or destroy particles. (Rest energy >> kinetic energy is another way, in a relativistic context, of describing what you are calling the "low velocity limit".)

Third, consider what happens if we remove the restriction I just described, and consider the possibility of interactions that can create and destroy particles. Now you cannot ignore rest energy, because it can turn into kinetic energy, and vice versa--for example in particle-antiparticle pair production and annihilation. This is why, in the relativistic case in general, you have to include rest energy in the energy. In such cases, of course, the concepts of "phase velocity" and "group velocity" don't work the way they do in the simple case of a free particle (in fact it's not clear to me that they're useful at all).

If you are disputing that a relativistic theory has to approach the classical theory in the low velocity limit, then I dont think this exchange is going to be productive.
In the relativistic case "energy" includes rest energy, and in the classical case, it doesn't, so the relativistic energy does not approach the classical energy in the low velocity limit. Sometimes you can get away with ignoring this and using kinetic energy. Sometimes you can't. So there can't possibly a "general principle" that says the relativistic theory has to approach the classical theory in all respects in the low velocity limit.

vanhees71
Gold Member
2019 Award
To clarify this issue, we can start from the relativistic formula for the energy, including the mass energy,
$$E=\sqrt{m^2 c^4+p^2 c^2}=m c^2 \sqrt{1+\frac{p^2}{m^2 c^2}}.$$
You get the non-relativistic limit if ##p \ll m c##. Then you can expand the square root,
$$E=m c^2 \left (1+\frac{p^2}{m^2 c^2} + \mathcal{O}(p^4/m^4 c^4) \right).$$
I.e., neglecting the higher orders you get
$$E=m c^2 + \frac{p^2}{2m},$$
which is completely agreeing with the kinetic energy in Newtonian physics up to an (irrelevant!) additive constant ##E_0=m c^2##.

I am well aware of everything everyone is telling me but it doesnt lead me to believe that my way of looking it it is wrong. Mostly I think it was a poor exercise that was never meant to be taken this seriously and that the phase velocity issue would never come up outside of a student exercise. One final point might settle the issue. In the classical case for a free particle you have

${\rm{ }}{v_{ph}} = \frac{w}{k} = \frac{{E\left( p \right)}}{{p}}{\rm{ }}\\ {\rm{ }}{v_g} = \frac{{dw}}{{dk}}{\rm{ = }}\frac{{dE\left( p \right)}}{{dp}}{\rm{ }}$

where E is Kinetic Energy.

In Relativistic QM for a free particle, do the above formulas involving w and k apply? Do you use Kinetic or Kinetic plus Rest Energy when calculating w and k?

PeterDonis
Mentor
2019 Award
In Relativistic QM for a free particle, do the above formulas involving w and k apply? Do you use Kinetic or Kinetic plus Rest Energy when calculating w and k?
You already know how to get the right answer, to the extent that it matters--as you yourself say, the issue wouldn't come up outside of a student exercise, so the conclusion that "it was a poor exercise, don't worry about it" seems obvious. (And as I've already pointed out, the only case where it makes a difference anyway is phase velocity.) I'm not sure what else you need to know.

vanhees71
Gold Member
2019 Award
I am well aware of everything everyone is telling me but it doesnt lead me to believe that my way of looking it it is wrong. Mostly I think it was a poor exercise that was never meant to be taken this seriously and that the phase velocity issue would never come up outside of a student exercise. One final point might settle the issue. In the classical case for a free particle you have

${\rm{ }}{v_{ph}} = \frac{w}{k} = \frac{{E\left( p \right)}}{{p}}{\rm{ }}\\ {\rm{ }}{v_g} = \frac{{dw}}{{dk}}{\rm{ = }}\frac{{dE\left( p \right)}}{{dp}}{\rm{ }}$

where E is Kinetic Energy.

In Relativistic QM for a free particle, do the above formulas involving w and k apply? Do you use Kinetic or Kinetic plus Rest Energy when calculating w and k?
I don't understand this recent hype about this non-issue either. There were some lengthy superfluous papers in EJP recently about it too. They are more confusing than helpful.

In relativistic physics one should use the covariant expressions, i.e., the four-vector with energy (including rest energy) as time and three-momentum as spatial components, because it's the natural way to write it.

Whether you include rest energy in the non-relavistic case or not is completely irrelevant since quantum states do not dependent on any additive constant to the energy. For the time evolution of state vectors it contributes only an overall phase factor, common to all states, and thus the corresponding state itself (which is represented by a unit ray in Hilbert space rather than a Hilbert-space vector) is independent of it.

I guess I am just having hard time getting used to the idea that phase velocity in QM is real enough that the author of the book prescribes precisely how to calculate it in the classical and relativistic cases, but it not real enough that the one has to be the low velocity limit of the other. That is a new concept for me. I will just have to deal with it. Thank for your help.