# Confusion in de Broglie wave and wave function

Recently I saw some questions related to the same topic and got to know that matter-wave functions(Ψ) are superpositions of multiple de Broglie waves corresponding to multiple momenta, which is governed by the Heisenberg uncertainty principle. If it is so, then how is the group velocity of de Broglie waves defined?

And while calculating group or phase velocities, if we consider those wave packets to be the superposition of de Broglie waves(i.e. Ψ itself), the values of phase and group velocities should match for both.

But referring to Concepts of Modern Physics by Arthur Beiser, we equate energy of de Broglie wave with total energy(kinetic+realtivistic) as

$$h\nu=\gamma mc^2$$ $$\nu=\frac{\gamma mc^2}h$$

so $$v_{p}=\frac{\omega}k=\nu \lambda$$ $$v_{p}=\nu\frac{h}p=\frac{\gamma mc^2}h \cdot \frac{h}{\gamma mv}$$ $$v_{p}=\frac{c^2}v$$

and group velocity is same as particle velocity so $$v_{g}=v$$

so we get the relation $$v_{p} \cdot v_{g}=c^2$$

And all this does not match with that of wave function packet, for which we know relation $$v_{p}=\frac{v_{g}}2$$.

So which relation is true?

Also, the relation $$v_{p}=\frac{v_{g}}2$$ for wave function is also a little doubtful, as $$v_{p}$$ is defined for individual waves and $$v_{g}$$ is defined for wave packet that is formed by combining all those waves. $$v_{p_{i}} = \frac{p_{i}}{2m}$$ can be written for individual waves which are momentum-pure states (value of momentum is exactly equal to $$p_{i}$$ in this one). and $$v_{g}=\frac{

}{m}$$

, where $$

$$

is the expected(average) momentum obtained by wave packet. So I think $$v_{p}=\frac{v_{g}}2$$ cannot be directly stated...

• I thought that $E=h\nu$ is only valid for photons. Commented Apr 28 at 10:44
• @Gorga The duality between matter waves and matter is very similar to that of EM waves and photons. So, I think we can apply the formula here as well. Commented Apr 28 at 11:58
• These relations are way too crude. We have the correct quantum wave equations, from which you can derive the correct relations. Commented Apr 28 at 12:21

This is because non-relativistic quantum mechanics (Schrödinger equation) uses for the matter wave the classical energy-momentum relation $$E=mv^2/2=p^2/(2m)$$ and the relativistic de Broglie wave uses the relativistic energy-momentum relation $$E^2 = (mc^2)^2+ (pc)^2$$ By inserting $$E=\hbar \omega$$ and momentum $$p=\hbar k$$ into both equations, one obtains different dispersion relations $$\omega(k)$$ for the non-relativistic and relativistic case, even in the low velocity limit, because in the relativistic equation the relativistic energy of the mass (first term RHS of second equation) is retained. In general, this gives different group $$v_g=d \omega/dk$$ and phase $$v_p=\omega/k$$ velocities for the non-relativistic and relativistic de Broglie wave. In the low velocity limit, the group velocities become the same, but the phase velocity of the non-relativistic equation is $$v_p=v_g/2$$ and of the relativistic equation is $$v_p=c^2/v_g$$.
• @Nihal Popat In principle both ways are correct. What really matters, is the group velocity, which corresponds to the particle velocity. It is the same for both the classical and the relativistic energy relation in the limit of small velocities. In this limit, the rest mass energy can be disregarded because it only adds a constant additional frequency, which doesn't change the group velocity. The phase velocity has no physical significance. Also the wave vector k is given in both cases by $p=\hbar k$. Note: Both the group and the phase velocity are defined at a given k. Commented Apr 28 at 17:10