Question 1.10.12: Calculate the group and phase velocities for the wave packet...

Recall that the energy and momentum of a relativistic particle are given by

E=mc^2=\frac{m_0c^2}{\sqrt{1-\upsilon ^2/c^2} },       p=m\upsilon =\frac{m_0\upsilon }{\sqrt{1-\upsilon ^2/c^2} } ,            (1.206)

where m_0  is the rest mass of the particle and c is the speed of light in a vacuum. Squaring and adding the expressions of E and p, we obtain E^2=p^2c^2+m^2_0c^4;  hence

E=c\sqrt{p^2+m^2_0c^2}.              (1.207)

Using this relation along with p^2+m^2_0c^2=m^2_0c^2/(1-\upsilon ^2/c^2)   and (1.122) \upsilon _g =\frac{dE(p)}{dp} ,\upsilon _{ph}=\frac{E(p)}{p}, ,  we can show that the group velocity is given as follows:

\upsilon _g =\frac{dE}{dp} =\frac{d}{dp}\left(c\sqrt{p^2+m^2_0c^2} \right)=\frac{pc}{\sqrt{p^2+m^2_0c^2} } =\upsilon .                 (1.208)

The group velocity is thus equal to the speed of the particle, \upsilon _g=\upsilon .

The phase velocity can be found from (1.122) and (1.207): \upsilon _{ph}=E/p=c\sqrt{1+m^2_0c^2/p^2}   which, when combined with p=m_0\upsilon /\sqrt{1-\upsilon ^2/c^2},   leads to \sqrt{1+m^2_0c^2/p^2 } =c/\upsilon ;   hence

\upsilon _{ph}=\frac{E}{p}=c\sqrt{1+\frac{m^2_0c^2}{p^2} }=\frac{c^2}{\upsilon }.                (1.209)

This shows that the phase velocity of the wave corresponding to a relativistic particle with m_0\neq 0   is larger than the speed of light, \upsilon _{ph}=c^2/\upsilon \gt c.   This is indeed unphysical. The result \upsilon _{ph} \gt c   seems to violate the special theory of relativity, which states that the speed of material particles cannot exceed c. In fact, this principle is not violated because \upsilon _{ph}   does not represent the velocity of the particle; the velocity of the particle is represented by the group velocity (1.208). As a result, the phase speed of a relativistic particle has no meaningful physical significance.

Finally, the product of the group and phase velocities is equal to c^2,i.e.,\upsilon _g\upsilon _{ph}=c^2.