Question

Dispersion relation for relativistic particle is
E = vp2c2 + m 2c4
The ratio of phase velocity Vo to group
velocity V, is:​

Answer 1

Answer:

Phase and Group Velocity of Matter Waves—C.E. Mungan, Spring 2017

Consider a beam of particles traveling in free space in the same direction with nonrelativistic

speed υ. Find their quantum-mechanical phase velocity υp and group velocity υg.

The phase speed of a wave is υp = ω / k where ω = 2π f is the angular frequency and

k = 2π / λ is the angular wavenumber (with f and λ the frequency and wavelength). Assume the

energy E and linear momentum p of the particles are given by the Einstein and de Broglie

relations as E = !ω = hf and p = !k = h / λ in terms of Planck’s constant. (If you wish, you

could take these as definitions of the frequency and wavelength of the matter wave associated

with the beam of particles.) In the nonrelativistic limit such that

γ ≡ 1

1−υ2 / c2 ≈1 (1)

to lowest order, then p = γ mυ ≈ mυ as expected classically where m is the (rest) mass of the

particles. So far we have

υp ≈ E

mυ

(2)

but there is now a choice of two ways to proceed.

Classically E is the sum of the kinetic and potential energies. In free space there is no

potential energy and thus

υp ≈ K

mυ =

1

2 mυ2

mυ = 1

2υ (3)

which is a strange result, in that there is nothing physically traveling at half the speed of the

particles. Nevertheless Griffiths1 and Mundarain2 argue that Eq. (3) is correct because it accords

with the Schrödinger equation i!∂ψ / ∂t = −(!2 / 2m)∂2

ψ / ∂x2 for a plane wave

ψ (x,t) = Aexp(ikx − iωt).

Another approach is to start from special relativity to get E = γ mc2 ≈ mc2 in which case

υp ≈ mc2

mυ = c

υ

c (4)

which is also strange in that it is superluminal. This answer is valid even for relativistic particles

because

υp = E

p = γ mc2

γ mυ

(