For a free electron gas the magnitude of phase velocitt and group velocity are such that
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The wave variables are the angular frequency ω and the wavenumber k. Both are related to the frequency ν and the wavelength λ as follows:
ω = 2πν, k = 2π/λ. (0)
If we now state the “Planck- de Broglie hypothesis”, which associates to each quantum particle an angular frequency, the relation of the latter to the total energy E of the particle is
E = ħω. (1)
On the other hand, the momentum p of this particle is
p = ħk. (2)
The wave associated to the above particle has a “phase velocity”
vf ≡ λ/T = ω/k, (3)
where T is the period of the corresponding oscillation. Substituting eqs.(1-2) into eq.(3), we obtain
vf = E/p. (4)
The wave also has a “group velocity”, defined as
vg ≡ dω/dk. (5)
The essence of the wave-particle duality is about relating the speed v of the particle with the phase and group velocities of the associated wave. At least for vg , we see that we cannot do this unless we know the functional relation between ω and k (called “dispersion relation”). This relation is obtained from the dynamical expression of the energy E; to this we now turn.
A particle’s particlelike properties are its energy E and its momentum p. We will consider two cases.
1) In general, we may take the particle to be a relativistic one, and write
E = mc2/√[1 –(v/c)2] , (6 a)
p = mv/√[1 – (v/c)2]. (6b)
As a consequence of Eqs.(6 a-b), we have
E2 = (cp)2 + (mc2)2 . (7)
2) As a special case when β ≡ v/c < < 1, we may write
E = mc2 + [p2/(2m)] ≈ mc2 , (8 a)
p = mv. (8b)
Having thus separated---as far as possible---particlelike from wavelike properties, we proceed to combine them.
1) Relativistic ( R) particle.
In order to compute the phase velocity from Eq.(4), let us divide Eq.(7) by p2 ; employing Eq.(6b), we find
vf 2 = (E/p)2 = (c2/v)2 ,
from which
vf = c2/v > c. (9)
Substituting Eqs.(1-2) into Eq.(7), we have
(ħω)2 = (ħck)2 + (mc2)2 . (10)
Eq.(10) is the dispersion relation for the “wave-particle”. Differentiating Eq.(10) with respect to k, we obtain
vg = c2k/ω = c2/vf = v. (11)
From Eq.(11) we have
vfvg = c2 ; (11’)
that is, c is the geometric mean of vf and vg . Since, from Eq.(9), vf > c, we conclude that vg < c; this suggests that vg is a possible speed for the particle-wave. Indeed, Eq.(11) shows that vg is equal to the speed of the particle.
We have as a subcase that of a photon, with zero mass, whose energy is therefore
E = cp. (12)
Effecting the customary substitutions, we find the dispersion relation
ω = ck. (13)
From Eq.(13) we conclude that the phase and group velocities of a photon are equal:
vf = vg = c. (14)
2) Non-relativistic (NR) particle.
In this case, β ≡ v/c < < 1. Employing again Eq.(4), and substituting Eq.(8 a) into it---not yet effecting the last approximation---,we get (with an obvious notation for phase velocities)
vfNR = E/p = (c2/v) + [p/(2m)] = vfR + (v/2), (15)
where in the last equality we have used Eq.(8b). we see that the non-relativistic phase velocity vfNR is even greater than the relativistic one, vfR .
Substituting now Eqs.(1-2) into Eq.(8 a) without the last approximation, we obtain the dispersion relation
ħω = mc2 + [(ħk)2/(2m)]; (16)
differentiating Eq.(16) with respect to k:
vg ≡ dω/dk = ħk/m = p/m = v. (17)
Eq.(17) tells us that the group velocity coincides with the speed of the particle.
Approximating now E ≈ mc2, we find
vf = E/p = c2/v > c. (9’)
Eq.(9’) gives the same result as Eq.(9) Therefore, Eqs.(17-18) lead again to Eq.(11’).
Employing Eq.(1), and solving for ω, the dispersion is here
ω = mc2/ħ = const. (18)
Eq.(18) tells us that, in this case, a classical free particle corresponds to a monochromatic wave. Differentiating Eq.(18), we have
vg = 0. (19)
Eq.(19) looks surprising, but it only means that in this case there is no “group” or “packet” of waves.