Modern Physics: PHY-104 Spring Semester 2013

Recitation on Matter Waves Solution

1. An atom is moving in vacuum with a slow speed v ≪ c. A Gaussian wave packet pilots the electron as shown below.

x1 >x 2>x 3

v v v x 0 xx1 x2 x3 At some At some even later time later time

Show that the group velocity vg is equal to the physical speed of the particle. Show that the phase velocity is NOT the speed of the particle. Use the Einstein’s relationship E = ~ω and the classical energy E = p2/2m which is relevant for a slow particle. Answer 1: The group velocity is given by, ( )( ) dω dω dE v = = . (1) g dk dE dk According to Einstein’s energy equation,

E = ~ω. (2)

Classically, the energy of a slow moving particle is given by, 1 p2 E = mv2 = , (3) 2 2m where p = momentum of the particle = h/λ = (h/2π)(2π/λ) = ~k. Slow means v ≪ c, implying that we can use the non-relativistic expressions. Therefore equation (3) becomes, ~2k2 E = . (4) 2m From equation (2), dE = ~ dω dω 1 ⇒ = , (5) dE ~

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while from equation (4) we obtain, dE 2~2k ~2k = = . (6) dk 2m m Putting equations (5) and (6) into equation (1), 1 ~2k ~k p v = = = = v , as required. g ~ m m m particle Whereas the phase velocity is given by, ω E ~ E p2/2m p v v ======particle ≠ v . p k ~ p p p 2m 2 particle

This clearly shows that for the atom, vparticle is exactly reproduced by the group velocity of the associated matter wave, whereas the phase velocity is half the particle velocity. √ 2 2 2 2. Now use the relativistic expression for energy E = (m0c ) + (pc) to calculate the phase velocity and group velocity. Answer 2:

For convenience, let’s denote vparticle = v. The energy and momentum of a relativistic particle (in this case, electron) are given by, m c2 E = m c2 = √ o · (7) − v2 1 c2 m v p = m v = √ o · − v2 1 c2

The energy in terms of the relativistic momentum p and the rest mass mo can be obtained from the expression we have now encountered several times,

2 2 2 2 4 E = p c + mo c √ 2 2 2 E = c p + mo c . (8)

The corpuscular features (energy and momentum) of an electron are connected to its wave characteristics (wave frequency and number ) by the relations,

E = ~ ω and p = ~ k.

Therefore the group and phase velocities will become, dω dE v = = , and g dk dp E E v = = · ph p p

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2 2 2 2 2 mo c From Equation (7) and (8), we find that p + m c = 2 . So, the phase velocity is o (1− v ) c2 ( ) dE d √ v = = c p2 + m c2 g d p dp o p c = √ p2 + m2 c2 √o − v2 mo v c/ 1 c2 = E c m v c2 = √o E 1 − v2 c2 √ 2 2 1 − v √mo v c × c2 · = 2 − v2 mo c 1 c2

Hence vg = v.

This shows that the speed of relativistic particle is equal to its group velocity. Similarly, the phase velocity of relativistic particle can be calculated as

E vph = p√ c p2 + m2 c2 = o √ p m2 c2 = c 1 + o p2 √ m2 c2 v2 = c 1 + o × (1 − ) m2 v2 c2 √ o c2 v2 = c 1 + (1 − ) √ v2 c2 c2 = c . ( v)2 c Hence v = c . ph v

As c > v, this means vph > c, predicting that the phase velocity for the relativistic particle is greater than the speed of light c. This appears to be a violation of the postulates of special theory of relativity. Actually, the phase velocity does not represent the physical velocity of the particle, rather it is the group velocity which represents the

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speed of propagation of the particle. Hence, the result that vg, and not vp, represents particle speeds, holds both in the relativistic and non-relativistic scenarios. √ 2 2 2 2 3. Qualitatively discuss when can we use E = (m0c ) + (pc) , E = pc, E = p /2m or p2 E = 2m + V , where V is the potential energy. Answer 3: This question will be discussed by the instructor.

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