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Class 17: The Uncertainty Principle

Wave packets and uncertainty A natural question to ask is ‘if particles behave like , what is waving?’ The varying quantity is related to the probability of finding the particle at a particular location. A of a well-defined single will extend everywhere and is not localized. To localize a particle the wave must have a finite extent. This leads to the concept of a , which is shown schematically below.

Δx

The width of the wave packet, Δx, is a measure of the uncertainty in the position of the particle. As the wave packet moves, the wave inside also moves and not necessarily at the same speed as the envelope of the wave packet. Hence the number of maxima in length Δx will change. Suppose we try to determine the wavelength associated with the wave packet by counting the maxima. We find

∆x λ = . (17.1) n

The uncertainty in the wavelength from our counting is

∆x ∆λ = ∆ n. (17.2) n2

Typically ∆n ≈ 1. Using equation (17.1) to eliminate n, we get

∆λ ∆x ≈ λ 2. (17.3)

The de Broglie relations relate momentum to wavelength. Hence an uncertainty in the wavelength gives an uncertainty in the momentum

h h ∆=p ∆≈λ . (17.4) λ 2 ∆x

This can be written as

∆p ∆ x ≈ h . (17.5)

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By considering the relationship between energy and frequency, the uncertainty in energy of a wave packet localized to a time interval Δt is such that

∆E ∆ t ≈ h . (17.6)

Relations like that in equations (17.5) and (17.6) were first encountered by Werner Heisenberg in 1927, and hence are referred to as the Heisenberg uncertainty principle. From a more precise analysis, the uncertainty principle is

1 ∆p ∆ x ≥ ℏ, (17.7) 2 and

1 ∆E ∆ t ≥ ℏ. (17.8) 2

Consequences of the uncertainty principle Consider a particle of mass m confined in a 1-D box of length L. Since ∆x ≤ L , the momentum of the particle cannot be exactly zero. Hence the particle has a minimum kinetic energy

p2ℏ 2 K = ≥ . (17.9) 2m 8 mL 2

The mechanical energy of a harmonic oscillator of mass m and spring constant k is

1p2 1 E= + kx 2. (17.10) 2m 2

Heisenberg’s uncertainty principle sets a non-zero lower limit to the mechanical energy which can be found by making the change of variables

14− 14 p=() km qx, = () km y , (17.11)

so that

1 k  1 2 E=  () q2 + y 2 . (17.12) 2 m 

The constraint from the uncertainty principle is ℏ ∆q ∆ y ≥ . (17.13) 2

Assuming that p and x are normally distributed, we have

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1k12 1  k 12 E=() qy22 +=  () ∆+∆ qy 22 , (17.14) 2m 2  m

because the mean values of p and x are both zero. From equation (17.14), we see that a point (∆q, ∆ y ) of energy E lies on a circle. The constraint (17.13) requires the point to lie in a region bounded by a hyperbola. The minimum energy occurs when the circle just touches the hyperbola. It does this on the line ∆q = ∆ y . Hence the minimum allowed energy is

1 2 k  1ℏ 1 ℏ Emin =  = ω, (17.15) m  2 2

where ω is the angular frequency of the oscillator.

These examples show that a particle confined by a potential has a minimum zero-point energy . Some consequences of this zero-point energy are that a crystalline solid will melt if compressed to high density even at zero temperature, and cold fusion could occur if the density was sufficiently large that the zero point energy is high enough for penetration of the Coulomb barrier to occur.

More detailed treatment of wave packets A harmonic wave in 1D can be represented by the functional form

y= Re Ae i( kx−ω t ) , (17.16)

where k and ω take definite values. Clearly this wave function extends everywhere. Since a wave packet is to represent a localized particle, it cannot have a single wavelength. A general expression to describe a wave packet is

∞ i kx−ω() k t  y= Re∫ A() k e  dk . (17.17) −∞

The relationship between ω and k is called the relation. For example for waves on a string the is ω = kc , where c is the speed of propagation of the wave.

Suppose that the amplitude function A(k) is uniform over a narrow range of wave numbers (k0, k 0 + ∆ k ) and zero outside that range. Then

k0 +∆ k  i kx−ω() k t   y= Re Ae∫   dk  . (17.18) k0 

Let k= k 0 + κ, so that

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∆k  ikx0+κ x − ω() k 0 + κ t  y= Re Ae∫   d κ  . (17.19) 0 

Because ∆k is small, we can expand ω(k0 + κ ) in a Taylor series and discard second and higher order terms. We get

∆k  ikx0+κω x −()() kt 0 − ω′ k 0 κ t  yAe= Re ∫   d κ  0  (17.20) ∆k  ikx()()0−ω 0 t ix κ − ω 0′ t = ReAe∫ e d κ  , 0  where ω0= ω (k 0 ), and ω0′= ω ′ (k 0 ).

ω0  ikx()0−ω 0 t We see that the wave packet consists of a harmonic wave function fx− t  = Ae multiplied k0  ∆k iκ() x− ω 0′ t by an envelope function Fx()−ω0′ t = ∫ e d κ . 0

The harmonic wave propagates at speed

ω0 vp = . (17.21) k0

This is called the of the wave packet.

The envelope propagates at a speed

dω vg = () k 0 . (17.22) dk

This is called the of the wave packet.

For waves on a string, since ω = kc , we see that the phase velocity is equal to the group velocity.

Phase and group velocity for non-relativistic and relativistic free particles Assume that the appropriate expression to use for the energy in the de Broglie relation is the mechanical energy. For a free particle of mass m,

p2 E= K = . (17.23) 2m

Using the de Broglie relations, this becomes

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()ℏk 2 ℏω = . (17.24) 2m

The dispersion relation for the waves associated with a free particle is ℏ ω = k 2. (17.25) 2m

We see that the phase and group velocities are

ω ℏk p vp = = = , k2 m 2 m (17.26) dω ℏ k p vg = = = , dk m m

and that it is the group velocity that gives the classical relationship between momentum and velocity. This is consistent with the idea that it is the envelope of the wave packet that localizes the particle.

Next assume that the appropriate energy to use in the de Broglie relations is the total relativistic energy. For a free particle of mass m, we have

2 2 E2=() cp + ( mc 2 ) . (17.27)

The de Broglie relations give

2 2 2 ()()ℏω =c ℏ k + ( mc 2 ) . (17.28)

The dispersion relation is

2 mc 2  ω 2=k 2 c 2 +   . (17.29) ℏ 

The group velocity is

2 ℏ dω kd ω k2 k 2 cp vg == == ccc = . (17.30) dkω dk2 ωℏ ω E

Again, we see that it is the group velocity that is equal to the velocity of the particle.

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