Photons are the particles of light.  Matter is made of atoms, and atoms are made protons, neutrons and electrons.  These are not macroscopic particles.  Typical atomic dimensions are on the order of 10-10 m, nuclear dimensions are on the order of 10-15 m, and the electron seems to be a point particle with no size at all.  How do these particles behave?

If a wave equation describes the behavior of photons, maybe a wave equation also describes the behavior of other microscopic particles.

In 1924, Luis deBroglie (Nobel Prize in Physics in 1929) proposed that a wave function is associated with all particles.  Where this wave function has nonzero amplitude, we are likely to find the particle.  The standard interpretation is that the intensity of the wave function of a particle at any point is proportional to the probability of finding the particle at that point.  The wavelengths of the harmonic waves used to build the wave function let us calculate the most likely momentum of the particle and the uncertainty in the momentum.  The wave function for a material particle is often called a matter wave.

The relationship between momentum and wavelength for matter waves is given by p = h/λ, and the relationship energy and frequency is E = hf.  The wavelength λ = h/p is called the de Broglie wavelength, and the relations λ = h/p and f = E/h are called the de Broglie relations.  These are the same relations we have for the photon, but for particle E = (1/2)mv2 = p2/(2m), so E = ћ2k2/(2m), λ = h/√(2mE)The relationship between λ and E is different for particles than for photons.

A spread in wavelengths means an uncertainty in the momentum. The uncertainty principle also holds for material particles. The minimum value for the product ∆x ∆p is on the order of ħ.

∆x∆p ~ ħ.

For any particle, we cannot predict its position and momentum with absolute certainty. The product of the uncertainties is on the order of h/2π = ħ or greater.

A free particle of mass m moving with exactly determined velocity v in the positive x-direction has momentum p = mv, pointing into the positive x-direction and kinetic energy E = p2/(2m).  Its wave function, which often is denoted by ψ(x,t),  is a plane wave.

ψ(x,t) = Acos(kx - ωt + φ)

[The wave function is actually a complex function, and Acos(kx - ωt + φ) is the real part of this function.  We will learn about complex functions in the next section.]

The wave function has a well defined deBroglie wavelength λ = h/p which determines k = 2π/λ
The wave function also has a well defined
deBroglie frequency f = E/h, which determines the angular frequency ω = 2πf.

Problem:

• What is the deBroglie wavelength of an electron after being accelerated through a potential difference of 25 kV in a television set?
• Solution:
λ = h/p,  E = p2/(2m),  p = √(2mE), λ = h/√(2mE).
The energy of the electron is 25000 eV * 1.6*10-19 J/eV = 4*10-15 J.
λ = (6.626*10-34 Js)/√(2*9.1*10-31 kg*4*10-15 J) = 7.8*10-12 m.
This wavelength is approximately 100 times smaller than the typical size of an atom.

Problem:

• If you double the kinetic energy of a particle, how does the deBroglie wavelength change?
• Solution:
λ = h/p,  E = p2/(2m),  p is proportional to  √E,  l  is proportional to 1/√E.
λ21 = √(E1/E2) = 1/√2.
The deBroglie wavelength decreases by a factor of 1/√2.
λ2 = 0.707 λ1.

Problem:

• If you double the speed of a particle, how does it deBroglie wavelength change?
• Solution:
λ = h/p,  p = mv,  p is proportional to v,  λ is proportional to 1/v.
λ21 = (v1/v2) = 1/2.
The deBroglie wavelength decreases by a factor of 1/2.
λ2 = 0.5 λ1.

• An electron and a proton have the same speed.  Which has the greater deBroglie wavelength?

If a particle is described by a plane wave ψ(x,t) = Acos(kx - ωt + φ) we know the particles momentum and wavelength exactly, but its position is completely unknown.  The plane wave extends to infinity, the particle could be anywhere.  The wave function for a particle for which we have some position information is a wave packet, and the particle's momentum now has some uncertainty.

The more massive the particle, the larger is the magnitude of the momentum p of the particle when it is moving with speed v, and the smaller is the deBroglie wavelength λ.  Wave packets build from harmonic waves with smaller wavelength can be smaller in size.  The smallest wave packet we can build has a size on the order of the deBroglie wavelength λ of a free particle moving with the same speed v.

Example:

When we synthesized a square pulse with a width of 100 units using harmonic waves, the longest wavelength we used λ = 200 units.

We can know the position of more massive particles moving with approximate speed v more precisely than the position of less massive particle moving with the same speed since the more massive particles have shorter deBroglie wavelengths.

Problem:

• Assume we want to locate an object with an uncertainty on the order of 1 Å, (1 Å (Angstrom) = 10-10 m, the approximate size of an atom).
What is the uncertainty in the speed of the object if the object is
(a) an electron
(m = 9.1*10-31 kg),
(b) a proton (m =
1.67*10-27 kg),
(c) an uranium nucleus (m = 3.9*10-25 kg),
(d) a dust particle (m = 10-14 kg)?

• Solution:
∆x = 10-10 m.  ∆p = ħ/10-10 m = 10-24 kgm/s, ∆v = (10-24 kgm/s)/(mass of particle)

(a)  ∆v =  (10-24 kgm/s)/(9.1*10-31 kg) = 1.1 106 m/s
So if we want to put an electron into a box the size of an atom, the electron must be zipping around in that box, and its speed can be as large as ~1 million meters per second.  Electrons are confined in atoms.  The uncertainty principle tells us that they cannot be sitting still, but must be moving around with high speeds.
(b)  ∆v =  (10-24 kgm/s)/1.67*10-27kg) = 6*102 m/s
(b)  ∆v =  (10-24 kgm/s)/(3.9*10-25 kg) = 2.6 m/s
(d)  ∆v =  (10-24 kgm/s)/(10-14 kg) = 10-10 m/s
For a dust particle the uncertainty in the speed is approximately 1 Å/s or one atomic size/s.  It must be jittering around a bit, but there are few instruments capable of detecting this jitter.
As we increase the mass of the object further, the uncertainty in the speed of the object becomes completely negligible.
So while the uncertainty principle also holds for massive, macroscopic particles, it has no practical consequences.

Dispersion

Photons always move with the speed of light c.
Monochromatic EM plane wave are of the form E(x,t) = Emaxcos(kx - ωt + φ)Emaxcos((2π/λ)x - (2πf)t + φ).
They all have speed ω/k = 2πf/(2π/λ) = λf = c.  If we build a wave packet from plane waves of the above form, the speed of the wave packet is c.
This relationship between ω and k, ω = ck, is called the dispersion relation for photons in free space.

Particles can move with any speed.  The plane matter waves of a free particle moving with speed v has the form ψ(x,t) = Acos(kx - ωt + φ).
This wave function has a well defined wavelength
λ = h/p and a well defined frequency f = E/h.  We have p = h/λ = ħk and E = hf = ħω.
But for a free particle we also have E = mv2/2 = p2/(2m) = ħ2k2/(2m).
Equating the two expressions for E, we find
the relationship ω = ħk2/(2m). This relationship is called the dispersion relation for free particles.

For any wave λf = ω/k = vwave.  Using  the dispersion relation for a free particle we find  vwave = ω/k = ħk/(2m) = p/(2m) = mv/(2m) = v/2.

This may seem surprising.  Our classical intuition suggests that a wave describing a particle moving with speed v should move with speed v itself.  How else can it track the particle?  But remember!  If a particle is described by a plane wave ψ(x,t) = Acos(kx - ωt + φ) we know the particles momentum and wavelength exactly, but its position is completely unknown.  It is impossible to track a particle whose momentum is exactly known.

The dispersion relation for free particles implies that the plane matter waves of particles do not all have the same speed.  The wave function for a particle for which we have some position information is a wave packet.  We have to superimpose plane waves describing particles with slightly different momenta and energies, or with slightly different wavelengths and frequencies.  But these component waves now all move with slightly different speeds.  So what happens to the shape of a wave packet as time goes on?

Let us investigate using a spreadsheet.  We have previously built a square wave packets at t = 0 from sinusoidal waves with different wavelengths λl or different wave numbers k.

p = ħk, E = p2/2m = (ħk)2/(2m),  ω = E/ħ = ħk2/(2m)

We will now let each of these waves move with speed  ω/k = vwave.
(We change the expression for each wave from ψ(x,t) = Acos(kx) to ψ(x,t) = Acos(kx  - (ħk2/(2m))t) in the spreadsheet.  We choose units of 10-12 m = pm for length and 10-18 s for time.  For the mass of the particle we choose m = 10-30 kg, approximately the mass of the electron.