Having established that photons can have properties of both waves and particles, a physicist named Louis de Broglie postulated that the same was true for matter. That is, he postulated that electrons, protons, neutrons, and even entire atoms will exhibit wave-like properties under the proper circumstances.
For a photon, fl = c, E = hf, and p = E/c = hf/c = h/l. For matter, c = fl is almost certainly not the case, and so how do we determine the wavelength and frequency of a "matter wave"?
De Broglie suggested that the wavelength of a matter particle of momentum p = mv (non-relativistic particle) is
Calculate the de Broglie wavelength of a proton moving at (a) 2.00×104m/s; (b) 2.00×108m/s.
(a) The momentum of the proton is
p = mv = (1.67×10-27kg)(2.00×104m/s) = 3.34×10-23kg m/s.
The de Broglie wavelength (the wavelength of the matter wave) is
l = h/p = (6.63×10-34Js)/(3.34×10-23kg m/s) = 1.99×10-11m.
(b) This proton is moving at at close to the speed of light.
Let's calculate the momentum with the relativistic formula, just to be safe:
p = gmv = 4.48×10-19kg m/s
l = h/p = (6.63×10-34Js)/( 4.48×10-19kg m/s) = 1.48×10-15m.
The idea of particles behaving like waves caught on very quickly. Schrödinger put this idea to work in his wave equation for quantum mechanics, the Schrödinger equation. Schrödinger's equation specifies how the wave function, traditionally denoted by the greek letter psi, Y, behaves.
The wave function, Y, depends on location and time, and Schrödinger's equation tells us how to determine Y for a given situation. But, it doesn't tell us what Y means. To understand what it means, let's look at what we know about photons and electromagnetic waves.
Consider a double slit experiment with light. In the classical solution, we consider the light as a wave emerging from each of the two slits in phase. At the viewing screen, we add up the electric fields of the waves from each slit, giving us the resulting electric field. In some locations the electric fields add, giving bright fringes, and in others the electric fields subtract, giving dark fringes.
Now consider this from the photon perspective. A single photon passing through the slits hits the screen at some location. We can's predict ahead of time where it will hit, but we know that more photons must reach the area of a bright fringe than the area of a dark fringe. We assign a probability to the photon hitting a particular location, and the probability turns out to be proportional to the square of the electric field at that location, as predicted by the classical solution, probability is proportional to E².
The electric field is the wave function of the photon. The square of the wave function at a particular location is proportional to the probability of finding the photon at that location.
We extend this interpretation to the particle wave function: Y² at some location is proportional to the probability of finding the particle at that location (and time).
The uncertainty principle is another well known modern Physics result. In words, the uncertainty principle says that it is not possible to precisely know a particle's position and momentum (speed) at the same time. In a sense, measurement of one disturbs the other.
Expressed as a formula, the uncertainty principle is
A second uncertainty principle exists between energy and time:
In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.10nm (the diameter of the atom). If the speed of the electron is on the order of the uncertainty in the speed, how fast is the electron moving?
Using the uncertainty principle, we find
Dp = h / (4p Dx) = (6.63×10-34Js)/4p(1.0×10-10m) = 5.3×10-25kg m/s
Using the mass of the electron, we can find the uncertainty in the speed, which is approximately equal to the speed:
v = Dv = Dp/m = (5.3×10-25kg m/s) / (9.11×10-31kg) = 5.8×105m/s.
I used the non-relativistic expression for momentum, p=mv, and the fact that the resulting velocity is much less than the speed of light is an a posteriori justification.
(a) Show that the kinetic energy of a non-relativistic particle can be written in terms of its momentum as KE = p²/2m. (b) Use the result of (a) to find the minimum kinetic energy of a proton confined within a nucleus havin a diameter of 1.0×10-15m.
(a) Non-relativistic kinetic energy is KE = ½mv², and non-relativisitic momentum is p=mv.
p²/2m = (mv)²/2m = ½mv² = KE.
(b) Since the proton is confined to the nucleus, Dx = 1.0×10-15m, and from the uncertainty principle,
Dp = h/4p Dx = (6.63×10-34Js)/4p(1.0×10-15m) = 5.3×10-20kg m/s.
The minimum average momentum of the proton is roughly equal to the uncertainty in the momentum, therefore, the minimum kinetic energy is:
KEmin = p²min/2m = (Dp)²/2m = (5.3×10-20kg m/s)²/2(1.67×10-27kg) = 8.4×10-13J = 5.3×106eV = 5.3MeV.