Discussion of Pauli exclusion principle, and the connection of quantum mechanics to the periodic table and chemistry. (Notes not provided, but to summarize:)
Classical argument, given L, can calculate magnetic moment mu
mu= (-e/2m) L,
and thus the energy in a magnetic field is
U = - mu . B = (e hbar/2m)m B
(the m downstairs is mass, the quantity in parentheses is the Bohr magneton)
Expect odd numbers of energy levels. Anomalous Zeeman effect: e.g., sometimes see even numbers. (Pauli, 1923 "How can one look happy when he is thinking about the anomalous Zeeman effect?")
Stern Gerlach, 1921, free electron through B field with a gradient, gives a force from the gradient of mu.B. Observation: beam splits into two.
1925 Goudsmit and Uhlenbeck suggested that the electron has intrinsic angular momentum, called it spin, with (1/2) units of hbar angular momentum. (NOT classical, a classical ball of charge, of size of "classical radius of electron" gives an equatorial velocity >> c)
Complete set of hydrogen wave functions includes n,l,m and now s. Then, 1925: Pauli's exclusion principle: No 2 electrons in an atom can exist in the same state.
Generalizes: (fermions and bosons) Fermions obey the exclusion principle. Fermions are spin odd half integer, bosons are integer spin. Examples of fermions: proton, electron, Delta, neutrinos... Examples of bosons: photons, pions,... Comes from antisymmetry of fermion wave function, symmetry of boson wave function.
Consequence: periodic table! (Discussed the order of the elements, why the 2s get filled before 2p, due to angular momentum barrier and shielding of charge, etc)
(Review for final)
(Possible) special topic: An example of using wave functions, and an introduction to an important concept. - Perturbation theory.
Let's see how to modify the ground state energy of hydrogen, based on the knowledge that the nucleus is not really a point, it has a finite size. The Bohr radius is 10^-8 cm, while the radius of a nucleus is about 10^-13cm, so you expect this to be a tiny correction. The point is that our potential is not quite V=e/r, when you get really close to r=0.
We could try to start from scratch, with the modified potential, taking into account that the charge of the nucleus is not at a point. If you assume it's a shell of charge at radius r0, then V(r)=e/r0 when r<r0 (constant), and V(r)= e/r when r>r0. But, this would be a very hard differential equation!
Instead, we'll make a sequence of approximations, and get a very decent guess. The essence of this is example is perturbation theory.
We already have a good idea of what psi is, so we assume it won't change much when we add this small extra effect, or perturbation.
because r0 is so incredibly small, and the original wavefunction which looks like exp[-r/2a0] hardly varies between 0 and r0. (since r0/a0 = 10^-5) Now, since V(r)=e/r0 is constant for r<r0, we can estimate the expected potential energy of the electron inside the finite nucleus:
On the other hand, if the nucleus was truly pointlike, then
The difference in potential energy of the electron in the two cases is
This is a tiny shift, since (r0/a0)^2 is about 10^-10. The difference is positive, which means the energy is in reality a tiny bit higher than what we got before, by the amount given above.
More correspondences with Bohr: Transitions
In the Bohr model, an electron dropping from state m to n gives of E=h nu = Em-En. Now we can see why this is the case!
Recall that the time dependent solution for a wave function is
Let's begin by looking at an electron in a fixed quantum state, n.
What is <x>? (We can also look at <y>, and <z> too...)
The time dependent terms cancel exactly, so <x> is time independent. So in a certain sense, the electron is not vibrating. It's expected position is staying the same! Thus, it is also not radiating. This was demanded (by fiat) by Bohr, and we see now it is a consequence of the Schrodinger equation.
What if the electron, somehow, begins in a mixed state, that is,
So it's partly in state n, and partly in state m.
|a|^2 tells you the probability we're in state n;
|b|^2 tells you the probability we're in state m.
NOW look at <x> again:
But we know that Exp[i theta]+Exp[-i theta] = 2cos(theta)
So here we have something of the form
This means that <x> is now time dependent! The particle is "wiggling" back and forth, with a frequency nu = (En-Em)/h
So, quantum mechanics is predicting that if you start in such a state, you wiggle at the "Bohr frequency", and thus also radiate at this frequency!
This is all still admittedly a bit vague. If we stimulate the atom, tickling it at frequency nu, you might imagine this could stimulate some admixture "b" of the other state, and hence make this radiation. This is stimulated emission. But, how does spontaneous emission happen? One really needs more than plain quantum mechanics - QED provides the real explanation. Electric and magnetic fields are always fluctuating, sometimes at the right frequency to tickle (and hence admix some "b") which then feeds on itself as we saw above. These "vacuum fluctuations" of E and B are like the zero point motion of a harmonic oscillator. QED also shows that this vibration doesn't give ordinary electric dipole radiation, but rather makes a photon....
Back to the list of lectures