Class
calendar, week 9
Mon., 10/25:
# Students can bring one 8.5x11 page to the midterm with notes on it.
The exam is on material through
spin resonance.
# The radial 2d Schrodinger equation cannot be cast in the form of
a 1d Schrodinger equation, hence
we can not infer anything about the existence of bound states for circularly
symmetric shallow potentials
in 2d from the radial reduction.
# Potentials that fall to minus infinity faster than -1/r^2 have no
ground state (in the sense that, if we
regulate the potential by flattening it at r_0, and then take the limit
as r_0 goes to zero, the ground
state energy goes to minus infinity). The potential -a/r^2 has a gound
state if a is smaller than some critical value,
otherwise it does not. (See Landau & Lifshitz, section 35, "Fall
of a particle to the centre".)
# I repeated much of Friday's material, and completed the discussion
of the spherical square well.
See last friday's minutes, where I included some of what I said on
10/25.
# Coulomb potential: Initially solved by Pauli using Heisenberg's matrix
mechanics, i.e. using an
algebraic method. Then solved by Schrodinger using the Schrodinger
equation. Let's do the latter first:
Multiply the radial eqn by 2m/hbar^2:
[-(d^2/dr^2) + (l(l+1)/r^2) - 1/r_0 r + kappa^2] u(r) = 0,
where
r_0 = hbar^2/2me^2 = a_B/2 (half the Bohr radius), and
kappa = Sqrt[-2mE]/hbar.
Wed., 10/27:
# Normalizable solutions exist only for discrete values of kappa^2,
which yields the quantized. energy levels.
These can be found by series solution of the above radial eqn.
As r goes to infinity only the kappa^2 term matters, so u ~ exp(-kappa
r) .
(The exponentially growing solution is not normalizable.)
As r goes to zero, the 1/r^2 term dominates if l is not
zero, so u ~ r^(l+1).
Factor out these behaviors: u(r) = r^(l+1) exp(-kappa r) (c_k r^k)
(sum on k = 0 to infinity) and try to
solve for the series coefficients c_k. Plug into the radial eqn. There
are nine combinations of
the two derivatives. When they both hit r^(l+1) that cancels the l(l+1)/r^2
term (if l=0 the second derivative
kills r^1, and there is no 1/r^2 term to cancel). When they both hit
exp(-kappa r) that cancels the kappa^2 term.
Evaluating the remaining terms we get an eqn like [ ... ]_k r^k = 0
which, since true for all r, implies [ ... ]_k = 0.
This yields a recursion relation c_k/c_(k-1) = [2 kappa (k + l + 1)
- 1/r_0]/ k (k + 2l + 1).
At large k then c_k/c_k-1 ~ 2 kappa/k. This is like the seriesexpansion
of exp(2 kappa r), so it will make the
whole function u blow up like exp(kappa r), which is not normalizable.
The only way out is if the series terminates,
which is only possible if [2 kappa (k + l + 1) - 1/r_0] = 0 for some
k. This yields 2 kappa r_0 = 1/n, with n = k+l+1,
or E = -(me^4/2 hbar^2) (1/n^2).
# n = k + l + 1 is called the principal quantum number. For
a given l, the lowest level is at k=0, so k+1 is what we
previously called the radial quantum number. For clarity, let's now
call the radial quantum number n_l, which counts
the levels of the radial Hamiltonian H_l. (Sometimes people alternatively
call k the radial quantum number.)
k = n_l -1 is the number of nodes in the radial wavefunction.
# The Coulomb spectrum is maximally degenerate: We showed last
week that for a given radial quantum
number, higher angular momentum states have higher energy. The most
degeneracy we could therefore have is
if the nth level of H_l is degenerate with the (n+1)st level
of H_(l-1), which is exactly the situation for the Coulomb spectrum.
This is sometimes called ``accidental degeneracy", but it is no accident.
It is a result of an extra symmetry of the
Coulomb problem (which is related to the classical fact that the orbits
are closed in the Coulomb potential). This
is explained in the supplement, Lenz vector and the
Coulomb potential. This supplement first explains Pauli's algebraic
solution of the Coulomb problem, which uses the Lenz vector, and then
explains how the Lenz vector yields the degeneracy.
# Different labeling schemes exist for the levels of central potentials.
They all have the form N l, where l is the angular
momentum, but they differ in N. When there is no degeneracy, as for
the square well, the most natural scheme seems
to be to let N be the radial quantum number n_l. For the Coulomb potential,
it is more natural to use the principal quantum
number. Nevertheless, you will also see people use N = n_l + l to label
levels even when there is no degeneracy.
For example this is done for the square well in the handout from Principles
of Modern Physics by R.B. Leighton, page 603.
# I explained how to do the problem from the last homework, involving
bound states of the -delta(r - d) potential
subject to the condition that the momentum space wavefunction vanishes
for p<p_0. (After calss Kenton pointed out
that this condition should not properly be called a "boundary condition",
since it refers to all p<p_0, and since it
does NOT imply that the wavefunction vanishes as p approaches p_0 from
above. This discontinuity in the
momentum space wave function is allowed, since there are no derivatives
in the momentum space Schrodinger equation
for the delta shell potential.) I also explained that in this setting
where the states with p<p_0 are occupied by fermions
and hence unavailable due to the exclusion principle, a bound state
would be any state with E<p_0^2/2m, since all the
available propagating states have higher energy. Also, such a state
would be spatially localized, and hence normalizable.
Fri., Oct. 29:
Mid-term exam