PHYS 623
Introduction to Quantum Mechanics II
Spring 2018, University of Maryland
Prof. Ted Jacobson
Notes
Thursday Mar. 1:
Today:
1. The simple product of hydrogenic variational wavefunctions
for a proton with two electrons yields an upper bound -0.945
Ry to the ground state energy. Since this is not less than -1
Ry, it is higher that the ionized state, H + e-, so
it fails to reveal the existence of the hydrogen anion, H-.
Challenge: try to find a trial wavefunction that
establishes the existence of the bound state. It will
presumably need to incorporate correlations between the two
electrons, so that the electrons tend to be on opposite sides
of the proton...If you find one, please share it with the
class.
2. Excited states of helium, as covered in Schwabl. I harped
on the fact that while the exchange integral "K" can be shown
to be positive, it isn't manifestly positive, and I
conjectured that if the Coulomb potential were modified, it
might not be positive. For instance: suppose the denominator
is replaced by d + w, where d is the distance between the
electrons (the usual Coulomb denominator), and w is a constant
length. This would decrease the advantage of the two electrons
avoiding being in the same location, which leads me to suspect
that it might invalidate the positivity of the integral. Let
me know if you test this (for different w's)...
3. Multi-electron atoms: I tried to summarize the salient
points. Many more details, and in particular explanations of
the Hartree-Fock method of the self-consistent field
approximation, can be found in Schwabl or Littlejohn. I don't
plan to delve into those details, but wanted to emphasize a
conceptual point: the ground state, being an energy eigenstate
of a rotationally invariant Hamiltonian, must be an eigenstate
of F2 (unless it's degenerate, which it ain't
except wrt the mF value), hence has a definite
total angular momentum quantum number F. To the extent that
the hyperfine interaction can be neglected, it can also be
taken to be an eigenstate of J,L and S, but that is only a
(very good for many purposes) approximation.
4. I explained the notion of electron configurations and
shells, along the lines in Schwabl. I also explained the
meaning of the "term symbol", (2S+1)LJ,
and how Hund's rules allow you to infer --- usually correctly
--- the term symbol, given the configuration. I explained the
example of oxygen, showing it is 3P2.
I'd like to do another couple of examples here: iron
(Fe), and ytterbium (Yb).
Example: The configuration of Fe is [Ar]4s2 3d6.
The 4s shell is filled but the 3d shell has room for
2x5=10 electrons, and contains only 6. To maximize the
spin, we can distribute 5 electrons with spin up among
the 5 ml values, and place the 6th electron
with spin down in any one of the orbitals. That yields S
= 2. The maximal ml would be
obtained if we placed the 6th electron in
the ml=2
orbital, yielding a maximal
total ml=2, and therefore L = 2 is the maximal L
accessible
(given that S
was already
maximized).
Finally, since
the shell is
more than half-filled,
we should
maximize J
given S=2 and
L=2, which
yields J=4,
and therefore
the term
symbol is 5D4.
The states
with higher J
values but the
same L and S
are excited
states, and
it's
interesting to
inspect the
energy
differences.
You can see
them here: https://physics.nist.gov/PhysRefData/Handbook/Tables/irontable5.htm.
If you compute
the energy
differences
and divide by
J, for
adjacent
levels, [E(J)
- E(J-1)]/J,
you find that
they are
fairly
constant:
-104, -96,
-92, -90, in
units of
inverse cm.
This
near-constancy
follows from
the Landé
interval rule,
which in turn
follows from
the fact that, in first
order
perturbation
theory,
the net
effect of the
spin-orbit
coupling can
be seen to be
equal, for
each
configuration
and L and
S, to an
LS dependent
constant times
the
expectation
value of L.S
= (J^2 -L^2 -
S^2)/2 = [J(J+1) -L(L+1) - S(S+1)]/2. The
change of this
quantity
between J and
J-1, for fixed
L and S, is J.
The adjacent
energy
differences
are thus
proportional
to J, so if
you divide
them by J, the
result should
be J
independent.
As to the
magnitude of
the energy
differences, for
instance,
between the
ground state
(J=4) and the
next excited
state (J=3)
the energy
goes up by 416
inverse cm.
Now 1/cm is
about 0.12
meV, so 416/cm
is 50 meV.
Compared to 10
eV this is 50
10-4, which is ~ 10-3. This is ~ 10 times larger
than the
relative size
of
relativistic
corrections in
hydrogen. Apparently,
as the atomic
number grows,
spin-orbit
coupling
strengths do
increase, I
suppose
because the
participating
electrons are
moving faster
and/or they
are sometimes
seeing a
larger central
charge.
The final
example I want
to mention
here is Yb,
since a
student
mentioned
using that
atom in a
quantum
computer
setting in the
lab she works
in, and since
it is used in
the most
stable atomic
clock. (To
read about the
clock, see the
article linked
on the
Supplements
page.) The
ground state
configuration
is [Xe]4f146s2. The letter f denotes l=3, whose shell has 2x7=14
states, so it is a filled shell, which has S=0 and L=0,
and the term symbol of the ground state is 1S0. The two
6s electrons
are like in
helium, so
among the
excited states
is a spin
triplet (S=1)
(like orthohelium)
that
also has
orbital
angular
momentum L=1,
and the fine
structure of
this level
includes J =
0,1,2
sublevels. The
J=0 term has
term symbol 3P0.
In this state, the spin and orbital angular momenta are each
nonzero but they add to zero. The one photon transition from
this state down to the ground state is "completely forbidden"
as we'll see, because the ground state also has J=0, and
there is no J=0 to J=0 one photon transition. If that were
the whole story the lifetime of this state would be
extremely long (years?). That would be the case for one of
the I=0 isotopes of yttrium. However, the long lifetime also
means the transition line "width" is extremely narrow, so it
is too difficult to excite the transition to be practical.
The Yb-171 isotope, on the other hand, has nuclear spin
I=1/2, and therefore a magnetic dipole moment, which introduces
a hyperfine interaction. This means that the 3P0(F=1/2)
state mixes with the 3P1(F=1/2)
state, and the latter state has an allowed decay mode.
Although not as narrow, it's still a very narrow transition,
and correspondingly very long lived, since a spin flip
is needed to get from the triplet to the singlet spin
state, and since (I presume) the amount of
admixture of the latter state is small.
Tuesday Feb. 20:
For a pdf file of these notes is here.
Today I finished discussing the hyperfine splitting of
atomic energy levels. Since I approached the material rather
differently than does Schwabl or Littlejohn, I wanted to make a
brief synopsis here. The following main points were made:
1. Reviewed definition of gyromagnetic ratio and g-factor.
2. Hyperfine interactions are interactions of nuclear magnetic
dipole or electric quadrupole moments with the field produced by the
electrons, evaluated at the nucleus. One could also view this the
other way around: the influence of nulcear multipole fields on the
electron. Either way, it's the interaction energy that matters, and
we'll view it the first way.
3. An electric dipole moment (EDM) is forbidden unless the
Hamiltonian violates both time reversal symmetry and parity
symmetry. (See my Supplement on Nuclear moments.) Parity is strongly
violated by the weak interactions, and time reversal symmetry is
violated by the complex phases in the quark mixing matrix of the
standard model, but that would be a very small effect in nuclei, so
some very small EDM is expected, but none has been detected as yet.
The main search is for a neutron EDM. The present upper limit is
10^5 times larger than the value expected from the quark mixing
matrix, but other theoretical considerations suggest that there
should be a larger value, and finding it could be an important sign
of new physics. See the Supplement for a few more details.
4. The nuclear magneton e_0/(2m_p c) is smaller than the Bohr
magneton e_0/(2m_e c) by a factor m_e/m_p ~ 1/1836 ~ 10^-3.
5. The magnetic hyperfine interaction Hamiltonian is H_hf,mag = -
µ_nucleus.B_electrons, where the electronic magnetic field is
evaluated at the nuclear position. Using the Wigner-Eckart theorem,
I showed that the expectation value of this Hamiltonian in states of
definite |FIJm_F> is equal to the expectation value of I.J,
times two factors that depend only on I and J. (Here I
and J are the nuclear and electronic total angular momenta,
and F is the overall total angular momentum of the atom.)
The first factor is the nuclear gyromagnetic ratio. The second
factor depends on the electronic state. (The key observation
behind the argument is that µ_nucleus is a nuclear vector operator
and B_electrons is an electronic vector operator. Using this, we may
invoke the aspect of the Wigner-Eckart theorem that states that
matrix elements of any two tensor operators of the same rank are
proportional to each other.)
6. When treating the hyperfine interaction as a perturbation, we
face the fact that the states labeled by different m_I & m_J are
degenerate, so we must find the eigenvectors and eigenvalues of the
hyperfine Hamiltonian truncated to the degenerate (IJ) subspace. In
view of point 5., this is easy, because I.J = (F^2 -
I^2 - J^2)/2 = [F(F+1) - I(I+1) - J(J+1)]/2, i.e.,
I.J is already diagonal in the |FIJm_F> basis. The
level shifts thus have the form A(I,J)[F(F+1) - I(I+1) - J(J+1)].
7. The must fundamental example is the ground state of the hydrogen
atom. The unperturbed electronic orbital state is 1s, and the spin
state is arbitrary, while the nuclear spin state is also arbitrary.
Both the electron and the proton have spin-1/2, so I = 1/2 and J =
1/2, hence the possible values of F are 1 and 0, the triplet and
singlet. The corresponding values of
[F(F+1) - I(I+1) - J(J+1)] are 1/2 and -3/2. It turns out that the
coefficient A here is positive, so the ground state is the singlet,
which is depressed three times as much as the triplet is raised. The
frequency of this transition (i.e. the energy difference divided by
hbar) is around 1.4 GHz, and the wavelength is the famous "21 cm
line".
8. A super important example to physics is the ground state of
Cs-133. Cesium is an alkalai atom, with a 6s valence electron. All
the other electrons make up closed shells, which are exactly
rotationally invariant, and so contribute nothing to the magnetic
moment of the electrons. The nuclear spin is I = 7/2, so the
possible values of F are 4 and 3. The energy difference between
these is radiation with a frequency around 10 GHz. In fact the SI
unit of the second is *defined* to be exactly 9,192,631,770 periods
corresponding to this frequency.
9. And now what about electric quadrupole moments? See the
Supplement for the definition. The key thing is that the quadrupole
moment operator is a rank 2 irreducible tensor operator. Hence the
selection rule aspect of the Wigner-Eckart theorem tells us that a
spin-1/2 nucleus cannot have an electric quadrupole moment:
<1/2|Q_2|1/2> = 0, because 1/2 is not in 2 x 1/2. Any nuclear
spin greater than 1/2 however does support an electric quadrupole,
and in particular the Cs-133 nucleus does. So why don't we need to
consider how that affects the fine structure of the ground state of
Cs-133? Why can we conclude it is only split into a doublet,
corresponding to F = 4 and F = 3?
The answer, again, comes from the selection rule. To understand
this, we should first write down the interaction energy between the
nuclear quadrupole moment and the electronic field. I gave an only
partly true explanation of this in class, so let me try to fix it
here: The electrostatic interaction energy of a charge distribution
rho and an electrostatic field Phi is ∫ rho Phi dV. This leads to a
quadrupole interaction energy 1/2 Q^ij Phi,ij(0) (see below for a
derivation). Since Q^ij is traceless, the trace part of Phi,ij(0)
doesn't contribute here, so we may replace this by
1/2 Q^ij W^ij, where W^ij = (Phi,ij - 1/3 Phi,kk delta^ij)(0).
W^ij is an operator on the electronic Hilbert space, since Phi
is generated by the electron(s), and in fact it is an
irreducible tensor operator of rank 2. Hence its expectation
value vanishes in electronic states with J = 0 or J = 1/2. In
particular, in the case of Cs-133, <6s|W^ij|6s> = 0. All
of which is to say that, although the cesium-133 nucleus has a
reasonably large electric quadrupole moment, it can't "feel"
the electric field generated by the 6s electron!
Derivation of the quadrupolar
interaction energy:
Let's expand Phi around the origin, taken to be the center of
mass of the nucleus, viz,
Phi(x) = Phi(0) + Phi,i(0) x^i + 1/2 Phi,ij(0) x^i x^j + ....
The i and j indices are the cartesian coordiante indices, the
comma indicates partial derivative, and repeated indices are
summed over. Now we can carry out the interaction energy
integral with this expansion, and we get
(*) Q Phi(0) + p^i Phi,i(0) +
1/2 Q^ij Phi,ij(0) + 1/6 (∫ rho x^2 dV)Phi,ii(0) + ...
The last term is needed because in order to express ∫ rho x^i
x^j dV in terms of the quadrupole moment Q^ij we must subtract
and then add back in the trace part:
∫ rho x^i x^j dV = ∫ rho (x^i x^j - 1/3 x^2 delta^ij) dV + 1/3
delta^ij ∫ rho x^2 dV.
The first term in this expansion (*) corresponds to the
Coulomb interaction, the second term is the EDM discussed
above, which nearly vanishes, and the third term is the
electric quadrupole interaction.
The last term is a second moment of the charge
distribution, multiplied by the Laplacian of the
potential evaluated at the origin. According to
Maxwell's equations, the latter is proportional to
the electron charge density at the origin. This
term represents a nuclear finite size correction
to the Coulomb potential, and is a scalar, so it
doesn't break the degeneracy among the m_I m_J
states.