 On page 3 the differential equation and
recursion relation for
the polynomial coefficients of H() are given.
Consider the following example of the H(x) function
for a 10g Hatom wavefunction.
H(x)=240240  120120 x
+ 21840 x^{2}
 1820 x^{3} + 70 x^{4}  x^{5}
 What is l? n_{r}? ?
 Show that this H satisfies the H differential equation.
 Show that
the coefficients satisfy the recursion relation.
 This H is not quite a Laguerre polynomial...it is off by an overall
factor. What is that overall factor?
 At the end of page 3 there is a table showing
that at least for n=15 the total degeneracy is n^{2}.
Prove this result in general.
 On page 4 the method of normalizing the
Hatom wavefunction is described: it requires use of the Laguerre recursion
relation. Follow similar steps to find the expectation value
of r and r^{2}...i.e., show:
 <1/r>=1/n^{2}
 <r>=½[3n^{2}l(l+1)]
 <r^{2}>=½n^{2}
[5n^{2}+13l(l+1)]
For circular, large n orbits, find an expression for a
classical orbit radius consistent with the above two expressions.
For classical s orbits we expect r is less than twice
the same energy circular orbit. Explain this, and show that these
expressions are consistent with this result.
 With additional work one can show:
You may want to look up (e.g., in Szegö) how to write
L_{n}^{2l+1} in terms of a sum of
L_{k}^{2l}. However a faster proof is
via the FeynmanHellmann theorem.
 By the previous problem, ns have larger average size than
say np orbitals. Where is the probability density largest
for s orbitals? Explain how this can be consistent with
with the both the classical orbit and the result of a larger
average radius.
 Calculate the wavelength and energy of the following transitions.
Report what "type" of light is involved (e.g., , Xray,
UV, visible, IR, microwave, radio).
 Hatom: 2 1
 Hatom: 3 2
 Hatom: 4 3 ; 3
 Hatom: 5 4 ; 4
 Hatom: 11 10
 Hatom: 101 100
 Catom: 2 1
 Uatom: 2 1
 For what value of n would an Hatom ns orbital have a <r>
 the size of a "bucky ball" C_{60}: r=5Å
 the size of a big bacteria cell: r=0.5µm
 the size of a pin head: r=0.4mm
 The radius of a nucleus is approximately 1.2 (2Z)^{1/3} fm.
For what value of Z would <r> for 1s be
"only" 10 times the size of the nucleus. 100×?
 We can approximate the nucleus as a uniform ball of charge.
Find the electrostatic potential as a function of r inside
such a ball of charge. When we solved the Hatom problem we used
an exact 1/r, whereas we now know the potential deviates
from 1/r inside the nucleus. Using the difference between
the 1/r and the actual potential as a perturbation, find the
energy shift of 1s states as a function of Z.
 [Gasiorowicz] An electron is in the ground state of
tritium (a radioactive isotope of hydrogen whose nucleus has 2 neutrons
and 1 proton, i.e., H^{3}). A nuclear reaction instantaneously
changes the nucleus to He^{3} (a rare isotope of helium
whose nucleus has 1 neutron and 2 protons). Calculate the probability that
the electron remains in the 1s state.
 An electron is in the following superposition state:
=N(100>  2200> + i210> + 211>)
where 210> means the wavefunction has n=2, l=1, m=0.
 Find the normalization N.
 Find the expectation value of energy.
 Find the expectation value of r (use Mathematica).
 On page 5 probability densities for
a variety of wavefunctions with n_{r}=2 are displayed.
Why are the d wavefunctions nearly a factor of two larger than
the p wavefunctions? Describe how changing n_{r}
would change these pictures.
There are four "xy" graphs on this page. The first two
(with xaxis labels z & x) are very similar and
the next two (with xaxis labels x & z) are
very similar. Why are the pairs similar? why are they different?
Describe how changing n_{r}
would change these graphs.
 Review the material on WKB applied to 3d problems.
Applied to the Hatom, there is no need to rederive the
l(l+1)(l+½)^{2}
relation, simply change the potential from r^{2} to
1/r. Now find the WKB energies for the Hatom.
 On page 6 we used perturbation theory
to find the energy shift from electronelectron repulsion. The
states 2s & 2p (m=1,0,1) are degenerate,
yet we did not use degenerate perturbation theory. Why could we get
away with this error?
 Continue the "ltilting" calculations of page 6
for the 3s, 3p, and 3d cases.
 In excited twoelectron atoms (e.g., 1s2p) one expects that the 1s
shields the 2p electron, so that a better approximation to the
2p wavefunction would use a central charge of Z1 rather than Z.
We can do this if we make "unperturbed" Hamiltonian a sum:
an Hatom with Z for electron 1 and an Hatom with Z1 for electron 2.
Since the total Hamiltonian must add up to the total energy, the perturbation
must include a bit of the nuclear electrostatic potential for electron 2 and
the electronelectron repulsion. The result is called Heisenberg's
method and it is described starting in §24 of Quantum Mechanics of One and Two
Electron Atoms by Bethe and Salpeter. Reinvent this method, note
any odd problems you find along the way (there are several!) and find the
energy of the 1s2p state in He.
 The ground state energy of two electron atoms is an interesting problem for
the RayleighRitz (variational) method. The basic idea is to calculate
the expectation value of the Hamiltonian using a trial wavefunction,
i.e., a function with some variable parameters in it. We are guaranteed
that the expectation value of the Hamiltonian always lies above the
actual ground state energy. Using the variable parameters one can get the
best (lowest) such upper bound. Using our twoelectron Hamiltonian:
and the trial wavefunction: =
exp[a(r_{1}+r_{2})]
Here is the process of finding the expectation value as a function
of a and then finding the lowest such upper bound:
Justify each step in the above derivation. Some things to consider:
How did the derivative terms come about, e.g., what happened to
the angular part of the Laplacians. In line 1 we have three terms in
square brackets "[]", on the third line its just two...what happened?
The term that starts as the third square bracket term become an infinite sum
and then just the first term in that sum...what happened? What are the
P_{l} and what is the argument: _{12}?
Explain how each "dV" was eventually replaced and why.
Evaluate each integral with Mathematica and verify the
results.
 Let's apply the above formula to three two electrons atoms:
H^{}, He, and Li^{+}.
System  calculated energy  energy in (eV)
 System less one electron  energy of that ion (eV)  calculated ionization energy (eV)
 experimental ionization energy (eV)


H^{}  .473  12.86  H  13.61  .75  .75


He  .712  77.49  He^{+}  54.42  23.1  24.6


Li^{+}  .803  196.54  Li^{+2}  122.45  74.1  75.6


For comparison, ignoring the interaction, each 1s electron would give
a binding energy of ½, so for 1s^{2} we would have
an energy of 1. For large Z we approach that limit.
The variational calculation is fairly accurate, its estimated
energies are about 1½ eV too high. (It is, after all,
just an upper limit on the energy.) Unfortunately that 1½ eV
difference is critical for the existence of the ion: H^{}.
By our calculation H^{} is less strongly bound than H
and so should spontaneously dissociate. But it turns out that
H^{} is stable and is critical for understanding the opacity of the
atmosphere of the Sun. Furthermore, H^{} is difficult to make in
the lab so calculation was the basis for understanding this ion.
In fact, discrepancies between early calculations and
observational evidence on the Sun's radiation pointed out
the inaccuracies of the thenexisting wave functions for H^{}.
Thus improved calculations of H^{} properties were
very much on the agenda of physics in the 1950s. Our variational
calculation allows the electrons expand away from the nucleus
(as one would expect, given the partial shielding by the other electron)
[i.e., a has been reduced from the hydrogen value of 1].
One expects an additional effect: the electrons should avoid each other.
It turns out that we can to a bit better if we allow the electrons
to have different "orbital radii"...i.e., to use a trial wavefunction like:
exp[ar_{1}br_{2}] +
exp[br_{1}ar_{2}]
Use this wavefunction and show that the H^{} ionization energy
is positive (i.e., H^{} is more stable than H+e^{})
 Better yet is to use a trial wavefunction in which the
electron positions are (anti)correlated (so if one electron is at
r_{1} the other electron is unlikely to be found
near r_{1}). This requires a more complex trial wavefunction,
for example letting the wavefunction depend on the angle between
r_{1} and r_{2}. Three parameter
trial wavefunction like this are able to predict binding energies
to a fraction of an eV (and get the ionization energy of H^{} accurate
to about 5%). See Quantum Mechanics of One and Two
Electron Atoms by Bethe and Salpeter for details.
We consider here a trial wavefunction:
(exp[ar_{1}br_{2}] +
exp[br_{1}ar_{2}])
(1+cr_{2}r_{1})
It is a mess to find the energy expectation value, but here is the
result in Mathematica form (so you can cut and paste):
e=(2*a^11*b^2 + a^12*b^2  16*a^10*b^3 + 8*a^11*b^3  56*a^9*b^4 +
29*a^10*b^4  204*a^8*b^5 + 64*a^9*b^5  426*a^7*b^6 + 226*a^8*b^6 
426*a^6*b^7 + 368*a^7*b^7  204*a^5*b^8 + 226*a^6*b^8  56*a^4*b^9 +
64*a^5*b^9  16*a^3*b^10 + 29*a^4*b^10  2*a^2*b^11 + 8*a^3*b^11 +
a^2*b^12  6*a^11*b*c + 3*a^12*b*c  48*a^10*b^2*c + 24*a^11*b^2*c 
170*a^9*b^3*c + 85*a^10*b^3*c  360*a^8*b^4*c + 176*a^9*b^4*c 
1232*a^7*b^5*c + 165*a^8*b^5*c  2000*a^6*b^6*c + 1147*a^7*b^6*c 
1232*a^5*b^7*c + 1147*a^6*b^7*c  360*a^4*b^8*c + 165*a^5*b^8*c 
170*a^3*b^9*c + 176*a^4*b^9*c  48*a^2*b^10*c + 85*a^3*b^10*c 
6*a*b^11*c + 24*a^2*b^11*c + 3*a*b^12*c  6*a^11*c^2 + 3*a^12*c^2 
48*a^10*b*c^2 + 24*a^11*b*c^2  171*a^9*b^2*c^2 + 86*a^10*b^2*c^2 
364*a^8*b^3*c^2 + 184*a^9*b^3*c^2  536*a^7*b^4*c^2 + 269*a^8*b^4*c^2 
2363*a^6*b^5*c^2 + 48*a^7*b^5*c^2  2363*a^5*b^6*c^2 +
2868*a^6*b^6*c^2  536*a^4*b^7*c^2 + 48*a^5*b^7*c^2  364*a^3*b^8*c^2 +
269*a^4*b^8*c^2  171*a^2*b^9*c^2 + 184*a^3*b^9*c^2  48*a*b^10*c^2 +
86*a^2*b^10*c^2  6*b^11*c^2 + 24*a*b^11*c^2 + 3*b^12*c^2)/
(2*(a^10*b^2 + 8*a^9*b^3 + 28*a^8*b^4 + 120*a^7*b^5 + 198*a^6*b^6 +
120*a^5*b^7 + 28*a^4*b^8 + 8*a^3*b^9 + a^2*b^10 + 3*a^10*b*c +
24*a^9*b^2*c + 86*a^8*b^3*c + 184*a^7*b^4*c + 823*a^6*b^5*c +
823*a^5*b^6*c + 184*a^4*b^7*c + 86*a^3*b^8*c + 24*a^2*b^9*c +
3*a*b^10*c + 3*a^10*c^2 + 24*a^9*b*c^2 + 87*a^8*b^2*c^2 +
192*a^7*b^3*c^2 + 294*a^6*b^4*c^2 + 1872*a^5*b^5*c^2 +
294*a^4*b^6*c^2 + 192*a^3*b^7*c^2 + 87*a^2*b^8*c^2 + 24*a*b^9*c^2 +
3*b^10*c^2))
Find the minimum of this function. You should find a reasonable
value for the ionization potential of H^{} near 0.7 eV.
A 24 parameter trial wavefunction (published in 1956) gave
an ionization potential of: 0.7545 eV.

On page 9 there are two "oddities" in the
transition metals and 8 "oddities" at the end of the page. For each
oddity report what the electron configuration "should" have been. Can
you find any "rules" which would "explain" some of the oddities?

On page 10 I outline the approach for
calculating the Stark shifts for the n=3 orbitals.
Do this calculation; remember to check your resulting eigenenergies
using the given parabolic solution.
 How close would an ion have to get to a n=2 Hatom
to perturb the energy level enough so that the light produced in
the transition 21 is shifted
by 1 Å. (Approximate the coulomb field of the ion: assume
the ion's E field at the atom's nucleus is uniform throughout the
atom.)
Using the parabolic solution to the Stark Effect, find how close
an ion could come to a n=100 Hatom without the calculated
shift being large enough to ionize the atom.
 Carry out a crystal field calculation like that on page 11
but assume tetrahedral symmetry rather than octahedral. I.e., assume the
charges are at: (1,1,1), (1,1,1), (1,1,1), and (1,1,1).
 Carry out a crystal field calculation like that on page 11
but assume squareplanar symmetry rather than octahedral. I.e., assume the
charges are at: (1,0,0), (1,0,0), (0,1,0), and (0,1,0). (This problem
is more difficult than the previous problem.)
 Plot out the sp^{2} state Tri a.
 Calculate the dipole, i.e., <r> for a
sp^{1}, sp^{2}, and sp^{3}
state. (Use symmetry! You should know which direction <r>
points.) Which wavefunction reaches out the furthest?