PHY4605–Introduction to Quantum Mechanics II
Spring 2005
Test 2 SOLUTIONS
March 9, 2005
Short Answer. Must attempt (only) 3 of 6. Circle answers to be graded.
1. Identify or short answer:
(a) Born-Oppenheimer approx. Discuss the philosophy and its implementation
in the case of the H2 molecule or ion.
The B-O approx. is a method useful for problems where one has fast and
slow degrees of freedom. In the H2 molecule, the protons are slow and
the electrons are fast due to the difference in the masses. Thus the electrons may be assumed to a good accuracy to instantaneously adopt their
configuration of lowest energy around the 2 protons “placed” in space at
some separation rab . The protons feel an effective potential determined by
this electron cloud, and try to minimize this potential at their equilibrium
separation. The B-O method then fixes rab , factorizes the wave function
into an electron and proton part φψ, then keeps only the terms in the
proton motion which involve the large spatial gradients of the proton wave
function ψ. This gives an equation for the proton motion involving the
eigenvalue of the electronic part E (0) playing the role of an effective potential for the protons, (T + E (0) )ψ = Eψ, where T is the proton KE and
E (0) is the solution to H0 φ = E (0) φ. E (0) can be found variationally for a
given rab and therefore depends on rab . It has a minimum for the ground
◦
state of the electron system around r = 0.74A. Expanding E (0) around
its minimum gives the frequency of small oscillations, i.e. the vibrational
modes of the molecule.
(b) For Lithium, Z = 3, electron configuration 1s2 2s, standard tables give the
◦
radius of the atom as 1.5 A . If, on the other hand, one looks at the formula
for the radius of the n = 2 Bohr orbit for a hydrogenic 1-electron atom
with nuclear charge Ze, we find the Bohr radius a0 = 4π²0 h̄2 /(mZe2 ), and
the 2s radial function goes like e−r/2a0 , so we might have guessed that the
◦
◦
◦
“size” of the atom would be 2a0 = 2 ∗ (0.529A)/3 ' 0.35A (where 0.529A
is the Bohr radius for Hydrogen with Z = 1). What physical phenomenon
was neglected in our estimate which might lead to a larger orbit of the
valence electron if included? Explain.
The argument neglects screening, sometimes called shielding. The valence
(2s) electron is further from the nucleus, and therefore effectively “sees” a
nuclear charge closer to 3 − 2 = +1e rather than +3e because the two 1s
electrons are also with high probability inside the orbit. The orbit is thus
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a factor of three or so larger, not quite the “right” answer, but in the right
direction.
(c) Hyperfine structure of hydrogen. State its origin, give the magnitude of
the splitting of the H ground state either in terms of fraction of a Rydberg
using the fine structure constant, or the wavelength of the photon emitted
in the hyperfine transition.
Hyperfine structure arises from the dipole-dipole interaction of the proton and the electron spin. The Bohr energy of the ground state doesn’t
depend on electron spin, but including hyperfine coupling splits it into a
proton-electron spin triplet and spin singlet, with energy separation of order (m/mp )α2 R, where α is fine structure constant and R = 13.6eV . The
famous photon emitted has wavelength 21 cm.
(d) Give the electronic configuration à la chemistry for the Boron atom, with
Z = 5. Then enumerate the allowed term symbols for the ground state
from the point of view of allowed angular momenta alone.
Boron is 1s2 2s2 1p. Since the two 1s and two 2s electrons are in a singlet
state (S = 0) and have zero orbital angular momentum, the spin and
angular momentum are determine entirely by the 1p electron. The term
diagrams are therefore 2S+1 Lj = 2 P1/2 and 2 P3/2 , with j = 1/2 and 3/2
being the two values of j you can get from adding s = 1/2 and ` = 1. The
2
P1/2 turns out to be the ground state, but you can’t tell that from the
angular momentum arguments alone.
(e) You are told that an excited state ψn , n 6= 0 of a quantum system you
cannot solve is an eigenstate of Hermitian operator Q, Qψn = qn ψn , of
lowest possible energy (but not the ground state, which has a different
quantum number q0 ). You also know that [Q, H]=0. Show that any trial
wavefunction φ which obeys Qφ = qn φ can be used to provide an upper
bound to En .
Take a basis for Hilbert space χm,q , where q is the eigenvalue of Q, and
m represents all other eigenvalues. A general function may be expanded
P
u = m,q cmq χm,q . A function with definite eigenvalue qn is however φ =
P
m cmqn χmqn .
hφ|H|φi =
X
mm0
=
X
m,m0
=
X
m,m0
=
X
c∗mqn cm0 qn hχmqn |H|χm0 qn i
c∗mqn cm0 qn hχmqn |Em0 qn |χm0 qn i
c∗mqn cm0 qn δmm0 Em0 qn
|cmqn |2 Emqn ≥ En
X
|cmqn |2 = En
m
m
since En is by definition the lowest of the energies corresponding to qn , and
assuming φ is normalized.
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(f) State the “triangle rule” for addition of angular momenta: for given L, S,
what are the possible eigenvalues of the operator J 2 with J = L + S?
|` − s| ≤ j ≤ |` + s|, J 2 ψ = h̄2 j(j + 1)ψ.
2. Low-lying He excited states
Consider the diagram for the ground (not shown) and excited states of the
Helium atom.
(a) Write down an approximate wavefunction for the ground state of Helium,
using Z = 2 hydrogenic wavefunctions of the form ψ1s (r1 ), etc., and indicating the correct spin state of the two electrons. Make sure your final
expression for the full wave function obeys the Pauli principle. Why is
your wave function not exact?
Ψ(r1 , r2 ) = ψ1s (r1 )ψ1s (r2 )χs ,
√
where χs stands for the singlet S = 0 wavefunction (↑e ↓p − ↓e ↑p )/ 2. The
wavefunction is an exact solution (with lowest energy eigenvalue) for the
problem neglecting the interelectron Coulomb repulsiion e2 /(4π²0 r). When
the repulsion is added, it prevents an exact solution,since the Hamiltonian
is no longer the sum of one term corresponding to electron 1 and another
corresponding to electron 2 independently. So the best we can do is consider a wave function of this form as a variational guess.
(b) Examine the first two excited states in the diagram, labelled 2S (Ortho)
and 2S (Para). Explain what the symbols, as well as the terms ortho and
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para, mean. Explain qualitatively why the Pauli principle forces the parastates with the same quantum numbers to be higher than the ortho- ones.
Para=spin singlet, total spin S = 0, ortho means spin triplet, S = 1. In
the symbol 2S, S means total L = 0, and 2 means it’s the first excited level
of Helium in the approximation neglecting e2 /(4π²0 r), i.e. E = E1 + E2 ,
where En are the Bohr levels for Z = 2. Triplet states have a spatially
antisymmetric wavefunction, which vanishes when r1 = r2 , therefore the
probability of the two e− ’s being close is small. This lowers the interelectron Coulomb energy he2 /(4π²0 r)i relative to the singlet state.
(c) Call the eigenvalues of the Z = 2 hydrogenic problem En . Give perturbation theory expressions for the energy (not just the change in energy)
of all levels in the diagram labelled N L when one includes the interelectron Coulomb interaction, e2 /(4π²0 r12 ), in both the i) ortho and ii) para
case. You may assume that all states shown are combinations of a 1s electron and an n` electron. Do NOT attempt to evaluate the integrals you
encounter.
E1para = 2E1 + h1s1s|
e2
|1s1si
(4π²0 r12 )
e2
1
(h1sn`| + hn`1s|)
(|1sn`i + |n`1si)
2
(4π²0 r12 )
1
e2
= E1 + En + (h1sn`| − hn`1s|)
(|1sn`i − |n`1si)
2
(4π²0 r12 )
(1)
Enpara = E1 + En +
(2)
Enortho
(3)
(d) Next discuss the effect of spin-orbit coupling on the states shown. What
are the possible term symbols S LJ for the 2S states? For the 2P states?
Into how many states does the ortho 2P state split in the presence of
spin-orbit coupling (why)?
In general spin-orbit coupling (fine structure) will split the states shown.
For the 2S states, since a 1s and a 2s electron are being combined, the
only possible orbital angular momentum is ` = 0, but one can have spin 1
or 0. Therefore the possible terms are 3 S1 and 1 S0 . For the 2P states, we
have total ` = 1, still S=1 or 0, and using the triangle rule we get 3 P2 , 3 P1 ,
3
P0 , 1 P1 . So ortho (S = 1) 2P splits into three levels, with degeneracies 1
3
( P0 ), 3 (3 P1 ) and 5 (3 P2 ), respectively.
3. Particle on ring
(a) A particle of mass M moves freely on a ring of radius a. Write down the
energy eigenfunctions ψn and eigenvalues En . If you know them or can
guess them there is no need to derive them, but if you have
forgotten,
you
³
´
∂ψ
1 ∂
1 ∂2ψ
2
might need the Laplacian in polar coords., ∇ ψ = ρ ∂ρ ρ ∂ρ + ρ2 ∂φ2 .
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Eigenfunctions are ψn = einφ , n integer, e’values En = h̄2 n2 /(2ma2 ), n =
0, ±1, ±2....
(b) A potential is now applied to the ring of form V (φ) = A cos φ sin φ, with
A a constant. Calculate the matrix Vnn0 = hn|V |n0 i, n, n0 = ±1.
Z
dφ2π
0 cos φ sin φ = 0
h1|V |1i = A
Z
dφ2π
0 cos φ sin φ = 0
h−1|V | − 1i = A
Z
−2imφ
dφ2π
cos φ sin φ = −i
0 e
h1|V | − 1i = A
Z
π
2imφ
dφ2π
cos φ sin φ = i ,
0 e
2
h−1|V |1i = A
π
2
so


V =


0
iA/4
−iA/4 
0
.

(c) Calculate the first-order energy shifts of the state(s) with n = ±1 using
degenerate perturbation theory, and find the “good” eigenstates in which
V is diagonal.
The two states | ± 1i are degenerate for A = 0, and the matrix V is not
diagonal, so we have to use full-fledged degenerate perturbation theory,
and find the basis in which the matrix becomes diagonal. The eigenvalues
are found from the characteristic polynomial, δE (1) = ±A/4. To find the
eigenvectors you can use any method you like, e.g. consider the equations
V | ± 10 i = ±A/4| ± 10 i, and put |10 i = a|1i + b| − 1i, etc., then solve for
coefficients, remembering normalization of wave function. Find
1
|10 i = √ (| − 1i + i|1i)
2
1
| − 10 i = √ (|1i + i| − 1i)
2
(d) The potential is now removed (A = 0), and a second identical particle
added to the ring. Both particles have spin 1. Write down i) the Hamiltonian, ii) a valid wavefunction for the ground and first excited states of the
two particles, and iii) the energies of these states. If you could not do part
a), you can still express your answer in terms of the ψn which are solutions
to that part.
Hamiltonian is H = p21 /(2m) + p22 /(2m). Ground state is constant,
Ψ0 (φ1 , φ2 ) = 1 · χt , E = 0, where χt is triplet spin state of 2 spin-1
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particles. For 1st excited state we can put one particle in the n = ±1
state,
³
´
³
´
Ψ±1,+ (φ1 , φ2 ) = 1 · e±iφ2 + e±iφ1 · 1 χt
Ψ±1,− (φ1 , φ2 ) = 1 · e±iφ2 − e±iφ1 · 1 χs
overall symmetric under 1 ↔ 2 since spin-1 particles are bosons.
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