These LaTeX notes contain full solutions to the problems from the past papers for the Part B MPhys examination in B3: Quantum, Atomic and Molecular Physics. Features colour images of diagrams, as well as full explanations of the steps taken.
B3: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins
November 1, 2017
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
toby.adkins@merton.ox.ac.uk
2011
Question 1
The general Hamiltonian of an atom containing N electrons can be written as:
N N
~2 2 Ze2 e2
X X
H= − ∇i − +
2me 4π0 ri 4πrij
i=1 i6=j
where ri radial position of electron i, and rij = |ri − rj |. However, this is not separable due
to the electron-electron interaction term. Furthermore, we cannot treat it perturbatively,
as it is not in general small. Instead, we notice that the majority of the potential felt by
the electrons will be directed radially inwards, meaning that we can introduce some central
potential U (r).
This gives rise to the central
eld approximation, under which we decompose our Hamil-
tonian into:
N
~2 2
X
HCF = − ∇ + U (ri )
2me i
i=1
N N
e2 Ze2
X X
δHRE = − + U (ri )
4π0 rij 4π0 ri
i6=j i
where δHRE is the residual electrostatic Hamiltonian, to be treated perturbatively. U (r)
is some function
e2
U (r) = −Ze (r)
4π0 r
where Ze (r) is some eective charge satisfying the conditions that
Ze → Z, r→0
Ze → 1, r→∞
1
,Toby Adkins B3 Past Papers
Under the central
eld approximation, there is no coupling between electrons, and so the
system is characterised by the individual electron quantum numbers ni , `i , m`i , msi , with
the energy depending only on ni and `i . This means that we can write the state of the
atom in terms of electronic con
gurations:
n i `# #
i . . . nj `j
where 2(2`i + 1) is the maximum number of electrons in each orbital. When including the
residual electrostatic Hamiltonian, there is now coupling between electrons, so ni and `i
are no longer good quantum numbers. Instead, `i and si couple as
X X
L= Li , S= Si
i i
meaning that in the absence of external torques on the system, L2 , Lz , S2 and Sz are
constants of the motion. This means that we characterise our system by the eigenstates
|L, ML , S, MS i. This gives rise to the term:
n i `# #
i . . . nj `j
2S+1
L
where S is the total spin quantum number, and L is the total angular momentum quantum
number in spectroscopic notation. Finally, including the spin-orbit Hamiltonian:
X
δHSO = ζ(ri ) `i · si ∝ L · S
i
causes L and S to couple as J = L + S. Thought of in terms of the vector model, L and
S couple together and precess around the axis de
ned by J.
In this case, J2 , Jz , L2 and S2 commute with the Hamiltonian (and are thus constants
of the motion), meaning that we characterise out states by |J, MJ , L, Si, with the energy
levels being split by J . This gives rise to the levels:
n i `# #
i . . . nj `j
2S+1
LJ
Under the L-S coupling scheme, J = L + S, and our system can by characterised by the
eigenstates, |J, MJ , L, Si. This is valid given that hδHRE i
hδHSO i. Under this coupling
scheme, we want to
nd the projections of `i and si onto L and S respectively (according
to the vector model).
X X h`i · Li hsi · Si
ESO = hδHSO i = hζ(ri )i h`i · si i = hζ(ri )i L· S = βn,` hS · Li
L(L + 1) S(S + 1)
i i
Now,
1
S · L = (J2 − L2 − S2 )
2
so
1
ESO = βn,` [J(J + 1) − L(L + 1) − S(S + 1)]
2
2
,Toby Adkins B3 Past Papers
Then,
1
∆EJ,J−1 ∝ [J(J + 1) − J(J − 1)] ∝ J
2
This is known as the Landé-g interval rule; the splitting under the spin-orbit interaction is
proportional to the J of the upper state.
Singlet triplet splitting - Triplet state has an antisymmetric spatial wavefunction, meaning
that on average, the electrons are located at a greater spatial separation, which decreases
the contribution of the repulsive coulomb term to their energy. This means that the triplet
state is lower in energy than the singlet state.
4s2 0 1S
0 No Triplet
20335 3D
1
20349 3D
2 Triplet
3d4s
20371 3D
3
20850 1D
2 Singlet
The selection rules for the transitions are as follows:
∆` = ±1
∆L = 0, ±1
∆ML = 0, ±1
∆J = 0, ±1
∆MS = 0
∆MJ = 0, ±1
The energy level diagram is as follows:
3
, Toby Adkins B3 Past Papers
Question 2
The spin-orbit interaction is a relativistic correction to the energy of the system that arises
due to the fact that the electron is moving in the central potential of the nucleus U (r).
In the rest frame of the electron (from relativistic electrodynamics), the magnetic
eld
experienced is:
1 1
B=− 2
(v × E) = 2 (v × ∇U )
c c
The magnetic moment of the electron (that depends only on S as we are in its own rest
frame) couples to the magnetic
eld, causing the energy levels to be split according to
J , L and S (as J2 , Jz , L2 and S2 are all constants of the motion). We thus adopt the
eigenstates |J, MJ , L, Si. This is valid under the LS coupling scheme. Thought of in terms
of the vector model, L and S couple together, and precess around an axis de
ned by J.
This is valid assuming that:
hδHRE i
hδHSO i
which is the case for the lighter elements Z . 30.
The
rst order shift in the energy is given by
rst order perturbation theory:
ESO = hj, mj , `, s| HSO |j, mj , `, si
as J = j , MJ = mj , L = ` and S = s for a single electron atom. Assume a Coulomb
potential of the form:
e2 1
U (r) = −
4π0 r
such that
~2 1 e2 1 ~2 e2 ` · s
∆HSO = ` · s =
2m2e c2 r 4π0 r2 8π0 m2e c2 r3
Recalling the de
nition of the Bohr magneton, µB = e~/(2me ), so
µ2B ` · s µ0 µ2B ` · s
∆HSO = 2 3
=
2πc 0 r 2π r3
We then note that
1
h` · si = [j(j + 1) − `(` + 1) − s(s + 1)]
2
1 1
3
= 3
r (a0 n) `(` + 1)(` + 1/2)
4
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