Does instruction lead to learning?
A mini-quiz – 5 minutes
1. Write down the ground state wavefunction of the hydrogen atom?
2. What is the radius of the ground state of the hydrogen atom?
3. What is the radius of the n=120 state (L=119) of the hydrogen atom?
4. What is the radius of the n=120 state (L=119) of the 91-times ionized
uranium atom? (ignore relativity in this and the next question)
5. What is the radius of the ground state of neutral uranium ?
6. How does relativity change the answers to Q4 and Q5?
7. Draw an energy level diagram indicating binding energies of the lowest
10 states in neutral helium – as close to scale as possible (giving their
approximate binding energies)
They are all the SAME!
Atomic Physics
Hyperfine structure
Elementary version
Initial theory of hyperfine structure
(1) The nucleus also can have an intrinsic angular momentum –
The dipole form of this is called the nuclear spin.
Note – for later:
Higher order multipole distributions of the nuclear charge and “current” can
also occur.
(2) The electromagnetic field caused by the electronic wavefunction at the nucleus
leads to an interaction between the two:
#1 – the Coulomb interaction with the nuclear charge – done
#2 – the interaction between the nuclear spin and the electronic B-field
#3 – higher order multipole interactions – we will consider a more complete
theory later
#2 – The dipole interaction energy (Hamiltonian) is
W(hfs-dipole) = - μ (nuclear) . B (elect)
looks just like spin orbit interaction, except μ is less by m/M
Hyperfine structure
magnetic dipole term
We can define the magnetic dipole moment of a nucleus in terms of its
intrinsic spin (known as the nuclear spin) I.
μ = (μI/I) I
Where μ, and I are vectors, and μI is the nuclear magnetic moment
usually quoted in nuclear tables.
Important note: The Wigner-Eckart Theorem: The components of
any vector in a given subspace are proportional to the components of the
angular momentum vector of that subspace.
You just need to find one proportionality factor – as given here.
(usually use the z-direction mJ=J)
Hence, the magnetic field due to the electron wavefunction(s)
B = J B0/J ħ
where B0 is the magnetic field at the nucleus in the (e.g.) z-direction
(Jħ=mZ)
Hyperfine structure
magnetic dipole term
The interaction energy is
ΔE(hfs) = -μ ● B = A I ● J
A is a constant
= - <BZ> μ/(IJħ2) I ● J
= - <BZ> μ/(2IJħ2) {F(F+1)-I(I+1)-J(J+1)}
=a/2 {F(F+1)-I(I+1)-J(J+1)}
Note: A and a are fairly standard notation in text books
What is the magnetic field at the
nucleus? – s-states
S-states: The Fermi contact term
– only spin contributes to the magnetic field
We have a spherically symmetric charge distribution, with a non-zero
spin-magnetization density at r=0, P -> B=8πP/3 (classical)
=-8πμ0/3 | Ψ(0) |2.
(μ0 = vacuum permeability)
Collecting terms
as = (μ0/4π) ● (2μBμI/I) ● (8π|/3) |Ψ(0)|2
In hydrogenic systems, with nuclear charge Z, |Ψ(0)|2 is just the
normalization constant: Z3/(πa03n3)
In other 1-electron systems (e.g. alkalis),
just put in the shielded nuclear charge ….
The directions of the B-field (labeled H) and magnetic moments in
one electron S-states and P-states
What is the magnetic field at the
nucleus? – non-s-states (1)
Two parts – orbital motion and electron spin:
The magnetic field at the nucleus
B = (μ0/4πr3) ● {(e v x r) – μs + 3(μs●r )r/r2}
Where μs = 2μB S
e r x v =2 μB L
Define the resulting vector in {} brackets by N. 2μB gives us an interaction energy
proportional to I●N
Use Wigner-Eckart theorem to find component of N along J:
Then ΔE = (μ0/4π)● (2μBμI/I) ● <(N●J)>/J(J+1) ●<1/r3> proportional to <I●J>
= aJ/2 ● {F(F+1)-J(J+1)-I(I+1)} - definition of aJ
What is the magnetic field at the
nucleus? – non-s-states (2)
Calculate N●J ……
= { L – S + 3(S·r)r/r2) } ● (L + S)
= L2 – S2 + 3(S·r)(r·L)/r2) + 3(S·r)(r·S)/r2)
But r·L = r·(rxp) = 0 -> 3rd term is zero
The second term and the last term cancel!
-check by components and commutation relations for spin
-Thus
< N●J > = < L2 > a nice result!
-And
aJ = (μ0/4π) ● (2μBμI/I) ● <1/r3> L(L+1)/[J(J+1)]
Result for hydrogen ground state
The ground-state has L=0, S=1/s -> J=1/2
Spin of the proton I = ½
HFS can be calculated exactly – to give E = 1.42 GHZ
What are these 2 ground-state levels?
Result: experiment differs from theory by about 1 part in 103!
Why? – What is the problem (too much for relativity)?
Result for hydrogen ground state
The ground-state has L=0, S=1/s -> J=1/2
Spin of the proton I = ½
HFS can be calculated exactly – to give E = 1.42 GHZ
What are these 2 ground-state levels?
Result: experiment differs from theory by about 1 part in 103!
Why? –
Answer: - the anomalous magnetic moment of the electron…
gs = 2(1+ α/2π + higher order terms….)
The hydrogen maser uses this frequency
E = 1,420,405,751.7662 +- 0.0030 Hz
Example of hyperfine structure
What is the spin
of this nucleus?
Example of hyperfine structure
What is the spin
of this nucleus?
9/2 + I = 8, 7,…..1
->> I=7/2
Note: the regularity of the
separations, gradually decreasing, as
F gets smaller
ΔE{(F-1) – F)} = k. F
The HFS Electric quadrupole
interaction
We will first look at the next term in the electromagnetic interaction
between the nucleus and the electron(s) – this will show us how to
generalize the interaction to all interaction multipole moments…
The “and beyond” part will require the introduction of spherical tensors
and their manipulation…
References for the first part: (these notes should be self-contained and
independently understandable!)
G.K. Woodgate – Elementary atomic structure
H.B.G. Casimir - On the Interaction between Atomic Nuclei and
electrons
(Tweedegenootschap Prize – about 1935)
The electrostatic interaction is given by
H.
For a point nucleus, this reduces to the
Coulomb central field. For a finite
nucleus and several electrons H can be
expanded in specific multipole
contributions.
HFS E2 coordinates
HFS E2 expansion
We can expand the expression for H, using the coordinate system show
1st term – the Coulomb central field;
2nd term – zero - the nucleus has no electric dipole moment! Hence, in
general all “odd” terms should be zero.
3rd term – the electric quadrupole term.
HFS E2 separations
Now we use the following identity to expand in separated spherical harmo
The rank k and the projection q are the same in each angular
momentum space (nuclear and electronic).
Substituting in the quadrupole term, rearranging, and defining Q2 and
F2 gives us 5 terms 
HFS E2 nuclear definitions
We can now evaluate the expression for a state with given I,J,F |IJF>.
Since the interaction energy is very small relative to other electronic
energies, we can assume that I, J and F and their components are all
good quantum numbers.
It is conventional to define the Nuclear Quadrupole Moment in the MI=I
direction and have cylindrical symmetry for the nucleus – we can then
define this one component, and use the Wigner-Eckart theorem to
automatically define the other components
HFS E2 electronic definitions
Again, define the interaction in the z-direction MJ=J,
and use the W.E. theorem to evaluate components …..
Note: we have assumed that the second derivative is proportional
to 1/r3 at the origin.
HFS E2 evaluation
The resulting interaction energy is
Convention-1: define the electric
quadrupole interaction constant B - which is what gets measured in a
hyperfine structure experiment…
Convention-2:
Define K
HFS - M1 plus E2 results
Putting together our results for both the M1 and the E2 contributions:
The energy of any hyperfine level | nJ(LS)I F(IJ) > is
Example of measurements in 55Mn
Note that there is only a small variation from the interval rule
predicted by just the M1 interaction:
Units:
1mK = 10-3 cm-1.
The transition 3d54s2 6S5/2 – 3d54s4p 6P7/2 in neutral manganese