### CHM 4412 Chapter 13

```Lecture 22
Spin-orbit coupling
Spin-orbit coupling




Spin makes an electron act like a small magnet. An
electron orbiting around the nucleus also makes a
magnet. These two magnetic moments can interact
and, depending on the relative orientations of the
two moments, orbital energy can be slightly altered.
We use the so-called Na D line as a paradigm.
We use the first-order perturbation theory to
describe the shifts in orbital energies.
The spin-orbit interaction is a relativistic effect and
its derivation is beyond the scope of this course. We
treat it as a phenomenological effect explained in
analogy to two interacting magnets.
Na D line


The orange color of the sodium lamp is due to
the Na 3p→3s emission at ca. 17000 cm−1.
A close examination of this transition reveals
that the emission band consists of two bands
separated by 17 cm−1.
Public-domain image from Wikipedia
Spin-orbit coupling

Spin of an electron makes
it a magnet. Orbital motion
of the electron also makes
it a magnet. These two
magnetic moments can
interact or “couple” (spinorbit coupling) and cause
energy level splitting.
N N
S S
N S
S N
Sum of angular momenta


Each electron has two angular momenta (a
dual magnet): orbital angular momentum and
spin angular momentum.
The total momentum is the most naturally
j =l +s
Total
Orbital
Spin
s
l
j
Sum of angular momenta


j
must be (space) quantized.
So its total angular momentum
quantum number j is either a full
or half integer in the range:
j = jmin (0 or greater), jmin+1,…,
jmax–1, jmax
j =l +s
jmax = l + s = l + 12
jmin = l - s = l - 12
Examples

Identify the levels that may arise from the
configurations (a) (3p)1, (b) (3s)1.
Examples


(a) 3p orbital → l = 1. j = l ± ½ = 3/2 or 1/2.
(b) 3s orbital → l = 0. j = 0 + ½ = ½ (j = 0 – ½
is not allowed because j is non negative).
Spin-orbit coupling


Two magnets are the most
stable when they are
antiparallel and the least
stable when they are parallel.
In general, the energy due to
the interaction of spin and
orbital momenta should be
ESO µ s ×l = s × l cosq
l
θ
s
Spin-orbit coupling operator

The atomic Hamiltonian does not have this:
H =
2
Ñ +V
2
2m
This is because we do not have a counterpart
in the classical energy, from which the
Hamiltonian is derived. We add spin-orbit
interaction operator:
H =-
2
Ñ +V + ( hcA)
2
2m
sˆ × lˆ
2
Spin-orbit coupling operator

The spin-orbit interaction operator
H =-

2
Ñ +V + ( hcA)
2
2m
sˆ × lˆ
2
has the spin-orbit coupling constant A. It is
in units of cm−1, which is why hc is multiplied.
The value of A is extracted from experiment
(11.5 cm−1 for Na 3p from the splitting of 17
cm−1) or relativistic quantum mechanics.
Homework challenge #6



Study the special theory of relativity. One of
the best textbooks is “Special Theory for
Relativity for Beginners” by Jürgen Freund.
Study Dirac’s theory of relativistic quantum
mechanics and explain how it introduces the
concepts of spins and positrons from the first
principles.
Study the work of Pekka Pyykkö on the effect
of relativity on chemistry.
Spin-orbit coupling operator

The spin-orbit interaction operator
H =-

2
Ñ +V + ( hcA)
2
2m
sˆ × lˆ
2
makes the solution of the Schrödinger
equation difficult.
Since A is very small (0.001 of 3p-3s energy
difference), we use perturbation theory.
First-order perturbation theory
E
(1)
SO
( )Y
ˆ × lˆ
hcA
s
(0)*
= ò Y ls
2
(0)
ls
dt
j × j = (l + s ) ×(l + s ) = l × l + s × s + 2s × l
s ×l =
E
(1)
SO
1
2
( j × j - l ×l - s ×s)
( )Y
ˆ × lˆ
hcA
s
(0)*
= ò Y ls
= 12 hcA ò Y (0)*
ls
2
(0)
ls
dt
ˆj 2 - lˆ2 - sˆ2
2
Y (0)
dt
ls
= 12 hcA{ j( j + 1) - l(l + 1) - s(s + 1)}
Na D line
E3/ 2 
E1 / 2 
1
2
hcA  23  52  1  2 
1
2
hcA  12  23  1  2 
1
2

1
2
3
2


3
2
1
2
hcA
   hcA
Na D line
4-fold degenerate
3
2
hcA = 17 cm -1
A = 11.5 cm -1
2-fold degenerate
Spin-orbit coupling constants

The measured values of A:






Li: 0.23 cm–1
Na: 11.5 cm–1
K: 38.5 cm–1
Rb: 158 cm–1
Cs: 370 cm–1
Spin-orbit coupling arises from the special
theory of relativity and greater for the heavier
elements because the 1s electrons in high-Z
elements can go nearly as fast as the speed of
light.
Consequences of SO coupling





An electron in each orbital no longer has a well
defined spin (magnetic quantum number, α or
β).
States are no longer pure spin-singlet, doublet,
triplet, etc.
Radiative transitions between singlet and triplet,
between doublet and quartet, etc. become
weakly allowed (phosphorescence).
triplet, etc. become weakly allowed
(intersystem crossing).
These are more prominent in heavier elements.
Singlet and triplet states

The singlet and triplet states have different
spin and spin magnetic momenta. They are
orthogonal functions if it were not for the
SO interaction..
spatial
spin spin
ˆ
TDM = ò Y f zˆY i dt = ò Yspatial
z
Y
d
t
Y
f
i
ò f Y i dt
This is zero when final and initial
states have different spin
eigenfunctions, e.g., singlet and
triplet.
Separable because z
operator does not act on
spin part.
Fluorescence &
Phosphorescence
Fluorescence
Fluorescence is an emission of
light between the same spin
states (e.g., singlet to singlet).
Since this is an allowed
transition, it is intense and fast.
Intersystem crossing
Phosphorescence is
between different spin
states and mediated by
SO. It is “forbidden” and
it is weak and slow.
Both public-domain images from Wikipedia
Phosphorescence
Summary




Spin angular momentum as a magnet and
orbital angular momentum as another magnet
interact (spin-orbit coupling).
Spin-orbit coupling is a relativistic effect and
is greater for heavier elements.
It causes splitting of subshell states,
phosphorescence, and intersystem crossing.
The first-order perturbation theory describes
the coupling.
```