Download Lecture 17-PDF

Lecture 17 Title :Fine structure : Spin-orbit coupling
Page-0
In this lecture, we will concentrate on the fine structure of the one electron atoms.
The cause of this fine structure is the interaction between the orbital angular momentum
and the spin angular momentum.
We will review the origin for this interaction and the effect of the coupling between
orbital and spin on the spectral lines.
We will discuss here mainly hydrogen atom and also the other alkali atoms.
Page-1
 It is observed in the sodium spectrum (shown in the figure below) that D-line (yellow
emission) is split into two lines
D1 = 5895.93 Å or 589.59 nm, D2 = 5889.9 Å or 588.99 nm
 The reason of observing this doublet is that the energy levels split into two for the
terms except s-level (l = 0).
 Many lines of the other alkali atoms are also doublets.
 With high resolution spectrometer shows that even hydrogen atom lines are also
having this doublet nature.
 The Coulomb interactions between the nucleus and electron in the outermost orbit for
alkali atoms can not explain this observation
 Not only alkali atoms, but also in the other multielectron atoms the transitions between
the terms split into more number of transitions. This is known as Fine structure.
Emission from Sodium lamp
4000
3500
5895.93 Å
5889.96 Å
3000
2500
2000
Series1
1500
1000
500
0
580
585
590
Wavelength (nm) →
595
Page-2
In the previous lectures, we have discussed that the Hamiltonian for describing one
electron atoms (Alkali atoms) is
2 2
−
∇ ψ ( r , θ , φ ) + V ( r )ψ ( r , θ , φ ) =
Eψ ( r , θ , φ )
2µ
where V ( r ) is the Coulomb interaction between the nucleus and electron.
We have also discussed the Hamiltonian needed to explain multielectron atoms
=
H H * + H1
2 N 2 N
∑ ∇i + ∑U ( ri )
2me i 1 =i 1
=
where H * =−
where
N
Z e2
=
−
+
U
r
(
)
∑i =
∑
i
ri
i 1
e2
and=
H1 ∑ −
i < j rij
N
e2
∑
i < j rij
e2
∑
i < j rij
N
N
Non-Spherical part only
However, these descriptions are not enough to explain the observed splitting. We need to
include the spin-orbit coupling into the Hamiltonian.
In the following we will understand the origin of this spin orbit interaction.
Page-3
Orbital magnetic dipole moment:
Let us consider that an electron is moving with velocity V in a circular Bohr orbit of
radius r that produces a current. This current loop will produce a magnetic field with the
magnetic moment,
eω 2
1
µl =
iA =
πr =
−
− eω r 2
2π
2
eω
Where i = −
, ω is the angular velocity, A is the area.
2π
µl
v
r
-e
l
Magnitude of orbital angular momentum
=
l m=
mω r 2
eVr
Where, me is the mass of the electron. Substituting we get, the orbital magnetic moment
µl = −
e
l
2m
Now, the Bohr magneton is defined as the magnetic moment of the first Bohr orbital. So,
the Bohr magneton=
µB
e
=
 9.27x10-24 J/T
2m
The Orbital magnetic moment in terms of Bohr magneton is thus, in the vector form

µl = −
gl µ B 
µ 

l and putting gl = 1, µl = − B l


Substituting the value of the angular momentum

µ=
l
µB

l (l + 1)=
 µ B l (l + 1)
Along the Z-direction, the component of lZ = ml , so
µ
µ
µl =
− B lz =
− B ml  =
− µ B ml
z


ml 
z 
l

=
l
l (l + 1) 
Page-4
When this dipole moment is placed in an external magnetic field along the Z-direction, it
  
experiences a torque which can be expressed as τ= µ × B
The potential energy
∆E =− µˆ B ⋅ Bˆ
For a static magnetic moment, this torque tends to line up the magnetic moment with the
magnetic field B, so that it reaches its lowest energy configuration.
Here, the magnetic moment arises from the motion of an electron in orbit around a
nucleus and the magnetic moment is proportional to the angular momentum of the
electron.
The torque exerted then produces a change in angular momentum. This change is
perpendicular to that angular momentum, causing the magnetic moment to precess
around the direction of the magnetic field rather than settle down in the direction of the
magnetic field.
This precession is known as Larmor precession.
∆φ
l Sinθ
Z BZ

l
θ

µl = −
Figure-17.1
e 
l
2me
Page-5
When a torque is exerted perpendicular to the angular momentum l, it produces a change
in angular momentum ∆l which is perpendicular to l. Referring to figure-17.1, the torque
is given by,
τ
=
∆l l sin θ∆φ
=
= lω sin θ ………………………….(17.1)
∆t
∆t
And also


τ = µ × B = µl B sin θ =
µB

lB sin θ ……………………(17.2)
µB
So equating equations 17.1 and 17.2 we get

lB sin θ = lω sin θ
µ
=
> ω =B B

This is known as Larmor frequency.
Spin magnetic moment :
Similar to orbital dipole moment, electron also produces the magnetic moment due to the
spin angular momentum.
Z
The spin dipole moment, in terms of spin Lande g-factor g s .
gs µB 
s

µ sz = − g s µ B ms

µs = −
Where, µ SZ the component in the Z-direction

µS

s
Page-6
As we discussed above, that the fine-structure in atomic spectra cannot be explained by
Coulomb interaction between nucleus and electron.
We have to consider magnetic interaction between orbital magnetic moment and the
intrinsic spin magnetic moment. This is known as Spin-Orbit interaction.
Let us understand how this interaction takes place.
If we consider the reference frame of electron then nucleus moves about electron. This


current j = − ZeV produces magnetic field at the electron.
According to Ampere’s Law, the magnetic field at electron due to nucleus is


 µ0 j × r − Zeµ0 V × r
B =
=
r3
4π r 3
4π
 

 
Since l =
r × mV =
−mV × r


 − Zeµ0 mV × r Zeµ0 l
B =
=
r3
4π m
4π m r 3
If we take the average field, then
 Zeµ0 1 
B=
l ……………..(17.3)
4π m r 3
+ Ze

r
v
−e
Page-7
 
Now, the orientation potential energy of magnetic dipole moment is ∆ESpin −Orbit =
−µs ⋅ B
gµ 

We know that µs = − s B s

gs µB  
∆ESpin −Orbit =
s ⋅B

Transforming back to reference frame with nucleus, must include the factor of 2 due to
Thomas precession (Reference : Eisberg & Resnick):
1 gs µB  
s ⋅B
2 

This is the spin-orbit interaction energy. Substituting the value of B from equation 17.3,
we get
∆=
Eso
∆Eso
=
 
g s µ B Zeµ0 1  
⋅
=
s
l
A
l
SO ⋅ s …………………………..(17.4)
2 4π m r 3
Page-8
Let us first discuss about the alkali atoms. The Hamiltonian needed for calculating the
energy levels is
2 2
∇ + V ( r ) + H Spin −Orbit
H =−
2µ
= H 0 + H Spin −Orbit
We already know the energies calculated for H0 . We can treat H Spin −Orbit as perturbation
on the energies of H0.
But first let us see the effect of this interaction.
This interaction couples the spin and orbital
angular momentum to form the total angular
momentum J.
The figure 17.2 represents the vector diagram
and accordingly we write
  
j= l + s (for one electron atoms).
Because of this coupling, the one electron
wavefunctions l ml s ms which are the
eigenfunctions of H0 are no more the
eigenfunctions of the total Hamiltonian H.
The eigenfunctions of the H will be J mJ
which are the coupled wavefunctions and can be
derived from the uncoupled wavefunctions
l ml s ms , as described in previous lectures.
So the perturbation energy

∆ESO =
J mJ ASO l .s J mJ
  
Now, j= l + s

j 2 = l 2 + s 2 + 2l .s
 1 2 2 2
 j − l − s 
l=
.s
2
  
j= l + s

l

s
Figure-17.2
And thus, the energy correction due to the spin-orbit interaction is

ASO
=
∆ESO
J mJ ASO l .s J=
mJ
[ j ( j + 1) − l (l + 1) − s( s + 1)] ………………….(17.5)
2
Page-9
Let us first look at the Sodium energy levels.
The terms arising from the ground state configuration is 2 S .
So=
l 0,=
s 1 . According to coupling of angular momenta j = 1
2
2
2 s +1
The notation we will use is
LJ
According to this the ground state energy level is 2 S 1 . Using the relation given in
2
equation 17.5, we get
A1  1 1
2
( + 1) − 0(0 + 1) − 1 ( 1 + 1) 
∆ESO=
S1
2 2

2
2  2 2
=0
( )
Now for the first excited state, terms is 2 P
So=
l 1,=
s 1 . According to coupling of angular momenta j = 1 , 3
2
2 2
2
2
Now this terms will split into two energy levels P1 and P3
2
( )
2
∆ESO
=
P1
2
A2  1 1
( + 1) − 1(1 + 1) − 1 ( 1 + 1) 
2 2

2  2 2
= − A2
And
( )
2
∆ESO
=
P3
2
A2  3 3
( + 1) − 1(1 + 1) − 1 ( 1 + 1) 
2 2

2  2 2
A
= 2
2
2
Page-10
So the construction of the energy levels is
2
2
A2
P
A2
2
S
2
2
2
E0
P3
P1
2
5895.93 Å
(2) 2 S
5889.96 Å
(1)
1
2
Now let us calculate the spin orbit constant for 2P level.
For (1) the transition energy : ν 1 (cm −1 ) =
E0 +
A2
108
16978.04 cm −1
=
=
2 5889.96
For (2) the transition energy : ν 2 (cm −1 ) = E0 − A2 =
108
= 16960.85 cm −1
5895.93
Solving these two we get A2 = 11.46 cm −1 and E0 = 16972.31 cm −1
Page-11
Similar to sodium atom, Hydrogen atom also shows doublet.
Spectral lines of H found to be composed of closely spaced doublets. Splitting is due
to interactions between electron spin s and the orbital angular momentum l
Hα line is single line according to the Bohr or Schrödinger theory. occurs at 656.47
nm for Hydrogen and 656.29 nm for Deuterium (isotope shift, λ∆~0.2 nm).
Spin-orbit coupling produces fine-structure splitting of ~0.016 nm corresponds to an
internal magnetic field on the electron of about 0.4 Tesla.
Orbital and spin angular momenta couple together via the spin-orbit interaction
Internal magnetic field produces torque which results in precession of l and s about
their sum, the total angular momentum:
This kind of coupling is called L-S coupling or Russell-Saunders coupling
2
2
P3
P
2
10.2 eV
121.6 nm
2
2
P1
2
splitting
4.5×10-5 eV
S
2
S1
2
Page-12
Relativistic kinetic energy correction :
According to special relativity, the kinetic energy of an electron of mass m and velocity v
is:
p2
p4
where p is the momentum
T≈
−
2m 8m3c 2
The first term is the standard non-relativistic expression for kinetic energy. The second
term is the lowest-order relativistic correction to this energy.
Using perturbation theory, it can be show that
3 
Z 2α 4
 1
∆Erel =
− 3 mc 2 
− 
n
 2l + 1 8n 
This energy correction does not split the energy levels, it only produces an energy shift
comparable to spin-orbit effect.
So the total energy correction for the fine structure
∆EFS = ∆Eso + ∆Erel
As En = -Z2E0/n2, where E0 = 1/2α2mc2, we can write
EH − atom
Z 2 E0  Z 2α 2  1
3 
=
− 2 1 +
− 

n 
n  j + 1/ 2 4n  
Page-13
So we found out that
Energy correction only depends on j, which is of the order of α2 ~ 10-4 times smaller
that the principle energy splitting.
All levels are shifted down from the Bohr energies.
For every n>1 and l, there are two states corresponding to j = l ± 1/2.
States with same n and j but different l, have the same energies i.e., they are
degenerate.
L=l = 0
L=l = 1
L=l = 2
n=3
3 2S 1
∆EFS
2
n=2
2 2S 1
2
1 S1
2
P1
2
D3
2
2
2
∆EFS
2
2 2
∆EFS
n =1
2
2
P3
P3
P1
2
2
2
D5
2
Page-14
Here, we learnt that due to the orbital motion of the electron, the charge of the nucleus
creates a magnetic field at the electron and this interacts with the spin of the electron.
This interaction essentially couples the orbital and the spin angular momenta and the
effect is the splitting of energy levels.
This transition pattern is known as fine structure.
The coupled angular momentum J is the good quantum number when spin-orbit coupling
is introduced for energy level calculation.
Classical explanation of spin-orbit interaction is not enough for the level having l = 0.
The quantum explanation reveals that for describing the l = 0 level, the Fermi contact
term is needed which does not have any classical analogue.
Relativistic correction is needed to predict the accurate energies.