university of waterloo - chapter 7b – electron spin...

Winter 2013 Chem 356: Introductory Quantum Mechanics

Chapter 7b – Electron Spin and Spin-Orbit Coupling.................................................................................96

H-atom in a Magnetic Field: Electron Spin.............................................................................................96

Total Angular Momentum...................................................................................................................103

Chapter 7b – Electron Spin and Spin-Orbit Coupling

H-atom in a Magnetic Field: Electron Spin

If electron in orbital has angular momentum , one has a magnetic moment

This magnetic moment can interact with a magnetic field and the interaction energy is given by

is measured in Tesla 1T = 1 Newton/(ampere meter)

If we take to be in direction then

And the total Hamiltonian would be

The wavefunctions we obtained are also eigenfunctions of

including the magnetic field

each level splits in sublevels

This is what would be expected, from classical considerations. This is not what is observed!

96


Stern and Gerlach passed silver atoms (with electron in s-orbital) through inhomogeneous magnetic field.

1922!

They found this splits the beam into two. Classically one would expect lines for particle with

angular momentum . Here !!This was the first evidence that electrons need another quantum number: half-integer angular momentum.

Jumping to what we know now, we introduce an intrinsic angular momentum operator

This operator is postulated to have exactly the same combination relations as

Hence

From out general discussion we know the possible eigenstates as

Where I indicate the states as : Dirac bracket notation (page 159. MQ)

The splitting of silver atom beam indicates eigenstates

97


Everything we discussed on operators and angular momentum applies to spin, also the states , are orthonormal

What are the ‘coordinates’ that we integrate over? We do not have a clue!! But, also, we do not need it:

We know , are orthogonal because they are eigenstates of a Hermitian operator

with different eigenvalues . Hence they must be orthogonal! (See chapter 4).

We now have 2 sets of angular momentum operators

: orbital angular momentum

: spin angular momentum

The operators all commute, and they also commute with the Hamiltonian

, for the H-atom

Then we can characterize each eigenstate through

5 quantum numbersLet us facilitate the notation somewhat

Note: Later on this will require modification, as we are neglecting relativistic effects, which introduce so-called spin-orbit coupling.

98


We can indicate the -quantum number as s, p, d and get functions

, , We could then write, including spin

, ,

, , In what follows we will focus on one particular quantum number, which we can suppress. Moreover I

can indicate the -spin function by an overbar.

Then we get the 6 p-functions

, , , , , Or the d-functions

, , , , , , , , ,

This indicates the quantum numbers

In this way we can label the exact eigenstates of the (non-relativistic) Hamiltonian.

The splitting of the lines in a magnetic field would then be determined by the Hamiltonian

where

This factor determining the ratio between spin and orbital interactions with the magnetic field, can be calculated using relativistic quantum field theory. (Schwinger, Tomanaga, Feynman). Far beyond our aim.

It agrees to about 10 digits with the experimental value! (that is like measuring the distance from here to New York up to a millimeter!)

The , functions are eigenfunctions of this magnetic Hamiltonian, and we can easily calculate the energy splitting.

Unfortunately, this does not give correct results!!

The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity. They are of at least comparable importance, and cannot be neglected when discussing magnetic effects.

99


The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss.

One can approximate effects by including spin-orbit interaction in the Hamiltonian.

It is called coupling or spin-orbit coupling

Then the magnetic interaction can be included as

Let us first examine the energy levels for the non relativistic Hamiltonian

Let us denote and use , then

Let us consider the allowed emission lines: Normal Zeeman effect:

100


, only include include

Due to all -levels go up by all -levels go down by . Since transitions cannot change spin,

, I get same transitions as without spin!!

Our conclusion thus far.

If one does not consider spin levels split in a magnetic field using

3 equal spaced levels

5 equal spaced levels

If we include spin, for single electron states then all -states shift up by 1 unit of , all -levels shift

down by 1 unit of , and the transition energies are not affected by spin. Moreover all

multiples are split by the same amount . We would not see the effects of spin.

This is what was originally observed. It is called the ‘normal Zeeman effect’ and it was explained by Lorentz (two Dutch physicists).

However we do observe the effects of spin in emission spectra! The story is more complicated.

The complications occur already for the H-atom without a magnetic field. There is a substantial correction due to what is called spin-orbit interaction.

A good way to think about this is as follows:We usually think of the electron as wizzing around the nucleus. From the point of view of the

electron we can just as easily think that the nucleus is wizzing about the electron. This moving nucleus, with its angular momentum generates a magnetic field. This magnetic field interacts with the spin of the electron.

Compare the electron with your position on the spinning earth. The sun rises and sets from our standing still point of view, and moves with incredible velocities, in this frame. For a charged particle the magnetic force would be large.

101


It is called spin-orbit coupling

Spin-orbit coupling is usually said to be a relativistic effect. This is because it arises in a natural way from the fully relativistic Dirac equation. So does spin; it arises naturally. And so do particles and antiparticles, which also arise from the Dirac equation.

Let me say something more about spin. The spin operators are best represented by matrices.

, ,

These matrices satisfy the commutation relations of angular momentum

Moreover

Hence

matrices have only 2 eigenvectors. This is why we have only ,

The Dirac equation is a matrix equation and we get (2 spin 2 mass) solutions.

The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity. They are of at least comparable importance, and cannot be neglected when discussing magnetic effects.

The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss.

One can approximate the effects by including spin-orbit interaction in the Hamiltonian.

It is called coupling or spin-orbit coupling

102


Then the magnetic interaction can be included as

Our original spin-orbitals are however not eigenstates of the including coupling

We can classify the eigenstates of by doing a little more angular momentum theory.

Let me sketch the result, as this is, finally, an accurate description.

Total Angular Momentum

We have and

Moreover

This is because and act on different coordinates

Now we define total angular momentum

; ;

and satisfy the usual commutation relations:

Moreover

And

103


Likewise

But then also

And

Hence we can derive without too much trouble that the operators and all commute.

Moreover these operators commute with , and also with .

It then follows that the angular momentum operators , commute with the relativistic

Hamiltonian . And we can classify the states with quantum numbers

As for the non-relativistic case, the angular momentum problem, defining is independent from the radial equation, and can be solved once and for all.

These equations hold for the Hydrogen atom, but later on we will see that they are very similar for many-electron atoms, which also have spherical symmetry.

Can we deduce what the eigenstates of and might be?

First we note that eigenfunctinos of are easy.

Eg.

104


We also know that acting with on should give 0: the highest value in the multiplet.

It is easy to find the highest function: ,

Hence in the p-manifold , is the highest function.

In the d-manifold it is with

Acting with we can get the other function in the multiplet.

We get in functions

We are left with 2 functions in the p-manifold. This creates a multiplet.

, 2 functions.

How does this work, for the d- functions? highest em .1 522 2j j jm l s

52

j

5 3 1 1 3 5, , , , ,2 2 2 2 2 2jm

2 1J = 6 functions

In total we have 10 functions other multiplet has 4 functions

12

j l 1 322 2

105


For the Hydrogen atom we can construct always12

j l

12 1 2 22

l l

12

j l

12 1 22

l l

2 2 1l

2 2 1l is the total number of spin-orbitals of angular momentum l .

We label the final eigenstates as 2 1s

The j-multiplet have slightly different energies due to the 2 2 22L S J L S coupling, which for

equal ,L S is seen to depend on the J quantum number.

Finally, we are ready to discuss how these relativistic levels split in a magnetic field;

Within a multiplet J the levels split in 2 1J equally spaced levels due to a magnetic field

106


The splitting depends on , ,n l j

Allowed transitions: 1l , 0s , 0, 1Jm

All different energies. Tiny splitting, but this is how we can experimentally access degeneracies of energy levels.

The splitting of energy levels in a magnetic field is a complicated subject, because relativistic effects have to be considered at the same time.

Magnetic transitions using nuclei as in NMR are much simpler as we only need to consider angular momentum theory itself.

Better picture of transitions including spin-orbit:

0B 10 different transitions, 10 different frequencies

Selection rules

1l , 0s , 0, 1j , 0, 1jm

No 0 0J J transitions

2 22 3 1 1

2 2

P s3 transitions 0, 1jm

2 22 1 1

2 2

P s2 transitions

107


2 23 12 2

2 1P s

3 transitions

2 21 12 2

2 1P s

2 transitions10 transitions in total

To good approximation one can evaluate the shifts in magnetic field from 2 ˆˆ

2 2z ze e

e eL Sm m

Neglecting spin-orbit coupling.

108

university of waterloo - chapter 7b – electron spin...

Documents