Lecture notes on Hund’s rules
In the non-relativistic approximation the electrons in the atom move under the influence of the Coulomb field of the nucleus and that of the other electrons. The Hamiltonian
of an Z-electron atom, given by
H = −
Z
Z
h̄2 X
Ze2 X
e2 1 X
1
1
∇2i −
+
,
2m i=1
4π²0 i=1 |~ri |
4π²0 2 i6=j |~ri − ~rj |
(1)
is independent of the spin of the electrons. Strictly speaking, on should deal with a manyelectron wave function and not single electron wave functions. An approximate description
which works extremely well involves defining single electron states; these are defined in
terms of the solutions of the one-particle Schrödinger equation with an effective, centrally
symmetric field. The central field, however, is not the same for all electrons. Moreover,
the the wave function of one of the electrons determines the force on the others and vice
versa. So the fields have to be determined self-consistently.
Our discussion is based on the above picture. Since the effective field is centrally
symmetric each state of the electron is characterized by the same quantum numbers as
the electron in the hydrogen atom, n, `, and m. These one-electron states are called
orbitals. In contrast to the hydrogen atom the energies of the orbitals depend on the value
of `, in general. For example, the 2p state has higher energy than the 2s state, roughly
speaking, due to screening effects discussed in class. The set of 4` + 2 orbitals is called
a shell. So for a system of electrons we can specify the total orbital angular momentum
~ = PZ ~si since they are both separately conserved. The degeneracy
~ ≡ PZ ~`i and S
L
i=1
i=1
of a state given L and S is clearly (2L + 1)(2S + 1). When one includes relativistic effects
~ and S
~ are no longer separately conserved but
such as spin-orbit coupling are included L
~
~
~
the total angular momentum J = L + S is conserved. So the states of the system are
specified by {L, S, J, mJ }. We will ignore spin-orbit coupling first.
Using the same notation as for the hydrogen atom the electrons occupy successively
the states diagonally from the first to third quadrants as shown in class:
1s
2s
3s
4s
5s
6s
2p
3p
4p
5p
6p
3d
4d 4f
5d 5f · · ·
6d · · ·
(2)
The periodic table of Mendeleev is obtained by filling these levels taking into account the
effect of degeneracies: for a given n and ` we have a degeneracy of 2(2`+1) states and this
is referred to as a shell. The rule shown fails occasionally, for example, Gd, Gadolinium
with Z = 64.
1
Consider Silicon, Si, with 14 electrons; these occupy the 1s2 , 2s2 , 2p6 , 3s2 , 3p2 states.
This is referred to as a configuration. Except for the two electrons in the 2p state the
other levels are completely filled and are characterized by zero total orbital and total spin
angular momentum. Verify this!
Now the 2p state is 6-fold degenerate and the two electrons can occupy these oneelectron states in 6 × 5/2 = 15 ways. These 15 states do not have the same energy.
Which are the lowest states? Hund’s rules addresses the issue of which of the allowed
states(terms) for a given configuration has the lowest energy: for the case of Si, the two
electrons in the outermost shell both are labeled by the quantum numbers ` = 1, s = 1/2;
what is the lowest energy state?. The rules apply to most atoms except the heaviest
atoms (see later).
Hund’s rules which are empirical state (the first and second) that the term structure
with the maximum possible S and the largest possible L for the given S compatible with
the Pauli exclusion Principle has the lowest energy.
There is no simple proof of this rule; it has been verified experimentally and by numerical calculations. They can be justified crudely as follows. In the case of two electrons
the possible values of S are 0 or 1. The S = 0 state has an antisymmetric spin wave
function and consequently a symmetric spatial part of the wave function. Remember that
identical fermions must be described by an antisymmetric (total) wave function. Similarly
the S = 1 state is symmetric in spin space and antisymmetric in coordinate space. In the
spatially antisymmetric state (ψ(~r1 , ~r2 ) = −ψ(~r2 , ~r1 ) and ψ(~r, ~r) = 0 and therefore,) the
probability of finding the electrons close to each other is small and the repulsive electrostatic energy is reduced. In addition, when the electrons are farther apart they screen the
nucleus less increasing the attractive energy. Thus the lowest energy state has the maximum possible S. This argument can be generalized to more number of electrons since the
state with the highest spin is maximally symmetric and the corresponding spatial wave
function has to be totally antisymmetric.
Next it is reasonable that the electrons are farthest apart when the total orbital angular momentum is the largest. One way to see this within a classical picture for the state
with maximum mL = L. Clearly for a system of point masses rotating about the z-axis
the larger the angular momentum along z the farther apart they are. Now remember that
for a given L all the states with different mL values have the same energy and so our
argument for mL = L should be OK. This justifies the L part of the rule.
So for the 2p2 electrons clearly the lowest energy state has S = 1. What is the
value of L? Given `1 = 1 and `2 = 1 the total angular momentum assumes values
`1 + `2 , `1 + `2 − 1, ......|`1 − `2 which in this case yields L = 2, 1, 0. The trick is to
use mS and mL to guide our thinking remembering that the specific values
do not matter by spherical symmetry. Since S = 1 choose mS = 1, i.e, ms1 = 1/2
and ms2 = 1/2. Now both m`1 and m`2 cannot be the maximum value of 1 by Pauli
2
exclusion principle. Be sure of this! The highest allowed value of mL corresponds to
m`1 = 1 and m`2 = 0 or vice versa. Therefore, the maximum allowed value of L corresponds to 1. Thus the lowest energy state of Silicon has L = S = 1. Thus there are
(2L + 1)(2S + 1) = 3 × 3 = 9 states with the same energy. What are these states? They
are labeled by ML and MS . ML = 1, 0, −1 since L = 1 and similarly for S.
Given L = 1 and S = 1 the allowed values of the total angular momentum are
J = 2, 1, 0. Now since the shell is less than half-filled, the ground state corresponds to
the lowest value of J. (See below!)
The trick of using the maximum possible values of ML and MS is a useful way of
deducing the maximum L and S. So if we have three spins in the 3d orbital we have ` = 2
for each of the electrons. First we find the maximum S by finding the maximum MS
and then find the maximum allowed L consistent with the Pauli principle by considering
different m` values.
Table 1: Three 3d electrons
Quantum number
ms
ms
1
↑
2
2
↑
1
3
↑
0
Total
MS = 3/2
ML = 3
At the risk of confusing you this trick of focusing on the maximum MS and ML does
not mean that all the spins cannot point down. Of course they can; when we say S = 3/2,
MS can be ±3/2, ±1/2 and there are 4 possible states. This is a quick way finding the
maximum allowed S and L.
So we have decided which values of L and S are allowed, and in the jargon of spectroscopy we know the LS multiplet. Finally, we include the effect of spin-orbit coupling (fine structure arising from relativistic corrections) perturbatively. The degeneracy
((2L + 1)(2S + 1)) is partially lifted and the energy eigenstates are given in terms of
|L, S, J, mJ i rather than |L, S, mL , mS i. The different mJ values are degenerate at this
level.
Hund’s third rule (which applies for atoms or ions with a single unfilled shell) states
that if the unfilled shell is not more than half-filled the lowest value of J has the lowest
energy while if it is more than half-filled the largest value of J has the lowest energy.
So for Si the ground state corresponds to J = 0. the spectroscopic notation is arcane
and gives the L value in terms of a letter of the alphabet: S,P,D,F,G,H,I, .... for L =
0, 1, 2, 3, 4, 5, 6, 7... respectively. The S value is specified by giving the spin degeneracy,
2S + 1, as a left superscript and the J value as a right subscript. For Si the spectroscopic
notation is
3
P0 .
3
Look at Praseodymium, Pr or triply ionized Neodymium. These have three 4f electrons. You can verify that S = 3/2, L = 6, and J = 9/2. Thus the term structure
is
4
I9/2
4