Ph235
Autumn 2008
F. Merritt
Oct 16, 2008
Lecture 7 (October 13, 2008): (version 1.1; 11-Oct-2008::21:12)
Atomic systems and the periodic table. Part 3
Higher Z atoms (Z>4) and Hund’s Rules
In the last lecture we covered notation for electron (nl) subshells, and for the
standard spectroscopic term 2 S 1LJ . We also reviewed the periodic table, including
“Hund’s Rule Zero” which gives the ordering of subshells which governs the structure of
the periodic table:
We derived the ground state configuration for the first four elements. But to go further
than that, we need to use Hund’s Rules 1-3. That is what we will do today. The rules
are:
Hund #1: The lowest energy state will have the maximum value of S (total spin)
consistent with the electron configuration specified by Hund Zero.
Hund#2: The lowest state will have the maximum value of L (total orbital angular
momentum) consistent with the previous rules Hund#0 and Hund#1.
Hund#3: For shells that are half-filled or less, the lowest state is the one with
J  L  S ; for shells that are more than half filled, the minimum energy state
is the one with J  L  S .
Note that the first rule tells us how to determine S, the second how to determine L, and
the third how to determine J. The first element for which all 3 of these come into play is
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Oct 16, 2008
Carbon (Z=6), so I will use that as the primary example. But the rules will be derived
and extended to all members of the multiplet.
Before going through the rules in detail, let’s look at the results obtained by
applying these to first 10 elements, repeating the table of electron configurations and
extending it to include the spectroscopic term. That way, we can immediately see what
the rules predict and make sure that we understand how to apply the rules as we develop
them.
Z
1
Sym
Name
H Hydrogen
2
He
e configuration
(1s)
Spect. term
2
S1/ 2
Helium
(1s)2
1
2
S0
3
Li
Lithium
(1s) (2s)
2
4
Be
Beryllium
(1s)2 (2s)2
1
(1s)2 (2s)2 (2p)
2
P1/ 2
5
6
B
C
Boron
Carbon
S1/ 2
S0
2
2
2
3
P0
2
2
3
S3/ 2
(1s) (2s) (2p)
7
N
Nitrogen
(1s) (2s) (2p)
4
8
O
Oxygen
(1s)2 (2s)2 (2p)4
3
P2
(1s)2 (2s)2 (2p)5
2
P3/ 2
9
10
F
Ne
Flourine
Neon
2
2
6
(1s) (2s) (2p)
1
S0
The GSU website can help with understanding all of this, both with alternate explanations
and most of all with graphs and sketches to illustrate the physics. The Atomic Physics
site is:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/atomstructcon.html#c1
The most relevant items are “Many Electron Atoms”, “Term Symbols”, “Hund’s Rules”,
and also “Coulomb Interaction”, “Spin-spin interaction”, “Spin-orbit interaction”, and
“LS Coupling”. Some of these cover the same material and/or link to each other.
Rationale for Hund#1
2
2
Carbon has a filled 1s   2 s  core, and the two valence electrons must be in the
 2 p  subshell.
But the different m-states should have equal energies, so it’s not obvious
whether the ground state should have L=0, 1, or 2. The H1 rule tells us to maximize total
spin S, which seems strange because spin has no dynamical role in the Hamiltonian (yet);
but if we recall the difference between orthohelium and parahelium, the reason is clear.
We want the 2 p-electrons to be in a totally antisymmetric spatial wavefunction, since
this will minimize the ee interaction. Since the total wavefunction (space  spin) must be
antisymmetric under exchange, this means the spin wavefunction must be totally
symmetric. The way to insure this is to put both electrons in the same spin state (clearly
symmetric under exchange), and this automatically gives us the maximum value of S. In
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Autumn 2008
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Oct 16, 2008
the (2p) subshell, the ground state must have S=1 for 2 electrons, and S  32 for 3
electrons (corresponding to a half-filled p-subshell).
When the subshell is half filled, all the electrons in the ground state must have
different values of m  lz (=1, 0, and -1 for the p-subshell), since they have the same
value of ms (  12 ). But if the shell is half filled, that means that all of the m-states are
occupied, so when we add a fourth electron it must have the same m as one of the first
3 -- and that means it must have a different spin (spin-down). All of the additional
electrons in this subshell must have spin down, because all the spin-up states were filled
by the first 3. Therefore, after the subshell is half filled, each additional electron must
decrease the total spin by 12 , so that S=0 when the subshell is completely filled.
When we add a fourth electron, it must have ms   12 ; but the spin state must still be
totally symmetric, so we can write it as
1
2
         , which has a
total spin of S=1. (???? Prove this. Do we need to include spatial part? I think so.
[It perhaps is conceptually simpler to think of the more-than half-filled subshell as a
superposition of a filled subshell plus a subshell of “hole states”, where a hole state is
defined to be a state with a positively charged electron (like a positron). To minimize the
total energy, minimize the energy of the hole system – which means putting the hole
states in the same spin state, so their spin wf is symmetric and their space wf is
antisymmetric; this minimizes the hole-hole interaction energy, and so minimizes the
energy of the atomic system. So for subshells that are half-filled or less, use the
“electron” picture, but for subshells that are half-filled or more, use the “hole” picture.]
The generalization to higher l -subshells is obvious. S  12 for the first electron,
increases in half-integral steps with each new electron until the subshell is half filled, and
then decreases in half-integral steps until it is zero and the subshell is filled. This means
that the multiplicity in the spectroscopic term starts with the value 2, increases in unit
steps until the subshell is half filled, and then decreases in unit steps until it equals 2
again and the subshell is filled.
Rationale for Hund#2
Returning to Carbon as an example, we now know that in the ground state the two
electrons must be in the same spin state, and that the spatial part of the wf must be
antisymmetric under exchange. It seems plausible that the lowest energy state would
have the electrons orbiting in the same direction, so that they pass each other at a minimal
frequency. But the two electrons cannot have the same ml quantum numbers, because
the state must be antisymmetric (or equivalently, because of the Pauli Exclusion
Principle).
In general, we can write the antisymmetric spatial part of the state as
 m1 , m2  A  12  l , m1 l , m2  l , m2 l , m1 
(1.1)
Since we want to maximize L, let’s make M  LZ as great as possible, which means
m1  l , m2  l  1 . The M  2l 1 , so this state must have L  2l or L  2l 1. To find
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Oct 16, 2008
out which, apply L . If the state has L  2l , this will raise it from 2l , 2l 1 up to
2l , 2l ; but if the state had L  2l 1, then this operator will give zero. That is what
happens (make sure you understand why), so the state had L  2l 1, and since l  1 for
Carbon (p-subshell), this means the ground state has L  1
But how do we handle 3 valence electrons? Or more? How do we make a
wavefunction which is totally antisymmetric under exchange (and that means exchange
of any two electrons)? The way to do this is to write the state as a “Slater determinant”
(see Griffiths page 210, problem 5.7).
Let’s consider a general state in l -subshell with 3 electrons. Then if the subshell
is half filled or less, all of the electrons must have spin-up, so all of the m’s must be
different. Since we want to maximize L, choose them to be
(1.2)
a  l , l , b  l , l 1 , c  l , l  2
Then the state which is totally antisymmetric under exchange is uniquely defined as
1
1
ma , mb , mc 
ai b j ck   ijk 
 abc  bca  cab  acb  bac  cba  (1.3)

6 i , j ,k 1,2,3
6
where the first state in each term refers to particle 1, the second to 2, and the third to 3.
Check that this indeed swithches sign if you exchange any two particles. Now the total
LZ of this state is clearly  l    l  1   l  2  3l  3 . And if we apply L we
necessarily kill the state, because we will make 2 elements the same. Therefore, the total
L of the system is just the value of LZ , so L  3l  3 . For the case of Nitrogen (Z=7), this
gives L  0 .
This procedure can get a little complicated for more electrons, where we would
need to use  ijkl (which you have probably never encountered). A nice way of making
the wavefunction totally antisymmetric is to use the Slater determinant. For N particles
in N different states, draw a N  N matrix, with columns corresponding to states
(=a,b,c,… and rows corresponding to particles (=1,2,3,…). The the antisymmetric state
is just the normalized determinant of this matrix. For N=3, this becomes
 a(1) b(1) c(1) 
 a1 b1 c1 
1

 1


 
det  a(2) b(2) c(2)  
det  a2 b2 c2 
6
6
 a(3) b(3) c(3) 
a b c 
(1.4)


 3 3 3

1
a1b2c3  b1c2 a3  c1a2b3  a1c2b3  b1a2c3  c1b2c3 
6
which is the same as (1.3). This works because the determinant changes sign under the
interchange of any two rows, so the resulting wf is necessarily antisymmetric under
exchange.
But what happens when a subshell becomes more than half filled? Then all of the
electrons can’t be in the same spin state. That means that we can no longer factor out the
common spin wavefunction and deal with the spatial part alone. There are 3 ways to
handle this:
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F. Merritt
Oct 16, 2008
a) Note that when a subshell is exactly half filled, L=0 necessarily, since all of the
(2l+1) m-states are occupied. For the second half of the subshell, we just start over
(so far as L is concerned) are repeat the (2l+1) numbers we have written down for the
first half.
b) Note that a filled subshell necessarily has L=0 and S=0. Argue that a subshell of P
states and Q electrons is equivalent to a filled subshell with P-Q “hole” states. Think
of the holes as positively charged electrons with negative mass. The hole states also
carry angular momentum, so we simply maximize L for the hole states. For a
subshelll that is more than half filled, the hole states will all have spin up, so can
proceed exactly as we did for the first half. This gives the same (2l+1) values of L
that we found for the first half. (Note that the order is reversed this time, logically
speaking, but you will see that doesn’t matter much.)
c) Follow exactly the same formal procedure that we used for the first half, using Slater
determinants, but now include the correlation between spin and space wavefunctions.
For example, in the case of Nitrogen, with 4 electrons in the subshell, a possible state
would be  a   a   b   c   , with a,b,c defined as above in equation (1.2), with
l  1. This state has M L  1 and M S  1 ; but it is not a valid solution because it is not
totally antisymmetric. We make the totally antisymmetric state by constructing a
Slater determinant:
  a1    a1    b1    c1   

 a2    a2    b2    c2   
1

 A
 det
(1.5)
  a3    a3    b3    c3   
24


  a4    a4    b4    c4   
Multiplying this out gives 24 terms, and the state is of course totally antisymmetric
under exchange. It is also an eigenstate of L2 and S 2 with eigenvalues L  M  1 .
We can prove this easily by applying L or S  to the state, both of which destroy the
state because they make 2 of the states the same (and because of the antisymmetry
under exchange, this gives an eigenvalue of zero). So we conclude that this state has
L  1 , which is the same result our other methods gave.
Rationale for Hund#3
We now have determined both L and S for the ground state, and the wavefunction is
necessarily totally antisymmetric. But we are still free to adjust the relative orientation of
L and S. In general, we could have any value of J ranging from a maximum of L  S
down to a minimum of L  S . The ground state will be the one of these with lowest
total energy. Using the Hamiltonian that we have assumed up until now, all of these
would be degenerate. But we have left out a dynamical term involving electron spin.
Because the electrons are orbiting in a radial Coulomb field, in the instantaneous
rest frame of an electron there will be a magnetic field given by (Shankar , p.468)

  Ze 
Ze
Ze
(1.6)
B   E     2 rˆ    3   r 
L
c
c  r 
cr
mecr 3
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Ph235
Autumn 2008
F. Merritt
Oct 16, 2008
The B -field that an electron sees in its instantaneous rest frame is consequently
proportional to the angular momentum of its orbit!
Equation (1.6) is not exactly correct for 2 reasons. First, the instantaneous rest
frame of the electron is not an inertial frame, and correcting for this reduces the apparent
B -field by a factor of 2. Moreover, in a multielectron atom the E -field is not a pure
Coulomb field because of shielding effects from the other electrons. But it is
nevertheless a radial field, so equation (1.6) gives the right direction of the B -field and
the right order of magnitude.
We have already seen (in Ph234) that the interaction energy of an electron in a
magnetic field is given by (Shankar equation 14.4.19, page 389)
ge
H int    B 
Se B
(1.7)
2mc
where g=2 for the electron. Combining this with equation (1.6) gives the approximate
interaction energy of
Ze 2
Eso 
Se Le
(1.8)
2me2c 2 r 3
That is, the energy of the electron is minimized when its spin is antiparallel to its orbital
angular momentum.
Note that the S and L of equation (1.8) are those of the individual electron, rather
than the summed S and L that we have been dealing with. When we are dealing with a
subshell that is less than half filled, following the conventions we have used above with
each electron in a spin-up state, we always have SZ  0 and also SeZ  0 for each
electron that we add. The total interaction energy from (1.8) is consequently
Ze 2
Eso 
S L
(1.9)
2me2c 2 r 3
where S and L now refer to the total quantum numbers of the system. The lowest
energy state (the ground state) is consequently the one with S and L antiparallel,
so J  L  S .
For a subshell that is exactly half filled, M=0 and S  12 (2l  1) , so Eso  0 .
For a subshell that is more than half filled, the new electrons must have their spin
antiparallel to the total spin of the system. The interaction energy is lowest when each
new electron has its Le antiparallel to its own spin, and consequently parallel to the total
spin of the system. The total interaction energy in this case is
Ze2
Eso   2 2 3 S L
(1.10)
2me c r
That is, the energy is minimized when S and L are parallel, and consequently
J  LS .
One can get the same result, perhaps more simply, in terms of the “hole” multiplet
described above . Since the hole states have positive charge, the sign of their Eso term
is simply reversed from that of the “electron” states.
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Oct 16, 2008
This interaction energy is called the “spin-orbit” term, and is also described on the
GSU website under the “Spin-orbit interaction”. The diagrams for the sodium atom
(linked from that site) show that how the spectral lines are split by this effect. The total
magnitude of the splitting between 2 P1/ 2 and 2 P3/ 2 states in the (3p) subshell is .0021 eV.
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