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Lectures 10-11: Multi-electron atoms
o Schrödinger equation for
o Two-electron atoms.
o Multi-electron atoms.
o Helium-like atoms.
o Singlet and triplet states.
o Exchange energy.
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System of non-interacting particles
o
What is probability of simultaneously finding a particle 1 at (x1,y1,z1), particle 2 at (x2,y2,z2),
etc. => need joint probability distribution.
o
N-particle system is therefore a function of 3N coordinates:
(x1,y1,z1; x2,y2,z2; … xN,yN,zN)
o
Must solve
o
First consider two particles which do not interact with one
another, but move 
in potentials V1 and V2. The Hamiltonian is
(2)
Hˆ  Hˆ 1  Hˆ 2
 2 2
  2 2

 
1  V1( rˆ1 ) 
 2  V2 ( rˆ2 )
 2m1
  2m2

Hˆ (rˆ1, rˆ2,..., rˆN )  E(rˆ1, rˆ2,..., rˆN ) (1)
Vi  
2e 2
r1
x
i 1,2
o
The eigenfunctions of H1 and H2 can be written as the product:



e1
y
The electron-nucleus potential for helium is
4 0 ri
e2
r2
o

z
(rˆ1, rˆ2 )  1(rˆ1)2 (rˆ2 )
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System of non-interacting particles
Hˆ ( rˆ1, rˆ2 )  (Hˆ 1  Hˆ 2 )1 (rˆ1 ) 2 ( rˆ2 )
 (E1  E 2 )1 (rˆ1 ) 2 ( rˆ2 )
o
Using this and Eqns. 1 and 2,
o
ˆ (rˆ , rˆ )  E(rˆ , rˆ ) where E = E1+E2.
That is, H
1 2
1 2

o The product wavefunction is an eigenfunction of the complete Hamiltonian H, corresponding
to an eigenvalue E which is the sum of the energy eigenvalues of the two separate particles.
(rˆ1, rˆ2,...rˆN )  1(rˆ1)2 (rˆ2 ) N (rˆN )
o
For N-particles,
o
Eigenvalues of each particle’s Hamiltonian determine possible energies. Total energy is thus
N

E   Ei
i1
o
Can be used as a first approximation to two interacting particles. Can then use perturbation
theory to include interaction.

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Application to helium
o
o
Assuming each electron in helium is non-interacting, can assume each can be treated
independently with hydrogenic energy levels:
13.6Z 2
En  
n2
Total energy of two-electron system in ground
state (n(1) = n(2) = 1) is therefore

o
Observed
-50
-60
E  E1(1)  E1 (2)
 1
1 
 13.6Z 2 2 

n(2) 2 
n(1)
1 1 
 13.6(2) 2 2  2 
1 1 
 109 eV
Energy
(eV)
For first excited state, n(1) = 1, n(2) = 2 => E =-68 eV.
-70
-80
-90
-100
-110

Neglecting electron-electron interaction
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System of interacting particles
o
o

For He-like atoms can extend to include electron-electron
interaction:
 2 2
Ze 2   2 2
Ze 2 
e2
ˆ
H  
1 
2 
 

40 r1  2m
4 0 r2  40 r12
2m
z
e2
The final term represents electron-electron repulsion at a
distance r12.
e1
r12
r2
r1
y
o
Or for N-electrons, the Hamiltonian is:
x
N
Hˆ   Hˆ i
i1
 2 2
Ze 2  N e 2
 
i 
 
2m
4

r

0 i  i j 4 0 rij
i1
N
and the corresponding Schrödinger equation is again of the
form
Hˆ (rˆ1, rˆ2,..., rˆN )  E(rˆ1, rˆ2,..., rˆN )

N
where
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(r1, r2,..., rN )  1(r1)2 (r2 ) N (rN ) and E   E i

i1
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Wave function for system of interacting particles
o
The solutions to the equation
 2
Ze 2
e 2 
2
ˆ
H ii (rˆi )  
2m  i  4 r  4 r 
 i (rˆi )

0 i
0 ij 
 E i i ( rˆi )
can again be written in the form  i (rˆi , i ,  i )  Rn i li ( rˆi )li m i (i ,  i )
o
 are solutions to
The radial wave functions
d 2 Rn i li
dri

2
2 dRn i li 2 
e 2 

 2 E 
Rn i li  0
r dri
4

r

0 i 
and therefore have the same analytical form as for the hydrogenic one-electron atom.
o
 for
Allowable solutions again only exist
Z eff e 4
En  
(4 0 ) 2 2 2 n 2
2
where Zeff = Z - nl.
o
Zeff is the effective nuclear charge and nl is the shielding constant. This gives rise to the shell
model for multi-electron 
atoms.
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Atoms with two valence electrons
o
Includes He and Group II elements (e.g., Be, Mg, Ca, etc.). Valence electrons are
indistinguishable, i.e., not physically possible to assign unique positions simultaneously.
o
This means that multi-electron wave functions must have exchange symmetry:
| ( rˆ1, rˆ2 ,..., rˆK , rˆL ,..., rˆN ) |2 | (rˆ1, rˆ2,..., rˆL , rˆK ,..., rˆN ) |2
which will be satisfied if
o
(rˆ1, rˆ2,..., rˆK , rˆL ,..., rˆN )  (rˆ1, rˆ2,..., rˆL , rˆK ,..., rˆN )

That is, exchanging labels of pair of electrons has no effect on wave function.

o
The “+” sign applies if the particles are bosons. These are said to be symmetric with respect to
particle exchange. The “-” sign applies to fermions, which are anti-symmetric with respect to
particle exchange.
o
As electrons are fermions (spin 1/2), the wavefunction of a multi-electron atom must be antisymmetric with respect to particle exchange.
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Helium wave functions
o
He atom consists of a nucleus with Z = 2 and two electrons.
o
Must now include electron spins. Two-electron wave
function is therefore written as a product spatial and a spin
wave functions:
z
   spatial( rˆ1, rˆ2 ) spin
o
As electrons are indistinguishable =>  must be antisymmetric. See table for allowed symmetries of spatial and
functions.
spin wave
r12
e2
r2
y
e1
r1
Z=2
x
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Helium wave functions: spatial
o
State of atom is specified by configuration of two electrons. In ground state, both electrons
are is 1s shell, so we have a 1s2 configuration.
o
In excited state, one or both electrons will be in higher shell (e.g., 1s12s1). Configuration must
therefore be written in terms of particle #1 in a state defined by four quantum numbers (called
). State of particle #2 called .
o
Total wave function for a excited atom can therefore be written:    (rˆ1 )  (rˆ2 )
o
But, this does not take into account that electrons are indistinguishable. The following is
therefore equally valid:
    (rˆ1 ) (rˆ2 )

o
Because both these are solutions of Schrödinger equation, linear combination also a solutions:

where

1/ 2 is
1
( ( rˆ1)  ( rˆ2 )    ( rˆ1 ) ( rˆ2 ))
2
1
A 
( ( rˆ1)  ( rˆ2 )    ( rˆ1) ( rˆ2 ))
2
S 
Symmetric
Asymmetric
a normalisation factor.

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Helium wave functions: spin
o
There are two electrons => S = s1+ s2 = 0 or 1. S = 0 states are called singlets because they
only have one ms value. S = 1 states are called triplets as ms = +1, 0, -1.
o
There are four possible ways to combine the spins of the two electrons so that the total wave
function has exchange symmetry.
o
Only one possible anitsymmetric spin eigenfunction:
1
[(1/2,1/2)  (1/2,1/2)]
2
o
singlet
There are three possible symmetric spin eigenfunctions:

(1/2,1/2)
1
[(1/2,1/2)  (1/2,1/2)]
2
(1/2,1/2)

triplet
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Helium wave functions: spin
o
Table gives spin wave functions for a twoelectron system. The arrows indicate whether
the spin of the individual electrons is up or
down (i.e. +1/2 or -1/2).
o
The + sign in the symmetry column applies if
the wave function is symmetric with respect to
particle exchange, while the - sign indicates that
the wave function is anti-symmetric.
o
The Sz value is indicated by the quantum
number for ms, which is obtained by adding the
ms values of the two electrons together.
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Helium wave functions
o
Singlet and triplet states therefore have different spatial wave functions.
o
Surprising as spin and spatial wavefunctions are basically independent of each other.
o
This has a strong effect on the energies of the allowed states.
Singlet
Triplet
S
ms
spin
spatial
0
0
1/ 2(12 12 )
1
( ( rˆ1)  ( rˆ2 )    ( rˆ1) ( rˆ2 ))
2
1
+1
0

-1

12
1/ 2(12  12 )
12 
1
( ( rˆ1)  ( rˆ2 )    ( rˆ1) ( rˆ2 ))
2

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Singles and triplet states
o
Physical interpretation of singlet and triplet states can be obtained by evaluating the total spin
angular momentum (S), where
Sˆ  Sˆ1  Sˆ2
is the sum of the spin angular momenta of the two electrons.
o
The magnutude of the total spin and its z-component are quantised: S  s(s  1)

Sz  m s
where ms = -s, … +s and s = 0, 1.
o
o
If s1 = +1/2 and s2 = -1/2 => s = 0.
o Therefore ms = 0
(singlet state)
If s1 = +1/2 and s2 = +1/2 => s = 1.
o Therefore ms = -1, 0, +1 (triplet states)
singlet
state
s1=1/2
s2=-1/2

ms
+1
z
triplet
state
s1=1/2
s2=1/2
0
s=1
-1
s = 0, ms = 0
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Helium terms
o
Angular momenta of electrons are described by l1, l2, s1, s2.
o
As Z<30 for He, use LS or Russel Saunders coupling.
o
Consider ground state configuration of He: 1s2
o
Orbital angular momentum: l1=l2 = 0 => L = l1 + l2 = 0
o Gives rise to an S term.
o
Spin angular momentum: s1 = s2 = 1/2 => S = 0 or 1
o Multiplicity (2S+1) is therefore 2(0) + 1 = 1 (singlet) or 2(1) + 1 = 3 (triplet)
o
J = L + S, …, |L-S| => J = 1, 0.
o Therefore there are two states: 11S0 and 13S1
o
(also using n = 1)
But are they both allowed quantum mechanically?
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Helium terms
o
Must consider Pauli Exclusion principle: “In a multi-electron atom, there can never be more
that one electron in the same quantum state”; or equivalently, “No two electrons can have the
same set of quantum numbers”.
o
Consider the 11S0 state: L = 0, S = 0, J = 0
o n1 = 1, l1 = 0, ml1 = 0, s1 = 1/2, ms1 = +1/2
o n2 = 1, l2 = 0, ml2 = 0, s2 = 1/2, ms2 = -1/2
o
11S0 is therefore allowed by Pauli principle as ms quantum numbers differ.
o
Now consider the 13S1 state: L = 0, S = 1, J = 1
o n1 = 1, l1 = 0, ml1 = 0, s1 = 1/2, ms1 = +1/2
o n2 = 1, l2 = 0, ml2 = 0, s2 = 1/2, ms2 = +1/2
o
11S1 is therefore disallowed by Pauli principle as ms quantum numbers are the same.
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Helium terms
o
First excited state of He: 1s12p1
o
Orbital angular momentum: l1= 0, l2 = 1 => L = 1
o Gives rise to an P term.
o
Spin angular momentum: s1 = s2 = 1/2 => S = 0 or 1
o Multiplicity (2S+1) is therefore 2(0) + 1 = 1 or 2(1) + 1 = 3
o
For L = 1, S = 1 => J = L + S, …, |L-S| => J = 2, 1, 0
o Produces 3P3,2,1
o
Therefore have, n1 = 1, l1 = 0, s1 = 1/2
and
n2 = 2, l2 = 1, s1 = 1/2
o
For L = 1, S = 0 => J = 1
o Term is therefore 1P1
o Allowed from consideration of Pauli principle
No violation of Pauli principle
=> 3P3,2,1 are allowed terms
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Helium terms
o
Now consider excitation of both electrons from ground state to first excited state: gives a 2p 2
configuration.
o
Orbital angular momentum: l1 = l2=1 => L = 2, 1, 0
o Produces S, P and D terms
o
Spin angular momentum: s1 = s2= 1/2 = > S =1, 0 and multiplicity is 3 or 1
o
*Violate
L
S
J
Term
0
0
0
1S
0
1
0
1
1P
2
0
2
1D
2
0
1
1
*3S
1
1
2, 1, 0
2
1
3, 2, 1
3P
1
1
2,1,0
*3D
3,2,1
Pauli Exclusion Principle (See Eisberg & Resnick, Appendix P)
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Helium Grötrian diagram
o
Singlet states result when S = 0.
o Parahelium.
o
Triplet states result when S = 1
o Orthohelium.
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Exchange energy
o
Need to explain why triplet states are lower in energy that singlet states. Consider
 2 2
Ze 2   2 2
Ze 2 
e2
ˆ
H  
1 
2 
 

40 r1  2m
4 0 r2  40 r12
2m
 Hˆ  Hˆ  Hˆ
1
o
o
2
12
The expectation value of the Hamiltonian is
 
Hˆ spatiald 3rˆ1d 3rˆ2
*
spatial

The energy can be
split into three parts,
E = E1 + E2 + E12

where
*
E i    spatial
Hˆ i spatiald 3 rˆ1d 3 rˆ2
E12 
o
E 
 
Hˆ 12 spatiald 3 rˆ1d 3 rˆ2
*
spatial
The expectation value of the first two terms of the Hamiltonian is just

E  E1  E 2
 1
1 
 4 E R  2  2 
n1 n 2 
where ER = 13.6 eV is called the Rydberg energy.

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Exchange energy
o
o
The third term is the electron-electron Coulomb repulsion energy:
Evaluating this integral gives
E12 
 
*
spatial

 
*
spatial
Hˆ 12 spatiald 3 rˆ1d 3 rˆ2
e2
40 r12
 spatiald 3 rˆ1d 3 rˆ2
E12 = D  J

where the + sign is for singlets and the - sign for triplets and D is the direct Coulomb energy
and J is the exchange Coulomb energy:
D 
J 
o
e2
4 0
2
e
4 0
  (rˆ )  (rˆ ) r
 ( rˆ1)  ( rˆ2 )d 3 rˆ1d 3 rˆ2
  (rˆ )  (rˆ ) r
  ( rˆ1) ( rˆ2 )d 3 rˆ1d 3 rˆ2
*
1
*
1
2
12
*
1
*
1
2
12
The resulting energy is E12 ~ 2.5 ER. Note that in the exchange integral, we integrate the
expectation valueof 1/r12 with each electron in a different shell. See McMurry, Chapter 13.
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Exchange energy
o
The total energy is therefore
 1
1 
E  4 E R  2  2  D  J
n1 n 2 
where the + sign applies to singlet states (S = 0) and the -sign to triplets (S = 1).
o
Energies of the singlet and
triplet states differ by 2J. Splitting of spin states is direct
consequence of exchange symmetry.
o
We now have,
o
Compares to measure value of ground state energy, 78.98 eV.
o
Note:
o Exchange splitting is part of gross structure of He - not a small effect. The value of
2J is ~0.8 eV.
E1 + E2 = -8ER
and
E12 = 2.5ER => E = -5.5ER = -74.8 eV
o Exchange energy is sometimes written in the form E exchange  2J Sˆ1  Sˆ 2
which shows explicitly that the change of energy is related to the relative alignment of
the electron spins. If aligned = > energy goes up.

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Helium terms
o
Orthohelium states are lower in energy than the parahelium states. Explanation for this is:
1. Parallel spins make the spin part of the wavefunction symmetric.
2. Total wavefunction for electrons must be antisymmetric since electrons are fermions.
3. This forces space part of wavefunction to be antisymmetric.
4. Antisymmetric space wavefunction implies a larger average distance between electrons
than a symmetric function. Results as square of antisymmetric function must go to zero
at the origin => probability for small separations of the two electrons is smaller than for
a symmetric space wavefunction.
5. If electrons are on the average further apart, then there will be less shielding of the
nucleus by the ground state electron, and the excited state electron will therefore be more
exposed to the nucleus. This implies that it will be more tightly bound and of lower
energy.
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