Electron Spin
Electron spin hypothesis
Solution to H atom problem gave three quantum numbers, n, , m.
These apply to all atoms.
Experiments show not complete description. Something missing.
Alkali metals show splitting of spectral lines in absence of magnetic field.
s lines not split
p, d lines split
Na D-line (orange light emitted by excited Na) split by 17 cm-1
Many experiments not explained without electron "spin."
Stern-Gerlach experiment early, dramatic example.
Copyright – Michael D. Fayer, 2007
Stern-Gerlach experiment
glass
plate
Magnet
ms
oven baffles
Ag
Ag beam
N
+1/2
S
-1/2
Beam of silver atoms deflected by inhomogeneous magnetic field.
Single unpaired electron - should not give to lines on glass plate.
s-orbital. No orbital angular momentum. No magnetic moment.
Observed deflection corresponds to one Bohr magneton. Too big for proton.
Copyright – Michael D. Fayer, 2007
Assume: Electron has intrinsic angular momentum called "spin."
1
s
2
11

2
S s ms    1 
2 2 
square of
angular
momentum
operator
projection of
angular
momentum on
z-axis
S z s ms  
1
2
s

s ms
s ms
 s  e /2mc
s
2

e
mc

e
2mc
Electron has magnetic moment.
One Bohr magneton.
Charged particle with angular mom.
Ratio twice the ratio for
orbital angular momentum.
Copyright – Michael D. Fayer, 2007
Electronic wavefunction with spin
   ( x, y, z )
   ( x , y , z , ms )
ms  
with spin
1
2
spin angular momentum kets
s ms
 ( ms ) 

11
22

1 1

2 2
11
1 1
,

22
2 2
Copyright – Michael D. Fayer, 2007
Electronic States in a Central Field
Electron in state of orbital angular momentum 
Y m ( ,  )
m
ms
label as orbital angular momentum
quantum number
spin angular momentum quantum number
 (1/2),  (1/2)
  m ms   Y m
  ms 
product space of angular momentum (orbital and spin) eigenvectors
Copyright – Michael D. Fayer, 2007
  m ms 
Simultaneous eigenvectors of
L
L  
2

2
Lz
S
2
Sz
 1 
Lz   m 
constant - eigenvalues of S2 are s(s + 1), s = 1/2
S  
2
3

4
S z   ms 
2  2  1
linearly independent functions of form   m ms  .
Copyright – Michael D. Fayer, 2007
Example - p states of an electron
Orbital functions
=1
 ( r , , )  R( r )Y m ( , )
radial
spherical harmonics
angular momentum part
Y11  11
Y10  10
j1m1
j1m1
Y11  1 1
j1m1
spin functions

j1m1  m
11
22
j2m
1 1

2 2
j2m
2
2

j2 m2  s ms
Copyright – Michael D. Fayer, 2007
In the m1m2 representation
s m ms
m ms
1
1
1
Y11  1
1
 1
2
2
2
Y11   1
1
1
1
2
2
 1
Y10  1
1
2
0
 0
Y10   1
1
2
0
Y11  1
1
1
1
2
2
Y11   1
1
1
1
 1   1 
2
2
2
1
2
1
2
1
2
 0
 1
1
2
1
2
1
2
Each of these is multiplied by R(r).
j1 j2  s
Copyright – Michael D. Fayer, 2007
Total angular momentum - jm representation
Two states of total angular momentum
j  j1  j2   s 
3
2
j  j1  j2  1  j1  j2   s 
1
2
The jm kets are
33
22
31
22
3 1

2 2
3 3

2 2
11
22
1 1

2 2
Copyright – Michael D. Fayer, 2007
The jm kets can be obtained from the m1m2 kets using the
table of Clebsch-Gordon coefficients.
11
2
1
1 1


1

0
For example
22
3
2
3 2
jm
Table of Clebsch-Gordon
Coefficients
3
2
3
2
1
2
3
2
1
2
1
2
1
2
1
3
2
3
0
1
2
2
3
0
2
1
2
3
1
3
1
1
2
1
3
 2
j1 = 1
j 2 = 1/2
1
2
1
1

1
1  2
m1 m2
m1 m2
3
2

1
2
1
2

1
2

3
2
j
3
2
m
1

1
3
3
1
m1 m2
Copyright – Michael D. Fayer, 2007
Spin-orbit Coupling
An electron, with magnetic moment moves in the electric field
of rest of atom (or molecule).


W   E V  
classical energy of magnetic dipole 
moving in electric field E with velocity V
energy
 
eS
spin magnetic moment of electron
2mc
Then
W 
E V   S

2mc
e
Using


e E  grad 
Coulomb potential (usually called V, but V is velocity)
and
p  mV
Copyright – Michael D. Fayer, 2007


1
W
( grad   p)  S
2
2m c
Hydrogen like atoms
ze 2
Coulomb potential
 
4 0 r
unit vector


ze 2 r
ze 2
grad  

r
vector derivative of potential
2
3
4 0 r r 4 0 r
Substituting
ze 2
1
H so 
r p S
3
2
8 0 m c r

(r  p)
L
H so
operates on
radial part of
wavefunction

orbital angular momentum
ze 2
1

L S
3
2
8 0 m c r
operates on
orbital ang. mom.
part of wavefunction
operates on
spin ang. mom.
part of wavefunction
Copyright – Michael D. Fayer, 2007
One unpaired electron – central field
(Sodium outer electron)
  V (r )
 V (r ) 1
grad V ( r ) 
r
r r


H so 
result of operating on
radial part of wavefunction
1 1  V (r )
L  S  a(r ) L  S
2
2m c r  r
Many electron system
H so
 1  V ( ri ) 
1

Li  S i   ai ( r ) Li  S i

2 
2m c i  ri  ri 
i
Ignore terms involving, Li  S j , the orbital angular momentum of one electron
with the spin of another electron. Extremely small.
Copyright – Michael D. Fayer, 2007
Spin-orbit coupling piece of Hamiltonian
H so  a ( r ) L  S
H-like atom – spatial part of operator
ze 2
1
H so ( r ) 
8 0 m 2 c r 3

E (r ) 


 (r ) H so (r )d

Normalization constant (eq.7.63) contains
z 3 /2
 E ( r )  z  z 3 /2 z 3 /2
H – little S.O. coupling
Br – sizable S.O. coupling
a( r )  z 4
heavy atom effect
similar for all atoms
and molecules
Heavy atom effect and external heavy
atom effect important in many processes.
Copyright – Michael D. Fayer, 2007
H so  a ( r ) L  S
The a(r) part is independent of the angular momentum states.
Consider L  S
L  S  L x S x  L y S y  Lz S z
Rearranging the expressions for the angular momentum
raising and lowering operators
1
S x  S  S 
2
1
S y  S  S 
2i
1
L x   L  L 
2
1
L y   L  L 
2i
Substituting
L  S  Lz S z 
1
 L S   L S  
2
Copyright – Michael D. Fayer, 2007
Spin-Orbit Coupling in the m1m2 representation
L  S  Lz S z 
1
 L S   L S  
2
not diagonal in the m1m2 representation
States in the m1m2 representation
1
1
2
1
1
2
0
1
2
0
1
2
1
1
2
1 
1
2
The spin-orbit coupling Hamiltonian matrix in the m1m2 representation
is a 6  6.
Calculating the matrix elements
Kets are diagonal for the Lz S z piece of the Hamiltonian. For example
1
1 1
 1
2
2 2
m ms
m  ms
Lz S z 1
Lz brings out m and Sz brings out ms .
Copyright – Michael D. Fayer, 2007
1
 L S   L S  
2
Apply
not diagonal
1
 L S   L S   to the six kets.
2
Examples
1
1
 L S   L S   1  0
2
2
can’t raise above largest values
1
1
2 1
 L S   L S   1   0
2
2
2
2
Left multiply by 0
0
1
2
1 1
1
2
L
S

L
S
1


    
2 2
2
2
matrix element
Copyright – Michael D. Fayer, 2007
1


H SO  a( r )  L z S z   L S   L S   
2


11
2
Hso=
1
12
1
1 2
1
0
2
a(r) 1
0 2
1 1
2
1 1
2
1
1
2

1
2
01
2
put kets along top
0
1
2
1 1
2
1  1
2
1
2
put bras down left side
2 2
2 2
0
0
2 2
2 2

1
2
1
2
example
0
1 1
1
2
 L S   L S   1  
2 2
2
2
Matrix is block diagonal.
Copyright – Michael D. Fayer, 2007
Matrix – Block Diagonal
Each block is independent. Diagonalize separately.
Form determinant from block with  (eigenvalue) subtracted from
diagonal elements.
1
 
2
2 2
2 2
0

The two blocks are the same. Only need
to solve one of them.
The two 11 blocks
are diagonal – eigenvalues.
Expand the determinant. Solve for . Each block multiplied by a(r).
1/2   2  1/2  0
  1/4  3 / 4
  [1/ 2 or  1]
1
2
  a(r )
  1 a(r )
The two 22 blocks each give these results.
The two 11 blocks each give 1/2a(r).
Therefore,
4 eigenvalues of 1/2a(r).
2 eigenvalues of -1a(r).
Copyright – Michael D. Fayer, 2007
Energy level diagram
 four-fold degenerate
0.5a(r)
fluorescence
no spin-orbit coupling 3p
E = 16,980 cm-1
0
E
 two-fold degenerate
-1a(r)
Na D-line (3p to 3s transition) split by 17 cm-1.
17 cm-1 = 3/2 a(r).
a(r) = 11.3 cm-1.
Ratio
a(r )
 7  104
E
3s
Ratio small, but need spin-orbit coupling
to explain line splitting.
Copyright – Michael D. Fayer, 2007
Spin-orbit Coupling in the jm representation
The jm kets are
33
22
31
22
3 1

2 2
3 3

2 2
11
22
1 1

2 2
However
a( r ) L  S
is in the m1m2 representation.
Can’t operate HSO directly on the jm kets.
Use table of Clebsch-Gordan coefficient to take jm kets into m1m2 rep.,
operate, then convert back to jm rep.
Copyright – Michael D. Fayer, 2007
Table of Clebsch-Gordan Coefficients
3
2
3
2
1
2
3
2
1
2
1
2
1
2
1
3
2
3
0
1
2
2
3
0
2
1
2
3
1
1
2
1
3
j1 = 1
j 2 = 1/2
1
2
1
1

3
2

1
2
1
2

1
2

3
2
j
3
2
m
1

1
3
1
3

2
3
1
1  2
1
m1 m2
jm
Example
11

22
m1 m2
m1 m2
2
1
1 1
1

0
3
2
3 2
Copyright – Michael D. Fayer, 2007
Operate a( r ) L  S
a(r ) L  S
on 3 3
22
L  S  Lz S z 
1
 L S   L S  
2
33
1
1
1
 a(r ) L  S 1  a(r ) 1
2 2
2
2
2
Found already that this is diagonal.
Can’t raise above largest values.
Therefore,
a(r ) L  S
33
22
33
1
33
 a(r )
2 2
2
2 2
is an eigenket of a( r ) L  S with eigenvalue 1/2a(r)
33
1
 1
2 2
2
1
1
2
forms 11 block in matrix. Eigenvector
Copyright – Michael D. Fayer, 2007
Consider a( r ) L  S operating on
 2
11
1
1 1 

1 
0 
22
2
3 2 
 3
jm
m 1 m2
m1 m2
11
22
from Clebsch-Gordan table
1
1 1 
11
1

 2
1 
0 
a( r ) L  S
 a ( r )  L z S z   L S   L S    
2
3 2 
22
2

 3


1
1
1
2 1
1
2
1
1  2
0  0
  1 a( r )
1   0  a(r )  0  2
2
3
2
3 2
2
3
2


 2
11
1
1 1 
 1 a ( r ) 
1 
0   1 a ( r )
22
2
3 2 
 3
a( r ) L  S
11
11
 1 a ( r )
22
22
eigenvector with eigenvalue –1a(r)
Copyright – Michael D. Fayer, 2007
Applying HSO to all jm kets
Eigenkets
The jm representation kets are the eigenkets because HSO couples the
orbital and spin angular momenta.
Coupled representation eigenkets.
The jm kets
33
22
31
22
11
22
1 1

2 2
3 1

2 2
-1.0a(r)
eigenvalues - +0.5a(r)
eigenvalues - -1.0a(r)
+0.5a(r)
Na - 3p
unpaired
electron
3 3

2 2
33
22
31
22
11
22
1 1

2 2
3 1

2 2
3 3

2 2
0
Copyright – Michael D. Fayer, 2007
The jm representation kets will be the eigenkets
whenever a term in Hamiltonian couples
two types of angular momenta.
Examples:
Hyperfine Interaction - I  S
Coupling of electron spin and nuclear spin.
Small splitting in Na optical spectrum.
Structure in ESR spectra.
NMR - nuclear spins coupled to each other.
Electronic triplet states - coupling of two unpaired electrons.
Copyright – Michael D. Fayer, 2007
Have diagonalized HSO in m1m2 representation.
Have shown that jm kets are eigenkets.
A unitary transformation U
m m2
will take the non-diagonal matrix H so1
in the m1m2 representation
into the diagonal matrix H sojm in the jm representation.
H so  U H so1 2 U
jm
mm
1
U is the matrix of Clebsch-Gordan Coefficients.
Copyright – Michael D. Fayer, 2007
11
2
1
1 2
Both are block diagonal.
Only need to work with
corresponding blocks.
1
2

1
2
01
2
0
1
2
1
0 2
1 1
2
1 1
2
0
1 1
1 
2
1
2
2 2
2 2
m1m2 matrix
0
0
2 2
2 2

1
2
1
2
3
2
3
2
1
2
3
2
1
2
1
2
1
2
1
3
2
3
0
1
2
2
3
 1
0
 21
2
3
1
1
2
1
3
j1 = 1
j 2 = 1/2
1
2
1
1
1
2
1
2
1
12
Hso= a(r)
1

1  21
3
2

1
2
1
2

1
2

3
2
j
3
2
m
1
Table of
Clebsch-Gordan
coefficients
3
1
3

2
3
1
m1 m2
Copyright – Michael D. Fayer, 2007
Multiplying the 2×2 blocks
U






1
3






1
3
2
3
2
3
H SO
2  1
 
3  2
1  2


3  2
2  1 1

3  2 3
1  1 2


3  2 3
 1/ 2 0 
 0 1 


2 

2 

0 

U
1
3
2
3
1
2 

3 

1


3
2


3

1 

3 
diagonal matrix with
eigenvalues on the diagonal
Copyright – Michael D. Fayer, 2007
Electron Spin, Antisymmetrization of Wavefunctions, and the Pauli Principle
In treating the He atom, electron spin was not included.
Electron - particle with intrinsic angular momentum, spin
S  1/ 2
m s  1/ 2
1/2 1/2  
1/2 1/2  
Copyright – Michael D. Fayer, 2007
Excited States of He neglecting Spin
Ground State of He - perturbation theory
H'
e2
perturbation
4 0 r12
Zeroth order wavefunctions (any level)
Product of H atom orbitals
n
1 1 m1
(1)  n2
2 m2
(2)   n1 1m1 (1)  n2
2 m2
(2)
(1)
electron 1; coordinates ( r1 , 1 ,  1 )
(2)
electron 2; coordinates ( r2 , 2 ,  2 )
Copyright – Michael D. Fayer, 2007
Zeroth order energy
 1
1 
En01n2  4 Rhc  2  2 
 n1 n2 
 e4
R 2 3
8 0 h c
 - reduced mass of H atom
First excited state energy
zeroth order
E 0  5 Rhc
since
n1  1 and n2  2
n2  1 and n1  2
same energy
Copyright – Michael D. Fayer, 2007
First Excited Configuration
States corresponding to zeroth order energy - 8 fold degenerate
1 s  1 2 s  2 
1 s  1 2 p y  2 
2 s 1 1s  2 
2 p y  1 1 s  2 
1 s  1 2 p x  2 
1 s  1  2 pz  2 
2 p x  1 1 s  2 
2 pz  1  1 s  2 
All have same zeroth order energy.
All have
n1  1 and n2  2
or
n2  1 and n1  2
Copyright – Michael D. Fayer, 2007
Degenerate perturbation theory problem
System of equations (matrix)
Form determinant
1
1 1s(1)2s(2)
2 2s(1)1s(2)
3 1s(1)2p x (2)
4 2 p x (1)1s(2)
2
3
4
5
6
7
8
Js  E' Ks
Ks Js  E'
Jpx  E'
Kpx
Kpx
Jpx  E'
5 1s(1)2 p y (2)
6 2p y (1)1s(2)
=0
Jpy  E ' Kpy
Kpy
Jpy  E'
7 1s(1)2p z (2)
Jpz  E'
8 2 p z (1)1s(2)
Kpz
Kpz
Jpz  E'
E' - eigenvalues
J's - diagonal matrix elements
K's - off diagonal matrix elements
Copyright – Michael D. Fayer, 2007
J s    1s(1)2 s(2)
e2
4 o r12
K s    1s(1)2 s(2)
e2
4 o r12
J px    1s(1)2 p x (2)
py
2 s(1)1s(2) d 1d 2
e2
4 o r12
K px    1s(1)2 p x (2)
J
1s(1)2 s(2) d 1d 2
e2
4 o r12
1s(1)2 p x (2) d 1d 2
2 p x (1)1s(2) d 1d 2
, K p y , J p z , K p z  replace x with y and z

Copyright – Michael D. Fayer, 2007
J's - Coulomb Integrals
Classically represent average Coulomb interaction energy of
two electrons with probability distribution functions
1s(1)
2
for
2 s(2)
and
2
Js
K's - Exchange Integrals
No classical counter part.
Product basis set not correct zeroth order set of functions.
Exchange - Integrals differ by an Exchange of electrons.
K s    1s(1)2 s( 2)
e2
4 o r12
2 s(1)1s( 2) d 1d 2
Copyright – Michael D. Fayer, 2007
Other matrix elements are zero.
Example
e2
  1s(1)2s(2) 4 r
1s(1)2 pz (2) d 1d 2  0
o 12
odd function - changes sign on inversion
through origin
Other functions are even.
Integral of even function times odd function over all space = 0.
Copyright – Michael D. Fayer, 2007
Eigenvalues (E')
E  J s  K s
1) Set determinant = 0
2) Expand
Four 2×2 blocks
3 blocks for p orbitals
identical
Js  Ks
J pi  K pi
triple root, px, py, pz
J pi  K pi
triple root, px, py, pz
Energy level diagram
1P
triply degenerate
1S2P
first excited states
n1,n2 = 1,2
3P
1S2S
triply degenerate
1S
3S
no off-diagonal
elements
1S2
ground state
n1,n2 = 1,1
1S
S configuration more stable.
S orbital puts more electron
density close to nucleus greater Coulombic attraction.
Copyright – Michael D. Fayer, 2007
Eigenfunctions
E  J s  K s
1
1s(1)2s(2)  2 s(1)1s(2)
2
E  J s  K s
1
1s(1)2s(2)  2 s(1)1s(2)
2
E  J pi  K pi
E  J pi  K pi
1
1s(1)2 pi (2)  2 pi (1)1s(2)
2
i  x, y, z
1
1s(1)2 pi (2)  2 pi (1)1s(2)
2
Copyright – Michael D. Fayer, 2007
Symmetric and Antisymmetric Combinations
+
combination symmetric.
Symmetric with respect to the interchange of
electron coordinates (labels).
P  1
P is the permutation operator.
P interchanges the labels of two electrons (labels - coordinates)
Applying P to the wavefunction gives back identical function times 1.
Copyright – Michael D. Fayer, 2007
-
combination antisymmetric.
Antisymmetric with respect to the interchange of
electron coordinates (labels).
Switching coordinates of two electrons gives negative of function.
P P

1
1s(1)2 s(2)  2 s(1)1s(2)
2
1
1s(2)2 s(1)  2 s(2)1s(1)
2
 1
1
1s(1)2 s(2)  2 s(1)1s(2)
2
 1 
Copyright – Michael D. Fayer, 2007
Both + and - functions are
eigenfunctions of permutation operator.
Symmetric function
Antisymmetric function
eigenvalue +1
eigenvalue -1
All wavefunctions for a system containing two or more identical particles
are either symmetric or antisymmetric with respect to exchange of a pair
of particle labels.
Many electron wavefunctions must be eigenfunctions of the
permutation operator.
Copyright – Michael D. Fayer, 2007
Inclusion of electron spin in He atom problem
Two spin 1/2 particles
In m1m2 representation
4 possible states.
 (1) (2)
 (1)  (2)
 (1) (2)
 (1)  (2)
The two functions,  (1) (2) and  (2) (1)
are neither symmetric nor antisymmetric.
m1m2 representation not proper representation
for two (or more) electron system.
Copyright – Michael D. Fayer, 2007
Transform into jm representation
In jm representation
 (1) ( 2)
 11
1
 (1)  ( 2)   (1) ( 2)
2
 (1)  ( 2)
 10
1
 (1)  ( 2)   (1) ( 2)
2
symmetric
spin = 1
 1 1
 00
antisymmetric
spin = 0
Copyright – Michael D. Fayer, 2007
Eight spatial functions.
H independent of spin.
Each of 8 orbital functions can be multiplied by
the 4 spin functions.
32 total (spin×orbital) functions.
1s(1)2 s(2) (1) (2)
2 s(1)1s(2) (1) (2)
1s(1)2 p x (2) (1) (2)
1s(1)2 s(2) 
1
 (1)  (2)   (1) (2) 
2
Copyright – Michael D. Fayer, 2007
Secular determinant looks like
8
8
=0
Block diagonal same as before except the correct zeroth order functions
are obtained by multiplying each of the previous spatial functions by
the four spin functions.
Example 1s2s orbitals
Copyright – Michael D. Fayer, 2007
Totally Symmetric
1
1s(1)2 s(2)  2 s(1)1s(2)   1  2 
(1-3 spat. & spin sym.
2
4 spat. & spin antisym.)
1
1
1s(1)2 s(2)  2 s(1)1s(2)   1   2    1  2  
2
2
1
1s(1)2 s(2)  2 s(1)1s(2)   1   2 
2
1
1

1
s
(1)2
s
(2)

2
s
(1)1
s
(2)

  1   2    1  2  
2
2
Totally Antisymmetric
1
(5-7 spat. antisym; spin sym.
1s(1)2 s(2)  2 s(1)1s(2)   1  2 
2
8 spat. sym.; spin antisym.)
1
1
1
s
(1)2
s
(2)

2
s
(1)1
s
(2)


  1   2    1  2  
2
2
1
1s(1)2 s(2)  2 s(1)1s(2)   1   2 
2
1
1
1s(1)2 s(2)  2 s(1)1s(2)   1   2    1  2  
2
2
Copyright – Michael D. Fayer, 2007
Energy Level Diagram
Divided according to totally symmetric or totally antisymmetric.
Totally Symmetric
Totally Antisymmetric
1s2p
1s2p
1P
1s2p
1s2p
3P
1s2s
1s2s
1s2s
1s2s
1S
1s2
1s2
1S
symmetric
spin function
3S
antisymmetric
spin function
No perturbation can mix symmetric and antisymmetric states (H symmetric).
Copyright – Michael D. Fayer, 2007
Which functions occur in nature?
Totally symmetric or totally antisymmetric
Answer with experiments.
All experiments
States occuring in nature
are always
Totally Antisymmetric
(Example - ground state of He atoms not paramagnetic,
spins paired.)
Copyright – Michael D. Fayer, 2007
Assume:
The wavefunction representing an actual state of a system
containing two or more electrons must be totally
antisymmetric in the coordinates of the electrons; i.e.,
on interchanging the coordinates of any two electrons
the wavefunction must change sign.
Q.M. statement of the Pauli Exclusion Principle
Copyright – Michael D. Fayer, 2007
To see equivalence of Antisymmetrization and Pauli Principle
Antisymmetric functions can be written as determinants.
A(1) represents an orbital×spin function for one electron,
for example, 1s(1)(1),
and B, C, … N are others, then

A(1)
A(2)
B(1)
B(2)
N (1)
N (2)
A( N ) B( N )
N(N )
Totally antisymmetric because interchange of any two rows
changes the sign of the determinant.
Copyright – Michael D. Fayer, 2007
Example - He ground state
1s(1)  (1)  1s(1)
(no bar means α spin)
1s(1)  (1)  1s(1)
(bar means  spin)
1s(1)1s(1)
 1s(1)1s(2)  1s(2)1s(1)
1s(2) 1s(2)
 1s(1) (1)1s(2)  (2)  1s(2) (2) 1s(1)  (1)
 1s(1)1s(2)  1  (2)   (1)  2  
Correct antisymmetric ground state function.
Copyright – Michael D. Fayer, 2007
Another important property of determinants
Two Columns of Determinant Equal
Determinant Vanishes
Pauli Principle
For a given one electron orbital there are only two possible
orbital×spin functions,
i. e., those obtained by multiplying the orbital function by  or  spin functions.
Thus,
no more than two electrons can occupy the same orbital in an atom,
and the two must have their spins opposed,
that is,
no two electrons can have the same values of all four quantum numbers
n, , m, ms
otherwise two columns will be equal and the determinant,
the wavefunction, vanishes.
Copyright – Michael D. Fayer, 2007
Example - He ground state with both spins 
1s(1) 1s(1)

1s(2) 1s(2)
1s(1) 1s(1)
 1s(1) (1)1s(2) (2)  1s(2) (2)1s(1) (1)  0
1s(2) 1s(2)
Requirement of antisymmetric wavefunctions is the equivalent of
the Pauli Principle.
Copyright – Michael D. Fayer, 2007
Singlet and Triplet States
Totally Antisymmetric
symmetric
spin function
1s2p
1P
1s2p
3P
1s2s
1s2s
1S
1s2
1S
term symbols
3S
antisymmetric
spin function
singlet states, sym. orbital function × (single) antisym. spin function.
triplet states, antisym. orbital function × (three) sym. spin functions.
Copyright – Michael D. Fayer, 2007
For same orbital configuration
Triplet States lower in energy than singlet states
because of electron correlation.
Triplet
sym. spin. Therefore, antisym. orbital
 (1) (2)

 1
 T  [1s(1)2 s(2)  2 s(1)1s(2)]   ( (1)  (2)   (2)  (1))
 2
  (1)  (2)
For example:
Singlet
 11
 10
 1 1
antisym. spin. Therefore, sym. orbital
 S  [1s(1)2 s(2)  2 s(1)1s(2)] 
1
( (1)  (2)   (2)  (1))
2
 00
Copyright – Michael D. Fayer, 2007
Singlet - 2 electrons can be at the same place.
Consider a point with coordinates, q.
 S  [1s(1)2s(2)  2s(1)1s(2)]
 S  [1s(q)2s(q)  2s(q)1s(q)]
 S  2[1s(q)2s(q)]
Correlation Diagram
Fix electron 1.
Plot prob. of finding 2.
(1)
Prob. of finding electron (2) at
r given electron (1) is at q.
D
q
r
(schematic illustration)
Copyright – Michael D. Fayer, 2007
Triplet - 2 electrons cannot be at the same place.
2 electrons have node for being in same place.
Consider a point with coordinates, q (any q).
 T  [1s(1)2 s(2)  2 s(1)1s(2)]
 T  [1s(q )2 s(q )  2 s(q )1s(q )]
Correlation Diagram
Fix electron 1.
Plot prob. of finding 2.
Prob. of finding electron (2) at
r given electron (1) is at q.
(1)
T  0
(1)
D
D
q
r
r
q
(schematic illustrations)
Copyright – Michael D. Fayer, 2007
In triplet state
Electrons are anti-correlated.
Reduces electron-electron repulsion.
Lowers energy below singlet state of same orbital configuration.
Copyright – Michael D. Fayer, 2007