Electron Spin
Electron spin hypothesis
Solution to H atom problem gave three quantum numbers, n, , m.
These apply to all atoms.
E
Experiments
i
t show
h nott complete
l t description.
d
i ti
Something
S
thi missing.
i i
Alkali metals show splitting of spectral lines in absence of magnetic field.
s lines not split
p, d lines split
N D
Na
D-line
li (orange
(
light
li ht emitted
itt d by
b excited
it d Na)
N ) split
lit by
b 17 cm-1
Many experiments not explained without electron "spin."
Stern-Gerlach experiment early, dramatic example.
Copyright – Michael D. Fayer, 2009
Stern-Gerlach experiment
glass
plate
Magnet
ms
oven baffles
Ag
Ag
g beam
N
+1/2
S
-1/2
Beam off silver
B
il
atoms d
deflected
fl
db
by magnetic
i fi
field
ld gradient.
di
Single unpaired electron - should not give two lines on glass plate.
s-orbital. No orbital angular momentum. No magnetic moment.
Observed deflection corresponds to one Bohr magneton.
magneton
Copyright – Michael D. Fayer, 2009
Assume: Electron has intrinsic angular momentum called "spin."
1
s 
2
11

2
S s ms    1   2 s ms
2 2 
square of
angular
t
momentum
operator
projection of
angular
momentum on
z-axis
1
S z s ms    s ms
2
 s  e/2mc
s
s



e
mc

e
2mc
Electron
El
t
has
h magnetic
ti moment.
t
One Bohr magneton.
Charged particle with angular mom.
Ratio twice the ratio for
orbital angular
g
momentum.
Copyright – Michael D. Fayer, 2009
Electronic wavefunction with spin
   ( x, y, z )
   ( x , y, z , ms )
with spin
1
ms  
2
s ms
spin angular momentum kets
 ( ms ) 

11
22

1 1

2 2
11
1 1
,

22
2 2
Copyright – Michael D. Fayer, 2009
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 m s   Ym

  ms 
product space of angular momentum (orbital and spin) eigenvectors
Copyright – Michael D. Fayer, 2009
   m m s 
Simultaneous eigenvectors of
L
2
Lz
S
2
Sz
L       1 
2
L z   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 m s  .
Copyright – Michael D. Fayer, 2009
Example - p states of an electron
=1
Orbital functions
 ( r , , )  R( r )Ym ( , )
radial
di l
spherical
h i l harmonics
h
i
orbital angular momentum part
Y11  11
Y10  10
j1m1
Y11  1 1
j1m1
j1m1
11
22
1 1
 
2 2
j2m
spin functions

j 2m
j1 m1   m
2
2
j2 m2  s m s
Copyright – Michael D. Fayer, 2009
In the m1m2 representation
 s m m s
m m s
1
1
1
1
Y11  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, 2009
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, 2009
The jm kets can be obtained from the m1m2 kets using the
table of Clebsch-Gordon coefficients.
11
2
1
1 1



1
0
p
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
1
2
1
3
j1 = 1
j 2 = 1/2
1
2
1
1

1
1  2
m1m2
m1m2
3
2

1
2
1
2

1
2

3
2
j
3
2
m
1

1
3
1
3

2
3
1
m1 m2
Copyright – Michael D. Fayer, 2009
Spin-orbit Coupling
An electron, with magnetic moment moves in the electric field
of rest of atom (or molecule).
  
W   E V  


energy

 

 
classical energy
gy of magnetic
g  dipole
p
moving in electric field E with velocity V

eS
spin magnetic moment of electron
2mc
Then
W 
e
2mc

  
E V  S
Using



e E  grad 

Coulomb potential (usually called V, but V is velocity)
and


p  mV
Copyright – Michael D. Fayer, 2009


 
1
W 
( grad   p )  S
2
2m c
Hydrogen like atoms
ze 2
 
Coulomb potential
4 0 r
unit
it vector
t



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


L
 
( r  p)
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, 2009
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 ) L i  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, 2009
Spin-orbit coupling piece of Hamiltonian
H so
 
 a( r ) L  S
H-like atom – spatial part of operator
1
ze 2
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
littl S.O.
S O coupling
li
Br – sizable S.O. coupling
a(r )  z 4
heavy atom
h
t
effect
ff t
similar for all atoms
and molecules
Heavy atom effect and external heavy
atom effect important in many processes.
Copyright – Michael D. Fayer, 2009
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
Substituting
L

S

L
S

 L S   L S  
z
z
1
2
L x   L  L 
2
1
diagonal
off-diagonal
L y   L  L 
2i
Copyright – Michael D. Fayer, 2009
Spin-Orbit Coupling in the m1m2 representation
 
1
L  S  L z S z   L S   L S  
2
di
l in
i the
h m1m2 representation
i
not diagonal
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 L z 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, 2009
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
1
L
S

L
S


    
2 2
2
2
matrix element
Copyright – Michael D. Fayer, 2009
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
p
2 2
2 2
0
0
2 2
2 2

1
2
1
2
example
p
0
1 1
1
2
 L S   L S   1  
2 2
2
2
Matrix is block diagonal.
Copyright – Michael D. Fayer, 2009
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 bl
Th
blocks
k are the
h same. O
Only
l need
d
to solve one of them.
The two 11 blocks
are diagonal – eigenvalues.
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
  a( r )
2
  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, 2009
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
small, but need spin-orbit
spin orbit coupling
to explain line splitting.
Copyright – Michael D. Fayer, 2009
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
iis in
i the
th m1m2 representation.
t ti
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, 2009
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
E
Example
l
11

22
m1 m2
m1 m2
2
1
1 1
1

0
3
2
3 2
Copyright – Michael D. Fayer, 2009
 
Operate a ( r ) L  S
on 3 3
22
 
1
L  S  L z S z   L S   L S  
2
  33
  1
1
1
a(r ) L  S
 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.
Th f
Therefore,
  33
1
33
a(r ) L  S
 a( r )
2 2
2
2 2
33
22
 
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, 2009
 
11
Consider a ( r ) L  S operating on
22
 2
11
1
1 1 
1 
0 

22
2
3 2 
 3
jjm
m1m2
m1m2
from Clebsch-Gordan table
  11
1
1 1 
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
  11
11
 1 a ( r )
a(r ) L  S
22
22
eigenvector with eigenvalue –1a(r)
Copyright – Michael D. Fayer, 2009
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, 2009
The jm representation kets will be the eigenkets
whenever a term in Hamiltonian couples
t ttypes off angular
two
l momenta.
t
Examples:
 
Hyperfine Interaction - I  S
Coupling of electron spin and nuclear spin.
Small splitting in Na optical spectrum.
Structure in ESR spectra.
Electronic triplet states - coupling of two unpaired electrons.
Copyright – Michael D. Fayer, 2009
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.
j
jm
mm
H so  U H so1 2 U
1
U is
i the
th matrix
t i off Clebsch-Gordan
Cl b h G d Coefficients.
C ffi i t
Copyright – Michael D. Fayer, 2009
11
2
1
1 2
Both are block diagonal.
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 
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 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
Clebsch-Gordan
coefficients
3
1
3

2
3
1
m1 m2
Copyright – Michael D. Fayer, 2009
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

3  2
1  1


3  2
 1/ 2 0 
 0 1 


U
2 

2 

0 

1
3
2
3

1
3
2
3
1
2 

3 

1


3
2

3

1 

3 
diagonal matrix with
eigenvalues on the diagonal
Copyright – Michael D. Fayer, 2009
Electron Spin, Antisymmetrization of Wavefunctions, and the Pauli Principle
In treating the He atom, electron spin was not included.
Electron - p
particle with intrinsic angular
g
momentum,, spin
p
S  1/ 2
m s  1/ 2
1/2 1/2  
1/2 1/2  
Copyright – Michael D. Fayer, 2009
Excited States of He neglecting Spin
G
Ground
d State
St t off H
He - perturbation
t b ti th
theory
H'
e2
perturbation
4 0 r12
Zeroth order wavefunctions (any level)
Product of H atom orbitals
 n  m (1)  n  m (2)   n  m (1)  n  m (2)
1 1
1
2 2
2
1 1
1
2 2
2
(1)
electron 1; coordinates ( r1 , 1 ,1 )
(2)
electron 2; coordinates ( r2 , 2 , 2 )
Copyright – Michael D. Fayer, 2009
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, 2009
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, 2009
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
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
g
matrix elements
Copyright – Michael D. Fayer, 2009
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, 2009
J's - Coulomb Integrals
Classically represent average Coulomb interaction energy of
two electrons with probability distribution functions
1s(1)
2
f
for
2 s(2)
and
2
Js
K's - Exchange
g Integrals
g
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, 2009
Other matrix elements are zero.
p
Example
e2
  1s(1)2 s(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.
even
g of even function times odd function over all space
p
= 0.
Integral
Copyright – Michael D. Fayer, 2009
Eigenvalues (E')
E  Js  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,
root px, py, pz
J pi  K pi
triple root, px, py, pz
Energy level diagram
1P
t i l degenerate
triply
d
t
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.
puts more electron
S orbital p
density close to nucleus greater Coulombic attraction.
Copyright – Michael D. Fayer, 2009
Eigenfunctions
E  Js  K s
1
E  Js  K s
1
E  J pi  K pi
2
2
1s(1)2 s(2)  2 s(1)1s(2)
1s(1)2 s(2)  2 s(1)1s(2)
1
2
E  J pi  K pi
1s(1)2 pi (2)  2 pi (1)1s(2)
i  x, y, z
1
1s(1)2 pi (2)  2 pi (1)1s(2)
2
Copyright – Michael D. Fayer, 2009
Symmetric and Antisymmetric Combinations
+
combination symmetric
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, 2009
–
combination antisymmetric.
Antisymmetric with respect to the interchange of
electron coordinates (labels).
Switching coordinates of two electrons gives function back
multiplied by –1.
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, 2009
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, 2009
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
ep ese tat o not
ot proper
p ope representation
ep ese tat o
for two (or more) electron system.
Copyright – Michael D. Fayer, 2009
Transform into jm representation
I jm
In
j representation
i
 (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, 2009
Eight spatial functions.
H independent of spin.
p
by
y
Each of 8 orbital functions can be multiplied
the 4 spin functions.
32 total (spin×orbital) functions.
1s(1)2 s(2) (1) (2)
) s(2)
( ) ((1)) ((2))
2 s((1)1
1s(1)2 p x (2) (1) (2)

1s(1)2 s(2) 
1
 (1)  (2)   (1) (2)
2

Copyright – Michael D. Fayer, 2009
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
p functions.
Example 1s2s orbitals
Copyright – Michael D. Fayer, 2009
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1s(1)2 s(2)  2 s(1)1s(2)   1  2
1s(1)2 s(2)  2 s(1)1s(2) 
1
2
 1   2    1   2  
Totally Symmetric
(1-3 spat. & spin sym.
4 spat. & spin antisym.)
1s(1)2 s(2)  2 s(1)1s(2)   1   2
1s(1)2 s(2)  2 s(1)1s(2) 
1
2
 1   2    1   2  
1s(1)2 s(2)  2 s(1)1s(2)   1  2
1s(1)2 s(2)  2 s(1)1s(2) 
1
2
Totally Antisymmetric
(5-7 spat. antisym; spin sym.
8 spat. sym.; spin antisym.)
 1   2    1   2  
1s(1)2 s(2)  2 s(1)1s(2)   1   2
1s(1)2 s(2)  2 s(1)1s(2) 
1
2
 1   2    1   2  
Copyright – Michael D. Fayer, 2009
Energy Level Diagram
Divided according to totally symmetric or totally antisymmetric.
Totally Symmetric
Totally Antisymmetric
1s2p
1s2p
1P
12
1s2p
12
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, 2009
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, 2009
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, 2009
To see equivalence of Antisymmetrization and Pauli Principle
Antisymmetric
t sy
et c functions
u ct o s can
ca be written
w tte as dete
determinants.
a ts.
A(1) represents an orbital×spin function for one electron,
for example, 1s(1)(1),
and B, C, … N are others, then

A(1)
B(1)

N (1)
A(2)
B(2)

N (2)



A( N ) B( N )  N ( N )
Totally antisymmetric because interchange of any two rows
changes the sign of the determinant
determinant.
Copyright – Michael D. Fayer, 2009
Example - He ground state
1s(1)  (1)  1s(1)
(no bar means α spin)
1s(1)  (1)  1s(1)
(bar means  spin)
1s((1)1
) s((1))
1s(2) 1s(2)
 1s(1)1s(2)  1s(2)1s(1)
 1s(1) (1)1s(2)  (2)  1s(2) (2) 1s(1)  (1)
) s((2))  1  ((2))   ((1))  2  
 1s((1)1
Correct antisymmetric ground state function.
Copyright – Michael D. Fayer, 2009
Another important property of determinants
T Columns
Two
C l
off Determinant
D t
i
t Equal
E
l
D t
Determinant
i
t Vanishes
V ih
Pauli Principle
For a given orbital there are only two possible
orbital×spin functions,
p y g the orbital function by
y  or  spin
p functions.
i. e.,, those obtained byy multiplying
Thus,
no more than two electrons can occupy the same orbital in an atom,
p opposed,
pp
,
and the two must have their spins
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, 2009
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
R
i
t off antisymmetric
ti
t i wavefunctions
f
ti
is
i the
th equivalent
i l t off
the Pauli Principle.
Copyright – Michael D. Fayer, 2009
Singlet and Triplet States
T t ll Antisymmetric
Totally
A ti
ti
symmetric
spin function
1s2pp
1P
1s2p
3P
1s2s
12
1s2s
1S
1s2
1S
term
e
symbols
sy bo s
3S
antisymmetric
spin function
singlet
i l t states,
t t
sym. orbital
bit l function
f
ti × (single)
( i l ) antisym.
ti
spin
i function.
f
ti
triplet states, antisym. orbital function × (three) sym. spin functions.
Copyright – Michael D. Fayer, 2009
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, 2009
Singlet - 2 electrons can be at the same place.
Consider a point with coordinates,
coordinates q.
q
 S  [1s(1)2 s(2)  2 s(1)1s(2)]
 S  [1s(q )2 s(q )  2 s(q )1s(q )]
 S  2[1
[ s(q ))2 s(q )]
Correlation Diagram
Fix electron 1.
Pl t prob.
Plot
b off finding
fi di 2.
2
(1)
Prob. of finding electron (2) at
r given electron (1) is at q.
D
q
r
(schematic illustration)
Copyright – Michael D. Fayer, 2009
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, 2009
In triplet state
Electrons are anti-correlated.
Reduces electron-electron repulsion.
Lowers energy below singlet state of same orbital configuration.
Copyright – Michael D. Fayer, 2009