Multi-Electron Atoms
• Start with Helium: He+ - same as H but with Z=2
• He - 2 electrons. No exact solution of S.E. but can use H wave
functions and energy levels as starting point. We’ll use some aspects
of perturbation theory but skip Ritz variational technique (which sets
bounds on what the energy can be)
• nucleus screened and so Z(effective) is < 2
• “screening” is ~same as e-e repulsion (for He, we’ll look at e-e
repulsion. For higher Z, we’ll call it screening)
• electrons are identical particles. Will therefor obey Pauli exclusion
rule (can’t have the same quantum numbers). This turns out to be due
to the symmetry of the total wave function
P460 - Helium
1
Schrod. Eq. For He
• have kinetic energy term for both electrons (1+2)
 T 
 22 T  VT T  ET T


 
VT  V (r1 )  V (r2 )  V12 (r1  r2 )

2
2m
2
1
2
2m
• V12 is the e-e interaction. Let it be 0 for the first approximation, that is
for the base wavefunctions and then treat it as a (large) perturbation
• for the unperturbed potential, the solutions are in the form of separate
wavefunctions
 


 T ( r1 , r2 )   ( r1 ) ( r2 )
ET  E1  E2
P460 - Helium
2
Apply symmetry to He
• The total wave function must be antisymmetric
• but have both space and spin components and so 2 choices:
 He   space spin  either
 space  symmetric   spin  antisym
 space  antisym  spin  sym
• have 2 spin 1/2 particles. The total S is 0 or 1
OR
 spin
S ms ms1 ms 2
1  2
1
2
(1  2  1  2 )
1  2
1
2
(1  2  1  2 )
• S=1 is spin-symmetric
1
2
1
2
1
1
1
0 
1
2
1 1 
1
2
0
0 

1
2
1
2

1
2

1
2
S=0 is spin-antisymmetric
P460 - Helium
3
He spatial wave function
• There are symmetric and antisymmetric spatial wavefunctions which
go with the anti and sym spin functions. Note a,b are the spatial
quantum numbers n,l,m but not spin
 space ( sym ) 
 space ( asym ) 
1
2
( a (1) b ( 2)   a ( 2) b (1))
1
2
( a (1) b ( 2)   a ( 2) b (1))
• when two electrons are close, the antisymmetric state is suppressed
(goes to 0 if exactly the same point). symmetric state is enhanced
• “Exchange Force” S=1 has the electrons (on average) further apart
(as antisymmetric space). So smaller repulsive potential and so lower
energy
• if a=b, same space state, must have symmetric space and
antisymmetric spin S=0 (“prove” Pauli exclusion)
P460 - Helium
4
He Energy Levels
• V terms in Schrod. Eq.:
2e 2
2e 2
e2
V 


 
40r1 40r2 40 | r1  r2 |
• Oth approximation. Ignore e-e term.
ET  E1  E2  
4  13.6eV
4  13.6eV

n12
n22
n1  n2  1

 109 eV
n1  1( 2), n2  2(1)

 68 eV
n1  n2  2

 27 eV
• 0.5th approximation: Guess e-e term. Treat electrons as point objects with average
radius (for both n=1) a0/Z (Z=2). electrons on average are ~0.7ao apart with
repulsive energy:
e2
e2
 1.4  13.6 eV  1.4  19eV
  
40 | r1  r2 | 40 a0
P460 - Helium
5
He Energy Levels II
• V terms in Schrod. Eq.:
2e 2
2e 2
e2
V 


 
40 r1
40 r2
40 | r1  r2 |
• First approximation: look at expectation value of e-e term which will
depend on the quantum states (i,j) of the 2 electrons and if S=0 or 1
Eij  Eij  Vij
with
Vij
e2

 
40 | r1  r2 |
• Assume first order wavefunction for ground state has both electrons in
 


1S f100 (spin S=0)
u0 ( r1 , r2 )  f100 ( r1 )f100 ( r2 )
• expectation value
e2
3
3
V11   u (1,2)
  u (1,2)d r1d r2
40 | r1  r2 |


with u* (1,2)u (1,2) | f100 ( r1 ) |2 | f100 ( r2 ) |2
*
and
1
1
  
| r1  r2 | ( r12  r22  2 r1r2 cos 12 )1 / 2
P460 - Helium
6
He Energy Levels III
• do integral. Gets 34 eV. The <V11> is measured to be 30 eV 
E(ground)=-109+30=-79 eV
• For n1=1, n2=2. Can have L2=0,1. Either S=0 or S=1. The space
symmetrical states (S=0) have the electrons closer  larger <V12>
  S  0, sin glet
and larger E
  S  1, triplet
• formally V  
e

2
12
12


4 0 | r1  r2 |
12
 12  f100 (1)f2 l 0 ( 2)  f100 ( 2)f2 l 0 (1)
| 12 |2 | f100 (1) |2 | f2 l 0 ( 2) |2  | f100 ( 2) |2 | f2 l 0 (1) |2
*
*
 f100
( 2)f2*l 0 (1)f100 (1)f2 l 0 ( 2)  f100
(1)f2*l 0 ( 2)f100 ( 2)f2 l 0 (1)
• first two terms are identical ”Jnl” as are the second two “Knl”. Both
positive definite. Gives
V12  E2( ,Sl 0,1)  J 2 ,l  K 2 ,l
1 e in n=1 state
 En( S,l 0,1)  J n ,l  K n ,l
P460 - Helium
7
He Energy Levels III
• L=0 and L=1 have different radial wavefunctions. The n=2, L=1 has
more “overlap” with the n=1,L=0 state  electrons are closer 
larger <V12> and larger E
• usually larger effect than two spin states. Leads to Hund’s rules (holds
also in other atoms)
N=1 L=0
N=2 L=0
P(r)
N=2 L=1
r
P460 - Helium
8
He Energy Levels IV
N1=2,n2=2
S
-27
0
L1=1,L2=0
1
0
1
E
L1=0,L2=0
N1=1,n2=2
-70
-110
n1=1,n2=1 S=0
P460 - Helium
9
Multi-Electron Atoms
• can’t solve S.E. exactly  use approximations
• Hartree theory (central field model) for n-electron atoms. Need an
antisymmetric wave function (1,2,3 are positions; i,j,k…are quantum
states)
 (1,2,3....n ) 
 (1) ( 2) (3).....

i
j
k
allterms
  n!
terms / antisymmetrical
 a (1)  a ( 2)  a (3)
n  3  16  b (1)  b ( 2)  b (3)
 c (1)  c ( 2)  c (3)
• while it is properly antisymmetric for any 1  j practically only
need to worry about valence effects
P460 - Helium
10
Schrod. Eq. For multi-electron
• have kinetic energy term for all electrons

2
2m
 T 
2
1
2
2m
  T  ...... 
2
2
• and (nominally) a complete potential energy:
2
2m
2n T
 Ze 2
e2
VT  
 


40 ri
i
n
n 40 | rn  rn |
• Simplify if look at 1 electron and sum over all the others
•  ~spherically symmetric potential as filled subshells have no
(theta,phi) dependence. Central-Field Model. Assume:
Z e ( r )e 2
V (r )  
40 r
with Z e (0)  Z , Z e ()  1
Ylm ( , f ) " good " eigenfunctions
l , ml " good " eigenvalues
P460 - Helium
11
Multi-electron Prescription
• “ignore” e-e potential..that is include this in a guess at the effective Z
• energy levels can depend on both n and L
• wavefunctions are essentially hydrogen-like
1
2
3
4
guess at effective Z: Ze(r)
solve (numerically) wave equation
fill energy levels using Pauli principle
use wavefunctions to calculate electrons’ average radii (radial
distribution)
5 redetermine Ze(r)
6 go back to step 2
P460 - Helium
12
First guess
• electron probability P(r). Electrons will fill up “shells” (Bohr-like)
N=1
n=2
n=3
P(r)
radius
assume all electrons for given n are at the same r with a Vn for each n.
Know number of electrons for each n. One can then smooth out this
distribution to give a continuous Z(r)
Z-2
Z ne2
Vn ( r )  
r
Zn
Z-8
Z-18
radius
P460 - Helium
13
Argon Z=18
• First shell
n=1 1S
Z1 ~ 18-2 = 16
Second shell n=2 2S,2P Z2 ~ 18-2-8 = 8
Third shell n=3 3S,3P Z3 ~ 18-10-8/4 = 4
• for the outermost shell…good first guess is to assume half the
electrons are “screening”. Will depend on where you are in the
2
Periodic Table
Z


En   n   13.6eV
• guess energy levels.
 n 
E1  (16) 2 13.6  3500eV
E2  ( 82 ) 2 13.6  220eV
E3  ( 43 ) 2 13.6  24eV
Zn
16
8
4
Z(r)
radius
P460 - Helium
14
Atomic Energy Levels
After some iterations, get generalizations
A for the innermost shell: Z1  Z  2 
1
r1 
a0
Z 2
E1  ( Z  2) 2 E1 ( H )
B Outermost shell
Zn  n 
n2
rn 
a0  na0
Zn
En
Z n2

n2
E1 ( H )  E1 ( H )
• atoms grow very slowly in size. The energy levels of the outer/valence
electrons are all in the eV range
P460 - Helium
15
Comment on X-ray Spectra
• If one removes an electron from an inner orbit of a high Z material,
many transitions occur as other electrons “fall”
• as photon emitted, obey
m  0,1
l  1
• “large” energies so emit x-ray photons. Or need x-ray photons (or
large temperature or scotch tape) to knock one of the electrons out
E1S  ( Z  2) 2 13.6eV  90 keV ( Z  82)
T 
90 keV
 109 0K
k
• study of photon energy gives effective Z for inner shells
P460 - Helium
16
Spectroscopic States in Atoms
• Only valence electrons need to be considered (inner “shield” nucleus)
• Filled subshells have L=0 and S=0 states (like noble gases).
• Partially filled subshells: need to combine electrons, make
antisymmetric wavefunctions, and determine L and S. Energy then
depends on J/L/S
• if >1/2 filled subshell then ~same as treating “missing” electrons like
“holes”.
Example: 2P no. states=6
1 electron  5 electrons
2 electrons  4 electrons
3 electrons
6 electrons (filled)
(can prove by making totally antisymmetric wavefunctions)
P460 - Helium
17
Periodic Table
• Follow “rules”
• for a given n, outer subshell with the lowest L has lowest E (S<P<D).
Reason: lower L  smaller <r>  larger effective Z
P(r)
H
<r2P> < <r2S> for H
2P
but “bump” in 2S at
2S
low r gives smaller <r>
for higher Z (need
radius
to weight by effective Z)
• For a given L, lowest n has lowest energy (smaller n and smaller <r>)
• no rule if both n,L are different. Use Hartree or exp. Observation. Will
vary for different atoms
• the highest energy electron (the next state being filled) is not
necessarily the one at the largest radius (especially L=4 f-shells)
P460 - Helium
18
Alkalis
• Filled shell plus 1 electron. Effective Z 1-2 and energy levels similar
to Hydrogen
Z2
E
eff
2
n
EH
• have spin-orbit coupling like H
-2
with
Z eff ( n, l )
j  0,1
l  1
4P 3D
3S 3P 3D
3
4S
-3
E(eV)
-4
2
-5
-6
E 
13.6
n2
H

2P
3P
2S
3S
(1.2 ) 2
n2
13.6
Li
P460 - Helium

(1.7 ) 2
n2
13.6
Na
19
Helium Excited States
• Helium: 1s2 ground state. 1s2s 1s2p first two excited states. Then
1s3s 1s3p and 1s3d as one electron moves to a higher energy state.
• combine the L and S for the two electrons to get the total L,S,J for a
particular quantum state. Label with a spectroscopic notation
• Atoms “LS” coupling: (1) combine Li give total L (2) combine Si to
give total S; (3) combine L and S to give J
• (nuclei use JJ: first get Ji and then combine Js to get total J)
• as wavefunctions are different, average radius and ee separation will
be different  2s vs 2p will have different energy (P further away,
more shielding, higher energy)
• S=0 vs S=1 S=1 have larger ee separation and so lower energy
• LS coupling similar to H  Lower J, lower E
P460 - Helium
20
Helium Spectroscopic States
• D=degeneracy=number of states with different quantum numbers (like
Sz or Lz) in this multiplet
l1  0,1,2  l 2  0  L  0,1,2
s1 
2 s 1
1
2
 s2 
1
2
 S  0  S  1 ( d  1,3)
L j  lev el
L
S
1s1s
0
0
1
1s 2 p
1
0
1
1s 2 p
1
1
1s 2 s
0
1s 3s  1s 2 s
( d  1,3,5)
0
states
( d  1)
S0
( d  3)
P1
3
P0
P2 ( d  1,3,5)
3
3
P1
1
S 0 , 3S1
( d  1,3)
1s 3 p  1s 2 p ( higher E )
1s 3d
2
0
1s 3d
2
1
( d  5)
1
D2
3
D1
3
D2
P460 - Helium
D3 ( d  3,5,7 )
3
21
Helium:States, Energies and Transitions
• Transitions - First Order Selection Rules (time dep. Pert. Theory)
s  0  no spin flip
l  1  as photon S  1
j  0,1
no 0  0 ( 3S1 3P1 OK )
• Metastable 2S with S=1 state - long lifetime
3
2
E
1
S0
1
P1
1
D2
3
S1
3
P0,1, 2
3
D3, 2,1
Triplet S=1
1
Singlet S=0
P460 - Helium
22
Spectroscopic States if >1 Valence Electron
• Look at both allowed states (which obey Pauli Exclusion) and shifts in
energy
• Carbon: 1s22s22p2 ground state. Look at the two 2p highest energy
electrons. Both electrons have S=1/2
" #1"
" #2"
n
l
n
l
2
1
2
1
2 p2
2
1
3
0
2 p1 3s1
2
1
3
1
2 p1 3 p 1
• combine the L and S for the two electrons to get the total L,S,J for a
particular quantum state
P460 - Helium
23
States for Carbon II
• 2p13s1 different quantum numbers  no Pauli Ex.
12 states = 6 ( L=1, S=1/2) times 2 (L=0, S=1/2)
l1  1  l 2  0  L  1
s1 
1
2
 s2 
1
2
( d  3)
 S  0  S  1 ( d  1,3)
J  L  S  1  0  1;1  1  2  1  0
2 s 1
Lj 
S
states
1
0
1
1
1
L
( d  3)
P1
3
P0
3
P1
P2 ( d  1,3,5)
3
• where d=degeneracy = the number of separate quantum states in that
multiplet
P460 - Helium
24
• 2p13p1 different quantum numbers  no Pauli Ex.
36 states = 6 ( L=1, S=1/2) times 6 (L=01 S=1/2)
l1  1  l 2  1  L  0,1, 2
s1 
1
2
 s2 
1
2
( d  1,3,5)
 S  0  S  1 ( d  1,3)
J  L  S  0  0  0; 1  0  1; 2  0  2
0  1  1; 1  1  2  1  0; 2  1  3  2  1;
2 s 1
Lj 
L
0
1
S
0
0
0
1
2
1
1
• where d=degeneracy = 2j+1
( d  1)
1
0
2
1
states (deg eneracy )
S0
( d  3)
1
P1
( d  5)
1
D2
S1 ( d  3)
3
3
P0
3
D1
3
P1
3
P460 - Helium
P2 ( d  1,3,5)
3
D2
D3 ( d  3,5,7 )
3
25
• 2p2 state: same L+S+J combinations as the 2p13p1 but as both n=2
need to use Pauli Exclusion
reject states which have the same
3
D3
ml1 , ml2 , ms1 , ms1
mj  3 
ml1  ml2  1  ms1  ms2 
1
2
• not allowed. Once one member of a family is disallowed, all rejected
 no
3
D1, 2,3
states
• But many states are mixtures. Let:
S, Sz
0,0 
1,0 
 AS1 , S1 z
1
2
1
2
 12 , 12  
 12 , 12  
P460 - Helium
  BS 2 , S 2 z 
1
1
1


,

2
2
2
1
1
1

,

2
2 
2
26
• Need a stronger statement of Pauli Exclusion principle (which tells us
its source)
• Total wavefunction of n electrons (or n Fermions) must be
antisymmetric under the exchange of any two electron indices
• 2 electrons spin part of wavefunction:S=0 or S=1
S  1  symmetry 1  2
S  0  symmetry 1  2
• So need the spatial part of the wavefunction to have
S  1, need " L" have
 symmetry
1 2
S  0, need " L" have
 symmetry
1 2
D states L  2    1D2 ( S  0, deg  5)
P states L  1    3P2 ,1, 0 ( S  1, deg  5,3,1)
S states L  0    1S0 ( S  0, deg  1)
• other states not allowed (but are in in 2p3p)
P460 - Helium
27
Building Wave Functions
• Use 2p2 how to build up more complicated wave functions.
• All members of the same multiplet have the same symmetry
• redo adding two spin 1/2  S=0,1 . Start with state of maximal Sz.
Always symmetric (all in same state)
S  1, S z  1  S z1 
1
2
, Sz2 
use stepdown S   1,0 
1
2
1
2
(
1
2
,
1
2
 
must be symmetric.repeat  1,1  
1
2
1
2
,
,
1
2
)
1
2
• other states must be orthogonal (different Szi are orthogonal to each
other. But some states are combinations, like a rotation, in this case of
45 degrees)
1
1
1
1
1
S  0, S z  0 
2
(
2
,
2
 
2
,2 )
• S=0 is antisymmetric by inspection
P460 - Helium
28
Building Wave Functions
• Adding 2 L=1 is less trivial  2,1,0
• L=2. Values in front of each term are Clebsch-Gordon coeff. Given
by “eigenvalues” of stepdown operator
L  2, S z  2  Lz1  1, Lz 2  1
use stepdown L  2,1 
repeat  2,0 
1
6
 1,1
1
2
( 1,0  0,1 )
 2 0,0   1,1

A
• other states are orthogonal. Given by CG coefficients/inspection
L  1, Lz  1 
1,0

1
2
 1,1
1
2
( 1,0  0,1 )
  1,1

B
• L=1 is antisymmetric by inspection. Do stepdown from L=1 to L=0
L  0, Lz  0 
1
3
 1,1
 0,0   1,1

C
• L=0 symmetric by inspection. Also note orthogonality: A*C=1-2+1
and B*C=1-1
P460 - Helium
29
More Energy Levels in Atoms
• If all (both) electrons are at the same n. Use Hund’s Rules (in order of
importance)
• minimize ee repulsion by maximizing ee separation  “exchange
force” where antisymmetric spatial wavefunction has largest spatial
separation
•
 maximum S has minimum E
• different L  different wavefunctions  different effective Z (radial
distributions)  maximum L has minimum E
•  minimum J has minimum E unless subshell >1/2 filled and then
 minimum J has maximum E


L  S coupling ( S  0)
E
 Hydrogen
 K  j ( j  1)  l (l  1)  s ( s  1) 
P460 - Helium
30
3d4p S=0,1 L=1,2,3
L=1 J=1
S=0
L=2 J=2
J=2
L=3 J=3
J=1
L=1
J=0
J=3
L=2
S=1
J=2
L=3
J=1
J=4
J=3
J splitting
E ( j )  K ( j ( j  1)  l (l  1)  s ( s  1))
J=2
E ( j  1)  K (( j  1)( j  2)  l (l  1)  s ( s  1))
E ( j  1)  E ( j )  K (( j  1)( j  2  j ))  2 K ( j  1)
Lande interval rule
P460 - Helium
31