20_01fig_PChem.jpg Hydrogen Atom M m r Potential Energy 2 () 4 4 e n o o qq Ze Vr r r + Kinetic Energy ˆ ˆ ˆ n e K K K R C 2 2 2 2 ˆ 2 2 R r K M m 2 2 ˆ () 2 r Kr m M m m
Feb 18, 2016
20_01fig_PChem.jpg
Hydrogen Atom
M
m
r
Potential Energy2
( )4 4
e n
o o
q q ZeV rr r
+
Kinetic Energy
ˆ ˆ ˆn eK K K
R
C2 2
2 2ˆ2 2R rKM m
22ˆ ( )
2 rK rm
M m m
20_01fig_PChem.jpg
Hydrogen Atom2 2
2ˆ ˆ ˆ( ) ( )2 4r
o
ZeH K r V rm r
22 2
2 2 2 2 2
1 1 1sinsin sinr
d d d d drr dr dr r d d r d
22 2
2 2
1 1ˆ sinsin sin
d d dLd d d
2 2 22
2 2
ˆˆ2 2 4 o
d d L ZeH rmr dr dr mr r
Radial Angular Coulombic
20_01fig_PChem.jpg
Hydrogen Atom( , , )r will be an eigenfunction of 2ˆ ˆ ˆ, & zH L L
.( , , ) ( ) ( , )n l mr R r Y Separable
ˆ ( , , ) ( , , )H r E r
2 2 22
. .2 2
ˆ( ) ( , ) ( ) ( , )
2 2 4 n l m n n l mo
d d L Zer R r Y E R r Ymr dr dr mr r
2 22
. .2 2
( 1)( , ) ( ) ( , ) ( )2 2l m n l m n
d d l lY r R r Y R rmr dr dr mr
2
. .( , ) ( ) ( ) ( , )4l m n n n l m
o
ZeY R r E R r Yr
2 2 22
2 2
( 1) ( ) 02 2 4 n n
o
d d l l Zer E R rmr dr dr mr r
20_01fig_PChem.jpg
Hydrogen Atom2
22 2 2 2
21 ( 1) ( ) 02
nn
o
E md d l l Zmer R rr dr dr r r
2 22 2
2 2 2 2
1 1 22d d d d d dr r rr dr dr r dr dr r dr dr
Recall
2
2
4 0.0529ooa nm
me
Bohr Radius
2
2 2 2
22 ( 1) 2 ( ) 0nn
o
E md d l l Z R rr dr dr r a r
20_01fig_PChem.jpg
Hydrogen Atom
21
2 2 2
22 ( 1) 2 0r
o
E md d l l Z er dr dr r a r
Assume1( ) 0 asR r r
1( ) rR r e Let’s try
2 12 2
22 ( 1) 2 0r
o
E ml l Z er r a r
2 12 2
21 2 1( 1) 2 0o
E mZl lr a r
2 12
22( 1) 0; 2 0; & 0o
E mZl la
It is a ground state as it has no nodes
20_01fig_PChem.jpg
Hydrogen Atom
2 2
10; ; &2o
Zl Ea m
2 12
22( 1) 0; 2 0; & 0o
E mZl la
22 2 2 2 2
1 2 202 2 4 o
Z Z meEa m m
2 4
2 24 2o
Z me
0
1
0,( , , ) ( ) ( , ) ( )
Zral
n l mr R r Y CR r Ce
The ground state as it has no nodes n=1, and since l=0 and m = 0, the wavefunction will have no angular dependence
2 2
1 202
ZEa m
20_01fig_PChem.jpg
Hydrogen AtomIn general:
Laguerre Polynomials
11
12
33
1 23
33
55
1 0 ( ) 1
2 0 ( ) (2!)(2 )
1 ( ) (3!)
3 0 ( ) (3!) 3 3 0.5
1 ( ) (4!)(4 )
2 ( ) (5!)
n l L x
n l L x x
l L x
n l L x x x
l L x x
l L x
0
2Zrxna
0
32 1
4 3 30 0 0
4 ( 1)! 2 2( )[( )]
l Zrnal l
n n lZ n l Zr ZrR r e Ln a n l na na
2 1
0
2ln l
ZrLna
1S- 0 nodes
2S- 1 node
3S-2 nodes
Energies of the Hydrogen AtomIn general:
4 2
2 2 2
124
n
o
me ZEn
2 2
20
124 o
e Za n
2
22Zn
2
0
27.24H
o
eE eVa
Hartrees
kJ/mol
627.51 / 2625.5 /kcal mol kJ mol
Wave functions of the Hydrogen AtomIn general:
0
32 1
4 3 30 0 0
4 ( 1)! 2 2( )[( )]
l Zrnal l
n n lZ n l Zr ZrR r e Ln a n l na na
,1( , ) (cos( ))2
mm iml m l lY C P e
,( , , ) ( ) ( , )ln l mr R r Y
Z=1, n = 1, l = 0, and m = 0:
00 (cos( )) 1P 0
012
C
11
0
2 1rLa
01 0,0( , , ) ( ) ( , )r R r Y
001 3
0
2( )raR r e
a
0,0
1( , )2
Y
0 0
3 30 0
2 1 12
r ra ae e
a a
Z=1, n = 2, l = 0, and m = 0:
12
0 0
2! 2r rLa a
0,01( , )
2Y
02
300
122 2
rae r
aa
0202 3
00
1( ) 122
ra rR r e
aa
02 0,0( , , ) ( ) ( , )r R r Y
Wave functions of the Hydrogen Atom
Hydrogen AtomZ=1, n = 2, l = 1
022,1,0 3
0 0
1 2( , , ) cos8
rarr e
a a
m = 0: m = +1/-1:
022,1, 1 3
0 0
1 1( , , ) sin8
rarr e e
a a
022,1, 2,1, 1 2,1, 1 3
0 0
1 1 1( , , ) ( , , ) ( , , ) sin cos2 8
ra
xrr r r e
a a
022,1, 2,1, 1 2,1, 1 3
0 0
1 1 1( , , ) ( , , ) ( , , ) sin sin2 8
ra
yrr r r e
i a a
+
_
-+-+
+-
-+
+- -
+
20_06fig_PChem.jpg
* 2, , , ,( ) ( , , ) ( , , ) sinn l m n l mP R r r r drd d
* * 2, ,( ) ( , ) ( ) ( , ) sinl l
n l m n l mR r Y R r Y r drd d * 2 *
, ,( ) ( ) ( , ) ( , )sinl ln n l m l mR r R r r dr Y Y d d
For radial distribution functions we integrate over all angles only2
* 2 *, ,
0 0
( ) ( ) ( ) ( , ) ( , )sinl ln n l m l mP r R r R r r Y Y d d
* 2( ) ( )l ln nR r R r r Prob. density as a function of r.
Radial Distribution Functions
20_09fig_PChem.jpg
Radial Distribution Functions0
22
30
4 rar e
a
0* 0 21,0,0 1 1( ) ( ) ( )P r R r R r r00
1 30
2( )raR r e
a
0202 3
00
1( ) 122
ra rR r e
aa
0* 0 22,0,0 2 2( ) ( ) ( )P r R r R r r
0 3 42
30 0 02 4
rae r rra a a
20_08fig_PChem.jpg
* 2, , , ,( , , ) ( , , ) ( , , ) sinn l m n l mP r r r r
* 2 *, , , ,( ) ( ) ( , ) ( , )sinl l
n n l m l mR r R r r Y Y
, ,( ) ( , )n l l mP r P
X
Y
Z
Probability Distributions
0
42
2,1 1,0 50
( ) ( , ) cos sin32
rarP r P e
a
, ( )n lP r, ( , )l mY
20_12fig_PChem.jpg
Atomic UnitsSet:
2
2
4 1 . .ooa a u
me
2 4 2
2 22 2 24 2n
o
Z me ZEnn
2 2 22
2 2
ˆˆ2 2 4e e o
d d L ZeH rm r dr dr m r r
2
0
1, 1, & 14eem
Hartrees
22
2 2
ˆ12 2
d d L Zrr dr dr r r
2
2Zr
a.u.
Much simpler forms.
0
3 3
1 30
Zra Zr
sZ Ze ea
023 3 2
2 300
1 12 22 22 2
r ra
sZ e r Z e r
aa
AtomsPotential Energy
2
( )4 4
e nen i
o i o i
q q ZeV rr r
Kinetic Energy2 2
2 2ˆ2 2 iR r
i e
KM m
22ˆ ( )
2 ii ri i e
K rm
C
me
me
2
( )4 4
i jee
o ij o ij
q q eV rr r
=r12
M
2
2i
i
i
Zr
1
ijr
1
1( ) ( )en i ee iji i j i iji ij
ZV V r V rr r
Helium Atom
C
me
me
=r12
M
2 1ˆ ˆ ˆ2i
i i j ii ij
ZH K Vr r
2 12i
i i ji ij
Zr r
,
1ˆi
i i i j ij
Hr
1 212
1ˆ ˆH Hr
Cannot be separated!!!
2
2i
ii
ZHr
Hydrogen like 1 e’ Hamiltonian
i.e. r12 cannot be expressed as a function of just r1 or just r2
What kind of approximations can be made?
Ground State Energy of Helium Atom
Eo
E1
E2
I1 = 24.587 ev
Eo
E1
E2
I2 = 54.416 evIonization Energy of He
EFree
Eo=- 24.587 - 54.416 ev =- 79.003 ev =- 2.9033 Hartrees
Perturbation Theory 1 212
1ˆ ˆ ˆH H Hr
01 2
ˆ ˆ ˆH H H 1
12
1Hr
0 0 01 2 1 2( , ) ( ) ( )r r r r
0 0 01 1 1 1 1 1
ˆ ( ) ( )H r E r 101
1(1 ) rs e
1
20 11
12r ZH
r
2 2
0 11 2 2
1
2 22 2 1ZEn
Ground State Energy of Helium Atom0
1 2ˆ ˆ ˆH H H 0 0 0
1 2 1 2( , ) ( ) ( )r r r r
0 0 0 01 2 1 2 1 2
ˆ ˆ ˆ( , ) ( ) ( )H r r H H r r 0 0 0 0
1 1 2 2 1 2ˆ ˆ( ) ( ) ( ) ( )H r r H r r
0 0 0 02 1 1 1 2 1
ˆ ˆ( ) ( ) ( ) ( )r H r r H r
0 0 0 02 1 1 1 2 1( ) ( ) ( ) ( )r E r r E r
1 2 0 00 0 2 1( ) ( )E E r r
0 01 2( , )E r r
0 0 01 2 2 2 4E E E H
20 011 22
1
22ZE En
Not even close.Off by 1.1 H, or3000 kJ/mol
Therefore e’-e’ correlation, Vee, is very significant
Ground State Energy of Helium Atom0
12
1ˆ ˆH Hr
1
12
1Hr
0 0 01 2 1 2( , ) ( ) ( )r r r r
0 01 1 1 1 1 1
ˆ ( ) ( )H r E r
01 2
ˆ ˆ ˆH H H
0 1 0 0 0 1 01 2 1 2 1 2
ˆ, ,E E E E E r r H r r
1 2
0 1 0 0* 01 2 1 2 1 2 1 2 1 2
12
1ˆ, , , ,S S
r r H r r r r r r dV dVr
1 2 1 2ˆ ( , ) ( , )H r r E r r 0 1
1 2 1 2 1 2( , ) ( , ) ( , )r r r r r r
Ground State Energy of Helium Atom
30 (1 ) iZri
Zs e
1 2
1 0 1 0 0* 01 2 1 2 1 2 1 2 1 2
12
1ˆ, , , ,S S
E r r H r r r r r r dV dVr
1 2
0* 0* 0 01 2 1 2 1 2
12
1
S S
r r r r dV dVr
1 2 1 2
2 22 2
1 2 1 2 1 2 1 2 1 2120 0 0 0 0 0
2 2 2 2 1 2 2 2 2 sin sinZr Zr Zr Zre e e e r r dr dr d d d dr
12
1 5 51 (1)1 (2) 1 (1)1 (2)8 4Zs s s s
r
0 1 54 2.75H4
E E E Closer but still far off!!!
1
0
1.25 31.5%4
EE
Perturbation is too large for PT to be accurate, much higher corrections would be required
Variational Method
0ˆ
exact exactH E
i ii
c i i iH E
The wavefunction can be optimized to the system to make it more suitable
Consider a trail wavefunction t and exactIs the true wavefunction, where:
Then0
ˆt t
t t
HE
The exact energy is a lower bound
,n exactis a complete set
Assume the trial function can be expressed in terms of the exact functions
0 0ˆ ˆ 0t t t t t tH E H E
We need to show that
texact
t
Variational Method
0 0ˆ ˆ
t t i i j ji j
H E c H E c
*0
ˆi j i j
i j
c c H E *
0ˆ
i j i j i ji j
c c H E *
0i j i ij iji j
c c E E
*0 0i i i
i
c c E E *
00 & 0i i ic c E E
Since
0
ˆt t
t t
HE
Variational Energy
var
ˆ( ) ( )( )
( ) ( )t t
t t
HE
E0
Evar()
var ( ) 0d Ed
min
2
var2 ( ) 0d Ed
Variational Method For He Atom3
1,0,01 ( ) ( ) iZri
Zs i e
r
Let’s optimize the value of Z, since the presence of a second electrons shields the nucleus, effectively lowering its charge.
1 2 1 2var
1 2 1 2
ˆ( , ) ( , )( , ) ( , )
HE
r r r rr r r r
33
1,0,01 ( ) ( ) eff ieff Z ri
Zs i e
r 1 2( , ) 1 (1)1 (2)s s r r
1 2 1 2( , ) ( , ) 1 (1) 1 (1) 1 (2) 1 (2) 1s s s s r r r r
var 1 212
1ˆ ˆ1 (1)1 (2) 1 (1)1 (2)E s s H H s sr
Variational Method For He Atom
var 1 212
1ˆ ˆ1 (1)1 (2) 1 (1)1 (2)E s s H H s sr
1 2
12
ˆ ˆ1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2)
11 (1)1 (2) 1 (1)1 (2)
s s H s s s s H s s
s s s sr
1 2
12
ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (2) 1 (2) 1 (1) 1 (1)
11 (1)1 (2) 1 (1)1 (2)
s H s s s s H s s s
s s s sr
1 212
1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)s H s s H s s s s sr
Variational Method For He Atom
31
3
1 (1) effeff Z rZs e
var 1 212
1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)E s H s s H s s s s sr
1 1 11 1 1 1
ˆ ˆ ˆ1 (1) 1 (1) 1 (1) 1 (1) 1 (1) 1 (1)eff effZ ZZ Zs H s s K s s K sr r r r
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)effeff
Zs K s Z Z s s
r r
1 11
ˆ ˆ effZH K
r
2
1ˆ 1 (1) 1 (1)
2effZ
H s s 2
22eff
n
ZE
n
2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
Variational Method For He Atom2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
3 31 1
3 322
1 1 1 1 11 10 0 0
1 11 (1) 1 (1) sineff effeff effZ r Z rZ Zs s e e r dr d d
r r
31
3 22
1 1 1 1 10 0 0
sineffZ reffZre dr d d
3123
1 10
4 effZ reffZ re dr
2
00
1 ( 1)au auue du au ea
23 0
2
14 2 1 2 0 12
effZ aeff eff eff
eff
Z Z e Z eZ
32
14 14eff eff
eff
Z ZZ
Variational Method For He Atom2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
2
2eff
eff eff
ZZ Z Z
2
2ˆ1 (2) 1 (2)
2eff
eff eff
Zs H s Z Z Z
12
1 51 (1)2 (2) 1 (1)2 (2)8 effs s s s Z
r
Similarly
Recall from PT
var 1 212
1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)E s H s s H s s s s sr
2 2 52 2 8eff eff
eff eff eff eff eff
Z ZZ Z Z Z Z Z Z
Variational Method For He Atom2 2
var528eff eff eff effE Z Z ZZ Z
2 528eff eff effZ ZZ Z
var5 52 2 08 16eff eff
eff
d E Z Z Z ZdZ
5 27216 16effZ
2
var27 27 5 272(2) 2.8479H16 16 8 16
E Much closer to -2.9033 H
(D E= 0.055 H =144.4 kJ/mol error)
Variational Method For He Atom
27 1.6916effZ
3 27161 271 ( )
16ir
s i e
1 23 27 27
16 161 2
1 27( , ) 1 (1)1 (2)16
r rs s e e
r r
Optimized wavefunction
Variational Method For He Atom2716effZ
3 27161 271 ( )
16ir
s i e
1 23 27 27
16 161 2
1 27( , ) 1 (1)1 (2)16
r rs s e e
r r
Optimized wavefunction
1 2 1 2
32
1 2( , ) Z r Z r Z r Z rZ Ze e e e
r r
1.19 & 2.18Z Z var 2.8757HE
Other Trail Functions
(D E= 0.027 H =71.1 kJ/mol error)
Optimizes both nuclear charges simultaneously
Variational Method For He Atom
1 2( )1 2 12
1( , ) (1 )Z r re brN
r r
1.849 & 0.364Z b var 2.8920HE
Other Trail Functions
(D E= 0.011 H =29.7 kJ/mol error)
Z’, b are optimized. Accounts for dependence on r12.
In M.O. calculations the wavefunction used are designed to give the most accurate energies for the least computational effort required.
The more accurate the energy the more parameters that must be optimizedthe more demanding the calculation.
Variational Method For He Atom
In M.O. calculations the wavefunction used are designed to give the most accurate energies for the least computational effort required.
The more accurate the energy the more parameters that must be optimizedthe more demanding the calculation.
-2.862879 H
-2.862871 H-2.84885 H
Experimental -79.003 ev -2.9003 H