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Lecture 3, January 14, 2011
energetics
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
Course number: Ch120a
Hours: 2-3pm Monday, Wednesday, Friday
William A. Goddard, III, [email protected]
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Wei-Guang Liu <[email protected]>
Caitlin Scott <[email protected]>
Ch120a-Goddard-L03
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2010 William A. Goddard III, all rights reserved
1
Course schedule
Friday January 14: L3 and L4
Monday January 17: caltech holiday (MLKing)
Wednesday January 19: wag L5 and L6
Friday January 21: wag L7 and L8, caught up
Monday January 24: wag L9
Wednesday January 26: wag L10 and L11
Friday January 28: wag participates in a retreat for our
nanotechnology project with UCLA
Back on schedule
Monday January 31: wag L12
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2010 William A. Goddard III, all rights reserved
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Last time
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2010 William A. Goddard III, all rights reserved
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Molecular Orbitals: Alternative way to view states of H2
Valence bond: start with ground state at R=∞ and build
molecule by bonding atoms
Molecular orbitals (MO): start with optimum orbitals of one
electron molecule at R=Re and add electrons
u: antibonding
g: bonding
Put 2 electrons in 2
orbitals, get 4 twoelectron
states
Ch120a-Goddard-L03
1-electron
molecular
orbitals
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2010 William A. Goddard III, all rights reserved
4
Analyze gu
and ug states
in 2 electron
space
All four have one
nodal plane and
lead to same KE
and same PE
except for the
electron-electron
repulsion term
gu
gu-ug
ug
gu+ug
<Φ(1,2)|1/r12| Φ(1,2)>
Worst case is for
z1=z2, along diagonal Never have
Maximum at z1=z2
z1=z2 2011
great
EE
EE
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2010
William
A. Goddard III, allterrible
rights reserved
5
Ground and excited states from MO analysis
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Compare ground state MO and VB near Re
KE: best possible
KE: pretty good
EE: too much
ionic character
EE: excellent
Now consider large R:
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compare ground state from VB and MO
Ground state MO wavefunction
is half covalent and half ionic
OK for R=Requilbrium = Re, but not
for R = ∞
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8
Compare bond dissociation for VB and MO
MO limit half
covalent and half
ionic H- + H+
Covalent
limit H + H
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9
VB and MO for u excited states
Pure covalent, but
antibonding
Pure ionic, but
bonding
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10
Energies for H2 states based on atomic orbitals (z=1)
Ch120a-Goddard-L03
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Summary 2nd Postulate QM
EQM = KEQM + PEQM
where for a system with a potential energy function, V(x,y,z,t)
PEQM= < Φ| V|Φ>=∫Φ(x,y,z,t)*V(x,y,z,t)Φ(x,y,z,t)dxdydz
Just like Classical mechanics except weight V with P=|Φ|2
KEQM = (Ћ2/2me) <(Φ·Φ>
where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz
We have assumed a normalized wavefunction, <Φ|Φ> = 1
The stability of the H atom was explained by this KE (proportional
to the average square of the gradient of the wavefunction).
Also we used the preference of KE for smooth wavefunctions to
explain the bonding in H2+ and H2.
So far we have been able to use the above expressions to reason
qualitatively. However to actually solve for the wavefunctions
12
Ch120a-Goddard-L03
© copyright
2010which
2011
William A. we
Goddard
III, all rights
reserved
requires
the Schrodinger
Eqn.,
derive
next.
Alternative form for QM KE
One dimensional
KEQM = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>
integrate by parts:
Goddard form
∫(du/dx) (v) dx = -∫(u)(dv/dx)dx
if u,v  0 at boundaries
Let du/dx = (dΦ*/dx) and v = dΦ/dx then
KEQM = - (Ћ2/2me) ∫(Φ*)(d2Φ/dx2) dx = - (Ћ2/2me) <Φ|(d2/dx2)| Φ>
^ | Φ> standard form
KEQM = <Φ| - (Ћ2/2me) (d2/dx2)| Φ> = <Φ| KE
^ = - (Ћ2/2me) (d2/dx2)
Where KE operator is KE
Three dimensions
KEQM = (Ћ2/2me) <(Φ·Φ>
Goddard form
where <(Φ·Φ> ≡ ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz
integrating by parts:
^ | Φ> standard form
KEQM = <Φ| - (Ћ2/2me)2 | Φ> = <Φ| KE
Where KE operator is ^ = - (Ћ2/2me) 2
KE
2 + (dΦ/dx)2]
2Ch120a-Goddard-L03
= [(dΦ/dx)2 + (dΦ/dx)
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13
3rd Postulate of QM, the variational principle
The ground state wavefunction is the system, Φ, has the lowest
possible energy out of all possible wavefunctions.
Consider that Φex is the exact wavefunction with energy
Eex = <Φ’|Ĥ|Φ’>/<Φ’|Φ’> and that
Φap = Φex + dΦ is some other approximate wavefunction.
Then Eap = <Φap|Ĥ|Φap>/<Φap|Φap> ≥ Eex
This means that for sufficiently small
dΦ, dE = 0 for all possible changes,
dΦ
We write dE/dΦ = 0 for all dΦ
E
Eex
Eap
This is called the variational principle.
For the ground state, d2E/dΦ ≥ 0 for all
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possible
changes
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Summary deriviation of Schrödinger Equation
^ | Φ> + < Φ| V|Φ> = <Φ| Ĥ | Φ>
EQM = <Φ| KE
^ + V and KE
^ = - (Ћ2/2m )2
where the Hamiltonian is Ĥ ≡ KE
e
And we assume a normalized wavefunction, <Φ|Φ> = 1
V(x,y,z,t) is the (classical) potential energy for the system
Consider arbitrary Φap = Φex + dΦ and require that
dE= Eap – Eex = 0
Get <dΦ|Ĥ-Eex|Φex>] = ∫ dΦ*[(Ĥ-Eex)Φex] = 0 for all possible dΦ
This  [(Ĥ-Eex)Φex] = 0 or the Schrödinger equation
Ĥ Φex = EexΦex
The exact ground state wavefunction is a solution of this equation
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15
Excited states
The Schrödinger equation Ĥ Φk = EkΦk
Has an infinite number of solutions or eigenstates (German
for characteristic states), each corresponding to a possible
exact wavefunction for an excited state
For example H atom: 1s, 2s, 2px, 3dxy etc
Also the g and u states of H2+ and H2.
These states are orthogonal: <Φj|Φk> = djk= 1 if j=k
= 0 if j≠k
Note < Φj| Ĥ|Φk> = Ek < Φj|Φk> = Ek djk
We will denote the ground state as k=0
The set of all eigenstates of Ĥ is complete, thus any arbitrary
function Ө can be expanded as
Ө = Sk Ck Φk where <Φj| Ө>=Cj or Ө = Sk Φk <Φk| Ө>
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Coordinates
of H atom
The Hamiltonian for H2+
For H atom the Hamiltonian is
r
Ĥ = - (Ћ2/2me)2 – e2/r or
Ĥ = - ½ 2 – 1/r (in atomic units)
For H2+ molecule the Hamiltonian (in atomic units) is
Ĥ = - ½ 2 + V(r) where
1
1
1
Coordinates
of H2+
We will rewrite this as
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The Schrödinger Equation for H2+
The exact (electronic wavefunction of H2+ is obtained by solving
Here we can ignore the 1/R term (not
depend on electron coordinates) to write
where e is the electronic energy
Then the total energy E becomes
E= e + 1/R
R
Since v(r) depends on R, the
E(R)
wavefunction φ depends on R.
Thus for each R we solve for φ and
e and add to 1/R to get the total
energy E(R)
Born-Oppenheimer approximation
Ch120a-Goddard-L03
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18
Symmetry Considerations
The operation of inversion (denoted as ^
I ) through the origin
of a coordinate system changes the coordinates as
x  -x
y  -y
z  -z
Taking the origin of the coordinate system as the
bond midpoint, inversion changes the electronic
coordinates as illustrated.
After inversion the electron is remains a distance of ra from one of
the nuclei and rb from the other, but their identities are
transposed. Thus the potential energy, v(r) is unchanged by
inversion, v(-x,-y,-z) ≡ v( ^
I r) = v(r) where r is considered as the 3D
vector with components x,y,z
Thus
we say that v(r)© copyright
is invariant
under inversion
19
Ch120a-Goddard-L03
2010 William A. Goddard III, all rights reserved
2011
Considering symmetry
Under inversion the kinetic energy terms in the Hamiltonian are
also unchanged
Hence the full Hamiltonian is invariant under inversion
h(-x,-y,-z) ≡ h( ^
I r) = h(x,y,z) = h(r)
Now consider that we had solved
hφ=eφ
for the exact wavefunction φ and apply the inversion to both sides
^
I hφ= e^
Iφ
Which we can rewrite as
h(-r)φ(-r)=eφ(-r)
But h(-r) = h(r) because of inversion symmetry
20
Thus
h(r)φ(-r)=eφ(-r)© copyright 2011
Ch120a-Goddard-L03
2010 William A. Goddard III, all rights reserved
4th postulate of QM
Consider the exact eigenstate of a system
HΦ = EΦ
and multiply the Schrödinger equation by some CONSTANT
phase factor (independent of position and time)
exp(ia) = eia
eia HΦ = H (eia Φ) = E (eia Φ)
Thus Φ and (eia Φ) lead to identical properties and we
consider them to describe exactly the same state.
4th Postulate: wavefunctions differing only by a constant
phase factor describe the same state
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21
Continue with symmetry discussion
We just derived that because h(-r) = h(r)
Then for any eigenfunction φ(r) of h
h(r)φ(r) = eφ(r)
It must be that φ(-r) also is an eigenfunction of the same h with
the same energy, e.
φg(r)
h(r)φ(-r) = eφ(-r)
Thus for a system with inversion, each
nondegenerate eigenstate is of either g
or u inversion symmetry.
φg(r) = + φg(r) g for gerade or even
φu(r)
φu(r) = - φu(r) u for ungerade or odd
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Now consider symmetry for H2 molecule
For H2 we use the coordinate
system at the right.
Using the same conventions and
assumptions as for H2+ leads to
where 1/r12 is the interaction between the two electrons
and
Contains all terms depending only on the coordinates of
electron 1
inverting the coordinates of electron 1 leaves h(1)
invariant but not H(1,2)
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Invert coordinates of electron 1
As before h(1) is invariant
and so is h(2), but clearly something is different for H2.
The problem is 1/r12. Inverting electron 1 or 2 separately
does not preserve the r12 distance.
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For multielectron systems, the inversion symmetry
applies only if we invert all electron coordinates
simultaneously
Thus we define I to be
Which we write as
This leads to
Now we see that H(1,2) is
invariant under inversion.
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25
Symmetry for H2
Just as for H2+, inversion symmetry of H2 leads to the
result that every eigenstate of H2 is either g or u
Consider the VB wavefunctions
^
I
=
does NOT have symmetry
But
I[
I
] =
+
is
] =
-
is
Which is why we labeled them as such
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2010 William A. Goddard III, all rights reserved
26
Now consider inversion symmetry for MO wavefunctions
This is easy since each MO has inversion symmetry
Thus
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Permutation Symmetry
transposing the two electrons in H(1,2) must leave the
Hamiltonian invariant since the electrons are identical
1
2
H(2,1) = h(2) + h(1) + 1/r12 + 1/R = H(1,2)
We will denote this transposition as t where tΦ(1,2) = Φ(2,1)
Note that t2 = e, the einheit or identity operator t2Φ(1,2) = Φ(1,2)
Thus t is of order 2 and the previous arguments on inversion
apply equally to transposition
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Permutational symmetry of wavefunctions
Every exact two-electron wavefunction must be either
symmetric, s, or antisymmetric, a with respect to transposition
Applying this to the product MO wavefunctions leads to
No symmetry
Thus symmetry alone, would tell us that the ug and
gu product wavefunctions are wrong
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Permutational symmetry for the u MO states of H2
We saw previously that combining these wavefunctions leads
to a low lying covalent state and a high energy ionic state
Which lead properly to symmetric and antisymmetric
permutation states
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permutational symmetry for H2 wavefunctions
symmetric
antisymmetric
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31
Electron spin, 5th postulate QM
Consider application of a magnetic field
Our Hamiltonian has no terms dependent on the magnetic field.
Hence no effect.
But experimentally there is a huge effect. Namely
The ground state of H atom splits into two states
b
B=0
Increasing B
a
This leads to the 5th postulate of QM
In addition to the 3 spatial coordinates x,y,z each electron has
internal or spin coordinates that lead to a magnetic dipole aligned
either with the external magnetic field or opposite.
We label these as a for spin up and b for spin down. Thus the
ground states of H atom are φ(xyz)a(spin) and φ(xyz)b(spin)
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32
Electron spin
So far we have considered the electron as a
point particle with mass, me, and charge, -e.
In fact the electron has internal coordinates,
that we refer to as spin, with two possible
angular momenta.
+½ or a or up-spin and -½ or b or down-spin
But the only external manifestation is that this spin leads to a
magnetic moment that interacts with an external magnetic field
to splt into two states, one more stable and the other less
stable by an equal amount.
DE = -gBzsz
b
B=0
Increasing B
a
Now the wavefunction of an atom is written as a spinorbital ψ(r,s)
where r refers to the vector of 3 spatial coordinates, x,y,z
andCh120a-Goddard-L03
s refers to the internal
spin
coordinates
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2010 William
2011
A. Goddard III, all rights reserved
33
Spin states for 1 electron systems
Our Hamiltonian does not involve any terms dependent on the
spin, so without a magnetic field we have 2 degenerate states for
H atom.
φ(r)a, with up-spin,
ms = +1/2
φ(r)b, with down-spin, ms = -1/2
The electron is said to have a spin angular momentum of S=1/2
with projections along a polar axis (say the external magnetic
field) of +1/2 (spin up) or -1/2 (down spin). This explains the
observed splitting of the H atom into two states in a magnetic
field
Similarly for H2+ the ground state becomes
φg(r)a and φg(r)b
While the excited state becomes
φu(r)a and φu(r)b
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2010 William A. Goddard III, all rights reserved
34
Spin states for 2-electron systems
Since each electron can have up or down spin, any two-electron
system, such as H2 molecule will lead to 4 possible spin states
each with the same energy
Φ(1,2) a(1) a(2)
Φ(1,2) a(1) b(2)
Symmetric spin
Φ(1,2) b(1) a(2)
Neither symmetric
nor antisymmetric
Φ(1,2) b(1) b(2)
Symmetric spin
This immediately raises an issue with permutational symmetry
Since the Hamiltonian is invariant under interchange of the spin
for electron 1 and the spin for electron 2, the two-electron spin
functions must be symmetric or antisymmetric with respect to
interchange of the spin coordinates, s1 and s2
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35
Spin states for 2 electron systems
Combining the two-electron spin functions to form symmetric and
antisymmetric combinations leads to
M
S
Φ(1,2) a(1) a(2)
Φ(1,2) [a(1) b(2) + b(1) a(2)]
+1
Symmetric spin
Φ(1,2) b(1) b(2)
Φ(1,2) [a(1) b(2) - b(1) a(2)]
0
-1
Antisymmetric spin
0
Adding the spin quantum numbers, ms, to obtain the total spin
projection, MS = ms1 + ms2 leads to the numbers above.
The three symmetric spin states are considered to have spin S=1
with components +1.0,-1, which are referred to as a triplet state
(since it leads to 3 levels in a magnetic field)
The antisymmetric state is considered to have spin S=0 with just
© copyright
2010
A. Goddard
III, all rights reserved
oneCh120a-Goddard-L03
component, 0. It is
called2011
aWilliam
singlet
state.
36
Spinorbitals
The Hamiltonian does not depend on spin the spatial and spin
coordinates are independent. Hence the total wavefunction can
be written as a product of
a spatial wavefunction, φ(s), called an orbital, and
a spin function, х(s) = a or b.
We refer to the composite as a spinorbital
ψ(r,s) = φ(s) х(s)
where r refers to the vector of 3 spatial coordinates, x,y,z
while s to the internal spin coordinates.
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37
spinorbitals for two-electron systems
Thus for a two-electron system with independent
electrons, the wavefunction becomes
Ψ(1,2) = Ψ(r1,s1,r2,s2) = ψa(r1,s1) ψb(r2,s2)
= φa(r1) хa(s1) φb(r2) хb(s2)
=[φa(r1) φb(r2)][хa(s1) хb(s2)]
Where the last term factors the total wavefunction
into space and spin parts
Ch120a-Goddard-L03
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2010 William A. Goddard III, all rights reserved
38
Permutational symmetry again
For a two-electron system the Hamiltonian is invariant
(unchanged) upon transposition of the electrons (changing both
spatial and spin coordinates)
H(2,1) = H(1,2)
Again the simultaneous transposition of space-spin coordinats of
two electrons is of order 2
Thus for every eigenstate of the Hamiltonian we obtain either
Ψs(1,2) = +1 Ψs(1,2)
Ψa(1,2) = -1 Ψa(1,2)
Here the transposition interchanges both spin and space
components of the wavefunction simultaneously
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39
Permutational symmetry continued
Factoring the wave function as spatial and spin coordinates
Ψ(1,2) = (1,2)(1,2)
We know that the H(1,2) is separately unchanged by transposing
either just the spacial coordinates or the spin coordinates
Thus
(2,1) = +(1,2) or (2,1) = -(1,2)
and either
(2,1) = +(1,2) or (2,1) = -(1,2)
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40
Permutational symmetry, summary
Our Hamiltonian for H2,
H(1,2) =h(1) + h(2) + 1/r12 + 1/R
Does not involve spin
This it is invariant under 3 kinds of permutations
Space only: r1  r2
Spin only: s1  s2
Space and spin simultaneously: (r1,s1)  (r2,s2)
Since doing any of these interchanges twice leads to the identity,
we know from previous arguments that
Ψ(2,1) =  Ψ(1,2) symmetry for transposing spin and space coord
Φ(2,1) =  Φ(1,2) symmetry for transposing space coord
Χ(2,1) =  Χ(1,2) symmetry for transposing spin coord
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41
Permutational symmetries for H2 and He
H2
Have 4 degenerate g
ground states for H2
Have 4 degenerate u
excited states for H2
He
Have 4 degenerate
ground state for He
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2010 William A. Goddard III, all rights reserved
42
Permutational symmetries for H2 and He
H2
Have 4 degenerate g
ground states for H2
Have 4 degenerate u
excited states for H2
He
Have 4 degenerate
ground state for He
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2010 William A. Goddard III, all rights reserved
43
Permutational symmetries for H2 and He
H2
He
Ch120a-Goddard-L03
the only states
observed are
those for
which the
wavefunction
changes sign
upon
transposing all
coordinates of
electron 1 and
2
Leads to the
6th postulate of
44
© copyright 2011
2010 William A. Goddard III, all rights reserved QM
The 6th postulate of QM: the Pauli Principle
For every eigenstate of an electronic system
H(1,2,…i…j…N)Ψ(1,2,…i…j…N) = EΨ(1,2,…i…j…N)
The electronic wavefunction Ψ(1,2,…i…j…N) changes
sign upon transposing the total (space and spin)
coordinates of any two electrons
Ψ(1,2,…j…i…N) = - Ψ(1,2,…i…j…N)
We can write this as
tij Ψ = - Ψ for all I and j
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45
Implications of the Pauli Principle
Consider two independent electrons,
1 on the earth described by ψe(1)
and 2 on the moon described by ψm(2)
Ψ(1,2)= ψe(1) ψm(2)
And test whether this satisfies the Pauli Principle
Ψ(2,1)= ψm(1) ψe(2) ≠ - ψe(1) ψm(2)
Thus the Pauli Principle does NOT allow
the simple product wavefunction for two
independent electrons
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46
Quick fix to satisfy the Pauli Principle
Combine the product wavefunctions to form a symmetric
combination
Ψs(1,2)= ψe(1) ψm(2) + ψm(1) ψe(2)
And an antisymmetric combination
Ψa(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
We see that
t12 Ψs(1,2) = Ψs(2,1) = Ψs(1,2) (boson symmetry)
t12 Ψa(1,2) = Ψa(2,1) = -Ψa(1,2) (Fermion symmetry)
Thus for electrons, the Pauli Principle only allows the
antisymmetric combination for two independent
electrons
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47
Consider some simple cases: identical spinorbitals
Ψ(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
Identical spinorbitals: assume that ψm = ψe
Then Ψ(1,2)= ψe(1) ψe(2) - ψe(1) ψe(2) = 0
Thus two electrons cannot be in identical spinorbitals
Note that if ψm = eia ψe where a is a constant phase
factor, we still get zero
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48
Consider some simple cases: orthogonality
Consider the wavefunction
Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
where the spinorbitals ψm and ψe are orthogonal
hence <ψm|ψe> = 0
Define a new spinorbital θm = ψm + l ψe (ignore normalization)
That is NOT orthogonal to ψe. Then
Ψnew(1,2)= ψe(1) θm(2) - θm(1) ψe(2) =
ψe(1) θm(2) + l ψe(1) ψe(2) - θm(1) ψe(2) - l ψe(1) ψe(2)
= ψe(1) ψm(2) - ψm(1) ψe(2) =Ψold(1,2)
Thus the Pauli Principle leads to orthogonality of
spinorbitals for different electrons, <ψi|ψj> = dij = 1 if i=j
=0 if i≠j
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Consider some simple cases: nonuniqueness
Starting with the wavefunction
ψm
Ψold(1,2)= ψe(1) ψm(2) - ψm(1) ψe(2)
Consider the new spinorbitals θm and θe where
θm = (cosa) ψm + (sina) ψe
Note that <θi|θj> = dij
a
Then Ψnew(1,2)= θe(1) θm(2) - θm(1) θe(2) =
+(cosa)2 ψ (1)ψ (2) +(cosa)(sina) ψ (1)ψ (2) θm
ψe
θe = (cosa) ψe - (sina) ψm
e
m
e
e
-(sina)(cosa) ψm(1) ψm(2) - (sina)2 ψm(1) ψe(2)
θe
a
-(cosa)2 ψm(1) ψe(2) +(cosa)(sina) ψm(1) ψm(2)
-(sina)(cosa) ψe(1) ψe(2) +(sina)2 ψe(1) ψm(2)
[(cosa)2+(sina)2] [ψe(1)ψm(2) - ψm(1) ψe(2)] =Ψold(1,2)
Thus
linear combinations
of the
spinorbitals do not change Ψ(1,2)50
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Determinants
The determinant of a matrix is defined as
The determinant is zero if any two columns (or rows) are identical
Adding some amount of any one column to any other column
leaves the determinant unchanged.
Thus each column can be made orthogonal to all other
columns.(and the same for rows)
The above properties are just those of the Pauli Principle
Thus
we will take determinants
of our wavefunctions.
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The antisymmetrized wavefunction
Now put the spinorbitals into the matrix and take the deteminant
Where the antisymmetrizer
determinant operator.
can be thought of as the
Similarly starting with the 3!=6 product wavefunctions of the form
The only combination satisfying the Pauil Principle is
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Example:
Interchanging electrons 1 and 3 leads to
From the properties of determinants we know that interchanging
any two columns (or rows) that is interchanging any two
spinorbitals, merely changes the sign of the wavefunction
Guaranteeing that the Pauli Principle is always satisfied
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Historical Note
The idea of combining functions to form say a
determinant in the same way we would combine
numbers algebraically goes back to a German
Mathematician, Stäckel in ~1905
Later Heisenberg introduced the concept of an
antisymmetrized product of orbitals in discussing
magnetic systems in ~1925
For some reason they are now called Slater
determinants, I believe this is because Slater showed
that the simplest allowed product wavefunction would
have this form, in ~1928
Warning: these dates are from remembering reading these papers
when I was a graduate student, they may be a bit off
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New material L3
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55
Energy for 2 electron product wavefunction
Consider the product wavefunction
Ψ(1,2) = ψa(1) ψb(2)
And the Hamiltonian
H(1,2) = h(1) + h(2) +1/r12 + 1/R
In the details slides next, we derive
E = < Ψ(1,2)| H(1,2)|Ψ(1,2)>/ <Ψ(1,2)|Ψ(1,2)>
E = haa + hbb + Jab + 1/R
where haa =<a|h|a>, hbb =<b|h|b>
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12
Represent the total Coulomb interaction between the electron
density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2
Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1
and 2, the integral is positive, Jab > 0
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Details in deriving energy: normalization
First, the normalization term is
<Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)>
Which from now on we will write as
<Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi are normalized
Here our convention is that a two-electron function such as
<Ψ(1,2)|Ψ(1,2)> is always over both electrons so we need not put
in the (1,2) while one-electron functions such as <ψa(1)|ψa(1)> or
<ψb(2) ψb(2)> are assumed to be over just one electron and we
ignore the labels 1 or 2
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Details of deriving energy: one electron terms
Using H(1,2) = h(1) + h(2) +1/r12 + 1/R
We partition the energy E = <Ψ| H|Ψ> as
E = <Ψ|h(1)|Ψ> + <Ψ|h(2)|Ψ> + <Ψ|1/R|Ψ> + <Ψ|1/r12|Ψ>
Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant
<Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> =
= <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> =
≡ haa
Where haa≡ <a|h|a> ≡ <ψa|h|ψa>
Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> =
= <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> =
≡ hbb
The remaining term we denote as
Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy is
E =Ch120a-Goddard-L03
haa + hbb + Jab + 1/R
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The energy for an antisymmetrized product, A ψaψb
The total energy is that of the product plus the exchange term
which is negative with 4 parts
Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa >
- < ψaψb|1/r12|ψb ψa >
The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+
<ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R
But <ψb|ψa>=0
Thus all are zero
Thus the only nonzero term is the 4th term:
-Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy
(or the 2-electron exchange) since it arises from the exchange
term due to the antisymmetrizer.
Summarizing, the energy of the Aψaψb wavefunction for H2 is
E = haa + hbb + (Jab –Kab) + 1/R
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The energy of the antisymmetrized wavefunction
The total electron-electron repulsion part of the energy for any
wavefunction Ψ(1,2) must be positive
Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0
This follows since the integrand is positive for all positions of r1
and r2 then
We derived that the energy of the A ψa ψb wavefunction is
E = haa + hbb + (Jab –Kab) + 1/R
Where the Eee = (Jab –Kab) > 0
Since we have already established that Jab > 0 we can conclude
that
Jab > Kab > 0
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Separate the spinorbital into orbital and spin parts
Since the Hamiltonian does not contain spin the spinorbitals can
be factored into spatial and spin terms. For 2 electrons there are
two possibilities:
Both electrons have the same spin
ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)]
So that the antisymmetrized wavefunction is
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)]
Also, similar results for both spins down
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]=
=[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)]
Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb>
We see that the spatial orbitals for same spin must be orthogonal
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Energy for 2 electrons with same spin
The total energy becomes
E = haa + hbb + (Jab –Kab) + 1/R
where haa ≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb>
where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>
We derived the exchange term for spin orbitals with same spin as
follows
Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>
`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)>
≡ Kab
where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>
Involves only spatial coordinates.
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Energy for 2 electrons with opposite spin
Now consider the exchange term for spin orbitals with opposite
spin
Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)>
`````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)>
=0
Since <a(1)|b(1)> = 0.
Thus the total energy is
Eab = haa + hbb + Jab + 1/R
With no exchange term unless the spins are the same
Since <ψa|ψb>= 0 = < Φa| Φb><a|b>
There is no orthogonality condition of the spatial orbitals for
opposite spin electrons
In general < Φa| Φb> =S, where the overlap S ≠ 0
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Summarizing: Energy for 2 electrons
When the spinorbitals have the same spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]
The total energy is
Eaa = haa + hbb + (Jab –Kab) + 1/R
But when the spinorbitals have the opposite spin,
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=
The total energy is
Eab = haa + hbb + Jab + 1/R
With no exchange term
Thus exchange energies arise only for the case in
which both electrons have the same spin
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Consider further the case for spinorbtials with opposite spin
Neither of these terms has the correct permutation symmetry
separately for space or spin. But they can be combined
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]=
A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)]
Which describes the Ms=0 component of the triplet state
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]=
A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(2)][a(1)b(2)]
Which describes the Ms=0 component of the singlet state
Thus for the ab case, two Slater determinants must be combined
to obtain the correct spin and space permutational symmetry
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65
Consider further the case for spinorbtials with opposite spin
The wavefunction
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]
Leads directly to
3E
ab = haa + hbb + (Jab –Kab) + 1/R
Exactly the same as for
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)]
[Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)]
These three states are collectively referred to as the triplet state
and denoted as having spin S=1
The other combination leads to one state, referred to as the
singlet state and denoted as having spin S=0
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]
We will analyze the energy for this wavefunction next.
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Consider the energy of the singlet wavefunction
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba)
The next few slides show that
1E
= {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)
Where the terms with S or Kab come for the exchange
\
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energy of the singlet wavefunction - details
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba)
1E = numerator/ denominator
Where
numerator =<(ab+ba)(ab-ba)|H|(ab+ba)(ab-ba)> =
=<(ab+ba)|H|(ab+ba)><(ab-ba)|(ab-ba)>
denominator = <(ab+ba)(ab-ba)|(ab+ba)(ab-ba)>
Since
<(ab-ba)|(ab-ba)>= 2 <ab|(ab-ba)>=
2[<a|a><b|b>-<a|b><b|a>]=2
We obtain
numerator =<(ab+ba)|H|(ab+ba)> = 2 <ab|H|(ab+ba)>
denominator = <(ab+ba)|(ab+ba)>=2 <ab|(ab+ba)>
Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>
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energy of the singlet wavefunction - details
1E
= <ab|H|(ab+ba)>/<ab|(ab+ba)>
Consider first the denominator
<ab|(ab+ba)> = <a|a><b|b> + <a|b><b|a> = 1 + S2
Where S= <a|b>=<b|a> is the overlap
The numerator becomes
<ab|(ab+ba)> = <a|h|a><b|b> + <a|h|b><b|a> +
+ <a|a><b|h|b> + <a|b><b|h|a> +
+ <ab|1/r12|(ab+ba)> + (1 + S2)/R
Thus the total energy is
1E
= {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)
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69
Re-examine the energy for H2+
For H2+ the VB wavefunctions were
Φg = (хL + хR) and
Φu = (хL - хR) (ignoring normalization)
where H = h + 1/R. This leads to the energy for the bonding state
eg = <L+R|H|L+R>/ <L+R|L+R> = 2 <L|H|L+R>/ 2<L|L+R>
= (hLL + hLR)/(1+S) + 1/R
And for the antibonding state
eu = (hLL - hLR)/(1-S) + 1/R
We find it convenient to rewrite as
eg = (hLL + 1/R) + t/(1+S)
eu = (hLL + 1/R) - t/(1-S)
where t = (hLR - ShLL) includes the terms that dominate the
bonding and antibonding character of these 2 states
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The VB interference or resonance energy for H2+
The VB wavefunctions for H2+
Φg = (хL + хR) and Φu = (хL - хR) lead to
eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx
eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux
where t = (hLR - ShLL) is the VB interference
or resonance energy and
ecl = (hLL + 1/R) is the classical energy
As shown here the t dominates the bonding
and antibonding of these states
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71
Analysis of classical and interference energies
The classical energy, ecl = (hLL + 1/R), is the total energy of the
system if the wavefunction is forced to remain an atomic orbital
as R is decreased.
The exchange part of the energy is the change in the energy
due to QM interference of хL and хR, that is the exchange of
electrons between orbitals on the L and R nuclei
The figure shows that ecl is weakly antibonding with little
change down to 3 bohr whereas the exchange terms start
splitting the g and u states starting at ~ 7 bohr.
Here the bonding of the g state arises solely from the
exchange term, egx = t/(1+S) where t is strongly negative,
while the exchange term makes the u state hugely repulsive,
eux = -t/(1-S)
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Analysis of classical and interference energies
egx = t/(1+S) while eux = -t/(1-S)
Consider first very long R, where S~0
Then egx = t while eux = -t
so that the bonding and antibonding effects are similar.
Now consider a distance R=2.5 bohr =
1.32 A near equilbrium
Here S= 0.4583
t= -0.0542 hartree leading to
egx = -0.0372 hartree while
eux = + 0.10470 hartree
ecl = 0.00472 hartree
Where the 1-S term in the denominator
makes the u state 3 times as
antibonding
as the g ©state
is2011
bonding.
Ch120a-Goddard-L03
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73
Analytic results - details
Explicit calculations (see appendix A of chapter 2) leads to
S = [1+R+ R2/3] exp(-R)
ecl = - ½ + (1 + 1/R) exp(-2R)
t = -[2R/3 – 1/R] exp(-R) – S(1+1/R) exp(-2R)
t ~ -[2R/3 – 1/R] exp(-R) neglecting terms of order exp(-3R)
Thus for long R, t ~ -2S/R
That is, the quantity in t dominating the bond in H2+ is
proportional to the overlap between the atomic orbitals.
At long R this leads to a bond energy of the form
t~ -(2/3) R exp(-R)
That is the bond strength decreases exponentially with R.
t has a minimum at ~ R=2 bohr, which is the optimum R.
But S continues to increase until S=1 at R=0.
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74
Contragradience
The above discussions show that the interference or
exchange part of KE dominates the bonding.
This decrease in the KE due to overlapping orbitals is
dominated by
tx = ½ [< (хL). ((хR)> - S [< (хL)2>
хL
Which is large and
negative in the
shaded region
between atoms,
where the L and R
orbitals have
opposite slope
(congragradience)
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хR
75
Analysis of classical and exchange energies for H2
For H2 the VB energy for the bonding state (g, singlet) is
1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>
g
1E = {(h
2
2
g
aa + hbb + (hab + hba) S + Jab + Kab + (1+S )/R}/(1 + S )
Similary for the VB triplet we obtain
3E = <ab|H|(ab-ba)>/<ab|(ab-ba)>
u
3E = {(h
2
2
u
aa + hbb - (hab + hba) S + Jab - Kab + (1-S )/R}/(1 - S )
As with H2+, we find it useful to define a classical energy, with no
exchange or interference or resonance
Ecl = <ab|H|ab>/<ab|ab> = haa + hbb + Jab + 1/R
Again we can define the exchange energy as
1E = Ecl + E x
g
g
3E = Ecl + E x
u
u
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76
The VB exchange energies for H2
For H2, the classical energy is slightly attractive, but again the
difference between bonding (g) and anti bonding (u) is
essentially all due to the exchange term.
1E
g
3E
u
= Ecl + Egx
= Ecl + Eux
Each energy
is referenced
to the value
at R=∞,
which is -1
for Ecl, Eu, Eg
and 0 for Exu
and Exg
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Analysis of the exchange energies
Egx = {(hab + hba) S + Kab –EclS2}/(1 + S2) = Ex/(1 + S2)
Eux = -{(hab + hba) S + Kab –EclS2}/(1 - S2) = - Ex/(1 - S2)
where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2
Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St contains the 1e part
T2 = {Kab –S2Jab} contains the 2e part
Clearly the Ex is
dominated by T1
and clearly T1 is
dominated by the
kinetic part, T1t.
Thus we can
understand
bonding by
analyzing just the
KE Ch120a-Goddard-L03
part
T2
Ex
T1
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78
Analysis of the exchange energies
The one electron exchange
for H2 leads to
Eg1x ~ +2St /(1 + S2)
Eu1x ~ -2St /(1 - S2)
which can be compared to
the H2+ case
egx ~ +t/(1 + S)
eux ~ -t/(1 - S)
For R=1.6bohr (near Re), S=0.7
Eg1x ~ 0.94t vs. egx ~ 0.67t
Eu1x ~ -2.75t vs. eux ~ -3.33t
For R=4 bohr, S=0.1
Eg1x ~ 0.20t vs. egx ~ 0.91t
Eu1x ~ -0.20t vs. eux ~ -1.11t
E(hartree)
Eu1x
Consider a very small R
with S=1. Then
Eg1x ~ 2t vs. egx ~ t/2
so that the 2e bond is twice
as strong as the 1e bond
1x
E
g
but at long R, the 1e bond is
R(bohr)
79
stronger
than the 2e bond
Ch120a-Goddard-L03
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