Chem 327
Multielectron Wavefunctions and Bond Formation
The word ‘orbital’ is a synonym for one-electron wavefunction.
A molecular orbital is usually expressed as a linear combination of atomic orbitals
located on different atom, say for instance hydrogen atom A and hydrogen atom B of the
H2 molecule.
an example of such a molecular orbital is:
 1s  1s A  1s B
If I want to explicitly include the label of the one electron in this ‘orbital’ then I can:
 1s 1  1s A  1s B 1  1s A 1  1s B 1
Now, if I want to write a two-electron wavefunction for H2, then I simply write
something like this where I’ve been sure to include an antisymmetric spin wavefunction
to ensure in conjunction with the symmetric spatial part that the overall wavefunction
antisymmetric with respect to interchange of electron labels 1 and 2. Because there is
only one spin function associated with this spatial function (and because these spins are
paired) this is called a singlet.
normalization constant
 MO  N  1 2  1  2  a2 1
1
spatial part (symmetric here)
spin part (antisymmetric here)
.
If I need to represent an excited state, I can do so in a couple of different ways. First, I
have a singlet excited state resulting from the distribution of one electron in a 
molecular orbital and a second in a * MO. This singlet corresponds to a symmetric
spatial part and an antisymmetric spin part.
spin part (antisymmetric)
 MO  N  1 * 2   2 * 1  1  2   1 2
1
spatial part (symmetric here)
normalization constant
If instead I had wanted to use an antisymmetric spatial part I could have used a
symmetric spin wavefunction for which there are three alternatives and is therefore called
a triplet
 MO
3
 1 2



 N  1 * 2   2 * 1  1 2   2 1


 1 2
Problem
You have an expression for a ground state wavefunction for the H2 molecule
above. Write down the spatial part of this wavefunction (don’t worry about the spin
part) use the definition of the molecular orbital for each electron (1) and (2) and FOIL
it out. If you do this properly, you will get four term (expressed in terms of products of
1sA or 1sB, etc.) in the expansion of this ground state wavefunction. Can you find a
meaning to each of these terms.
The valence bond wavefunction for H2 (as opposed to the molecular orbital wavefunction
that you have already seen) is simply another, different way of expression a molecular
wavefunction. The VB wavefunction for H2 is as follows
 VB  N 1s A 11sB 2  1s A 21sB 1 1 2   1 2
Compare this (VB) wavefunction to the MO wavefunction for H2, what are their
similarities and differences?

Figure 1. A representation of the radial dependence of the H 1s orbital.
Naturally, the the actual orbital is spherical but this diagram simply depicts the
radial decay of the orbital at increasing distance from the nucleus.


+
+
A

B
R

+

A
node
-
B
Figure 2. A depiction of the  and * molecular orbitals of H2 separated by distance
R. These are respectively the constructive (additive) and destructive (subtractive)
linear combinations of atomic orbitals.
overlap integral
 1s
A
 1s
 1s
1s B d  S
A
1
1s A
rB
A
1
1s B
rB
electron –other nucleus attraction integral
quantum mechanical ‘resonance’ integral
Figure 3. Names for and interpretations of commonly found integrals in quantum
mechanical descriptions of bonding.
Potential energy surface for H2+
Here we will examine the potential energy surface of the hydrogen molecule ion,
+
H2 . This is a one-electron system so we can actually write down a relatively simple
expression for the expectation value for total energy of the molecule. Still, you can
unambiguously see that there is a bond between the two H and the point of this
illustration is to examine the origin of the molecular (ionic in this case) potential energy
surface. That is, to examine how each of the energy expressions that show up in the total
energy (expectation value) expression gives rise to the overall character of the PES.
The wavefunction
Again, because this is a one-electron problem, the form of the wavefunction is
particularly simple. We say that the electron occupies a  bonding molecular orbital
(defined above). We can multiply this spatial orbital by either one of two spin functions
for one electron ( or )- so this state is a doublet.
  1 
  1 
  N 1s A 1  1s B 1

  1
  1
Where the normalization constant N is given below.
1
N
1
2  2S  2
   1
Prove that the one-electron (approximate) wavefunction above is normalized.
Energy Expression
The expression we need to evaluate in order to generate the potential energy
surface is constructed just like it is for any expectation value evaluation. The operator for
total energy, the Hamiltonian operator H, is given below. The denominator of this
expression is simply the square of the wavefunction which we know is normalized to
unity so we don’t really need to worry about it here.
E 
 * Hˆ  d   * Hˆ  d   * Hˆ  d

1
 * d
The Hamiltonian
The Hamiltonian operator is relatively simple as well, notice that it has only four
terms. The first is simply the kinetic energy operator, the second and third terms are
operators that represent the energy of attraction between the negatively charged electron
with each of the two (equivalent) positively charged nuclei. The final term simply
represents the mutual repulsion of the two positively charged nuclei
internuclear
repulsion
between two
nuclei
1
1
1
Hˆ  Tˆk 


rA
rB R AB
kinetic
energy
operator
ator
attraction of
electron to
nucleus A
attraction of
electron to
nucleus B
Evaluating the energy expression
Notice that in the next equation I have simplified the energy formula already.
First, I have dropped the spin part of the wavefunction. I can do this because there are no
terms in the Hamiltonian operator that depend directly on spin, so the spin functions do
not contribute to the energy and therefore can be ignored.
Editors Note: You probably recall from our discussions in class that given the same
spatial components (orbitals) of molecular wavefunctions that triplets tend to be lower in
energy than singlets: doesn’t this argue against my position that spin doesn’t matter.
Well, actually no, spin still does not directly affect energy. However, there is an indirect
effect. If you could write out an energy expression for a two (or more) electron molecule
or atom, you would find that the number an types of terms you write out would be
different for singlet and triplet wavefunctions. Among other things, the energy
expectation value expression for the triplet wavefunction has more of the mysterious
quantum mechanical “exchange” integrals which lower the energy.
E   N 1s A  1s B  Hˆ N 1s A  1s B  d
FOILing this expression out we get
E  N2
1s Hˆ 1s
A
A
d   1s A Hˆ 1s B d   1s B Hˆ 1s A   1s B Hˆ 1s B d

Now, we make use of the fact that because nuclei A and B are identical the following two
relationships between integrals hold:
 1s
 1s
A
B
Hˆ 1s A d   1s B Hˆ 1s B d
Hˆ 1s A d   1s A Hˆ 1s B d
Using this these facts we can simplify the energy expression to get


2
E  N 2  1sA Hˆ 1sA d  2  1sA Hˆ 1sB d 
 


1 1
1 
1 1
1 


 2N 2  1sA 
Tk   
1sA d   1sA 
Tk   
1sB d 






rA rB RAB 
rA rB R AB 

 

Where in the second expression above I have simply inserted the explicit form of the
Hamiltonian operator
I can rearrange these terms any way I like, after all addition is commutative. I choose
this way and you’ll see that I do this for a reason.



1
1 
1
1s A d   1s A  1s A d   1s A
E  2 N 2   1s A  Tk 
1s A d 
rA 
R AB

 rB 





1
1
1
 2 N 2   1s A  Tk  1s B   1s A  1s B   1s A
1s B 
rB 
R AB

 rA 


The reason I have chosen this way to arrange the terms in this way is that it leads
straightforwardly to

1
E  2 N 2   1s A E1s 1s A  1s A 
 rB


 1
1s A   1s A 

 R AB
 
1s A 
 

 1 
 1  




 2N 1sA E1s1sB   1sA  1sB   1sA 
R 
1sB

rB 
 AB  


2
Notice that what is different is the first term on each line. How can I do this? Well the
1
sum of terms Tˆk  is simply the Hamiltonian operator for the hydrogen atom and
rA
because the atomic contribution to the molecular orbital is 1sA, the eigenvalue E1s is the
energy of the hydrogen atomic 1s orbital. (atom B is the same as atom A so the same
argument holds).
Because E1s is simply a number, I can write.

1
E 2 N 2  E1s  1s A 1s A  1s A 
 rB



1
1s A 
1s A 1s A 

R AB




1 
1


 2N 2 E1s  1sA1sB   1sA 
1s

1s
1s
r  B R  A B 





B
AB
Furthermore, because the atomic wavefunction ,1sA, is normalized the integral of its
square is just 1. In addition, RAB is independent of electron coordinate ( ie the electronic
wavefunction) so we can pull it out of the integral in the third term on each line.
Similarly, we use the definition of overlap S to simplify further

1 
1 


 2N 2 E1s   1sA 
1s

r  A R 

 B 

AB 

 1 
1 


 2N 2 E1s S   1sA 
1s

r  B R S 

 B 


AB
I can now collect terms and rearrange to get

 


  
1 
E  1  1s  1 1s  E  1 S  1s  1 1s 
1s
 A  1s R 
 A rB  B 
1 S 
RAB  A 
rB 


AB 
energy due to
1s electron
binding to
hydrogen
nucleus
internuclear
repulsion
energy
attraction
energy of
electron around
one nucleus to
"other" nucleus
screening of
internuclear
repulsion by
electron
quantum
mechanical
energy from
“exchange”
attraction
Problem
Write a similar energy expectation value expression for the H2+ ion in its excited
electronic state, *(1). What kind of PES does this give rise to? Step one is to find the
normalization constant N, it is different here than it was for (1)- but not too much
different.
Energy
E(H1s)
RAB
equilibrium bond length
Figure 3. Potential energy surface for the H2+ ion.