Chemistry 243: Homework 2
Due by 2 pm on Monday, March 7, 2011
Problems
Problem 1: A H2+ toy model.
In this problem we are going to study a minimal H2+ toy model. We are
going to use Gaussians as atomic orbitals (that is, we are using a Gaussian
basis set). The reason is that the product of two Gaussians is still a Gaussian,
and therefore all the relevant integrals work out nicely. As a matter of fact, you
are given all the integrals in the “help” section and as a Mathematica file. We
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work with unnormalized Gaussians e−αr (we are working in atomic units). We
are going to fix the protons 1.4a0 apart.
(a) Set α = 0.417 and plot the resulting normalized Gaussian and a Slater
atomic orbital with parameter 1. They should be similar. (A 1D plot will
do.)
Define and plot the following MOs: first Gaussian AO, second Gaussian
AO, gerade, and ungerade. The first two are of course not good candidates
for MO as they are not symmetric enough to be considered eigenstates of
the Hamiltonian, but that’s OK.
(b) Find the energy for the states of the previous section (this could be a good
time to start familiarizing yourself with the relevant matrices and tensors,
as hinted in the help section). Which one is the bonding MO?
(c) You have probably heard that the bonding orbital bonds because the electron density in the middle is greater, and so it attracts both nuclei. Let’s
check that. Find the values for the kinetic energy, electron-nuclear interaction and total energy of the gerade and ungerade MOs. Calculate the
difference in kinetic energy and electron-nuclear interaction. What seems
to be the biggest contribution to the bonding so far?
Problem 2: A H2 toy model.
We add a second electron to the previous problem. The energy of any state
|Ψi is given by
X
1X
dpqrs gpqrs ,
E=
Dqp hpq +
2 pqrs
pq
1
where, as given in class,
!
Z
X ZJ
1
hpq = dr1 ψp∗ (r1 ) − ∇2 −
ψq (r1 )
2
rJ
J
Z
ψp∗ r1 ψr∗ (r2 )ψq (r1 )ψs (r2 )
gpqrs = dr1 dr2
,
r12
and the density matrix elements are
X
Dpq =
hΨ|a†pσ aqσ |Ψi
σ
dpqrs
X
=
hΨ|a†pσ a†rτ asτ aqσ |Ψi .
στ
(a) Calculate the electron-electron interaction for the bonding and anti-bonding
closed-shell state of two electrons. Calculate the total Coulomb interaction (electron-electron + electron-nuclear) for the two states, and the total
kinetic energy. Where do you find the biggest difference for the energy
between the two states?
(b) Let the bonding and anti-bonding states be |1σg2 i and |1σu2 i. Find their
total energy.
Consider the state
|g(θ)i = cos θ|1σg2 i + sin θ|1σu2 i .
Find the density matrix elements of |g(θ)i and the value of θ which minimizes the energy.
Problem 3: Toy HF.
We are going to implement a very basic HF solver using the SCF procedure
for H2 . Please note that we only need to solve for one MO (we are in the closed
shell model with only two electrons), which simplifies the implementation. That
is, the Fock matrix depends on a single vector of coefficients.
(a) Define a Mathematica function which takes a vector of coefficients and returns the Fock matrix.
(b) Check that the gerade MO is an eigenstate of S −1 F , where S is the overlap
matrix and F the corresponding Fock matrix.
Let ε be the eigenvalue (orbital energy). Check the following expression for
the energy in Hartree Fock (Szabo and Ostlund Eq. 3.183, but in our case
we have only one MO)
N/2
E0 =
X
(haa + εa ) .
a
(c) Apply the SCF procedure to solve the H2 problem starting with a MO
where both electrons are on the same proton. (Again, remember that you
only have a single vector of coefficients.)
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Help
You will need the following integrals, which you can find in Szabo and Ostlund
App. A.
Z
2
2
(A|B) ≡ dr e−α|r−RA | e−α|r−RB |
π 3/2 α
2
e− 2 |RA −RB |
Z 2α
2
2
1
1
(A| − ∇|B) ≡ dr e−α|r−RA | − ∇2 e−α|r−RB |
2
2
π 3/2 α
2
α
= (3 − α|RA − RB |2 )
e− 2 |RA −RB |
2α
Z2
2
ZC
ZC
−α|r−RA |2
(A| −
|B) ≡ dr e
e−α|r−RB |
−
|r − RC |
|r − RC |
2
π α
= − e− 2 |RA −RB | ZC f0 (2α|RP − RC |2 )
α
RA + RB
where RP =
2
Z
−α|r1 −RA |2 −α|r1 −RB |2 −α|r2 −RC |2 −α|r2 −RD |2
e
e
e
e
(AB|CD) ≡ dr1 dr2
r12
π 5/2 − α |RA −RB |2 − α |RC −RD |2
2
e 2
f0 (α|RP − RQ |2 )
=
4α5/2
RA + RB
RC + RD
where RP =
RQ =
,
2
2
=
and f0 is a normalized Err function
√
1
f0 (t) = √
t
Z
t
2
dy e−y .
0
You can download all this in a Mathematica file from the class website.
Notice the use of the “chemist” convention in the naming of the integrals.
For the homework, it also helps to define matrices and higher rank tensors.
For instance, if you have a function f [i, j, k.l], you can define a four rank tensor
ttt = Table[f[i, j, k, l], {i, 1, 2}, {j, 1, 2}, {k, 1, 2}, {l, 1, 2}]
Now, if you want to contract the last two indexes with a vector of coefficients
C1 (for the last index) and C2 , you can use Mathematica’s dot operator:
ttt.C1 .C2
Finally, you can change the order of the indexes with Mathematica’s transpose
function. Look at Mathematica’s help for that.
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