The helium atom
Gaussian basis functions
Objective: Program for calculating the helium ground state using a BO electronic Hamiltonian and by restricting
the electronic wave function to a simple uncorrelated antisymmetric form written in terms of Gaussian functions.
1
Gaussian functions
Gaussian functions of the type:
2
e−α(r−r1 )
(1)
are frequently used as basis functions, especially for atomic and molecular calculations. They are known as
GTO: Gaussian-Type Orbitals. An important feature of GTOs is that the product of two Gaussian functions,
centered at different centeres, can be written as a single Gaussian:
2
2
2
e−α(r−r1 ) e−β(r−r2 ) = e−(α+β)(r−r0 ) e
αβ
− α+β
(r1 −r2 )2
,
r0 =
αr1 + βr2
α+β
(2)
Some useful integrals involving GTOs:
Z ∞
0
Z ∞
1 π
2
e−αx dx = ( )1/2 ,
2 α
1
2α
(3)
(2n − 1)!π 1/2
2n+1 αn+1/2
(4)
2
xe−αx dx =
0
From which one derives:
Z ∞
2
e−αx x2n dx = (−1)n
0
Z ∞
∂n
∂αn
2
e−αx x2n+1 dx = (−1)n
0
2
Z ∞
2
e−αx dx =
0
∂n
∂αn
Z ∞
2
xe−αx dx =
0
n!
2αn+1
(5)
Procedure
We write the antisymmetric wave function as:
1
Ψ(r1 , s1 ; r2 , s2 ) = φ(r1 )φ(r2 ) √ [α(s1 )β(s2 ) − α(s2 )β(s1 )]
2
(6)
where α(s) and β(s) denotes the spin up and spin down wave function and φ(r) is an orbital (depending on the
spatial orbital only) which is shared by the two electrons. We take φ(r) as a linear combination of four fixed,
real basis functions (both Cp and χp are real):
φ(r) =
4
X
Cp χp (r)
(7)
p=1
The wave function we use is called uncorrelated because of the fact that the probability P (r1 , r2 ) for finding
an electron at r1 and another at r2 is uncorrelated, i.e. it can be written as a product of two one-electron
probabilities:
P (r1 , r2 ) = p(r1 )p(r2 )
The electronic Schrödinger equation for helium reads:
1
1
2
1
1
2
ĤBO = − ∇21 − ∇22 +
−
−
2
2
|r1 − r2 | r1 r2
(8)
The action of this Hamiltonian on the proposed wave function (Eq. 7) leads to (since the Hamiltonian does
not act on the spin, the spin-dependent part drops out):
1
2
1
1
2
− ∇21 − ∇22 +
−
φ(r1 )φ(r2 ) = Eφ(r1 )φ(r2 )
−
2
2
|r1 − r2 | r1 r2
(9)
This means that the ground state of the Helium atom can be described with a non-antisymmetric Hartree wave
function. In fact, in the ground state the two electrons have opposite spins and occupy the same spherically
symmetric orbital, i.e. the orbital part is the same.
In order to arrive to a simpler equation we remove the φ(r2 )-dependence by multiplying both sides from the
left side by φ∗ (r2 ) and by integrating over r2 . We then arrive at:
1
2
− ∇21 −
+
2
r1
Z
d3 r2 |φ(r2 )|2
1
φ(r1 ) = E 0 φ(r1 )
|r1 − r2 |
(10)
where several integrals yielding a constant (not dependent on r1 are absorbed in E 0 .
The parameterization (Eq. 7) leads to:

4
X
1
2
− ∇2 −
+
Cr Cs
2 1 r1 r,s=1

Z
4
4
X
X
1
0

Cq χq (r1 ) = E
Cq χq (r1 )
d r2 χr (r2 )χs (r2 )
|r1 − r2 | q=1
q=1
3
(11)
Multiplying Eq. 11 from the left side by χp (r1 ) and integrating over r1 leads to:
!
∀p :
X
hpq +
q
X
Cr Cs Qprqs Cq = E 0
rs
X
Spq Cq
(12)
q
with
1
2
hpq = hχp | − ∇21 − |χq i
2
r1
Z
Qprqs =
(13)
d3 r1 d3 r2 χp (r1 )χr (r2 )
1
χq (r1 )χs (r2 )
|r1 − r2 |
Spq = hχp |χq i
(14)
(15)
Equation 12 has the form of a self consistent problem. However, Eq.12 is not a generalised eigenvalue equation
because of the presence of the variables Cr and Cs , but can be solved by ’fixing’ Cr and Cs and resolving the
resulting generalized eigenvalue problem for Cq . The new values of Cr and Cs are used to restart the procedure
until convergence is achieved.
We need now to evaluate the matrix elements (Eqs. 13,14 and 15)
We will use Gaussian l=0 basis function (s-functions) for the parameterized wave function (Eq. 7):
χp = e−αp r
2
2
(16)
with the following ’optimized’ values of α:
α1
α2
α3
α4
2.1
=
=
=
=
0.298073
1.242567
5.782948
38.474970
Matrix elements
Matrix elements of the kinetic energy
Tpq
1
1
= hχp | − ∇2 |χq i = −
2
2
3
Z
−αp r2
3
d re
2 −αq r2
∇ e
αp αq π 2
=3
5
(17)
(αp + αq ) 2
Matrix elements of the Coulomb energy
2
Vpq = hχp | − |χq i = −2
r
Z
d3 re−αp r
2
1 −αq r2
4π
=−
e
r
αp + αq
(18)
Matrix elements of the overlap matrix
Z
Spq =
r2
d3 re−αp e−αq
r2
=
π
αp + αq
!3
2
(19)
Matrix elements Qprqs (two electrons integral)
1
χq (r1 )χs (r2 )
|r1 − r2 |
Z
2
2 1
2
2
=
d3 r1 d3 r2 e−αp r1 e−αr r2
e−αq r1 e−αs r2
r12
Z
Qprqs =
d3 r1 d3 r2 χp (r1 )χr (r2 )
(20)
(21)
5
=
2.2
2π 2
√
(αp + αq )(αr + αs ) αp + αq + αr + αs
(22)
Program pseudo-code
Finally, the program can be constructed as follows:
1. Construct the 4×4 matrices hpq , Spq , and the 4×4×4×4 array Qprqs (these depend only on αi ).
2. Choose initial values of Cp . These should always be normalized to unity via the overlap matrix:
4
X
Cp Spq Cq = 1
(23)
p,q=1
3. Use the chosen C values to construct the matrix:
Fpq = hpq +
X
Qprqs Cr Cs
(24)
rs
4. Solve the generalized eigenvalue problem
FC = E 0 SC
For the ground state the vector C is the one corresponding to the lowest eigenvalue.
3
(25)
5. The ground state energy is not simply E’ (see Eqs. 9 and 10). The ground state energy is given by the
expectation value of the Hamiltonian for the ground state just obtained:
E=2
X
Cp Cq hpq +
pq
X
Qprqs Cp Cq Cr Cs
(26)
pqrs
where the normalized eigenvector C result from the last diagonalization of F.
6. The new values of C are used to build the matrix F and so on until convergence is achieved. The exact
ground state energy is -2.903 a.u. (including correlation, which is neglected in this example).
4