The document discusses solving the ground state energy of a helium atom. It involves computing the Hamiltonian and overlap matrices (H and S) of the system by representing the wavefunction as a linear combination of basis functions. Computing the entries of H and S requires evaluating triple integrals over the internal coordinates of the atom. The main work is to derive a general closed form for these integrals. This involves repeatedly using integration by parts to reduce the exponents in the integrands, yielding sums of terms that can be directly evaluated or fed into computational software for further analysis. Solving these integrals is the crucial step to enable determining the ground state energy by solving the eigenvalue problem Hc = λSc.

1. Solving the energy problem of Helium
Yihsin Ma
Advisor: Henryk A. Witek
2018 Jenuary
Abstract
The goal of the project is to compute the ground state of a helium given its hamiltonian, which
is the sum of kinetic operator and potential operator of the atomic system. We can find all of its
energy by solving the general eigenvalue problem Hc = λSc, where the ground energy λmin is the
smallest eigenvalue of matrices H and S. The most arduous part to derive λmin is to compute H
and S, where H and S are matrices related to the basis functions we choose. Each entries of the
two matrices are triple integral over internal interparticle coordinates (r1, r2, r3). Since these triple
integrals are complicated and both H, S are of large scale, it is impossible to directly feed these
integrals to mathematical softwares.
One solution to overcome the obstacle is to derive the general form of each entry, which means
solving the triple integral. The main work of the project is calculating this general form by paper
and pen before feeding the general form into Maple Software for further computations.
1 Introduction
The explicit form of the Hamiltonian operator of a helium system is given by[2]
H := T + V (1)
where the kinetic operator T and potential operator V is
T := −
1
2r2
1
∂
∂r1
r2
1
∂
∂r1
−
1
2r2
2
∂
∂r2
r2
2
∂
∂r2
−
1
r2
3
∂
∂r3
r2
3
∂
∂r3
−
r2
3 + r2
1 − r2
2
2r1r3
∂2
∂r1∂r3
−
r2
3 − r2
1 + r2
2
2r2r3
∂2
∂r2∂r3
(2)
V :=
q0q1
r1
+
q0q2
r2
+
q1q2
r3
(3)
and the internal interparticle coordinate (r1, r2, r3) which are the distances between atomic nucleus and
two electrons. (See Fig(1))
1

2. Figure 1: Internal interparticle coordinates of the helium atom. In this figure, (q0, q1, q2) in (3) is equal
to (+Z, −e, −e)
We assume that the wave function φ lies in a space spanned by a set of basis functions.
Basis := {e−αr1−βr2−γr3
rm
1 rn
2 rs
3|m, n, s ∈ N, m + n + s ≤ N} (4)
In (4), the number N ∈ N controls the number of basis function we choose. The wave function φ can
be expressed as a linear combination of these basis functions.
φ(r1, r2, r3) :=
m,n,s
cm,n,s · e−αr1−βr2−γr3
rm
1 rn
2 rs
3 (5)
As we represent φ as a vector c = [cm,n,s] in the basis, the problem of finding eigenvalue/eigenfunction
of H becomes solving Hc = λSc, where H, S are both matrices of dimension |Basis| × |Basis| and is
defined as
Si,j :=
∞
0
∞
0
r1+r2
|r1−r2|
r1r2r3b(i)b(j)dr3dr2dr1
Hi,j :=
∞
0
∞
0
r1+r2
|r1−r2|
r1r2r3b(i)H[b(j)]dr3dr2dr1
(6)
(7)
Here, we simply denote b(i) := e−αr1−βr2−γr3
rm
1 rn
2 rs
3 for the i-th basis function which corresponds with
some (m, n, s).
2

3. 2 Calculation
Si,j =
∞
0
∞
0
r1+r2
|r1−r2|
r1r2r3b(i)b(j)dr3dr2dr1
=
∞
0
∞
0
r1+r2
|r1−r2|
rm1+m2+1
1 rn1+n2+1
2 rs1+s2+1
3 e−2αr1−2βr2−2γr3
dr3dr2dr1
=
∞
0
∞
0
r1+r2
|r1−r2|
ra
1 rb
2rc
3e−yr1
e−zr2
e−xr3
dr3dr2dr1
=
∞
0
∞
0
ra
1 rb
2e−yr1
e−zr2
r1+r2
|r1−r2|
rc
3e−xr3
dr3 dr2dr1
(8)
(9)
(10)
(11)
For simplicity, in (9) let
x = 2γ y = 2α z = 2β
a = m1 + m2 + 1 b = n1 + n2 + 1 c = s1 + s2 + 1
Doing integral by parts with respect to r3
r1+r2
|r1−r2|
rc
3e−xr3
dr3 = −
1
x
rc
3e−xr3
r1+r2
|r1−r2|
+
r1+r2
|r1−r2|
c
x
rc−1
3 e−xr3
dr3 = A1 + B1 (12)
where A1, B1 are the first and second terms respectively.
Doing the integral by parts on B1
B1 =
r1+r2
|r1−r2|
c
x
rc−1
3 e−xr3
dr3
= −
c
x2
rc−1
3 e−xr3
r1+r2
|r1−r2|
+
r1+r2
|r1−r2|
c(c − 1)
x2
rc−2
3 e−xr3
dr3 = A2 + B2
(13)
(14)
where A2, B2 are the first and second terms respectively.
Each time we do integral by part, we can reduce the exponent of r3 in the integrand by one.
After doing c times integral by part, we can expand the integral in (12) as
r1+r2
|r1−r2|
rc
3e−xr3
dr3 =
c
k=1
Ak + Bc (15)
where
Ak = −
(c)k−1
xk
rc−k+1
3 e−xr3
r1+r2
|r1−r2|
Bc =
−c!
xc+1
e−xr3
r1+r2
|r1−r2|
(16)
(17)
(c)k is defined as c(c − 1)(c − 2)...(c − k + 1).
3

4. Back to (11), we let PartA =
r1+r2
|r1−r2|
rc
3e−xr3
dr3
Si,j =
∞
0
∞
0
ra
1 rb
2e−yr1
e−zr2
PartA · dr2dr1
=
∞
0
r2
0
ra
1 rb
2e−yr1
e−zr2
PartA · dr1dr2 +
∞
0
r1
0
ra
1 rb
2e−yr1
e−zr2
PartA · dr2dr1
=
∞
0
rb
2e−zr2
r2
0
ra
1 e−yr1
PartA · dr1 dr2 +
∞
0
ra
1 e−yr1
r1
0
rb
2e−zr2
PartA · dr2 dr1
= PartS1 + PartS2
(18)
(19)
(20)
(21)
We first investigate the parentheses in the first term
PartB :=
r2
0
ra
1 e−yr1
PartA · dr1
=
c
k=1
r2
0
ra
1 e−yr1
Ak · dr1 +
r2
0
ra
1 e−yr1
Bc · dr1
=
c
k=1
r2
0
−ra
1 e−yr1
(c)k−1
xk
rc−k+1
3 e−xr3
r1+r2
|r1−r2|
· dr1 +
r2
0
ra
1 e−yr1
−c!
xc+1
e−xr3
r1+r2
|r1−r2|
· dr1
(22)
(23)
(24)
For the first term of PartB
PartB1 :=
c
k=1
r2
0
−ra
1 e−yr1
(c)k−1
xk
rc−k+1
3 e−xr3
r1+r2
|r1−r2|
· dr1
=
c
k=1
−
(c)k−1
xk
r2
0
ra
1 e−yr1
(r1 + r2)c−k+1
e−x(r1+r2)
· dr1
+
(c)k−1
xk
r2
0
ra
1 e−yr1
(r2 − r1)c−k+1
e−x(r2−r1)
· dr1
=
c
k=1
−
(c)k−1
xk
e−xr2
r2
0
ra
1
c−k+1
l=0
Cc−k+1
l rl
1rc−k+1−l
2 e−(x+y)r1
· dr1
+
(c)k−1
xk
e−xr2
r2
0
ra
1
c−k+1
l=0
(−1)l
Cc−k+1
l rl
1rc−k+1−l
2 e−(y−x)r1
· dr1
(25)
(26)
(27)
=
c
k=1
c−k+1
l=0
−Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
r2
0
ra+l
1 e−(x+y)r1
· dr1
+
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
r2
0
ra+l
1 e−(y−x)r1
· dr1 (28)
4

5. where
r2
0
ra+l
1 e−(x+y)r1
· dr1 =
a+l
i=1
−
(a + l)i−1
(x + y)i
ra+l−i+1
2 e−(x+y)r2
−
(a + l)!e−(x+y)r2
(x + y)a+l+1
+
(a + l)!
(x + y)a+l+1
(29)
r2
0
ra+l
1 e−(y−x)r1
· dr1 =



ra+l+1
2
a + l + 1
,if x = y
a+l
i=1
−
(a + l)i−1
(y − x)i
ra+l−i+1
2 e−(y−x)r2
−
(a + l)!e−(y−x)r2
(y − x)a+l+1
+
(a + l)!
(y − x)a+l+1
,if x = y
(30)
by simply doing integral by parts iteratively as we have done in (12).
The second part of PartB
PartB2 :=
r2
0
ra
1 e−yr1
−c!
xc+1
e−xr3
r1+r2
|r1−r2|
· dr1
=
r2
0
ra
1 e−yr1
−c!
xc+1
e−x(r1+r2)
· dr1 −
r2
0
ra
1 e−yr1
−c!
xc+1
e−x(r2−r1)
· dr1
=
−c!
xc+1
e−xr2
r2
0
ra
1 e−(x+y)r1
· dr1 +
c!
xc+1
e−xr2
r2
0
ra
1 e−(y−x)r1
· dr1
(31)
(32)
(33)
where
r2
0
ra
1 e−(x+y)r1
· dr1 =
a
i=1
−(a)i−1
(x + y)i
ra−i+1
2 e−(x+y)r2
−
a!
(x + y)a+1
e−(x+y)r2
+
a!
(x + y)a+1
r2
0
ra
1 e−(y−x)r1
· dr1 =



ra+1
2
a + 1
,if x = y
a
i=1
−(a)i−1
(y − x)i
ra−i+1
2 e−(y−x)r2
−
a!
(y − x)a+1
e−(y−x)r2
+
a!
(y − x)a+1
,if x = y
(34)
(35)
5

6. So far, we have solved the two inner integrals in the first part of (20). Assume x = y, then we have
∞
0
rb
2e−zr2
r2
0
ra
1 e−yr1
PartA · dr1 dr2 =
∞
0
rb
2e−zr2
PartB · dr2 (36)
=
∞
0
rb
2e−zr2
(PartB1 + PartB2) · dr2
=
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
−Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
a+l
i=1
−
(a + l)i−1
(x + y)i
ra+l−i+1
2 e−(x+y)r2
dr2
+
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
−Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
−
(a + l)!e−(x+y)r2
(x + y)a+l+1
dr2
+
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
−Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
(a + l)!
(x + y)a+l+1
dr2
+
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
a+l
i=1
−
(a + l)i−1
(y − x)i
ra+l−i+1
2 e−(y−x)r2
dr2
+
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
−
(a + l)!e−(y−x)r2
(y − x)a+l+1
dr2
+
∞
0
rb
2e−zr2
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
(a + l)!
(y − x)a+l+1
dr2
+
∞
0
rb
2e−zr2
c!
xc+1
e−xr2
a
i=1
(a)i−1
(x + y)i
ra−i+1
2 e−(x+y)r2
dr2
+
∞
0
rb
2e−zr2
c!
xc+1
e−xr2
a!e−(x+y)r2
(x + y)a+1
dr2 +
∞
0
rb
2e−zr2
−c!
xc+1
e−xr2
−a!
(x + y)a+1
dr2
+
∞
0
rb
2e−zr2
−c!
xc+1
e−xr2
a
i=1
(a)i−1
(y − x)i
ra−i+1
2 e−(y−x)r2
dr2
+
∞
0
rb
2e−zr2
−c!
xc+1
e−xr2
a!e−(y−x)r2
(y − x)a+1
dr2 +
∞
0
rb
2e−zr2
c!
xc+1
e−xr2
a!
(y − x)a+1
dr2
(37)
(38)
(39)
(40)
(41)
(42)
(43)
(44)
(45)
(46)
(47)
In order to make reviewers easily to check, I put the terms contributed by PartB1 in the first curly
brackets and PartB2 for the second. The next step is to solve the outermost integral in each term.
6

7. For Part1 :=(38)
Part1 =
c
k=1
c−k+1
l=0
a+l
i=1
Cc−k+1
l
(c)k−1(a + l)i−1
xk(x + y)i
η!
(2x + y + z)η+1
,where η = a + b + c − k − i + 2.
(48)
For Part2 :=(39)+(40)
Part2 =
c
k=1
c−k+1
l=0
Cc−k+1
l
(c)k−1(a + l)!
xk(x + y)a+l+1
η1!
(2x + y + z)η1+1
−
η1!
(x + z)η1+1
,where η1 = b + c − k − l + 1.
(49)
For Part3 :=(41)
Part3 =
c
k=1
c−k+1
l=0
a+l
i=1
(−1)l+1
Cc−k+1
l
(c)k−1(a + l)i−1
xk(y − x)i
η2!
(y − x)η2+1
,where η2 = a + b + c − k − i + 2.
(50)
For Part4 :=(42)+(43)
Part4 =
c
k=1
c−k+1
l=0
(−1)l+1
Cc−k+1
l
(c)k−1(a + l)!
xk(y − x)a+k+1
η3!
(y + z)η3+1
−
η3!
(x + z)η3+1
,where η3 = b + c − k − l + 1.
(51)
For Part5 :=(44)
Part5 =
a
i=1
(a)i−1c!
xc+1(x + y)i
η4!
(2x + y + z)η4+1
,where η4 = a + b − i + 1.
(52)
For Part6 :=(45)
Part6 =
a
i=1
a!c!
xc+1(x + y)a+1
b!
(2x + y + z)b+1
−
b!
(x + z)b+1
(53)
7

8. For Part7 :=(46)
Part7 =
a
i=1
−(a)i−1c!
xc+1(y − x)i
η5!
(y + z)η5+1
,where η5 = a + b − i + 1.
(54)
For Part8 :=(47)
Part8 = −
a!c!
xc+1(y − x)a+1
b!
(y + z)b+1
−
b!
(x + z)b+1
(55)
Note that the above calculation is under the assumption that x = y. If we consider x = y case, then
Part3(41) + Part4(42, 43) should be replaced by Part9 and Part7(46) + Part8(47) should be replaced
by Part10. (The reason of the replacement is (30) and (35).)
We defined Part9 and Part10 below.
Back to (28), the second term of it when x = y is
PrePart9 :=
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
r2
0
ra+l
1 e−(y−x)r1
· dr1
=
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
xk
rc−k+1−l
2 e−xr2
ra+l+1
2
a + l + 1
(56)
(57)
Hence we define
Part9 :=
∞
0
rb
2e−zr2
PrePart9 · dr2 =
c
k=1
c−k+1
l=0
(−1)l
Cc−k+1
l
(c)k−1
(a + l + 1)xk
η6!
(x + z)η6+1
,where η6 = a + b + c − k + 2.
(58)
Similarly, the last term of (33) when x = y is
PrePart10 :=
c!
xc+1
e−xr2
r2
0
ra
1 e−(y−x)r1
· dr1 =
c!
xc+1
e−xr2
ra+1
2
a + 1
(59)
We define
Part10 :=
∞
0
rb
2e−zr2
PrePart10 · dr2 =
(a + b + 1)!c!
(a + 1)xc+1(x + z)a+b+2
(60)
We have now obtained the general form of the first term of Si,j(20). Note that the two terms in (20) is
symmetric by exchanging (r1, a, y) and (r2, b, z). Hence there is no need of further paper-pen calculation.
8

9. (1.2)(1.2)
(1.4)(1.4)
(1.1)(1.1)
(1.3)(1.3)
Maple Code
Problem formulation
Generating the basis functions

10. (1.4)(1.4)
(1.5)(1.5)
nl denotes number of basis functions. Our matrices and are both nl*nl dimensions.
The General Form of Entries of S
Explanation
In the previous work, we have obtained the general form of
by solving the triple integral step by step.
Without loss of generality, we consider and
.
Hence
.
We have made the form simpler by letting
, , ,
x= , , .
and further we can treat as a function of
Code

11. Computation of S and H
Explanation
S matrix

12. (3.3)(3.3)
(3.1)(3.1)
(3.2)(3.2)
(3.4)(3.4)
(3.5)(3.5)
To compute , we first defined as the factor of the integrand in the triple integral .
For example we have
The integrand of is
Hence, in the round , is
Hence , , . We can then compute by simply feed the input into
.
H matrix
Recall that each entry of is defined as
.
Although the operator that acts on a basis function seems complicated. The calculation isn't as
hard as it seems.
Let's consider again and
where in can be written as
We can thus rewrite .
then becomes

13. (3.7)(3.7)
(3.6)(3.6)
SEntry( , , , , , ).
where corresponds to .
We define to be
.
the last eqution is just substitute
, , and having new coefficients .
Take and for example, The integrand of is
which can be written as .
where is now
Code Design
We are now ready to compute each entry and . In the following code, we run over
to compute S[i,j] and H[i,j].
For the round, the corresponding is computed. Then we scan over all .
Since for and we have
, we scan over from 0 to .
Code

14. (4.1)(4.1)
Ground State Energy
Finally, we compute the ground state energy by solving the generalized eigenvalue problem .
Code

15. 3 Result
We run the above Maple code for N = 2, 3, 4, 5 and the resulting ground energies are
λN=2
min = −2.836841080 λN=3
min = −2.889678191 λN=4
min = −2.900900995 λN=5
min = −2.903199555
According to [1], the ground state energy of helium is −79.005151042(eV ) = −2.90338583(a.u.) which
is very close to our result.
References
[1] In: NIST Atomic Spectra Database Ionization Energies Data ().
[2] Hylleraas. In: E.A.Z. Phys 48 (1928), pp. 469–494.
9