CITE THIS NOTEBOOK: Ground State Energy of Helium via Perturbation by Paul Abbott. Wolfram Community April 10 2023.

Schrödinger equation for Helium

In atomic units (
ℏ=c=
m
e
=1
) the Schrödinger equation for a two-electron atom (with nuclear charge
Z
) reads
ℋΨ(r)=
-
1
2
(
Δ
1
+
Δ
2
)
T
Ψ(r)+
-
Z
r
1
-
Z
r
2
+
1
r
12
V
Ψ(r)=ℰΨ(r),
(
1
)
where
ℋ
is the Hamiltonian,

is the kinetic energy operator, and

is the potential energy. The interparticle coordinates—
r
1
,
r
2
, and
r
12
—are defined below:
Show[Graphics[{Arrow[{{0,0},{1,0.1}}],Arrow[{{0,0},{0.7,1}}],Arrow[{{0.7,1},{1,0.1}}],AbsolutePointSize[40],Point[{0,0}],AbsolutePointSize[20],Point[{1,0.1}],Point[{0.7,1}],Text[
r
1
,{0.3,0.55}],Text[
r
2
,{0.5,0}],Text[
r
12
,{0.9,0.55}],GrayLevel[1],Text[1,{0.7,1}],Text[2,{1,0.1}],Text[
+

,{0,0}]}],AspectRatioAutomatic,PlotRange{{-0.1,1.1},{-0.1,1.1}}]
Δ
i
is the three-dimensional Laplacian operator for electron
i
. In spherical polar coordinates,
r=(r,θ,ϕ)
, the Laplacian reads
Δ=
1
2
r
∂
2
r
∂
∂r

∂r
-
1
2
r
2
l
(θ,ϕ),
(
2
)
where the angular part,
2
l
(θ,ϕ)
, is the square of the (total) angular momentum operator,
2
l
(θ,ϕ)=-
1
sin(θ)
∂sin(θ)
∂
∂θ

∂θ
+
1
2
sin
(θ)
2
∂
∂
2
ϕ
.
(
3
)
For spherically symmetric states (s-states), one has
Δ
1
+
Δ
2
=
1
2
r
1
∂
2
r
1
∂
∂
r
1
∂
r
1
+
1
2
r
2
∂
2
r
2
∂
∂
r
2
∂
r
2
-
1
2
r
1
+
1
2
r
2
2
l
(θ),
(
4
)
where
r
1
·
r
2
=
r
1
r
2
cos(θ)
(i.e.,
θ
is the angle between
r
1
and
r
2
) and
2
l
(θ)=-
1
sin(θ)
∂sin(θ)
∂
∂θ

∂θ
.
(
5
)

Perturbation Calculation

Considering the inter-electron potential,
1/
r
12
, as a perturbation, the Hamiltonian can be written as
ℋ
-
1
2
(
Δ
1
+
Δ
2
)-
Z
r
1
-
Z
r
2
(0)
ℋ
+
1
r
12
︸
′
ℋ
,
(
6
)
where
(0)
ℋ

-
1
2
Δ
1
-
Z
r
1
ℋ
1
-
1
2
Δ
2
-
Z
r
2
ℋ
2
,
(
7
)
and
′
ℋ

1
r
12
·
(
8
)
To zeroth order, the Hamiltonian is simply separated into the sum of two single-particle (hydrogenic) operators,
(0)
ℋ
=
ℋ
1
+
ℋ
2
. Hence the (unnormalized) ground state of the a two-electron atom can be approximated by a product of hydrogenic functions (with nuclear charge
Z
),
(0)
ψ
=
-Z(
r
1
+
r
2
)

,
(
9
)
and the energy is the sum of the two hydrogenic (ground-state) energies, i.e.,
(0)
ℋ
(0)
ψ
=
(0)
ℰ
(0)
ψ
where
(0)
ℰ
=-
2
Z
2
-
2
Z
2
=-
2
Z
.
(
10
)
1. Show that
(0)
ψ
is an eigenfunction of
(0)
ℋ
with corresponding zeroth order energy,
(0)
ℰ
=-
2
Z
. [2]

2. What is the helium ground state energy of the zeroth order approximation? Compare this to the exact value of -2.904... [2]


2
.
1
. First-order perturbation energy

The first-order perturbation energy,
(1)
ℰ
, is
(1)
ℰ
=
(0)
ℰ
+Δℰ
, where the energy shift is given by
Δℰ=

(0)
ψ
1
r
12
(0)
ψ


(0)
ψ

(0)
ψ

.
(
11
)

2
.
2
. “Standard method” for computing
Δℰ

The usual textbook computation of
Δℰ
is as follows. We need to compute

(0)
ψ
1
r
12
(0)
ψ


(0)
ψ

(0)
ψ

=

-Z(
r
1
+
r
2
)

1
r
12
-Z(
r
1
+
r
2
)


r
1

r
2
∫∫
-2Z(
r
1
+
r
2
)


r
1

r
2
.
(
12
)

Generating Function

Since
2
r
12
=
2
r
1
+
2
r
2
-2
r
1
r
2
cos(θ)
, we can expand the perturbation using the generating function for Legendre polynomials:
1
r
12
=
1
2
r
1
+
2
r
2
-2
r
1
r
2
cos(θ)
=
1
r
>
1+
2

r
<
r
>

-2
r
<
r
>
cos(θ)
=
1
r
>
∞
∑
l=0
l
r
<
r
>
P
l
(cos(θ)),
(
13
)
where
r
>
=max(
r
1
,
r
2
)
,
r
<
=min(
r
1
,
r
2
)
and
cos(θ)
is the angle between
r
1
and
r
2
.
Since
r
1
·
r
2

r
1
r
2
cos(θ)
u
1
·
u
2
cos(θ)
it is easy to show that
cos(θ)=cos(
θ
1
)cos(
θ
2
)+sin(
θ
1
)sin(
θ
2
)cos(
ϕ
2
-
ϕ
1
).
(
14
)

Addition Theorem

Using the addition theorem
P
l
(cos(θ))=
4π
2l+1
l
∑
m=-l
*
m
Y
l
(
θ
1
,
ϕ
1
)
m
Y
l
(
θ
2
,
ϕ
2
),
(
15
)
and the orthonormality of the spherical harmonics,
m
Y
l
(θ,ϕ)
, the angular integrals are trivial
2π
∫
0
π
∫
0
2π
∫
0
π
∫
0
P
l
(cos(θ))sin(
θ
1
)
θ
1

ϕ
1
sin(
θ
2
)
θ
2

ϕ
2
​​=
4π
2l+1
l
∑
m=-l
2π
∫
0
π
∫
0
*
m
Y
l
(
θ
1
,
ϕ
1
)
sin(
θ
1
)
θ
1

ϕ
1
2
π
δ
l,0
δ
m,0
2π
∫
0
π
∫
0
m
Y
l
(
θ
2
,
ϕ
2
)sin(
θ
2
)
θ
2

ϕ
2
2
π
δ
l,0
δ
m,0
=
2
(4π)
δ
l,0
.
(
16
)
We have used the fact that
0
Y
0
(θ,ϕ)
is a constant,
In[]:=
0
Y
0
(θ,ϕ)
Out[]=
1
2
π
and introduced this term into both integrals so that the orthonormality condition can be applied.
3. Verify the addition theorem for
0<=l<=6
. [2]


Perturbation Energy

Hence the six-fold integration reduces to a two-dimensional integral, with the inner integral split into two cases.

(0)
ψ
1
r
12
(0)
ψ
=
2
(4π)
∞
∫
0
-2Z
r
1

1
r
1
r
1
∫
0
-2Z
r
2

2
r
2

r
2
+
∞
∫
r
1
-2Z
r
2

1
r
2
2
r
2

r
2
2
r
1

r
1
.
(
17
)
4. Compute
ℐ=
(0)
ψ
1
r
12
(0)
ψ

. [1]

5. Compute the normalization integral
=
(0)
ψ

(0)
ψ

. [1]

6. Determine Δℰ. [1]


2
.
3
. Perimetric Coordinates

The matrix element of a general operator,

, can be computed from the integral
〈ψψ〉=∫∫
*
ψ
ψ
r
12

r
1

r
2
(
18
)
In interparticle coordinates this reads
〈ψψ〉=8
2
π
∞
∫
0
∞
∫
0
r
1
+
r
2
∫

r
1
-
r
2

*
ψ
ψ
r
1
r
2
r
12

r
12

r
2

r
1
≡8
2
π

∞
∫
0
r
1
∫
0
r
1
+
r
2
∫
r
1
-
r
2
*
ψ
ψ
r
1
r
2
r
12

r
12

r
2

r
1
+
∞
∫
0
∞
∫
r
1
r
1
+
r
2
∫
r
2
-
r
1
*
ψ
ψ
r
1
r
2
r
12

r
12

r
2

r
1

(
19
)
There is a simpler way: interparticle coordinates suffer from the fact that their variables satisfy triangular inequalities,

r
1
-
r
2
≤
r
12
<=
r
1
+
r
2
, complicating the integration. This can be avoided by introducing perimetric coordinates (Coolidge and James 1937), which treat all three particles on an equal footing:
with inverse:
In particular, with the exponential ansatz,
the general matrix element is trivial to compute:
Indeed, computing this matrix element integral is immediate:
Observe that
and, in general
Define this general matrix element (after moving the power of 2 to the left-hand side of the equation) as

Normalisation

Electron-electron interaction

Total energy