Technische Universität Graz
Institut für Festkörperphysik
Student project
Variational Wavefunction
for the Helium Atom
Molecular and Solid State Physics 513.001
submitted on:
30. November 2009
by:
Markus Krammer
Contents
1 Problem
3
2 Optimization of a given wavefunction
3
3 Calculating the energy E (α)
3.1 hx|Ĥ|ψi . . . . . . . . . . . . . . . . . . . . .
3.2 hψ|Ĥ|ψi . . . . . . . . . . . . . . . . . . . . .
3.3 hψ|Ĥ1 |ψi . . . . . . . . . . . . . . . . . . . .
3.4 hψ|ψi . . . . . . . . . . . . . . . . . . . . . .
3.5 hψ|Ĥ2 |ψi . . . . . . . . . . . . . . . . . . . .
3.5.1 Transforming to spherical coordinates
3.5.2 Determinant of the Jacobi matrix . . .
3.5.3 One more transformation . . . . . . .
3.5.4 Solving the integral . . . . . . . . . .
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4 Optimizing the Energy
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3
3
4
4
5
5
5
7
9
9
10
2
1
Problem
The Hamitonian for the helium atom is
Ĥ = −
~2
2e2
2e2
e2
∇21 + ∇22 −
−
+
2m
4π0 r1
4π0 r2
4π0 r12
(1)
where r12 is the distance between the two electrons.
There is no analytical solution for the Schrödinger equation for this problem. So in a first step we neglect the
electron electron interaction (term 5 in equation 1) to get an idea of how the real wavefunction could look like.
Solving the Schrödinger equation for this simplified problem leads to the following wavefunction for the ground
state:
r1 + r2
1
ψ (r1 , r2 ) = 3 exp −2
(2)
a0
a0
This approximation is, of course, not very good. But it gives us an idea of how a better approximation could
look like. So we try a variational wavefunction:
1
r1 + r2
(3)
ψ (r1 , r2 , α) = 3 exp −α
a0
a0
Now, as we have made a decision of how our approximate wavefunction for the ground state should look like,
we can optimize it.
2
Optimization of a given wavefunction
To optimize a wavefunction, we have to calculate the energy
E (α) =
hψ|Ĥ|ψi
hψ|ψi
(4)
and find the lowest energy depending on our parameter. If we would have more than one Parameter (e.g.
coefficients of a polynomial that we multiply with our wavefunction) we would have to find the global minimum
of our energy. But since we have just one parameter, finding the minimum energy is pretty easy.
dE (α)
=0
dα
(5)
Equation 5 gives us the optimized parameter α (Energy must be the global minimum for the best α, so we have
to check whether this energy is the global minimum or not).
3
3.1
Calculating the energy E (α)
hx|Ĥ|ψi
The Laplace operator in spherical coordinates looks like this:
1 ∂ 2 ∂
1
∂
1
∂
∂2
2
∇i = 2
ri
+ 2
sin θi
+ 2 2
ri ∂ri ∂ri
ri sin θi ∂θi
∂θi
ri sin θi ∂ϕ2i
Now we let the Laplace operator sink on the wavefunction:
1 ∂ 2 ∂ 1
r1 + r2
exp
−α
∇2i ψ = 2
ri
ri ∂ri ∂ri a30
a0
∇2i ψ = −
α 1 ∂ 2
r1 + r2
r
exp
−α
a40 ri2 ∂ri i
a0
∇2i ψ = −
α 1
a40 ri2
r1 + r2
α
r1 + r2
2ri exp −α
− ri2 exp −α
a0
a0
a0
3
(6)
α
a40
∇2i ψ =
α
2
−
a0
ri
r1 + r2
exp −α
a0
(7)
Putting this into hx|Ĥ|ψi returns:
2 X
hx|Ĥ|ψi =
i=1
~2 α
−
2m a0
α
2
−
a0
ri
2e2
−
4π0 ri
e2
+
4π0 r12
!
r1 + r2
1
exp
−α
a30
a0
with
e2
~2
=
ma0
4π0
(8)
this leads to
2 X
α2
2−α
1
−
−
+
2a0
ri
r12
i=1
~2
hx|Ĥ|ψi =
ma40
!
r1 + r2
exp −α
a0
(9)
hψ|Ĥ|ψi
3.2
With equation 9 it is not difficult to write down hψ|Ĥ|ψi:
Z
hψ|Ĥ|ψi =
~2
···
dx1 dy1 dz1 dx2 dy2 dz2
ma0
R6
Z
2 X
i=1
α2
2−α
−
−
2a0
ri
1
+
r12
!
r1 + r2
1
exp
−2α
(10)
a60
a0
The first four terms in this integral are no problem to solve. Only the last term with the r12 is a bit harder to
evaluate. So we split it up into two parts:
!
Z
Z
2 X
α2
2−α
r1 + r2
~2
−
hψ|Ĥ1 |ψi = · · ·
dx1 dy1 dz1 dx2 dy2 dz2
−
exp −2α
(11)
ma70 i=1
2a0
ri
a0
R6
and
Z
hψ|Ĥ2 |ψi =
~2 1
r1 + r2
···
dx1 dy1 dz1 dx2 dy2 dz2
exp −2α
ma70 r12
a0
R6
Z
(12)
hψ|Ĥ1 |ψi
3.3
The integrals
Z
Z
α2
2−α
r1 + r2
~2
···
dx1 dy1 dz1 dx2 dy2 dz2
−
−
exp
−2α
ma70
2a0
r1
a0
R6
and
Z
Z
···
dx1 dy1 dz1 dx2 dy2 dz2
R6
~2
ma70
−
α2
2−α
−
2a0
r2
r1 + r2
exp −2α
a0
are exactly the same, so we can write
Z
Z
~2
α2
2−α
r1 + r2
hψ|Ĥ1 |ψi = · · ·
dx1 dy1 dz1 dx2 dy2 dz2
−
−2
exp −2α
ma70
a0
r1
a0
R6
(13)
Now we tansform it into spherical coordinates and with the Jacobi determinant of the transformation we get:
dx1 dy1 dz1 dx2 dy2 dz2 = r12 sin θ1 r22 sin θ2 dr1 dϕ1 dθ1 dr2 dϕ2 dθ2
Our integral does not depend on ϕi and θi , so we can integrate these 4 variables seperately:
Z 2π
Z π
Z 2π
Z π
2
dϕ1
dθ1
dϕ2
dθ2 sin θ1 sin θ2 = (4π)
0
0
0
0
4
(14)
(15)
~2 (4π)
hψ|Ĥ1 |ψi =
ma70
2
Z
0
∞
Z ∞
α2 r12
r1
r2
2
dr1 −
dr2 r2 exp −2α
− 2 (2 − α) r1 exp −2α
a0
a0
a0
0
This integrals are all of the following form:
Z ∞
Z ∞
d2
d2
1
2
2
dx · x exp (−λx) =
dx · exp (−λx) =
= 3
2
2
dλ
dλ
λ
λ
0
0
Z ∞
Z ∞
d
d 1
1
dx · x exp (−λx) = −
dx · exp (−λx) = −
= 2
dλ
dλ
λ
λ
0
0
(16)
(17)
So the result is:

hψ|Ĥ1 |ψi =

2
2
2
2α
~ (4π) 
− 3
ma70
2α
a0
hψ|Ĥ1 |ψi = −
3.4
a0
− 2 (2 − α) 1
2α
a0
 2
2  3
2α
a0
~2 π 2 (4 − α)
ma20
α5
(18)
hψ|ψi
For calculating the energy (see equation 4) we need to know hψ|ψi.
Z
Z
hψ|ψi = · · ·
dx1 dy1 dz1 dx2 dy2 dz2 ψ ∗ ψ
(19)
R6
Z
hψ|ψi =
Z
···
dx1 dy1 dz1 dx2 dy2 dz2
R6
r1 + r2
1
exp
−2α
a60
a0
There is again no ϕi and θi dependence in this term, so with equation 14 and 15 we get
2
hψ|ψi =
(4π)
a60
Z
0
∞
Z ∞
r1
r2
dr1 r12 exp −2α
dr2 r22 exp −2α
a0
a0
0
and with equation 16
2

2
hψ|ψi =
(4π)  2 
 3 
a60
2α
a0
hψ|ψi =
3.5
π2
α6
(20)
hψ|Ĥ2 |ψi
As we see in equation 12 this integral is not that easy. The term
to transform the integral clever.
3.5.1
1
r12
makes it a bit more difficult. So we need
Transforming to spherical coordinates
In a first step we transform the integral into spherical coordinates. But not like we did it in equation 14, because
this would not make the integral easier. So we choose a different transformation. We transform x1 , y1 and z1
like before:
x1
=
r1 cos ϕ1 sin θ1
(21)
y1
=
r1 sin ϕ1 sin θ1
(22)
z1
=
r1 cos θ1
(23)
For the other coordinates we choose a different transformation. We use the r1 direction as z̃ axis and the x̃ axis
is in the zz̃ plane. So the ỹ axis is in the xy plane (see figure 1). In this coordinates we now transform x̃2 , ỹ2
5
Figure 1: Transformation, the z̃ axis has the same direction as r~1
and z̃2 as we are used to.
x̃2
=
r2 cos ϕ12 sin θ12
(24)
ỹ2
=
r2 sin ϕ12 sin θ12
(25)
z̃2
=
r2 cos θ12
(26)
So now we need to know the relation between x2 , y2 , z2 and x̃2 , ỹ2 , z̃2 . For this we write down the basis vectors:
êz̃ =
r~1
r1


cos ϕ1 sin θ1
êz̃ =  sin ϕ1 sin θ1 
cos θ1
êỹ =
(27)
êz̃ × êz
|êz̃ × êz |

  
cos ϕ1 sin θ1
0
1
 sin ϕ1 sin θ1  ×  0 
êỹ =
|êz̃ × êz |
cos θ1
1


sin ϕ1 sin θ1
1
 − cos ϕ1 sin θ1 
êỹ =
|êz̃ × êz |
0
6
v
 

u
u
sin ϕ1 sin θ1
sin ϕ1 sin θ1
u
|êz̃ × êz | = t − cos ϕ1 sin θ1  ·  − cos ϕ1 sin θ1 
0
0
|êz̃ × êz | =
q
sin2 ϕ1 sin2 θ1 + cos2 ϕ1 sin2 θ1
|êz̃ × êz | = sin θ1


sin ϕ1
êỹ =  − cos ϕ1 
0
(28)
êx̃ = êỹ × êz̃

 

sin ϕ1
cos ϕ1 sin θ1
êx̃ =  − cos ϕ1  ×  sin ϕ1 sin θ1 
0
cos θ1


− cos ϕ1 cos θ1
êx̃ =  − sin ϕ1 cos θ1 
sin θ1
(29)
x · êx + y · êy + z · êz = x̃ · êx̃ + ỹ · êỹ + z̃ · êz̃
(30)
With
and the equations 27, 28 and 29 we get

 
x2
− cos ϕ1 cos θ1
sin ϕ1
 y2  =  − sin ϕ1 cos θ1 − cos ϕ1
z2
sin θ1
0


cos ϕ1 sin θ1
x̃2
sin ϕ1 sin θ1   ỹ2 
cos θ1
z̃2
Now our transformation is complete. Equations 21 to 26 and 31 are all we wanted to know.
3.5.2
Determinant of the Jacobi matrix
For the integral tranformation we need the Jacobi determinant. The

cos ϕ1 sin θ1 −r1 sin ϕ1 sin θ1 r1 cos ϕ1 cos θ1
M̃1
 sin ϕ1 sin θ1 r1 cos ϕ1 sin θ1 r1 sin ϕ1 cos θ1
M̃2


cos
θ
0
−r
sin
θ
M̃3
1
1
1
J =

0
0
0


0
0
0
M
0
0
0
7
Jacobi matrix has the form








(31)
where M̃1 , M̃2 , M̃3 and M are the more complicate parts of the
ϕ12 and θ12 .
r1 cos ϕ1 sin θ1 r1 sin ϕ1 cos θ1
0
−r1 sin θ1
|J| = cos ϕ1 sin θ1 0
0
0
0
0
0
−r1 sin ϕ1 sin θ1 r1 cos ϕ1 cos θ1
0
−r1 sin θ1
− sin ϕ1 sin θ1 0
0
0
0
0
0
−r1 sin ϕ1 sin θ1 r1 cos ϕ1 cos θ1
r1 cos ϕ1 sin θ1 r1 sin ϕ1 cos θ1
+ cos θ1
0
0
0
0
0
0
r1 cos2 ϕ1 sin2 θ1
|J| =
r1 sin2 ϕ1 sin2 θ1
−r1 sin ϕ1 sin θ1 cos θ1
−r1 cos ϕ1 sin θ1 cos θ1
−r1 sin θ1
0
0
0
M̃3
−r1 sin θ1
0
0
0
M̃3
M
M
Jacobi matrix with the transformation to r2 ,
M̃2
M̃3
M
M̃1
M̃3
M
M̃1
M̃2
M
r1 sin ϕ1 cos θ1
0
0
0
M̃2
r1 cos ϕ1 cos θ1
0
0
0
M̃1
M
M
|J| = −r12 sin3 θ1 |M | − r12 sin2 ϕ1 sin θ1 cos2 θ1 |M | − r12 cos2 ϕ1 sin θ1 cos2 θ1 |M |
For the integral transformation we need the absolute value of the determinant of the Jacobi matrix.
||J|| = r12 sin θ1 ||M ||
(32)
So now we just need to calculate the absolute value of the determinant of M:


x2
M =  y2  · ∂r∂ 2 ∂ϕ∂12 ∂θ∂12
z2

− cos ϕ1 cos θ1
M =  − sin ϕ1 cos θ1
sin θ1
sin ϕ1
− cos ϕ1
0


cos ϕ1 sin θ1
x̃2
sin ϕ1 sin θ1   ỹ2  ·
cos θ1
z̃2
∂
∂r2
∂
∂ϕ12
∂
∂θ12
As the transformation matrix has no dependence on r2 , ϕ12 and θ12 we can easily get the determinante:



− cos ϕ1 cos θ1
sin ϕ1
cos ϕ1 sin θ1
x̃2
|M | =  − sin ϕ1 cos θ1 − cos ϕ1 sin ϕ1 sin θ1   ỹ2  · ∂r∂ 2 ∂ϕ∂12 ∂θ∂12 sin θ1
0
cos θ1
z̃2
8
− cos ϕ1 cos θ1
|M | = − sin ϕ1 cos θ1
sin θ1
sin ϕ1
− cos ϕ1
0
cos ϕ1 sin θ1
sin ϕ1 sin θ1
cos θ1


x̃2
 ỹ2  ·
z̃2
∂
∂r2
∂
∂ϕ12
∂
∂θ12
The determinante of the second part is the same as we were calculating many times bevor:


x̃2
∂
∂
∂
 ỹ2  ·
= −r22 sin θ12
∂r2
∂ϕ12
∂θ12
z̃2
And the first part is also not really difficult:
− cos ϕ1 cos θ1
sin ϕ1
cos ϕ1 sin θ1 − sin ϕ1 cos θ1 − cos ϕ1 sin ϕ1 sin θ1 = cos2 ϕ1 cos2 θ1 +sin2 ϕ1 sin2 θ1 +cos2 ϕ1 sin2 θ1 +sin2 ϕ1 cos2 θ1 = 1
sin θ1
0
cos θ1
So the absolute value of the determinante of the Jacobi matrix is:
||J|| = r12 r22 sin θ1 sin θ12
3.5.3
(33)
One more transformation
With the transformation done before the integral is still not easy to solve. So we need one more transformation.
We transform θ12 to r12 :
2
r12
= r12 + r22 − 2r1 r2 cos θ12
The Jacobi matrix for this transformation is:
1
0 0
0
0
1
0
0
0
0
1
0
|J| = 0
0
0
1
0
0
0
0
r12
r12
0
0
r1 −r2 cos θ12
r2 −r1 cos θ12
||J|| =
0
0
0
0
1
0
0
0
0
0
0
r12
r1 r2 sin θ12
r12
r1 r2 sin θ12
(34)
We need to take a look at the integration limits. In spherical coordinates they were the same as we are used
to. But now we have to get the correct limits for r12 . The limits for θ12 were 0 and π, so r12 now needs to go
from |r1 − r2 | to r1 + r2 .
3.5.4
Solving the integral
With all these transformations (equations 33 and 34) the integral finally has a form that is easy to solve:
Z ∞
Z ∞
Z r1 +r2
Z 2π
Z 2π
Z π
r12
hψ|Ĥ2 |ψi =
dr1
dr2
dr12
dϕ1
dϕ12
dθ1 r12 r22 sin θ1 sin θ12
r
r
1 2 sin θ12
0
0
|r1 −r2 |
0
0
0
2
~
1 1
r1 + r2
exp −2α
6
ma0 r12 a0
a0
Z ∞
Z ∞
Z r1 +r2
2
~
r1 + r2
2
=
2
(2π)
dr
dr
dr
r
r
exp
−2α
1
2
12 1 2
ma70
a0
0
0
|r1 −r2 |
Z
Z
Z ∞
Z r1 +r2
Z
∞
r1
r1 +r2
2
r1 + r2
~
2
dr1
dr2
dr12 +
dr2
dr12 r1 r2 exp −2α
=
2 (2π)
ma70
a0
0
0
r1 −r2
r1
r2 −r1
Z r1
Z ∞
Z ∞
~2
r
+
r
1
2
2
=
2 (2π)
dr1
dr2 2r2 +
dr2 2r1 r1 r2 exp −2α
ma70
a0
0
0
r1
9
Z
r1
dr2 r22
0
∂2
exp (−λr2 ) =
∂λ2
= exp (−λr1 ) −r1
Z
r1
Z
0
∂2
dr2 exp (−λr2 ) =
∂λ2
r1
1
+ 2
λ
λ
∂
∂λ
Z
∞
dr2 r2 exp (−λr2 ) = −
r1
r1
2
− 2− 3
λ
λ
+
exp (−λr1 )
1
−
+
λ
λ
∞
dr2 exp (−λr2 ) = −
r1
~
2
(4π)
ma70
hψ|Ĥ2 |ψi =
∞
Z
0
∂
=
exp (−λr1 )
∂λ
∂
∂λ
r2
1
2α

dr1 − exp − r1
a0
exp (−λr1 )
λ
2α
a0
= exp (−λr1 )
~2
2
(4π)
ma70
Z
∞
0
2
r1
1
+ 2 + 3
λ
λ
λ
r1
1
+ 2
λ
λ
2

+
2
+ 2r1 2α
a0
r1
2 
+ 3  r1 exp −2α
3
a0
2α
2α
a0

+
2
r12 λ2 + 2r1 λ + 2
2
=
−
exp
(−λr
)
+ 3
1
λ3
λ3
λ

2
a0


2α
r1
1 

 r1
dr1 r1 exp − r1  2α + 2  r1 exp −2α
a0
a0
2α
a0
a0

2
~
2
(4π)
ma70
=
∞
Z
0
r1 2α
a0

dr1 − +2
4α
3 exp − r1
a0
2α
a0

2
r1 
+ 3 exp −2α
 r1
a
0
2α
a0
Z ∞
~2 a0 3
4α
2α
2
2 2α
+
2r
r
r
(4π)
dr
exp
−
+
2r
exp
−
−
r
1
1
1
1
1
1
ma70 2α
a0
a0
a0
0
=
With equations 16 and 17 this leads to:
 


2 3
~
a0
2 2α
1 
1 
2 
hψ|Ĥ2 |ψi =
(4π) −  3
+ 2 2  + 2 2 
ma70 2α
a
0
4α
4α
2α
a0
2
~
ma70
=
a 5
0
2α
(4π)
2
2 2
− − +2
8 4
a0
a0
So finally we solved the integral:
hψ|Ĥ2 |ψi =
4
~2 π 2 5
ma20 α5 8
(35)
Optimizing the Energy
Now we calculated all terms to get the energy (see equation 4):
E (α)
=
hψ|Ĥ|ψi
hψ|ψi
=
hψ|Ĥ1 |ψi + hψ|Ĥ2 |ψi
hψ|ψi
With equation 18, 20 and 35 we get the result:
2
E (α) =
~ π
− ma
2
0
~2
E (α) = −
ma20
2
(4−α)
α5
π2
α6
+
~2 π 2 5
ma20 α5 8
27
α − α2
8
(36)
10
As already explained in equation 5 the first derivative of the energy must be zero
dE (α)
27
d
~2
2
= 0 =
−
α−α
dα
dα
ma20 8
27
~2
0 =−
− 2α
ma20 8
27
− 2α
0 =
8
So the optimized value for α is:
α=
27
16
(37)
As we can easily see from the energy E (α) this is the global minimum. The corresponding energy is (with
equation 36):
27
Eopt = E
16
2 !
~2
27 27
27
= −
−
2
ma0 8 16
16
2
~
729 729
= −
−
ma20 128 256
= −
Eopt
~2 729
ma20 256
(38)
(39)
The hartree Eh is the atomic unit of energy (see also equation 8):
Eh =
~2
e2
=
2
ma0
4π0 a0
(40)
So in hartree the energy would be
729
hartree
256
= −2.8477hartree
= −
Eopt
(41)
(42)
Putting the values
~
=
1.054571628 · 10−34 Js
−31
(43)
m
=
9.10938215 · 10
a0
=
5.291772186 · 10−11 m
(45)
−19
(46)
e
= −1.602176487 · 10
kg
(44)
C
into equation 39 results in:
Eopt
= −1.24151 · 10−17 J
(47)
= −77.4887eV
(48)
11