7.2 The ground state of helium
Our task:
To calculate the ground–state energy by using the
Variational Principle
experimental
ev 975
. 78
− =
g
E
Theoretically reproduce the value :
Figure 7.2:the helium atom
2
r
1 2
r r
−
e −
e −
2
r
2e +
The Hamiltonian for the helium atom system ignoring fine structure and small correction is
2 2
2 2
1 2
1 2
1 2
2 2
1 2
4 e
H m
r r
r r
πε
= − ∇ + ∇
− +
− −
2
1 2
1 4
e e
e V
r r
π ε
= −
Let
1 2
2 1
2 100
1 100
2 3
8 ,
r r a
r r r
r e
a
π
− +
Ψ ≡ Ψ
Ψ =
Consequently ,the energy that goes with this simplified If we ignore the electron-electron repulsion is, the
ground-state wave function is just
ee
V
where is hydrogen-like wave function with .
100
Ψ
2 =
Z
picture is see Chapter 5 .
1
8 109 ev
E = −
1
8
ee
H E
V Ψ =
+ Ψ
In the following we will apply the variational principle , using the as the trial wave function. The eigenfunction of
Hamiltonian is: Ψ
Thus
1
8
ee
H E
V =
+
where
1 2
2 4
2 3
3 1
2 3
1 2
8 4
r r
a ee
e e
V d r d r
a r
r
πε π
− +
= −
To get the above integral value conveniently, we do the
2
r
integral first and orient the coordinate system so that the
polar axis lies along see Figure 7.3.
1
r
2
r
By the law of cosines,
2 2
1 2
1 2
1 2 2
2 cos
r r
r r
r r
θ
− =
+ −
and hence
2 2
4 4
3 2
2 2
2 2
2 2
2 2
2 1
2 1
2 1 2
2
sin 2
cos
r a r a
e e
I d r
r dr d d
r r
r r
r r
θ θ φ
θ
− −
≡ =
− +
−
The integral is trivial ;the integral is:
2
φ
1
φ
π
2
Figure 7.3: choice of coordinate for the integral
2
r
2 2
2 2
1 2
1 2 2
sin 2
cos d
r r
r r
π
θ θ
θ
+ −
2 2
1 2
1 2 2
1 2
2 cos
r r
r r r r
θ
+ −
=
π
2 2
2 2
1 2
1 2 1
2 1 2
1 2
1 2
1 2
1 2
1 2
2
1 r
r rr
r r
rr rr
r r
r r
rr =
+ +
− +
−
= +
− −
{
1 2
1 2
2 1
2 , if
2 , if
r r
r r
r r
=
Thus
1 2
2 1
4 4
2 2
2 2
2 2
1 3
1 4
r r a
r a r
I e
r dr e
r dr r
π
π
∞ −
−
= +
1
3 4
1 1
2 1
1 8
r a
r a
e r
a
π
−
= −
+
It follows that is equal to
ee
V
1 1
2 4
4 1
1 1
1 1
1 3
2 8
1 1
sin 4
r a r a
r e
e e
r dr d d
a a
θ θ φ
πε π
− −
− +
2 2
4 8
2 5
128
r a r a
r a
re r
e dr
a
∞ −
−
− +
=
2 1
5 5
34 4
4 2
ee
e V
E ev
a
πε
= = −
=
Finally , then, The angular integals are easy ,and the integal becomes
π
4
1
r
4 4
2 a
πε
109 34
75 H
ev ev
ev = −
+ = −
And therefore
Not bad , but we
can do better
Can we think of a more realistic trial function
than
Ψ
??
We try the product function
1 2
3 1
1 2
3
,
Z r r
a
Z r r
e a
π
− +
Ψ ≡
and treat Z as a variable rather than setting it equal to 2.
The idea is that as each electron shields the nuclear charge seen by the other ,the effective Z is less than 2.
In the following, we will treat Z as a variational parameter,
picking the value that minimizes . H
Rewrite in the following form:
H
2 2
2 2
2 1
2 1
2 1
2 1
2
2 2
1 2
4 4
Z Z
e Z
Z e
H m
r r
r r
r r
πε πε
− −
= − ∇ + ∇
− +
+ +
+ −
The expectation value of is evidently
H
2
1 e
2 2
1
1 2
2 2
4
ee
e H
Z E Z
V r
πε
= +
− +
r 1
1 r
Here is the expectation value of in the one-particle hydrogenic ground state but with nuclear charge Z.
100
Ψ
And according to Chapter 6, we know
1 a
r Z
=
The expection value of is the same as before ,except that instead of ,we now want arbitrary Z—so we
multiply by :
2 =
Z
Z
a
ee
V
multiply by :
2 Z
a
2 1
5 5
8 4
4
ee
Z e
Z V
E a
πε
= = −
Putting all this together, we find
2 2
1 1
2 4
2 5 4 2
27 4 H
Z Z Z
Z E Z
Z E =
− −
= − +
The lowest upper bound occurs when is minimized:
H
1
4 27 4
d H
Z E
dZ = −
+ =
from which it follows that
2 7 1 .6 9
Z =
= 1 .6 9 1 6
Z =
=
Putting in this value for Z, we find
6 1
1 3 77.5
2 2 H
E ev
= = −
Much nearer to experimental value
7.3 The hydrogen molecule ion
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