Minimize the energy.
〈ψ|H|ψ〉=∫
d^3 r 1 d^3 r 2 φ∗ 100 (~r 1 )φ∗ 100 (~r 2 )[
p^21
2 m−
Ze^2
r 1+
p^22
2 m−
Ze^2
r 2+
e^2
|~r 1 −~r 2 |]
φ 100 (~r 1 )φ 100 (~r 2 )We can recycle our previous work to do these integrals. First, replace theZinH 1 with aZ∗and
put in a correction term. This makes theH 1 part just a hydrogen energy. The correction term is
just a constant overrso we can also write that in terms of the hydrogen ground state energy.
x =∫
d^3 r 1 φ∗ 100(
p^21
2 m−
Ze^2
r 1)
φ 100=
∫
d^3 r 1 φ∗ 100(
p^21
2 m−
Z∗e^2
r 1+
(Z∗−Z)e^2
r 1)
φ 100= Z∗^2 (− 13 .6 eV) + (Z∗−Z)e^2∫
d^3 r 1 |φ 100 |^21
r 1= Z∗^2 (− 13 .6 eV) + (Z∗−Z)e^2Z∗
a 0= −Z∗^21
2
α^2 mc^2 +Z∗(Z∗−Z)α^2 mc^2= α^2 mc^2(
Z∗(Z∗−Z)−
1
2
Z∗^2
)
Then we reuse the perturbation theory calculation to get theV term.
〈ψ|H|ψ〉 = 2[x] +5
4
Z∗
(
1
2
α^2 mc^2)
= −
1
2
α^2 mc^2[
2 Z∗^2 − 4 Z∗(Z∗−Z)−
5
4
Z∗
]
= −
1
2
α^2 mc^2[
− 2 Z∗^2 + 4ZZ∗−
5
4
Z∗
]
Use the variational principle to determine the bestZ∗.
∂〈ψ|H|ψ〉
∂Z∗= 0 ⇒ − 4 Z∗+ 4Z−
5
4
= 0
Z∗=Z−
5
16
Putting these together we get our estimate of the ground state energy.
〈ψ|H|ψ〉 = −1
2
α^2 mc^2 Z∗[
− 2 Z∗+ 4Z−
5
4
]
= −
1
2
α^2 mc^2 (Z−5
16
)
[
− 2 Z+
5
8
+ 4Z−
5
4
]
= −
1
2
α^2 mc^2[
2
(
Z−
5
16
) 2 ]
=− 77 .38 eV(really− 78. 975 eV).Now we are within a few percent. We could use more parameters for better results.