Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

For Z = 2 the correction is equal to 2.5 Ry and yields E = −8 + 2.5 = −5.5 Ry.
The experimental value is −5.8074 Ry. The perturbative approximation is not
accurate but provides a reasonable estimate of the correction, even if the “perturbation”,
i.e. the Coulomb repulsion between electrons, is of the same order
of magnitude of all other interactions. Moreover, he ground state assumed in
perturbation theory is usually qualitatively correct: the exact wave function for
He will be close to a product of two 1s functions.
D.2 Variational treatment for Helium atom
The Helium atom provides a simple example of application of the variational
method. The independent-electron solution, Eq.(D.10), is missing the phenomenon
of screening: each electron will ”feel” a nucleus with partially screened
charge, due to the presence of the other electron. In order to account for this
phenomenon, we may take as our trial wave function an expression like the
one of Eq.(D.10), with the true charge of the nucleus Z replaced by an “effective
charge” Z e , presumably smaller than Z. Let us find the optimal Z e
variationally, i.e. by minimizing the energy. We assume
ψ(r 1 , r 2 ; Z e ) = Z3 e
π e−Ze(r 1+r 2 )
and we re-write the Hamiltonian as:
[
H = −¯h2 ∇ 2 1
− Zq2 e
− ¯h2 ∇ 2 ]
2
− Zq2 e
+
2m e r 1 2m e r 2
[
− (Z − Z e)q 2 e
r 1
− (Z − Z e)q 2 e
r 2
We now calculate
∫
E(Z e ) = ψ ∗ (r 1 , r 2 ; Z e )Hψ(r 1 , r 2 ; Z e ) d 3 r 1 d 3 r 2
(D.15)
]
+ q2 e
r 12
(D.16)
(D.17)
The contribution to the energy due to the first square bracket in Eq.(D.16) is
−2Ze 2 a.u.: this is in fact a Hydrogen-like problem for a nucleus with charge Z e ,
for two non-interacting electrons. By expanding the remaining integrals and
using symmetry we find
∫
E(Z e ) = −2Ze 2 −
(in a.u.) with
|ψ| 2 4(Z − Z ∫
e)
d 3 r 1 d 3 r 2 +
r 1
|ψ| 2 = Z6 e
π 2 e−2Ze(r 1+r 2 )
Integrals can be easily calculated and the result is
|ψ| 2 2
r 12
d 3 r 1 d 3 r 2
(D.18)
(D.19)
E(Z e ) = −2Z 2 e − 4(Z − Z e )Z e + 2 5 8 Z e = 2Z 2 e − 27
4 Z e (D.20)
where we explicitly set Z = 2.
immediately leads to
Minimization of E(Z e ) with respect to Z e
Z e = 27 = 1.6875 (D.21)
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