130_notes.dvi

25 The Helium Atom

Hydrogen has been a great laboratory for Quantum Mechanics. After Hydrogen, Helium is the
simplest atom we can use to begin to study atomic physics. Helium has two protons in the nucleus
(Z= 2), usually two neutrons (A= 4), and two electrons bound to the nucleus.

This material is covered inGasiorowicz Chapters 18,inCohen-Tannoudji et al. Comple-
mentBXIV,and briefly in Griffiths Chapter 7.

25.1 General Features of Helium States

We can use the hydrogenic states to begin to understand Helium. The Hamiltonian has the same
terms as Hydrogen but has a large perturbation due to the repulsion between the two electrons.

H=

p^21

2 m

+

p^22

2 m

−

Ze^2

r 1

−

Ze^2

r 2

+

e^2

|~r 1 −~r 2 |

We can write this in terms of the (Z= 2) Hydrogen Hamiltonian for each electron plus a perturba-
tion,
H=H 1 +H 2 +V

whereV(~r 1 ,~r 2 ) = e

2
|~r 1 −~r 2 |. Note thatV is about the same size as the the rest of the Hamiltonian so
first order perturbation theory is unlikely to be accurate.

For our zeroth order energy eigenstates, we will useproduct states of Hydrogen wavefunctions.

u(~r 1 ,~r 2 ) =φn 1 ℓ 1 m 1 (~r 1 )φn 2 ℓ 2 m 2 (~r 2 )

These are not eigenfunctions of H because ofV, the electron coulomb repulsion term. IgnoringV,
the problem separates into the energy for electron 1 and the energy for electron 2 and we can solve
the problem exactly.
(H 1 +H 2 )u=Eu

We can write thesezeroth order energiesin terms of the principal quantum numbers of the two
electrons,n 1 andn 2. Recalling that there is a factor ofZ^2 = 4 in these energies compared to
hydrogen, we get

E=En 1 +En 2 =−

1

2

Z^2 α^2 mec^2

(

1

n^21

+

1

n^22

)

=− 54 .4 eV

(

1

n^21

+

1

n^22

)

.

E 11 =Egs=− 108 .8 eV

E 12 =E 1 st=− 68 .0 eV

E 1 ∞=Eionization=− 54 .4 eV

E 22 =− 27 .2eV

Note thatE 22 is above ionization energy, so the state can decay rapidly by ejecting an electron.