7The hydrogen atom

IB Quantum Mechanics



7.1 Introduction
Consider an electron moving in a Coulomb potential
V (r) =
e
2
4πε
0
1
r
.
This potential is due to a proton stationary at
r
= 0. We follow results from
the last section of the last chapter, and set the mass
µ
=
m
e
, the electron mass.
The joint energy eigenstates of H, L
2
and L
3
are of the form
φ(x) = R(r)Y
`m
(θ, ϕ)
for ` = 0, 1, ··· and m = 0, ±1, ··· , ±m.
The radial part of the Schr¨odinger equation can be written
R
00
+
2
r
R
0
`(` + 1)
r
2
R +
2λ
r
R = κ
2
R, ()
with
λ =
m
e
e
2
4πε
0
~
2
, E =
~
2
κ
2
2m
e
.
Note that here we work directly with
R
instead of
χ
, as this turns out to be
easier later on.
The goal of this chapter is to understand all the (normalizable) solutions to
this equation ().
As in the case of the harmonic oscillator, the trick to solve this is see what
happens for large
r
, and “guess” a common factor of the solutions. In the case of
the harmonic oscillator, we guessed the solution should have
e
y
2
/2
as a factor.
Here, for large r, we get
R
00
κ
2
R.
This implies R e
κr
for large R.
For small
r
, we by assumption know that
R
is finite, while
R
0
and
R
00
could
potentially go crazy. So we multiply by
r
2
and discard the
rR
and
r
2
R
terms to
get
r
2
R
00
+ 2rR
0
`(` + 1)R 0.
This gives the solution R r
`
.
We’ll be bold and try a solution of the form
R(r) = Cr
`
e
κr
.
When we substitute this in, we will get three kinds of terms. The
r
`
e
κr
terms
match, and so do the terms of the form r
`2
e
κr
. Finally, we see the r
`1
e
κe
terms match if and only if
2(`r
`1
)(κe
κr
) + 2(r
`1
)(κe
κr
) + 2λr
`1
e
κe
= 0.
When we simplify this mess, we see this holds if and only if
(` + 1)κ = λ.
Hence, for any integer
n
=
`
+ 1 = 1
,
2
,
3
, ···
, there are bound states with
energies
E
n
=
~
2
2m
e
λ
2
n
2
=
1
2
m
e
e
2
4πε
0
~
2
1
n
2
.
These are the energy levels of the Bohr model, derived within the framework of
the Bohr model. However, there is a slight difference. In our model, the total
angular momentum eigenvalue is
~
2
`(` + 1) = ~
2
n(n 1),
which is not what the Bohr model predicted.
Nevertheless, this is not the full solution. For each energy level, this only
gives one possible angular momentum, but we know that there can be many
possible angular momentums for each energy level. So there is more work to be
done.