7The hydrogen atom
IB Quantum Mechanics
7.2 General solution
We guessed our solution
r
`
e
−κr
above by looking at the asymptotic behaviour
at large and small
r
. We then managed to show that this is one solution of the
hydrogen atom. We are greedy, and we want all solutions.
Similar to what we did for the harmonic oscillator, we guess that our general
solution is of the form
R(r) = e
−κr
f(r).
Putting it in, we obtain
f
00
+
2
r
f
0
−
`(` + 1)
r
2
f = 2
κf
0
+ (κ − λ)
f
r
.
We immediately see an advantage of this substitution — now each side of the
equality is equidimensional, and equidimensionality makes our life much easier
when seeking series solution. This equation is regular singular at
r
= 0, and
hence we guess a solution of the form
f(r) =
∞
X
p=0
a
p
r
p+σ
, a
0
6= 0.
Then substitution gives
X
p≥0
((p + σ)(p + σ − 1) −`(` + 1))a
p
r
p+σ−2
=
X
p≥0
2(κ(p + σ + 1) −λ)a
p
r
p+σ−1
.
The lowest term gives us the indicial equation
σ(σ + 1) − `(` + 1) = (σ − `)(σ + ` + 1) = 0.
So either
σ
=
`
or
σ
=
−
(
`
+ 1). We discard the
σ
=
−
(
`
+ 1) solution since this
would make f and hence R singular at r = 0. So we have σ = `.
Given this, the coefficients are then determined by
a
p
=
2(κ(p + `) − λ)
p(p + 2` + 1)
a
p−1
, p ≥ 1.
Similar to the harmonic oscillator, we now observe that, unless the series termi-
nates, we have
a
p
a
p−1
∼
2κ
p
as
p → ∞
, which matches the behaviour of
r
α
e
2κr
(for some
α
). So
R
(
r
) is
normalizable only if the series terminates. Hence the possible values of λ are
κn = λ
for some
n ≥ `
+ 1. So the resulting energy levels are exactly those we found
before:
E
n
= −
~
2
2m
e
κ
2
= −
~
2
2m
e
λ
2
n
2
= −
1
2
m
e
e
2
4πε
0
~
2
1
n
2
.
for n = 1, 2, 3, ···. This n is called the principle quantum number.
For any given n, the possible angular momentum quantum numbers are
` = 0, 1, 2, 3, ··· , n − 1
m = 0, ±1, ±2, ··· , ±`.
The simultaneous eigenstates are then
ψ
n`m
(x) = R
n`
(r)Y
`m
(θ, ϕ),
with
R
n`
(r) = r
`
g
n`
(r)e
−λr/n
,
where g
n`
(r) are (proportional to) the associated Laguerre polynomials.
In general, the “shape” of probability distribution for any electron state
depends on
r
and
θ, ϕ
mostly through
Y
`m
. For
`
= 0, we have a spherically
symmetric solutions
ψ
n00
(x) = g
n0
(r)e
−λr/n
.
This is very different from the Bohr model, which says the energy levels depend
only on the angular momentum and nothing else. Here we can have many
different angular momentums for each energy level, and can even have no angular
momentum at all.
The degeneracy of each energy level E
n
is
n−1
X
`=0
`
X
m=−`
1 =
n−1
X
`=0
(2` + 1) = n
2
.
If you study quantum mechanics more in-depth, you will find that the degeneracy
of energy eigenstates reflects the symmetries in the Coulomb potential. Moreover,
the fact that we have
n
2
degenerate states implies that there is a hidden symmetry,
in addition to the obvious
SO
(3) rotational symmetry, since just
SO
(3) itself
should give rise to much fewer degenerate states.
So. We have solved the hydrogen atom.