2Some examples in one dimension
IB Quantum Mechanics
2.5 The harmonic oscillator
So far in our examples, the quantization (mathematically) comes from us requiring
continuity at the boundaries. In the harmonic oscillator, it arises in a different
way.
x
V
This is a harmonic oscillator of mass m with
V (x) =
1
2
mω
2
x
2
.
Classically, this has a motion of
x
=
A cos ω
(
t − t
0
), which is something we
(hopefully) know well too.
This is a really important example. First of all, we can solve it, which is a
good thing. More importantly, any smooth potential can be approximated by a
harmonic oscillator near an equilibrium x
0
, since
V (x) = V (x
0
) +
1
2
V
00
(x
0
)(x − x
0
)
2
+ ··· .
Systems with many degrees like crystals can also be treated as collections of
independent oscillators by considering the normal modes. If we apply this to
the electromagnetic field, we get photons! So it is very important to understand
the quantum mechanical oscillator.
We are going to seek all normalizable solutions to the time-independent
Schr¨odinger equation
Hψ = −
~
2
2m
ψ
00
+
1
2
mω
2
x
2
ψ = Eψ.
So simplify constants, we define
y =
mω
~
1
2
x, E =
2E
~ω
,
both of which is dimensionless. Then we are left with
−
d
2
ψ
dy
2
+ y
2
ψ = Eψ.
We can consider the behaviour for
y
2
E
. For large
y
, the
y
2
ψ
term will
be large, and so we want the
ψ
00
term to offset it. We might want to try the
Gaussian
e
−y
2
/2
, and when we differentiate it twice, we would have brought
down a factor of y
2
. So we can wlog set
ψ = f(y)e
−
1
2
y
2
.
Then the Schr¨odinger equation gives
d
2
f
dy
2
− 2y
df
dy
+ (E − 1) = 0.
This is known as Hermite’s equation. We try a series solution
f(y) =
X
r≥0
a
r
y
r
,
and substitute in to get
X
r≥0
(r + 2)(r + 1)a
n+2
+ (E − 1 −2r)a
r
y
r
= 0.
This holds if and only if
a
r+2
=
2r + 1 − E
(r + 2)(r + 1)
a
r
, r ≥ 0.
We can choose
a
0
and
a
1
independently, and can get two linearly independent
solutions. Each solution involves either all even or all odd powers.
However, we have a problem. We want normalizable solutions. So we want
to make sure our function does not explode at large
y
. Note that it is okay if
f
(
y
) is quite large, since our
ψ
is suppressed by the
e
−
1
2
y
2
terms, but we cannot
grow too big.
We look at these two solutions individually. To examine the behaviour of
f(y) when y is large, observe that unless the coefficients vanish, we get
a
p
/a
p−2
∼
1
p
.
This matches the coefficients of
y
α
e
y
2
for some power
α
(e.g.
P
p≥0
y
2p
p!
). This
is bad, since our ψ will then grow as e
1
2
y
2
, and cannot be normalized.
Hence, we get normalizable
ψ
if and only if the series for
f
terminates to
give a polynomial. This occurs iff
E
= 2
n
+ 1 for some
n
. Note that for each
n
,
only one of the two independent solutions is normalizable. So for each
E
, we get
exactly one solution.
So for n even, we have
a
r+2
=
2r − 2n
(r + 2)(r + 1)
a
r
for r even, and a
r
= 0 for r odd, and the other way round when n is odd.
The solutions are thus
f
(
y
) =
h
n
(
y
), where
h
n
is a polynomial of degree
n
with h
n
(−y) = (−1)
n
h
n
(y).
For example, we have
h
0
(y) = a
0
h
1
(y) = a
1
y
h
2
(y) = a
0
(1 − 2y
2
)
h
3
(y) = a
1
y −
2
3
y
3
.
These are known as the Hermite polynomials. We have now solved our harmonic
oscillator. With the constant restored, the possible energy eigenvalues are
E
n
= ~ω
n +
1
2
,
for n = 0, 1, 2, ···.
The wavefunctions are
ψ
n
(x) = h
n
mω
~
1
2
x
exp
−
1
2
mω
~
x
2
,
where normalization fixes a
0
and a
1
.
As we said in the beginning, harmonic oscillators are everywhere. It turns
out quantised electromagnetic fields correspond to sums of quantised harmonic
oscillators, with
E
n
− E
0
= n~ω
This is equivalent to saying the
n
th state contains
n
photons, each of energy
~ω
.