6Quantum mechanics in three dimensions
IB Quantum Mechanics
6.2 Separable eigenstate solutions
How do we solve the Schr¨odinger equation in three dimensions? Things are
more complicated, since we have partial derivatives. Often, those we can solve
are those we can reduce to one-dimensional problems, using the symmetry of
the system. For example, we will solve the hydrogen atom exploiting the fact
that the potential is spherically symmetric. To do so, we will use the method of
separation of variables, as we’ve seen in the IB Methods course.
Consider the simpler case where we only have two dimensions. The time-
independent Schr¨odinger equation then gives
Hψ = −
~
2
2m
∂
2
∂x
2
1
+
∂
2
∂x
2
2
ψ + V (x
1
, x
2
)ψ = Eψ.
We are going to consider potentials of the form
V (x
1
, x
2
) = U
1
(x
1
) + U
2
(x
2
).
The Hamiltonian then splits into
H = H
1
+ H
2
,
where
H
i
= −
~
2
2m
∂
2
∂x
2
i
+ U
i
(x
i
).
We look for separable solutions of the form
ψ = χ
1
(x
1
)χ
2
(x
2
).
The Schr¨odinger equation then gives (upon division by ψ = χ
1
χ
2
) gives
−
~
2
2m
χ
00
1
χ
1
+ U
1
+
−
~
2
2m
χ
00
2
χ
2
+ U
2
= E.
Since each term is independent of x
2
and x
1
respectively, we have
H
1
χ
1
= E
1
χ
1
, H
2
χ
2
= E
2
χ
2
,
with
E
1
+ E
2
= E.
This is the usual separation of variables, but here we can interpret this physically
— in this scenario, the two dimensions are de-coupled, and we can treat them
separately. The individual
E
1
and
E
2
are just the contributions from each
component to the energy. Thus the process of separation variables can be seen
as looking for joint or simultaneous eigenstates for
H
1
and
H
2
, noting the fact
that [H
1
, H
2
] = 0.
Example. Consider the harmonic oscillator with
U
i
(x
i
) =
1
2
ω
2
x
2
i
.
This corresponds to saying
V =
1
2
mω
2
kxk
2
=
1
2
mω
2
(x
2
1
+ x
2
2
).
Now we have
H
i
= H
0
(ˆx
i
, ˆp
i
),
with H
0
the usual harmonic oscillator Hamiltonian.
Using the previous results for the one-dimensional harmonic oscillator, we
have
χ
1
= ψ
n
1
(x
1
), χ
2
= ψ
n
2
(x
2
),
where
ψ
i
is the
i
th eigenstate of the one-dimensional harmonic oscillator, and
n
1
, n
2
= 0, 1, 2, ···.
The corresponding energies are
E
i
= ~ω
n
i
+
1
2
.
The energy eigenvalues of the two-dimensional oscillator are thus
E = ~ω (n
i
+ n
2
+ 1)
for
ψ(x
1
, x
2
) = ψ
n
1
(x
1
)ψ
n
2
(x
2
).
We have the following energies and states:
State Energy Possible states
Ground state E = ~ω ψ = ψ
0
(x
1
)ψ
0
(x
2
)
1st excited state E = 2~ω
ψ = ψ
1
(x
1
)ψ
0
(x
2
)
ψ = ψ
0
(x
1
)ψ
1
(x
2
)
We see there is a degeneracy of 2 for the first excited state.
Separable solutions also arise naturally when
H
(i.e. the potential
V
) has
some symmetry. For example, if the potential is spherically symmetric, we can
find solutions
ψ(x) = R(r)Y (θ, φ),
where r, θ, φ are the spherical polars.