6Quantum mechanics in three dimensions
IB Quantum Mechanics
6.4 Joint eigenstates for a spherically symmetric potential
Unfortunately, it is completely standard to use
m
to indicate the eigenvalue of
L
3
, as we did above. Hence, here we shall consider a particle of mass
µ
in a
potential V (r) with spherical symmetry. The Hamiltonian is
H =
1
2µ
ˆp
2
+ V = −
~
2
2µ
∇
2
+ V (r).
We have seen that we can write H as
H = −
~
2
2µ
1
r
∂
2
∂r
2
r +
1
2µ
1
r
2
L
2
+ V (r).
The first thing we want to check is that
[L
i
, H] = [L
2
, H] = 0.
This implies we can use the eigenvalues of
H
,
L
2
and
L
3
to label our solutions
to the equation.
We check this using Cartesian coordinates. The kinetic term is
[L
i
,
ˆ
p
2
] = [L
i
, ˆp
j
ˆp
j
]
= [L
i
, ˆp
j
]ˆp
j
+ ˆp
j
[L
i
, ˆp
j
]
= i~ε
ijk
(ˆp
k
ˆp
j
+ ˆp
j
ˆp
k
)
= 0
since we are contracting an antisymmetric tensor with a symmetric term. We
can also compute the commutator with the potential term
[L
i
, V (r)] = −i~ε
ijk
x
j
∂V
∂x
k
= −i~ε
ijk
x
j
x
k
r
V
0
(r)
= 0,
using the fact that
∂r
∂x
i
=
x
i
r
.
Now that
H
,
L
2
and
L
3
are a commuting set of observables, we have the joint
eigenstates
ψ(x) = R(r)Y
`m
(θ, ϕ),
and we have
L
2
Y
`m
= ~
2
`(` + 1)Y
`m
` = 0, 1, 2, ···
L
3
Y
`m
= ~mY
`m
m = 0, ±1, ±2, ··· , ±`.
The numbers
`
and
m
are usually known as the angular momentum quantum
numbers. Note that
`
= 0 is the special case where we have a spherically
symmetric solution.
Finally, we solve the Schr¨odinger equation
Hψ = Eψ
to obtain
−
~
2
2µ
1
r
d
2
dr
2
(rR) +
~
2
2µr
2
`(` + 1)R + V R = ER.
This is now an ordinary differential equation in
R
. We can interpret the terms
as the radial kinetic energy, angular kinetic energy, the potential energy and
the energy eigenvalue respectively. Note that similar to what we did in classical
dynamics, under a spherically symmetric potential, we can replace the angular
part of the motion with an “effective potential”
~
2
2µr
2
`(` + 1).
We often call
R
(
r
) the radial part of the wavefunction, defined on
r ≥
0.
Often, it is convenient to work with
χ
(
r
) =
rR
(
r
), which is sometimes called
the radial wavefunction. Multiplying the original Schr¨odinger equation by
r
, we
obtain
−
~
2
2µ
χ
00
+
~
2
`(` + 1)
2µr
2
χ + V χ = Eχ.
This is known as the radial Schr¨odinger equation.
This has to obey some boundary conditions. Since we want
R
to be finite as
r →
0, we must have
χ
= 0 at
r
= 0. Moreover, the normalization condition is
now
1 =
Z
|ψ(x)|
2
d
3
x =
Z
|R(r)|
2
r
2
dr
Z
|Y
`m
(θ, ϕ)|
2
sin θ dθ dϕ.
Hence, ψ is normalizable if and only if
Z
∞
0
|R(r)|
2
r
2
dr < ∞.
Alternatively, this requires
Z
∞
0
|χ(r)|
2
dr < ∞.
Example
(Three-dimensional well)
.
We now plot our potential as a function of
r:
r
V
−U
a
This is described by
V (r) =
(
0 r ≥ a
−U r < a
,
where U > 0 is a constant.
We now look for bound state solutions to the Schr¨odinger equation with
−U < E < 0, with total angular momentum quantum number `.
For r < a, our radial wavefunction χ obeys
χ
00
−
`(` + 1)
r
2
χ + k
2
χ = 0,
where k is a new constant obeying
U + E =
~
2
k
2
2µ
.
For r ≥ a, we have
χ
00
−
`(` + 1)
r
2
χ − κ
2
χ = 0,
with κ obeying
E = −
~
2
κ
2
2µ
.
We can solve in each region and match
χ, χ
0
at
r
=
a
, with boundary condition
χ
(0) = 0. Note that given this boundary condition, solving this is equivalent to
solving it for the whole R but requiring the solution to be odd.
Solving this for general
`
is slightly complicated. So we shall look at some
particular examples.
For
`
= 0, we have no angular term, and we have done this before. The
general solution is
χ(r) =
(
A sin kr r < a
Be
−κr
r > a
Matching the values at x = a determines the values of k, κ and hence E.
For ` = 1, it turns out the solution is just
χ(r) =
(
A
cos kr −
1
kr
sin kr
r < a
B
1 +
1
kr
e
−κr
r > a
.
After matching, the solution is
ψ(r) = R(r)Y
1m
(θ, ϕ) =
χ(r)
r
Y
1m
(θ, ϕ),
where m can take values m = 0, ±1.
Solution for general
`
involves spherical Bessel functions, and are studied
more in-depth in the IID Applications of Quantum Mechanics course.