5Axioms for quantum mechanics
IB Quantum Mechanics
5.5 Degeneracy and simultaneous measurements
Degeneracy
Definition
(Degeneracy)
.
For any observable
Q
, the number of linearly inde-
pendent eigenstates with eigenvalue
λ
is the degeneracy of the eigenvalue. In
other words, the degeneracy is the dimension of the eigenspace
V
λ
= {ψ : Qψ = λψ}.
An eigenvalue is non-degenerate if the degeneracy is exactly 1, and is degenerate
if the degeneracy is more than 1.
We say two states are degenerate if they have the same eigenvalue.
In one dimension, the energy bound states are always non-degenerate, as we
have seen in example sheet 2. However, in three dimensions, energy bound states
may be degenerate. If
λ
is degenerate, then there is a large freedom in choosing
an orthonormal basis for
V
λ
. Physically, we cannot distinguish degenerate states
by measuring
Q
alone. So we would like to measure something else as well to
distinguish these eigenstates. When can we do this? We have previously seen
that we cannot simultaneously measure position and momentum. It turns out
the criterion for whether we can do this is simple.
Commuting observables
Recall that after performing a measurement, the state is forced into the corre-
sponding eigenstate. Hence, to simultaneously measure two observables
A
and
B
, we need the state to be a simultaneously an eigenstate of
A
and
B
. In other
words, simultaneous measurement is possible if and only if there is a basis for
V
consisting of simultaneous or joint eigenstates χ
n
with
Aχ
n
= λ
n
χ
n
, Bχ
n
= µ
n
χ
n
.
Our measurement axioms imply that if the state is in
χ
n
, then measuring
A
,
B
in rapid succession, in any order, will give definite results
λ
n
and
µ
n
respectively
(assuming the time interval between each pair of measurements is short enough
that we can neglect the evolution of state in time).
From IB Linear Algebra, a necessary and sufficient condition for
A
and
B
to
be simultaneously measurable (i.e. simultaneously diagonalizable) is for
A
and
B to commute, i.e.
[A, B] = AB − BA = 0.
This is consistent with a “generalized uncertainty relation”
(∆A)
ψ
(∆B)
ψ
≥
1
2
|h[A, B]i
ψ
|,
since if we if have a state that is simultaneously an eigenstate for
A
and
B
, then
the uncertainties on the left would vanish. So
h
[
A, B
]
i
ψ
= 0. The proof of this
relation is on example sheet 3.
This will be a technique we will use to tackle degeneracy in general. If we have
a system where, say, the energy is degenerate, we will attempt to find another
operator that commutes with
H
, and try to further classify the underlying states.
When dealing with the hydrogen atom later, we will use the angular momentum
to separate the degenerate energy eigenstates.