5Axioms for quantum mechanics

IB Quantum Mechanics



5.2 Measurements
Key results for Hermitian operators
Our next set of axioms relate Hermitian operators with physical measurements.
Before we state them, we first look at some purely mathematical properties of
Hermitian operators. We first prove some purely mathematical results about
Hermitian operators.
Proposition. Let Q be Hermitian (an observable), i.e. Q
= Q. Then
(i) Eigenvalues of Q are real.
(ii)
Eigenstates of
Q
with different eigenvalues are orthogonal (with respect to
the complex inner product).
(iii)
Any state can be written as a (possibly infinite) linear combination of
eigenstates of
Q
, i.e. eigenstates of
Q
provide a basis for
V
. Alternatively,
the set of eigenstates is complete.
Note that the last property is not actually necessarily true. For example,
the position and momentum operators do not have eigenstates at all, since the
“eigenstates” are not normalizable. However, for many of the operators we care
about, this is indeed true, and everything we say in this section only applies to
operators for which this holds.
Proof.
(i) Since Q is Hermitian, we have, by definition,
(χ, ) = (Qχ, χ).
Let χ be an eigenvector with eigenvalue λ, i.e. = λχ. Then we have
(χ, λχ) = (λχ, χ).
So we get
λ(χ, χ) = λ
(χ, χ).
Since (χ, χ) 6= 0, we must have λ = λ
. So λ is real.
(ii) Let = λχ and = µφ. Then we have
(φ, ) = (Qφ, χ).
So we have
(φ, λχ) = (µφ, χ).
In other words,
λ(φ, χ) = µ
(φ, χ) = µ(φ, χ).
Since λ 6= µ by assumption, we know that (φ, χ) = 0.
(iii) We will not attempt to justify this, or discuss issues of convergence.
Measurement axioms
Consider an observable
Q
with discrete spectrum (i.e. eigenvalues) and normalized
eigenstates. From the results we just had, any given state can be written
ψ =
X
n
α
n
χ
n
,
with
n
= λ
n
χ
n
, and (χ
m
, χ
n
) = δ
mn
. This means we can calculate α
n
by
α
n
= (χ
n
, ψ).
We can assume also that
α
n
6
= 0 since we can just pretend the terms with
coefficient 0 do not exist, and that the corresponding
λ
n
are distinct by choosing
an appropriate basis. Then
The outcome of a measurement is some eigenvalue of Q.
The probability of obtaining λ
n
is
P
n
= |α
n
|
2
,
where α
n
= (χ
n
, ψ) is the amplitude.
The measurement is instantaneous and forces the system into the state
χ
n
.
This is the new state immediately after the measurement is made.
The last statement is rather weird. However, if you think hard, this must be
the case. If we measure the state of the system, and get, say, 3, then if we
measure it immediately afterwards, we would expect to get the result 3 again with
certainty, instead of being randomly distributed like the original state. So after
a measurement, the system must be forced into the corresponding eigenstate.
Note that these axioms are consistent in the following sense. If
ψ
is normalized,
then
(ψ, ψ) =
X
α
m
χ
m
,
X
α
n
χ
n
=
X
m,n
α
m
α
n
(χ
m
, χ
n
)
=
X
m,n
α
m
α
n
δ
mn
=
X
n
|α
n
|
2
=
X
n
P
n
= 1
So if the state is normalized, then the sum of probabilities is indeed 1.
Example.
Consider the harmonic oscillator, and consider the operator
Q
=
H
with eigenfunctions χ
n
= ψ
n
and eigenvalues
λ
n
= E
n
= ~ω
n +
1
2
.
Suppose we have prepared our system in the state
ψ =
1
6
(ψ
0
+ 2ψ
1
4
).
Then the coefficients are
α
0
=
1
6
, α
1
=
2
6
, α
4
=
i
6
.
This is normalized since
kψk
2
= |α
0
|
2
+ |α
1
|
2
+ |α
4
|
2
= 1.
Measuring the energy gives the following probabilities:
Energy Probability
E
0
=
1
2
~ω P
0
=
1
6
E
1
=
3
2
~ω P
1
=
2
3
E
4
=
9
2
~ω P
4
=
1
6
If a measurement gives
E
1
, then
ψ
1
is the new state immediately after measure-
ment.
Expressions for expectation and uncertainty
We can use these probability amplitudes to express the expectation and uncer-
tainty of a state in a more familiar way.
Proposition.
(i) The expectation value of Q in state ψ is
hQi
ψ
= (ψ, ) =
X
λ
n
P
n
,
with notation as in the previous part.
(ii) The uncertainty (δQ)
ψ
is given by
(∆Q)
2
ψ
= h(Q hQi
ψ
)
2
i
ψ
= hQ
2
i
ψ
hQi
2
ψ
=
X
n
(λ
n
hQi
ψ
)
2
P
n
.
From this, we see that
ψ
is an eigenstate of
Q
with eigenvalue
λ
if and only
if hQi
ψ
= λ and (∆Q)
ψ
= 0.
Proof.
(i) Let ψ =
P
α
n
χ
n
. Then
=
X
α
n
λ
n
χ
n
.
So we have
(ψ, ) =
X
n,m
(α
m
, χ
n
, α
n
λ
n
χ
n
) =
X
n
α
n
α
n
λ
n
=
X
λ
n
P
n
.
(ii) This is a direct check using the first expression.
Example.
Consider the harmonic oscillator as we had in the previous example.
Then the expectation value is
hHi
ψ
=
X
n
E
n
P
n
=
1
2
·
1
6
+
3
2
·
2
3
+
9
2
·
1
6
~ω =
11
6
~ω.
Note that the measurement axiom tells us that after measurement, the
system is then forced into the eigenstate. So when we said that we interpret the
expectation value as the “average result for many measurements”, we do not
mean measuring a single system many many times. Instead, we prepare a lot of
copies of the system in state ψ, and measure each of them once.