3Expectation and uncertainty
IB Quantum Mechanics
3.3 Heisenberg’s uncertainty principle
We will show that, in fact, quantum mechanics is not like classical mechanics.
Statement
The statement of the uncertainty principle (or relation) is
Theorem
(Heisenberg’s uncertainty principle)
.
If
ψ
is any normalized state (at
any fixed time), then
(∆x)
ψ
(∆p)
ψ
≥
~
2
.
Example. Consider the normalized Gaussian
ψ(x) =
1
απ
1
4
e
−x
2
/α
.
We find that
hˆxi
ψ
= hˆpi
ψ
= 0,
and also
(∆x)
2
ψ
=
α
2
, (∆p)
2
ψ
=
~
2
2α
.
So we get
(∆x)
ψ
(∆p)
ψ
=
~
2
.
We see that a small
α
corresponds to
ψ
sharply peaked around
x
= 0, i.e. it has
a rather definite position, but has a large uncertainty in momentum. On the
other hand, if
α
is large,
ψ
is spread out in position but has a small uncertainty
in momentum.
Recall that for
α
=
~
mω
, this is the lowest energy eigenstate for harmonic
oscillator with
H =
1
2m
ˆp +
1
2
mω
2
ˆx
2
,
with eigenvalue
1
2
~ω
. We can use the uncertainty principle to understand why
we have a minimum energy of
1
2
~ω
instead of 0. If we had a really small energy,
then we would just sit at the bottom of the potential well and do nothing, with
both a small (and definite) momentum and position. Hence for uncertainty to
occur, we need a non-zero ground state energy.
To understand where the uncertainty principle come from, we have to under-
stand commutation relations.
Commutation relations
We are going to define a weird thing called the commutator. At first sight, this
seems like a weird definition, but this turns out to be a really important concept
in quantum mechanics.
Definition
(Commutator)
.
Let
Q
and
S
be operators. Then the commutator
is denoted and defined by
[Q, S] = QS − SQ.
This is a measure of the lack of commutativity of the two operators.
In particular, the commutator of position and momentum is
[ˆx, ˆp] = ˆxˆp − ˆpˆx = i~.
This relation results from a simple application of the product rule:
(ˆxˆp − ˆpˆx)ψ = −i~ψ
0
− (−i~(xψ)
0
= i~ψ.
Note that if α and β are any real constants, then the operators
X = ˆx − α, P = ˆp − β
also obey
[X, P ] = i~.
This is something we will use when proving the uncertainty principle.
Recall that when we proved Ehrenfest’s theorem, the last step was to calculate
the commutator:
[ˆx, H] = ˆxH −H ˆx =
i~
m
ˆp
[ˆp, H] = ˆpH −H ˆp = −i~V
0
(ˆx).
Since
H
is defined in terms of
ˆp
and
ˆx
, we can indeed find these relations just
using the commutator relations between
ˆx
and
ˆp
(plus a few more basic facts
about commutators).
Commutator relations are important in quantum mechanics. When we first
defined the momentum operator
ˆp
as
−i~
∂
∂x
, you might have wondered where
this weird definition came from. This definition naturally comes up if we require
that
ˆx
is “multiply by
x
” (so that the delta function
δ
(
x − x
0
) is the eigenstate
with definition position
x
0
), and that
ˆx
and
ˆp
satisfies this commutator relation.
With this requirements, ˆp must be defined as that derivative.
Then one might ask, why would we want
ˆx
and
ˆp
to satisfy this commutator
relation? It turns out that in classical dynamics, there is something similar
known as the Poisson bracket
{·, ·}
, where we have
{x, p}
= 1. To get from
classical dynamics to quantum mechanics, we just have to promote our Poisson
brackets to commutator brackets, and multiply by i~.
Proof of uncertainty principle
Proof of uncertainty principle. Choose α = hˆxi
ψ
and β = hˆpi
ψ
, and define
X = ˆx − α, P = ˆp − β.
Then we have
(∆x)
2
ψ
= (ψ, X
2
ψ) = (Xψ, Xψ) = kXψk
2
(∆p)
2
ψ
= (ψ, P
2
ψ) = (P ψ, P ψ) = kP ψk
2
Then we have
(∆x)
ψ
(∆p)
ψ
= kXψkkP ψk
≥ |(Xψ, P ψ)|
≥ |im(Xψ, P ψ)|
≥
1
2i
h
(Xψ, P ψ) −(P ψ, Xψ)
i
=
1
2i
h
(ψ, XP ψ) −(ψ, P Xψ)
i
=
1
2i
(ψ, [X, P ]ψ)
=
~
2
(ψ, ψ)
=
~
2
.
So done.