The mathematical uncertainty principle
The speeding physicist
A physicist is stopped for speeding. ``No, Officer, I don't know how fast
I was going. But I do know exactly where I am.''
The joke is built on the implicit reference to Heisenberg's
uncertainty principle, a bedrock of modern Physics:
the position and the momentum of an object cannot both be measured exactly,
at the same time.
Quantitatively, Δ x Δ p ≥ h ⁄ 4π,
where Δ x
is the uncertainty in the measurement of position x,
Δ p the uncertainty in the measurement of momentum p,
and h is Planck's constant. In MKS units,
h ⁄ 4π works
out to be approximately 0.5 10^{34}kg m^{2}⁄s.
The speeding electron
For example, the mass m of the electron is close to 10^{30}
kg.
So any simultaneous measurements of its positon x and
velocity v are constrained by
Δ x Δ v ≥ 0.5 10^{4}
m^{2}⁄s,
using p = mv. If in some experiment we
can measure the position of an electron
to within one micron, or 10^{6} m,
then the uncertainty Δ v must be greater than
0.5 10^{28}/10^{30} = 50 m/s:
we can only know its velocity to within plus or minus 50 meters/sec.
(I checked with our Physics Department and was told not to worry:
an electron in an orbit
of that diameter typically travels at around 30,000 m/s.)
Singing into the piano
There is a purely mathematical phenomenon that closely parallels the
Heisenberg uncertainty principle.
Qualitatively, it can be understood as the problem of ascribing to
a burst of sound both
a location t in time and a frequency λ. It is clearly hard
to pinpoint the frequency of a very short burst: if Δ t
is small, then the indeterminacy Δ λ will be large.
Conversely, sound at one exact frequency (Δ λ = 0)
would correspond to
a perfect sine wave, with no beginning or end (Δ t = ∞).
To make this more precise we use the mathematical formalism
of the Fourier transform. Here again the analogy with sound is
helpful. If you sing a note into a piano while holding down the
damper pedal, so all the strings are free to vibrate, you will
hear the piano duplicate the sound of your voice. What is happening?
Each of the piano strings has its own sound, which does not sound
like your voice at all. But when the strings are excited by your voice,
each one resonates to the extent that its frequency is present
in your voice. When you stop singing the strings continue to vibrate,
each with the appropriate amplitude, and your voice is perpetuated.
The Fourier transform acts like a piano with infinitely many
strings, one giving each frequency λ between 0 and ∞.
The analogue of a sound burst (which could be the pressurewave due
to your voice) is a timevarying function f(t).
The amplitude picked up by the string with natural frequency
λ becomes
This is a finite integral for any value of
λ if you only sing for a finite time,
or if, more generally, the absolute value of f has
a finite integral from ∞ to ∞. We'll call such
a function "absolutely integrable."
The mathematical analogue to the piano's prolonging your voice
after you stop singing is the equation
The string with fundamental frequency λ is vibrating with
amplitude a(λ); the integral sums up all those vibrations
to reconstitute your voice.
What about phases?
To reconstruct an arbitrary f(t) exactly requires some more
information, since advancing or retarding the phase of a(λ)
cos(λt),
one of the harmonic constituents, gives a different function (although
it sounds the same to the ear). If that component is switched on
δ seconds earlier, it is represented by a(λ)
cos(λ(t + δ)). The trigonometric identity for
cos(a+b) rewrites this as
a(λ)[
cos(λt)cos(λδ) 
sin(λt)sin(λδ)]
which we can regroup as
a(λ)cos(λδ) cos(λt) 
a(λ)sin(λδ) sin(λt) = A(λ)cos(λt)
+ B(λ)sin(λt),
so the exact synthesis in general will require sines as well as cosines.
To avoid this slight complication, we will restrict attention to
even functions f (t):
those satisfying f (t)
= f (t); for these,
cosines are sufficient.
The Fourier Transform for even functions
Theorem. If f (t) is an even, absolutely integrable function
then
where
The theorem holds as stated for a continuous function
f (t); if the function has a
finite number of jump discontinuities t_{1}, ...,
t_{p} (i.e. points where the lefthand
limit f (t_{i})
is different from the righthand limit
f (t_{i}+)
, then the dλ integral
at one of those points splits the difference and gives
(1⁄2)[f (t_{i}+) +
f (t_{i}) ].
Some simple examples of even functions and their Fourier transforms
The most elementary functions with a finite integral on the whole line are
constant on a finite interval and zero everywhere else. They will be even
functions if that interval is centered about t = 0. Let us consider the
family f_{h}
of functions defined by
f_{h}(t) =

0 if t < h
1/2h if h ≤ t ≤ h
0 if h < t

The graphs of f_{
h} for h = 0.5 (blue), 1 (green), 2 (red).
The horizontal and vertical axes are scaled differently,
but these graphs all enclose the same area. They are
analogous to a family of
sound bursts of varying lengths, but of the same total power.
The Fourier transform of f_{h} can be
calculated by antidifferentiation:
The graphs of a_{h} for h = 0.5 (blue), 1 (green), 2 (red), colored the same as the graphs of the functions they transform.
Note that the width
of the transform is inversely proportional to the width of the function.
(The direct reconstruction of f_{h} from
a_{h}, which follows from the Theorem, can
also be achieved by a
fairly elementary but roundabout method. See, for
example, David L. Powers,
Boundary Value Problems, 4th Ed., Section 1.9).
Uncertainty principle for the Fourier transform
The f_{h} and their transforms
a_{h} show the
uncertainty principle for the Fourier transform
at work. Roughly, the
more tightly localized the f (t) signal is
(the shorter the duration
of the sound burst), the less tightly localized the a(λ)
distribution
must
be (the larger the spread in frequencies); conversely, the tighter
you cluster the frequencies, the wider the f (t)
distribution must be. This
principle has
very precise and natural formulation for normal probability distributions.
The normal distribution function.
A probability distribution function f is a positive function
with total integral equal to 1. (The f_{h}
defined above are probability distribution functions). One very
important example is the normal distribution with mean zero
and standard deviation 1:
an even function.
The Fourier transform
of this function is essentially the same function:
(This
calculation is not too difficult, using the identity
cos(λ)=(1 ⁄ 2)(e^{iλ} +
e^{iλ}) and
completing the squares in the resulting exponents.)
The uncertainty principle for normal distribution functions
with mean zero
The uncertainty principle takes a sharp form when
we adjust the normal distribution to have variance σ^{2}
(equivalently, standard deviation
σ), by letting
Then the same calculation as above yields
In words, the
Fourier transform of
a normal distribution with mean zero and variance σ^{2}
is a normal distribution with mean zero and variance
1 ⁄ σ^{2}. Let's use the
notation Ff for the
Fourier transform of f (the function we have been calling a),
and write D for variance. Then the relation between the
variances of a
normal probability
distribution function with mean zero and its Fourier transform can
be expressed in "uncertainty principle" form as
D(Ff ) D(f ) = 1.
And more generally ...
This property does not generalize to arbitrary probability distribution
functions (p.d.f.'s) because Ff may not be a p.d.f. at all
(it was not
for our f_{h} examples) and its variance may not
be defined. On the other hand, the Fourier transform
preserves the integral of the square of a function
(this is the Plancherel Theorem).
So if f ^{2} is a p.d.f., then
(Ff )^{2}, which is automatically nonnegative, will
also have total integral equal to 1: it
is also a p.d.f. Now our uncertainty relation does generalize, and becomes:
D((Ff )^{2}) D(f ^{2}) ≥ 1/2.
The resemblance between this statement and Heisenberg's uncertainty
principle is not coincidental, but is due to the relation between
position and momentum in quantum mechanics. They are
conjugate variables, and
the corresponding probability amplitudes are therefore
related just as f ^{2} and (Ff )^{2},
except with the scale change that produces
the extra h/2π.