Fourier Transforms and
Uncertainty Relations
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The function
exp(-x2) has no simple closed-form indefinite integral, but the
related function x exp(-x2) does have a simple integral,
namely,
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This identity can be used to
evaluate the definite integral of exp(-x2) from x = -¥ to +¥. Letting Q denote the value of this definite integral,
we can write
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In terms of polar coordinates
on the x,y plane we have x = r cos(q) and y = r sin(q), and therefore x2 + y2 =
r2. The Jacobian of the transformation is
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so the incremental area element
is
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Hence the double integral
expressed in terms of r,q coordinates is
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Making use of equation (1) to
evaluate the interior definite integral, we have
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Therefore the definite integral
of exp(-x2) from -¥ to ¥ is
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More generally, if we replace
the exponent -x2 with -ax2 + bx - c, we can define the parameter
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in terms of which we can
write
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This definite integral is
particularly useful when considering the Fourier transform of a normal density
distribution. Recall that for any function f(x) we can define another function
F(y) that satisfies the relations
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These two functions are a
Fourier transform pair, i.e., each of them is the Fourier transform of
the other. Now, if f(x) is the normal probability density
function
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then the Fourier transform
is
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so the exponent in the integral
is of the form -ax2 + bx + c with a = 1/2s2, b = iy +
m/s2, and c =
-m2/(2s2). Hence the
Fourier transform of the normal density distribution is
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Choosing our scales so that the
mean of f is zero, i.e., so that m = 0, the above expression reduces
to
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In other words, the Fourier
transform of the normal distribution with mean zero and standard deviation
s is also a normal distribution with mean zero, but with
standard deviation 1/s. This shows that the variances of the f and F
distributions satisfy the "uncertainty relation" var(f) var(F) = 1. This
equality is the limiting case of a general inequality on the product of
variances of Fourier transform pairs. In general, if f is an arbitrary
probability density distribution and F is its Fourier transform,
then
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Notice that f and F are just
two different ways of characterizing the same distribution, one in the amplitude
domain and the other in the frequency domain. Given either of these
distributions, the other is completely determined.
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The relation between conjugate
variables (such as position and momentum) in quantum mechanics can be expressed
in terms of the relation between Fourier transform pairs. Consider a physical
system with just one degree of freedom, represented by the operator q,
and let p denote the operator for the corresponding momentum (i.e., the
generalized momentum of q in the usual Hamiltonian formulation). The
basic commutation relation between these operators is
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Notice the symmetry between the
p and q operators if we replace i with -i. The usual Schrodinger
representation of this system takes the observable q as a "diagonal"
operator, i.e., with the eigenvalues specified explicitly, and then the
corresponding momentum operator is defined as
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However, we could (in theory)
just as well take p as a diagonal operator, and then q would be given
by
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Dirac referred to this as the
momentum representation of the system. This again shows the symmetry between
q and p under an exchange of i and -i. Indeed if we let
<q|S> and <p|S> for any given state S denote
the probability amplitudes that measurements corresponding to the operators
q and p will return the eigenvalues q and p respectively, then we
find
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In other words, the probability
amplitude distributions of two conjugate variables are simply the (suitably
scaled) Fourier transforms of each other. We saw previously that the dispersions
(variances) of two density distributions that comprise a Fourier transform pair
satisfy the inequality (2), so the variances of the probability amplitude
distributions of conjugate observables in quantum mechanics satisfy such an
inequality. Thus Heisenberg's uncertainty principle for conjugate pairs of
observables follows directly from the fact that those observables are
essentially the Fourier transforms of each other.
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Of course, this attribute of
Fourier transform pairs is purely mathematical, and has no a priori
applicability to pairs of observables such as position and momentum, or time and
energy. The physical content of quantum mechanics is based on the two
relations
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where E is energy, p is
momentum (in one dimension), h is Planck's (reduced) constant, w is the frequency with units second-1, and k
is the wave number with units meter-1. These relations were
introduced in the early 1900's by Planck, Einstein, and deBroglie to account for
non-classical phenomena such as cavity radiation and the photo-electric effect,
both of which depend on the particle-like behavior of entities that had
previously been modeled as waves, as well as phenomena involving wave-like
behavior of material particles. These are the relations that associate the
familiar observables of energy, momentum, space, and time, with the frequency
domain. Indeed in terms of the characteristic time t = 1/w and distance D = 1/k the above relations can be written
as
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which already clearly reveals
the conjugacy of time and energy, and of distance and momentum. In view of this,
it isn't surprising to find that the product of the dispersions of two conjugate
observables (such as position and momentum) cannot be less than one quanta of
action, represented by h.
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In a sense, there is also a
conjugacy between space and time - two observable that had been regarded as
disjoint and independent prior to the early 1900s. In special relativity the
inertial space and time intervals dx and dt between two events are components of
a single invariant spacetime interval ds between those events. These intervals
are related according to the Minkowski metric, which can be written in the
form
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This can be regarded as an
"uncertainty relation" for space and time. In general, physics was based, prior
to 1900, on the premise that h and 1/c2 were both zero. With the advent of
quantum mechanics and special relativity, it was realized that they both have
non-zero values, although they are extremely small in terms of ordinary
units.
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Friday, February 8, 2013
Fourier transform of the normal distribution with mean zero and standard deviation s is also a normal distribution with mean zero, but with standard deviation 1/s.
Fourier transform of the normal distribution with mean zero and standard
deviation s is also a normal distribution with mean zero, but with
standard deviation 1/s.
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