Generally speaking, the more concentrated f (x) is, the more spread out its Fourier transform  (ξ) must be. In particular, the scaling property of the Fourier transform may be seen as saying: if we squeeze a function in x, its Fourier transform stretches out in ξ. It is not possible to arbitrarily concentrate both a function and its Fourier transform.

The trade-off between the compaction of a function and its Fourier transform can be formalized in the form of an uncertainty principle by viewing a function and its Fourier transform as conjugate variables with respect to the symplectic form on the time–frequency domain: from the point of view of the linear canonical transformation, the Fourier transform is rotation by 90° in the time–frequency domain, and preserves the symplectic form.

Suppose f (x) is an integrable and square-integrable function. Without loss of generality, assume that f (x) is normalized:

$ \int_{-\infty}^\infty |f(x)|^2 \,dx=1. $

It follows from the Plancherel theorem that  (ξ) is also normalized.

The spread around x = 0 may be measured by the dispersion about zero[1] defined by

$ D_0(f)=\int_{-\infty}^\infty x^2|f(x)|^2\,dx. $

In probability terms, this is the second moment of $ |f(x)^2| $ about zero.

The Uncertainty principle states that, if f (x) is absolutely continuous and the functions x·f (x) and f ′(x) are square integrable, then[2]

$ D_0(f)D_0\left(\hat{f}\right) \geq \frac{1}{16\pi^2} $.

The equality is attained only in the case

$ \begin{align} f(x) &= C_1 \, e^{-\pi \frac{x^2}{\sigma^2} }\\ \therefore \hat{f}(\xi) &= \sigma C_1 \, e^{-\pi\sigma^2\xi^2} \end{align} $

where σ > 0 is arbitrary and $ \mathbf{C_1} = \frac{\sqrt[4]{2}}{\sqrt{σ}} $ so that f is L2-normalized.[2] In other words, where f is a (normalized) Gaussian function with variance σ2, centered at zero, and its Fourier transform is a Gaussian function with variance σ−2.

In fact, this inequality implies that:

$ \left(\int_{-\infty}^\infty (x-x_0)^2|f(x)|^2\,dx\right)\left(\int_{-\infty}^\infty(\xi-\xi_0)^2\left|\hat{f}(\xi)\right|^2\,d\xi\right)\geq \frac{1}{16\pi^2} $

for any x0, ξ0.[3]

In quantum mechanics, the momentum and position wave functions are Fourier transform pairs, to within a factor of Planck's constant. With this constant properly taken into account, the inequality above becomes the statement of the Heisenberg uncertainty principle.[4]

A stronger uncertainty principle is the Hirschman uncertainty principle, which is expressed as:

$ H\left(\left|f\right|^2\right)+H\left(\left|\hat{f}\right|^2\right)\ge \log\left(\frac{e}{2}\right) $

where H(p) is the differential entropy of the probability density function p(x):

$ H(p) = -\int_{-\infty}^\infty p(x)\log\bigl(p(x)\bigr) \, dx $

where the logarithms may be in any base that is consistent. The equality is attained for a Gaussian, as in the previous case.


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