Generally speaking, the more concentrated f (x) is, the more spread out its Fourier transform f̂ (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 f̂ (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 L^{2}-normalized.^{[2]} In other words, where f is a (normalized) Gaussian function with variance σ^{2}, 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 x_{0}, ξ_{0} ∈ ℝ.^{[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.
References
- ↑ .
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