uncertainty theorem
The uncertainty principle, as first formulated by Heisenberg, states that the product of the standard deviations of two conjugated variables^{}, cannot be less than some minimum. This statement has been generalized to a precise mathematical theorem, in the frame of wavelet^{} theory.
1 THE UNCERTAINTY THEOREM
Let $f(t)$ be a real function of the real variable, satisfying the ${L}^{2}$ condition (see below), and $F(\omega )$ its Fourier transform^{}. The standard deviation ${\mathrm{\Delta}}_{t}$ and ${\mathrm{\Delta}}_{\omega}$ of $t$ and $\omega $ respectively, satisfy the following inequality:
$${\mathrm{\Delta}}_{t}{\mathrm{\Delta}}_{\omega}\ge \frac{1}{2}$$ 
For this formula to make sense, ${\mathrm{\Delta}}_{t}$ and ${\mathrm{\Delta}}_{\omega}$ must be precisely defined.
2 THE ${\bm{L}}^{\mathrm{\U0001d7d0}}$ CONDITION
A real function $f(t)$ of the real variable $t$ will be said to satisfy the
${L}^{2}$ condition if $f(t)$, $tf(t)$ and the derivative ${f}^{\prime}(t)$ are all in ${L}^{2}$.
If $F(\omega )$ is its Fourier transform, $i\omega F(\omega )$ is the transform of ${f}^{\prime}(t)$. All the following functions^{} belong to ${L}^{2}$:
$$f(t),{f}^{\prime}(t),tf(t),F(\omega ),\omega F(\omega )$$ 
The first three functions are just the definition and the two last ones result from Parseval’s identity^{}, recalled here in its integral form:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\overline{U(\omega )}V(\omega )\mathit{d}\omega =2\pi {\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\overline{u(t)}v(t)\mathit{d}t$$ 
$U(\omega )$ and $V(\omega )$ are the Fourier transforms of $u(t)$ and $v(t)$.
3 DEFINITIONS
$f(t)$ and $F(\omega )$ being in ${L}^{2}$, we may define their finite norms:
$${\mid \mid f\mid \mid}^{2}={\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\mid f(t)\mid}^{2}\mathit{d}t\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}}{\mid \mid F\mid \mid}^{2}={\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\mid F(\omega )\mid}^{2}\mathit{d}\omega $$ 
By Parseval’s identity, they are related:
$${\mid \mid f\mid \mid}^{2}=\frac{1}{2\pi}{\mid \mid F\mid \mid}^{2}$$ 
We are now able to define the probability distributions $T(t)$ and $\mathrm{\Omega}(\omega )$ for the ”random” variables $t$ and $\omega $:
$$T(t)=\frac{{\mid f(t)\mid}^{2}}{{\mid \mid f\mid \mid}^{2}}\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}}\mathrm{\Omega}(\omega )=\frac{{\mid F(\omega )\mid}^{2}}{{\mid \mid F\mid \mid}^{2}}$$ 
Since the ${L}^{2}$ integrals of $T$ and $\mathrm{\Omega}$ are 1, they are proper probability distributions. The mean value ${t}_{0}$ of $t$ is defined the usual way:
$${t}_{0}={\int}_{\mathrm{\infty}}^{\mathrm{\infty}}tT(t)\mathit{d}t$$ 
Note that $\omega $’s mean value is always 0 because $f(t)$ is a real function. Finally, we have the standard deviations for the uncertainty theorem:
$${\mathrm{\Delta}}_{t}^{2}={\int}_{\mathrm{\infty}}^{\mathrm{\infty}}T(t){(t{t}_{0})}^{2}\mathit{d}t\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}}{\mathrm{\Delta}}_{\omega}^{2}={\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\mathrm{\Omega}(\omega ){(t{t}_{0})}^{2}\mathit{d}\omega $$ 
4 PROOF OF THE THEOREM
The heart of the proof is the CauchySchwarz inequality in the ${L}^{2}$ Hilbert space^{}: the product of the norms of two functions $u(t)$ and $v(t)$ is greater than, or equal to, the norm of their scalar product^{}:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\mid u(t)\mid}^{2}\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\mid v(t)\mid}^{2}\mathit{d}t\ge {\mid {\int}_{\mathrm{\infty}}^{\mathrm{\infty}}\overline{u(t)}v(t)\mathit{d}t\mid}^{2}$$ 
Equality occurs if, and only if, one of the functions is proportional to the other. For the two functions $u(t)=(t{t}_{0})f(t)$ and $v(t)={f}^{\prime}(t)$, we have therefore:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{(t{t}_{0})}^{2}f{(t)}^{2}\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{f}^{\prime}{(t)}^{2}\mathit{d}t\ge {\mid {\int}_{\mathrm{\infty}}^{\mathrm{\infty}}(t{t}_{0})f(t){f}^{\prime}(t)\mathit{d}t\mid}^{2}$$ 
The integral at the right hand side can be integrated by parts. Using the definition of $\mid \mid f\mid \mid $:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{(t{t}_{0})}^{2}f{(t)}^{2}\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{f}^{\prime}{(t)}^{2}\mathit{d}t\ge \frac{1}{4}{\mid \mid f\mid \mid}^{4}$$ 
But $\mid \mid f\mid \mid $ and $\mid \mid F\mid \mid $ are related by a $2\pi $ factor, so:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{(t{t}_{0})}^{2}f{(t)}^{2}\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{f}^{\prime}{(t)}^{2}\mathit{d}t\ge \frac{1}{8\pi}{\mid \mid f\mid \mid}^{2}{\mid \mid F\mid \mid}^{2}$$ 
Applying Parseval’s identity to the second integral of the left hand side, we get:
$$\frac{1}{2\pi}{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{(t{t}_{0})}^{2}f{(t)}^{2}\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\omega}^{2}F{(\omega )}^{2}\mathit{d}\omega \ge \frac{1}{8\pi}{\mid \mid f\mid \mid}^{2}{\mid \mid F\mid \mid}^{2}$$ 
We have used the fact that the Fourier transform of ${f}^{\prime}(t)$ if $i\omega F(\omega )$.
Now, dividing both sides by the norms, and simplifying by the $2\pi $ factor, we get exactly the uncertainty theorem:
$${\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{(t{t}_{0})}^{2}T(t)\mathit{d}t{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{\omega}^{2}F{(\omega )}^{2}\mathit{d}\omega \ge \frac{1}{4}$$ 
5 THE GAUSSIAN FUNCTION
The CauchySchwarz inequality becomes an equality if, and only if, one of the functions is proportional to the other. In our case, this condition is expressed by ${f}^{\prime}(t)=\lambda (t{t}_{0})f(t)$ where $\lambda $ is a constant. This differential equation is readily solved: $f(t)=k{e}^{\lambda {(t{t}_{0})}^{2}}$. $f(t)$ must be in ${L}^{2}$ so that $\lambda $ must be negative. Defining $\lambda =\frac{1}{2{\sigma}^{2}}$, we get the gaussian function in its traditional form:
$$f(t)={e}^{\frac{{(t{t}_{0})}^{2}}{2{\sigma}^{2}}}$$ 
The constant $k$ has been omitted because it cancels anyway in the probability distributions. The standard deviations are easily computed from their definitions:
$${\mathrm{\Delta}}_{t}=\frac{\sigma}{\sqrt{2}}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\u2006}{\mathrm{\Delta}}_{\omega}=\frac{1}{\sigma \sqrt{2}}$$ 
Their product is $\frac{1}{2}$, independent of $\sigma $. There is no other function with this property.
References

1
Roberto Celi TimeFrequency visualization of helicopter noise
http://celi.umd.edu/Jour/NoisePaperColor.pdf
Despite its frightening title, this paper is mostly theoretical and it is the only place where I saw the uncertainty theorem clearly stated. 
2
Robi Polikar The wavelet tutorial
http://users.rowan.edu/ polikar/wavelets/wttutorial.html
Title  uncertainty theorem 

Canonical name  UncertaintyTheorem 
Date of creation  20130322 18:50:37 
Last modified on  20130322 18:50:37 
Owner  dh2718 (16929) 
Last modified by  dh2718 (16929) 
Numerical id  14 
Author  dh2718 (16929) 
Entry type  Theorem 
Classification  msc 42C40 
Related topic  UncertaintyPrinciple 