# 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 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 $\Delta_{t}$ and $\Delta_{\omega}$ of $t$ and $\omega$ respectively, satisfy the following inequality:

 $\Delta_{t}\Delta_{\omega}\geq\frac{1}{2}$

For this formula to make sense, $\Delta_{t}$ and $\Delta_{\omega}$ must be precisely defined.

## 2 THE $L^{2}$ 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:

 $\int_{-\infty}^{\infty}\overline{U(\omega)}V(\omega)d\omega~{}=~{}2\pi\int_{-% \infty}^{\infty}\overline{u(t)}v(t)dt$

$U(\omega)$ 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:

 $\mid\mid f\mid\mid^{2}=\int_{-\infty}^{\infty}\mid f(t)\mid^{2}dt~{}~{}~{}~{}~% {}~{}~{}~{}\mid\mid F\mid\mid^{2}=\int_{-\infty}^{\infty}\mid F(\omega)\mid^{2% }d\omega$

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 $\Omega(\omega)$ for the ”random” variables $t$ and $\omega$:

 $T(t)=\frac{\mid\ f(t)\mid^{2}}{\mid\mid f\mid\mid^{2}}~{}~{}~{}~{}~{}~{}~{}~{}% \Omega(\omega)=\frac{\mid\ F(\omega)\mid^{2}}{\mid\mid F\mid\mid^{2}}$

Since the $L^{2}$ integrals of $T$ and $\Omega$ are 1, they are proper probability distributions. The mean value $t_{0}$ of $t$ is defined the usual way:

 $t_{0}=\int_{-\infty}^{\infty}tT(t)dt$

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:

 $\Delta_{t}^{2}=\int_{-\infty}^{\infty}T(t)(t-t_{0})^{2}dt~{}~{}~{}~{}~{}~{}~{}% ~{}\Delta_{\omega}^{2}=\int_{-\infty}^{\infty}\Omega(\omega)(t-t_{0})^{2}d\omega$

## 4 PROOF OF THE THEOREM

The heart of the proof is the Cauchy-Schwarz 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:

 $\int_{-\infty}^{\infty}\mid u(t)\mid^{2}dt\int_{-\infty}^{\infty}\mid v(t)\mid% ^{2}dt~{}~{}~{}\geq~{}~{}~{}\mid\int_{-\infty}^{\infty}\overline{u(t)}v(t)dt% \mid^{2}$

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:

 $\int_{-\infty}^{\infty}(t-t_{0})^{2}f(t)^{2}dt\int_{-\infty}^{\infty}f^{\prime% }(t)^{2}dt~{}~{}~{}\geq~{}~{}~{}\mid\int_{-\infty}^{\infty}(t-t_{0})f(t)f^{% \prime}(t)dt\mid^{2}$

The integral at the right hand side can be integrated by parts. Using the definition of $\mid\mid f\mid\mid$:

 $\int_{-\infty}^{\infty}(t-t_{0})^{2}f(t)^{2}dt\int_{-\infty}^{\infty}f^{\prime% }(t)^{2}dt~{}~{}~{}\geq~{}~{}~{}\frac{1}{4}\mid\mid f\mid\mid^{4}$

But $\mid\mid f\mid\mid$ and $\mid\mid F\mid\mid$ are related by a $2\pi$ factor, so:

 $\int_{-\infty}^{\infty}(t-t_{0})^{2}f(t)^{2}dt\int_{-\infty}^{\infty}f^{\prime% }(t)^{2}dt~{}~{}~{}\geq~{}~{}~{}\frac{1}{8\pi}\mid\mid f\mid\mid^{2}\mid\mid F% \mid\mid^{2}$

Applying Parseval’s identity to the second integral of the left hand side, we get:

 $\frac{1}{2\pi}\int_{-\infty}^{\infty}(t-t_{0})^{2}f(t)^{2}dt\int_{-\infty}^{% \infty}\omega^{2}F(\omega)^{2}d\omega~{}~{}~{}\geq~{}~{}~{}\frac{1}{8\pi}\mid% \mid f\mid\mid^{2}\mid\mid F\mid\mid^{2}$

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:

 $\int_{-\infty}^{\infty}(t-t_{0})^{2}T(t)dt\int_{-\infty}^{\infty}\omega^{2}F(% \omega)^{2}d\omega~{}~{}~{}\geq~{}~{}~{}\frac{1}{4}$

## 5 THE GAUSSIAN FUNCTION

The Cauchy-Schwarz 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: $f(t)=ke^{\lambda(t-t_{0})^{2}}$. $f(t)$ 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:

 $\Delta_{t}=\frac{\sigma}{\sqrt{2}}\hskip 32.0pt\Delta_{\omega}=\frac{1}{\sigma% \sqrt{2}}$

Their product is $\frac{1}{2}$, independent of $\sigma$. There is no other function with this property.

## References

• 1 Roberto Celi Time-Frequency 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 UncertaintyTheorem 2013-03-22 18:50:37 2013-03-22 18:50:37 dh2718 (16929) dh2718 (16929) 14 dh2718 (16929) Theorem msc 42C40 UncertaintyPrinciple