# 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