We know that Heisenberg uncertainty Principle states that $$\Delta f \Delta t \geq \frac{1}{4 \pi}.$$

But (in many case for Morlet wavelet) I have seen that they changed the inequality to an equality. Now my question is when are we allowed to change the inequality to an equality: $$\Delta f \Delta t = \frac{1}{4 \pi} $$ why =

  • $\begingroup$ it seems very interesting $\endgroup$
    – user350
    Apr 13, 2014 at 17:44
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    $\begingroup$ as i know it is equal if gaussian distribution is optimal shape ,please see this book The Illustrated Wavelet Transform Handbook: Introductory Theory and Applications in Science, Engineering, Medicine and Finance $\endgroup$
    – user350
    Apr 13, 2014 at 17:46
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    $\begingroup$ the link is broken mate, would you either email the book or send another link please? my email: <[email protected]> thanks @datodatuashvili $\endgroup$
    – SAH
    Apr 13, 2014 at 18:16

3 Answers 3


It is important to define the time and frequency widths $\Delta_t$ and $\Delta_{\omega}$ of a signal before discussing any special forms of the uncertainty principle. There is no unique definition of these quantities. With appropriate definitions it can be shown that only the Gaussian signal satisfies the uncertainty principle with equality.

Consider a signal $f(t)$ with Fourier transform $F(\omega)$ satisfying

$$\int_{-\infty}^{\infty}f^2(t)dt=1\quad\textrm{(unit energy)}\\ \int_{-\infty}^{\infty}t|f(t)|^2dt=0\quad\textrm{(centered around }t=0)\\ \int_{-\infty}^{\infty}\omega|F(\omega)|^2d\omega=0\quad\textrm{(centered around }\omega=0)$$

None of these conditions is actually a restriction. They can all be satisfied (for signals with finite energy) by appropriate scaling, translation and modulation.

If we now define time and frequency widths as follows

$$\Delta_t^2=\int_{-\infty}^{\infty}t^2|f(t)|^2dt\\ \Delta_{\omega}^2=\int_{-\infty}^{\infty}\omega^2|F(\omega)|^2d\omega$$

then the uncertainty principle states that


(if $f(t)$ vanishes faster than $1/\sqrt{t}$ for $t\rightarrow\pm\infty$)

where the inequality is satisfied with equality for the Gaussian signal

$$f(t)=\sqrt{\frac{\alpha}{\pi}}e^{-\alpha t^2}\tag{2.6.3}$$

The equation numbers above correspond to the proof below which is from Wavelets and Subband Coding by Vetterli and Kovacevic (p.80):

enter image description here

  • $\begingroup$ thanks for the math, I will try to understand it. @matt-l $\endgroup$
    – SAH
    Apr 13, 2014 at 19:06
  • $\begingroup$ @Matt L.: Why do you define the time and frequency widths with a squared weightfactor? I saw in school that the ∆t and ∆w the variances are. Variances of distributions are with a linear weightfactor? What is this? So does this mean that this uncertainty principle does not talk about the variances of a function and the variance of its spectrum, but something else? $\endgroup$ Jan 15, 2015 at 20:05
  • $\begingroup$ @MartijnCourteaux: This is just one possible way of defining the width of a signal. When applied to a time function, it's often called RMS duration, and it's simply the second moment of $|f(t)|^2$. $\endgroup$
    – Matt L.
    Jan 16, 2015 at 6:10
  • $\begingroup$ Is it possible to state mathematically a Heisenberg uncertainty principle that involves the second moment of $f(t)$? I can understand that Heisenberg used $|f(x)|^2$, because that is the probability of a particlewavefunction. But, I'd like to know the Heisenberg principle in the context of signal processing. $\endgroup$ Jan 16, 2015 at 8:10
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    $\begingroup$ @MartijnCourteaux: This is the Uncertainty principle in the context of signal processing. The second moment of $f(t)$ does not have an interpretation as duration, because $f(t)$ can be positive and negative. Imagine an odd signal $f(t)$. Its second moment $\int_{-\infty}^{\infty}t^2f(t)dt$ is always zero (if the integral converges). $\endgroup$
    – Matt L.
    Jan 16, 2015 at 8:34

I cannot give you all the theory behind this (as it literally fills books), but it turns out that Heisenberg becomes an exact equality for precisely this family of signals:

$$ s_{t_0,\omega_0,\sigma,\phi,\gamma}(t) = \exp\left(-\left(\frac{t-t_0}{\sigma}\right)^2 + i \left(\phi + \omega_0 (t-t_0) + \gamma (t-t_0)^2\right)\right) $$

where all parameters are real numbers. This family is generated by quadratic symplectomorphisms in time-frequency from a single Gabor atom. These symplectomorphisms preserve the Heisenberg uncertainty relation.

Edit: Let me make this more precise and also in fact more correct. The signals I gave above minimise the time-frequency area, but not the time-frequency uncertainty product. If you want minimal $\Delta F \cdot \Delta T$ then $\gamma$ from above must vanish.

The notion of time frequency area can however be generalised to measure the area of shapes that are not aligned with the time and frequency axis. That means instead of the uncertainty product between F and T we measure the minimal uncertainty product of any two conjugate variables spanned by F and T. I'll spare you the details, but for this definition of time-frequency area the family of signals gives you the minimum.

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    $\begingroup$ Isn't it the Gabuor fijlter fuonctiuons?` $\endgroup$
    – Jean-Yves
    Apr 13, 2014 at 18:21
  • $\begingroup$ One reason why it "fills books" is that the many conditions required for the equality are precisely defined and limited (oft beyond any usefulness in any other context, such as the real world). $\endgroup$
    – hotpaw2
    Apr 13, 2014 at 18:52
  • $\begingroup$ The original context of the Heisenberg uncertainty principle was physics, specifically quantum mechanics where the conjugate variables in question are position and momentum. It is not confined to time/frequency analysis. $\endgroup$
    – user2718
    Apr 15, 2014 at 23:25
  • $\begingroup$ @BZ, you're preaching to the choir here. I'm a mathematical quantum physicist. However I don't quite see the point of your comment here or that in your own answer. $\endgroup$
    – Jazzmaniac
    Apr 16, 2014 at 7:30

The uncertainty principle sets up a theoretical bound for resolution, so it is never written as an equality.

The equality relationships you are encountering are for for a specific analysis context and analysis implementation. In this case the context is signal analysis so time/frequency are the conjugate variables of interest, and the implementation is the specific wavelet in use.

The equality relationship provides a way of comparing resolutions across different analysis implementations. Care must be taken when interpreting these relationships because the definition of resolution shouldn't, but may vary.

An equality relationship is appropriate once you have defined two things: 1) the mathematical meaning of resolution. 2) the method of analysis (in this case, choice of wavelet).

  • $\begingroup$ If you dig deeper then Heisenberg's principle becomes much more than a statement about resolution. It's deeply linked with time frequency geometry in a mathematical structure called symplectic non-commutative geometry. It delivers an information theoretic measure for time-frequency information and becomes precisely integrally quantised. You can even use it to generalise the Shannon theorem for reconstruction of arbitrary TF-regions. $\endgroup$
    – Jazzmaniac
    Apr 13, 2014 at 20:31
  • $\begingroup$ In quantum mechanics, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle known as complementary variables, such as position x and momentum p, can be known simultaneously. For instance, in 1927, Werner Heisenberg stated that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. [Wikipedia - but I learned this in Physics and visited it again in analysis classes] $\endgroup$
    – user2718
    Apr 15, 2014 at 23:21

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