Shannon entropy, in terms of position space wave functions, can be written as, \begin{equation} S= -\int^{\infty}_{-\infty} \vert \psi(x) \vert^2 \log \vert \psi(x) \vert^2 dx. \end{equation} In explanation of the above expression, it is written in one article that the information entropy measures the localization of a distribution.

In another article, it is explained that the position-space information entropy measures the uncertainty in the localization of the wave packet in space. So, the lower this entropy, the more concentrated the probability density $\vert \psi(x) \vert^2$ is at points, and the smaller the uncertainty, and higher the accuracy in predicting the localization of the particle.

What is meant by localization of a distribution? And what is meant by "more concentrated probability density"?

  • $\begingroup$ Which articles are you referring to? Can you provide a link or the names? $\endgroup$ Aug 12, 2021 at 17:04
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    $\begingroup$ Shannon is overkill here. This is plain distribution entropy. You should reconfigure your question after you test the definition on a Gaussian with varying width. The smaller the width of the Gaussian, the more localized the distribution is. $\endgroup$ Aug 12, 2021 at 17:20
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    $\begingroup$ @Lucas Baldo For a pure state, it is the von Neumann entropy that vanishes, not the semiclassical distribution entropy written above, which some (mis-)refer to as "Shannon entropy"... Indeed, the quantity written quantifies the narrowness of the probability distribution in coordinate space. $\endgroup$ Aug 12, 2021 at 20:06
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    $\begingroup$ @Lucas They couldn’t be more different! vN is fully quantum and hence smaller than Shannon, which has sacrificed quantum information. Don’t be fooled by the similar looking symbols… vN involves operators but Shannon is essentially classical, merely using probability densities originating in QM. $\endgroup$ Aug 13, 2021 at 2:50
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    $\begingroup$ @Rococo Yes, there is a standard hidden length normalization inside the logarithm, which shifts out to a constant additive "origin setup" for S. In phase-space distributions, it is a Planck h , which helps prodigiously with the classical limit. $\endgroup$ Aug 13, 2021 at 15:49

3 Answers 3


There is little Shannon or information theory in the quantity you are examining, which is just the 19th century definition of classical entropy of probability distributions for $f(x)=|\psi(x)|^2$. There is also little actual quantum mechanics, as all you need is the above definition of your probability distribution, without regard of where it came from, or reference to the interference quantum effects associated with ψ, etc.

As Clausius' name implies, entropy describes probability dispersal, disorder, or randomness (τροπή ~ change).

Nevertheless, information scientists have adopted/appropriated it as differential entropy for lack of information (randomness); and there is no point in splitting hairs about names; in QM uncertainty discussions the name has stuck. At the end of the day, the above-defined classical entropy S does help with QM quite a bit, after all.

A normalized probability distribution $f(x)$ may be broad or narrow, which is what the standard deviation σ (width), or variance (its square), describes. First look at a Gaussian/normal distribution, $$ f(x)= \frac{1}{\sigma\sqrt{2\pi}} e^{-(x/\sigma)^2/2} \qquad \leadsto \\ S= \ln ~(\sigma \sqrt{2\pi e}), $$ so S increases with the variance.

Small variance indicates localization, and the limit of $\sigma\to 0$ is a δ-function spike: ultra-localization; meaning no uncertainty in the position of your wave packet (but an infinite uncertainty in its corresponding momentum). (It's up to you how to interpret this infinite information/negentropy, if you are hung up on information theory.)

Conversely, huge σ (huge S) signals huge dispersal of the probability distribution, on its way to terminal delocalization. So far, so good.

But... You may easily imagine a probability distribution consisting of two Gaussians with their centers at some distance a from each other and small widths. σ will be more like this distance (ultimately a/2), rather than the smaller widths of each constituent Gaussian. (Proceed to calculate such examples.) You know this distribution is less ordinary than an unremarkable Gaussian, and has more information content, (and is more localized in space, since the probability vanishes everywhere except in the two narrow bumps a apart).

You'd like to have some measure of this different than σ ~ a/2, and to indicate your two-bump distribution is more localized than a single Gaussian with the same σ. Hopefully, you found that this S is less than the S of that single Gaussian: S is a better measure of delocalization than σ. In fact, there is a theorem.

You know random things end up in Gaussians, and for a given σ, the Gaussian is the maximum entropy distribution; the most delocalized distribution for a given σ. The normal distribution is where information goes to die.

You may have fun experimenting with comb distributions, whose teeth are Gaussians, of the same σ, and thus further lower S; this is even more extraordinary and ordered/non-random: more informative. Try the ones listed.

The takeaway is that S is a far better measure of fine detail and localization than the crude variance, and, in fact, as a result, it provides a more stringent uncertainty inequality for the uncertainty principle, entropic uncertainty, linked above.

The actual entropy of a quantum state, however, is not what you are discussing. The quantum information entropy for QM is the von Neumann entropy, whose definition in phase-space outranges this discussion. As mentioned in a comment by @Lucas, for a pure state, the von Neumann entropy actually vanishes: you have maximum information about that state, and you are in no doubt about it — except some observables which are mutually incompatible and simultaneously unknowable. But that is another, more exciting story.


What is meant by localization of a distribution? And what is meant by "more concentrated probability density"

It means what you would expect: a "concentrated probability density/distribution" is a probability distribution that is only nonzero in a "small interval" around some point, where what "small" means depends on the context. A probability density is "localised" if it is concentrated around some point.

Regarding your $S$: this is the Shannon entropy associated with the probability density obtained measuring the state in the position basis. This is a bad way to quantify how localised/concentrated the position probability density is. To see it, note that there is no notion of "closeness" of different coordinates $x$ in the expression. You can always find distributions which have completely different "concentration" features, but same values for $S$.

For example, let $\psi$ be some well-localised function. Say, something that approximates well enough a delta around $0$: $\psi(x)\sim \delta(x)$. Thinking in terms of its graph, you can imagine splitting the distribution in two halves, and move the two halves in opposite directions an arbitrary distance. You'll get a new distribution which is completely delocalised, but has the same value of $S$ (if you worry about this process breaking continuity/differentiability, don't, you can always fix this by sufficiently smoothing the edges of the graph in a way that has negligible impact on $S$).


Anything fluctuating around the square modulus. It's a local smearing of the amplitude probabily of the wave. When quantum mechanics emtangles two states, it manifests in a non local way.


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