I have heard several pseudoscientific explanations about the Heisenberg Uncertainty Principle and find them hard to believe.

As a mathematician mainly focusing on functional analysis, I have a considerable interest in this matter. Although I still have no idea about it, I aim to truly understand it one day. For now, the question is:

Can the Heisenberg Uncertainty Principle be explained so that a non-scientist can at least get a correct idea of what it says and what not?

One of the most heard explanations is the one that says naïvely that while trying to determine the position of a particle you send energy (light) so you modify its speed, therefore making it impossible to determine the speed. If I understand well this should only be a some sort of metaphor to explain it, right?

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    $\begingroup$ The last is not meant as metaphore (I feel like its due to Heisenberg himself originally), but is also not quite the same, because it reduces the fundamental uncertainty to a practical impossibility in measurement. $\endgroup$
    – Bort
    Jan 13, 2016 at 11:20
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    $\begingroup$ The uncertainty principle is closely related to a more intuitive (at least for me) defining feature of quantum mechanics, that goes as follows. A quantum mechanical particle cannot have a definite trajectory, as opposed to what happens in classical mechanics (a trajectory is a map $t\mapsto (x(t),p(t))$ on the phase space). This is achieved requiring that the underlying mathematical theory forbid the possibility of localizing, at any time, the particle in a single point of the phase space. $\endgroup$
    – yuggib
    Jan 13, 2016 at 11:58
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    $\begingroup$ This is in turn realized by requiring that the more you are able to localize the particle in space (momentum), e.g. by measurement, the less you are able to localize it in momentum (space). Luckily, variables related mathematically by Fourier duality satisfy this uncertainty requirement automatically (in a suitable sense), and therefore they are the ones chosen to become the canonical variables of quantum mechanics. $\endgroup$
    – yuggib
    Jan 13, 2016 at 12:05
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    $\begingroup$ The big problem with this question is that my intuition is not the same as yours. In other words, there is not a unique answer that will be intuitive to everyone. Thus, there is no 'answer' to this question. $\endgroup$
    – Jon Custer
    Jan 13, 2016 at 17:31
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    $\begingroup$ Speed is movement over time. Position is location at a single instant. To accurately measure speed your test needs to take time (think of a long exposure camera). To accurately measure position you need to know the location in zero time. IOW, you need to stop time to get a perfect measurement of location. If time is stopped, you have absolutely zero information about movement over time, time is stopped! Conversely, if your measurement takes a long time, you have no information about a single instant. The more precision you have on attribute, the less you have about the other. $\endgroup$
    – Shane
    Jan 13, 2016 at 22:08

18 Answers 18


The best intuitive analogy I've heard is with classical sound waves. Consider a musical instrument playing a pure sine wave of frequency $\nu$ and amplitude $A$, and no other harmonic frequencies at all. Graphing this in frequency-amplitude space ($x$-axis=frequency, $y$=amplitude) gives you a $\delta$-function-like point function with value $y=A$ at $x=\nu$, and zero everywhere else. That represents your exact knowledge of the note's frequency.

But at what time was the note played? A pure sine wave extends from $-\infty<t<\infty$. Any attempt to play a shorter note necessarily introduces additional components/harmonics in its Fourier decomposition. And the shorter the interval $t_0<t<t_1$ you want, the broader your frequency spectrum has to become. Indeed, imagine an instantaneous sound. Neither your ear, nor any apparatus, can say anything about its frequency at all -- you'd have to sense some finite portion of the waveform to analyse its shape/components, but "instantaneous" precludes that.

So, you can't simultaneously know both a note's frequency and the time it's played, due to the Fourier conjugate nature of frequency/time. The better you know one, the worse you know the other. And, as @annav mentioned, that's analogous to the nature of conjugate quantum observables.


to address @sanchises remark about some "crude MSPaint drawings"...

For simplicity (i.e., my own simplicity generating the following "crude drawings"), I'm illustrating an almost-square wave below, rather than a sine wave. Suppose you wanted to produce a sound wave with a one-cycle duration, looking something like,

So the "tails" are zero in both directions, indicating the sound's finite duration. But if we try generating that with just two fourier components, we can't get those zero-tails. Instead, it looks like,

As you see, we can't "localize" the sound's duration with just two frequencies. To get a better approximation, four components looks like,

And that still fails to accomplish much by way of "localization". Next, eight components looks like,

And that's beginning to exhibit the behavior we're looking for. Sixteen looks like,

And I could go on. The initial illustration above was generated with 99 components, and looks pretty much like the intended square wave.


you guys coincidentally stepped into one of my little programs when mentioning drawings. See http://www.forkosh.com/onedwaveeq.html for a discussion, although not about uncertainty. To get the above illustrations, I used the following parameters in that "Solver Box" at top,


Just change the ncoefs=99 to generate the corresponding drawings above.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jan 18, 2016 at 8:56
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    $\begingroup$ So it's a fundamental property of a wave then? $\endgroup$
    – Yogi DMT
    Sep 23, 2016 at 12:32
  • $\begingroup$ Well, I'd say it doesn't mathematically emerge from a point-like particle picture alone. But "fundamental" is kind of a metaphysically loaded word I'd be hesitant to use physically. $\endgroup$
    – user89220
    Sep 25, 2016 at 3:09
  • $\begingroup$ I meant mathematically $\endgroup$
    – Yogi DMT
    Nov 6, 2016 at 23:02
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    $\begingroup$ ... observables commute - what is the physical explanation for this. And the only answer I can come up with is that they simply don't - but many operations in the everyday world don't commute (the shoe and sock donning operators, so we shouldn't be too surprised if classical measurements and their commutativity don't always hold. $\endgroup$ Jul 31, 2017 at 4:42

The explanation you have heard, extended, goes as follows: suppose I want to find the position of a particle in a box. To do so, I shine light on it, and, in a very similar way to what happens in the macroscopic world, by how the light bounces off I understand where the object is. However, the particle is so small that the momentum of a photon can push it and change its momentum. So: if I use a low-energy, large-wavelength photon, it won't change the particle's momentum much (because of low energy) but also won't tell me its position with high precision (because of large wavelength). If I want a higher precision in the position you need a short wavelength photon, which is unfortunately a high-energy photon and will change the momentum of the particle in an unpredictable way. See Compton Scattering for the physical details.

This, however, is only an example of a consequence of the uncertainty principle. Heisenberg's uncertainty relation is actually far more general and holds in principle, in the same sense as the conservation of energy is not "proven" by explaining why a certain kind of endless energy source cannot work.

A more general statement would be any kind of measurement changes the state of a system. I can only explain this in an axiomatic way, I'm personally unable to convince you based on physical arguments. But there is a sound reason for this. No physical argument based on our intuition of physics can explain quantum uncertainty, because it is fundamentally different from our intuition of physics.

To a person ready to accept this change of paradigm you can explain that the concept of state is different in qm. As someone writes in a comment, position and momentum don't simultaneously exist in qm (as, by the way, the angular momenta along different axes). Some states can have definite position, some can have definite momentum, but not both.

As you are a mathematician, I can explain to you axiomatically why this happens. In the standard theory of QM, it is usually taken as true that:

  • States are vectors in a complex Hilbert space
  • Observable quantities such as position and momentum correspond to operators in this Hilbert space. Their explicit form depends on the basis you choose, but the important thing is they don't commute in the case of $x$ and $p$.
  • States with a definite value for an observable are eigenvectors of the corresponding operator. The definite value is the corresponding eigenvalue.

If two operators don't commute, maths shows they cannot have simultaneous eigenvector bases, and therefore the two physical quantities are never well defined simultaneously.

Another way of putting this mathematically is by showing that the wavefunction (whose modulus squared is the probability of finding the particle at a certain location) and the "wavefunction" in the momentum space are Fourier transforms of each other. You can easily show that if you choose a low-variance distribution on one side, the variance increases on the other and vice versa.

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    $\begingroup$ Yes, I agree, @Martin. Maybe it's not clear from my answer, but I never liked people 'proving' HUC using that compton scattering example. I changed "example" to "consequence". $\endgroup$
    – Martino
    Jan 13, 2016 at 14:18
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    $\begingroup$ very good, multi-layered answer now, very good edits, chapeaux! $\endgroup$
    – avalancha
    Jan 13, 2016 at 15:10
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    $\begingroup$ I'm not a big fan of the scattering example, since it is much more of an experimental problem rather than a conceptual one. If, accidentally, position and momentum commuted then it would be no longer true. $\endgroup$
    – gented
    Jan 13, 2016 at 16:43
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    $\begingroup$ @GennaroTedesco I'm not a big fan either. As I've tried to understand the HUP, I've encountered tons of cases where the scattering example is presented as if it were the HUP itself, when it's just one concrete example that we can easily conceptualize because photons are like bouncing balls. Right? (I've even seen it presented like this: God may know both a particle's speed and position, but there will never be a way for us to measure both [because throwing photons at it must be the only way?]) The HUP is not just a statement about the theoretical limits on (even perfect) measuring instruments. $\endgroup$
    – Jon Coombs
    Jan 14, 2016 at 2:15
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    $\begingroup$ Thanks for the only answer that took the "As a mathematician mainly focusing on functional analysis" part into account. But two operators that don't commute can share eigenvectors, they just can't be co-diagonalizable (ie share a basis of eigenvectors). $\endgroup$
    – Evpok
    Jan 15, 2016 at 16:53

I think: Yes there's an intuitive explanation for the Uncertainty principle. The explanation is the following:

The most important thing to convince the non-scientist listener is that particles in Quantum Mechanics ARE NOT OBJECTS! This is observed in interference experiments and is a fact that we are very certain about. So they are waves. Once they wrap their heads around this idea, things become much easier to explain.

Show them this picture or similar:

enter image description here

And tell them. Electrons look like the wave you see in this picture on top. Can you tell me what's the position of this electron? The listener will fail, and will start to understand that instrumentation errors has nothing to do with it. It's all about what these subatomic particle are. Then explain to him that scientists have a way to say where the electron may act as a particle most likely (which we call the probability of finding the position of the electron). This is defined by where the wave has a higher amplitude (or even naively, where it's further from the x-axis). Now if we would like to map that kind of position and make a position for the electron, the lower picture is how it will look like.

So from this, the listener learned:

  • Electrons are waves

  • The problem is mapping waves to particles.

  • Mapping waves to particles gives uncertainty, which is given by the Heisenberg uncertainty principle.

Good luck!

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jan 18, 2016 at 8:55
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    $\begingroup$ RE: "The most important thing to convince the non-scientist listener is that particles in Quantum Mechanics ARE NOT OBJECTS". It is funny to me that your answer to "Is there an intuitive explanation of this principle?" is "Yes, first make sure they buy into this other non-intuitive principle, then the other principle follows intuitively." While accurate, it does somewhat undermine your answer. I think the idea that particles are not objects is one of the most non-intuitive ideas in modern physics for most people $\endgroup$
    – Kevin
    Oct 18, 2016 at 21:09

In my experience non scientists tend to turn quantum mechanics into metaphysics. A non scientist would not know even what a measurement error is, which is inherent in all data.

For mathematically inclined people the Fourier transform uncertainties are directly related to the HUP. Heisenberg identified h_bar as the lowest limit for pairs of conjugate variables, within a system where the probability distributions are derived from the solutions of a quantum mechanical equation. That the square of the complex conjugate of the wave function gives a probability is a postulate of quantum mechanics.

If one starts by explaining measurement errors the lay person will already get the wrong impression that the HUP is about measurement errors and will be looking for deterministic reasons for the behavior.

I believe that a minimum of mathematical sophistication is necessary and a minimum background on what physics is about, i.e. observations and measurements fitted by mathematical models.

Edit after going through the other answers:

The basic problem lies in transferring in lay terms, intuitively, the correct concept of the wave aspect of quantum mechanical entities as a probability distribution that has sinusoidal dependence on space and time.I will try to explain for a non scientist the probability aspect of the quantum mechanical framework.

A probability distribution is a function over a variable,x, and it describes how often x will appear .

The most familiar probability curve, even if not visualized, is the curve of throwing a dice.

There are six numbers, the x in our example being discrete. The probability curve against x for a large number of throws is predicted a flat line unless the dice is biased.

          1/6   - - - - - -


(number of throws)/

(total throws)


                 1 2 3 4 5 6

                number on dice

For an elementary particle and the variable in space x, the probability distribution for a "throw", i.e. a measurement, to come up with the value x is given by a solution to the quantum mechanical equation with the boundary conditions of the problem.

In the double slit one electron at a time experiment nature solves the complicated equations for us in this figure:


This figure shows both the particle nature of the electron and the wave nature of the probability.

The upper photo shows single electrons thrown against the slits. Their x ( and y) looks random, and it is a dot which in classical mechanics is considered a point particle signature. Thus the electron is called a particle because when measured/(its footprint seen on the screen) it has a point signature within the experimental errors.

The photos of gradually accumulating throws show an interference pattern for the probability of finding the electron at x . This is a demonstration that a wave nature exists in the probabilities of describing electron interactions with slits.

I could not find a photo for a single slit single electron at a time experiment. Here is what the accumulation looks like for a single slit:

single slit

Again a diffraction pattern is evident, and is a manifestation of the figure given in another answer, but a probability distribution, not a manifestation of a single electron against the x variable .

Back to the HUP.

The Heisenberg uncertainty arises as a measure of the indeterminacy introduced by the fact that the electron is not really a particle in the classical sense, with a fixed trajectory defined in all cases by classical mechanics, its trajectory is controlled by a probability distribution that may have sinusoidal variations. The HUP is inherent in the quantum mechanical equations, and is a clear shorthand mathematical description of the quantum mechanical behavior of particles in the regime where the value of h_bar is commensurate with the value of the variables measured.

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    $\begingroup$ It was hard for me to understand HUP until I saw a web video that incidentally explained that the momentum wave (speed) and matter wave (position) were Fourier duals. Since I studied to be an engineer in college, I knew enough to understand the tradeoff! $\endgroup$
    – CTMacUser
    Jan 16, 2016 at 1:26

Here's a maths-free explanation which, i think, can be understood intuitively.

A lot of people think (incorrectly) that they understand Heisenberg's Uncertainty Principle: they think that it's a measurement problem: ie, that we can't measure the properties without interacting with it, and this interaction changes the property, so we don't actually know what it's doing: we just know what it was doing before we measured it. This is true, but it's not the principle. The principle is that it is a law of nature that we cannot know, like how we cannot travel faster than light.

Thinking of the HUP as a practical problem to do with measuring things without interacting with them is like thinking that the reason we can't travel faster than light is that we can't build a powerful enough engine at the moment. This happens to be true, but it's not the real reason: the real reason is that it's a law of nature (specifically, that we would need infinite power, which is impossible).

The best illustration i've seen, which is truly mind-blowing, of the HUP is this: The atomic nucleus is positive (overall) and electrons are negative. We all know that positive and negative attract each other, right? So, why don't the electrons all just fall into the nucleus? The answer is that doing so would violate the uncertainty principle!

We (a hypothetical omniscient observer) would know the position of the electron (in the nucleus), and we would know its velocity (practically zero as it's stuck in the nucleus). Knowing this much about the two paired properties (there are other paired properties as well) is forbidden, and that's why the electron doesn't go there, regardless of who is trying to interact with it. In fact, the electron maintains a certain minimum distance from the nucleus, and this distance corresponds exactly with what the HUP predicts, based on the minimum allowed degree of uncertainty: so it's like the electron "wants" to be in the nucleus (or a lower shell), and it gets as close as the principle allows it to get.

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    $\begingroup$ I think it's a bad idea to say "law of nature" in any lay explanation. The problem is that that sounds too much like something that can be avoided by clever tricks or outright broken when noöne is looking. Instead, it's a fundamental property of the universe as we know it - just like the speed of light. $\endgroup$
    – Luaan
    Jan 14, 2016 at 12:27
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    $\begingroup$ To me, and to many people, "law of nature" and "fundamental property of the universe" mean exactly the same thing. I don't think many people would think of "laws of nature" as being like the laws of a country, that you could break if the police aren't watching. $\endgroup$ Jan 14, 2016 at 15:17
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    $\begingroup$ An illustration why this doesn't apply to the protons/neutrons might be useful (higher binding energies allowing more uncertainty in velocity?) $\endgroup$
    – Yakk
    Jan 14, 2016 at 18:52
  • $\begingroup$ @Yakk go for it $\endgroup$ Jan 15, 2016 at 13:36
  • $\begingroup$ @MaxWilliams hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c3 and hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c4 are first steps if anyone is curious. $\endgroup$
    – Yakk
    Jan 15, 2016 at 15:17

No, there is no intuitive explanation for the Heisenberg Uncertainty Principle, or most of other QM. Feynman was rumored to say

Anyone who claims to understand quantum theory is either lying or crazy.

To answer your second question, the HUP states the product of the uncertainties of two measurements on a system has a lower bound, provided those measurements are related in a special way (the most commonly seen are time/energy and position/momentum).

  • $\begingroup$ I understand quantum theory is unintuitive. $\endgroup$
    – D1X
    Jan 13, 2016 at 12:46
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    $\begingroup$ @D1X: QM and its old pal, Relativity, are notoriously counter-intuitive. Our intuition has been programmed by millions of years of experience of metre-scale objects with masses of kilos interacting on timescales of seconds. We have no way of comprehending scales on which QM or R act, unless guided by advanced mathematics. $\endgroup$ Jan 13, 2016 at 12:50
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    $\begingroup$ @OwenBoyle I agree that "Our intuition has been programmed by millions of years of experience of metre-scale objects with masses of kilos interacting on timescales of seconds." and I disagree that "We have no way of comprehending scales on which [relativity] act[s], unless guided by advanced mathematics." Even though I have a math degree, I have a non-mathematical yet still useful understanding of general relativity that I gained from Einstein's own book that he wrote for non-mathematicians. $\endgroup$ Jan 13, 2016 at 19:32
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    $\begingroup$ You know, Feynman said this in 1965. I think he'd be quite amused by people taking it verbatim as if it were word of God: "YOU SHALL NOT UNDERSTAND QUANTUM THEORY, OR I WILL SMITE THEE WITH FURY AND VENGEANCE!". It's not that quantum physics is fundamentally hard to understand - it's just that our "intuitive" models of the world are so unbelievably wrong. It's more like non-inertial mechanics - sure, your mammal's brain evolution didn't have to deal with it a lot, so it's unintuitive... but we all had to adapt when we started driving cars. $\endgroup$
    – Luaan
    Jan 14, 2016 at 12:32
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    $\begingroup$ @JonCoombs Eh, not really. If you spend twenty years of your life studying something, it better start feeling intuitive - if it doesn't, you probably haven't been studying all that well. It has nothing to do with how intuitive or not the underlying material is - if you want an example, look at any user interface in the history of the world. "Intuitive" only really means "familiar" - it has nothing to do with how simple something is. The more familiar you are with one approach, the less familiar others look in comparison - you see this all the time, and Einstein was no exception. $\endgroup$
    – Luaan
    Jan 14, 2016 at 12:37

I find it to be difficult to explain the Heisenberg uncertainty principle intuitively in one step. I find it helpful to split it into two halves. The first half explains why uncertainty like behaviors appear in wave mechanics. The second half argues why you have to consider wave mechanics when dealing with small particles.

For wave mechanics, I like to explain it using a wave which is more familiar to people: a violin string (or any other vibrating string). Pluck the violin string in the center. We'll ignore everything except the fundamental harmonic (this might have involved a particularly clever plucking scheme, or just involving hand waving to make our lives less complicated). Most people are comfortable with the idea that this wave has an amplitude, which can be determined from the maximum deflection of the string, and a phase, which is roughly "where in the oscilation it is," whether it is at an extreme position (maximum deflection), an extreme velocity (minimum deflection), or anywhere inbetween.

To make this a useful model for explaining QM, we don't get to gather any information about the plucked string except through our observation instrument: a camera. Everything we are going to learn about this wave we're going to learn by taking pictures, and looking at the results. We do get to adjust the shutter speed. Most are comfortable with the idea that a slow shutter speed induces motion blur, and a fast shutter speed creates a very sharp image.

If we take a very fast image, we can freeze the string in place. We can see exactly where the string is, but we have very little information about where the string is going. It could be on its way up, it could be on its way down. By contrast, if we take a long exposure, we can easily see the full extent of the oscilations, as it blurs together. However, we lost track of the phase information, because the string may have warbled a long distance during that image, and we don't know exactly how far.

From this we can see that amplitude and phase information share a connection. You can't know the amplitude and phase of a wave at the same time, using an observation from this camera. If you take a fast picture, you know exactly where the string is, but you don't know its phase, so you can't figure out the maximum amplitude. If you take a slow picture, you know the amplitude, but its really hard to tell what phase the string was at. You have a tradeoff.

Now there's a workaround here: take multiple pictures really fast, and use the extra information to figure out all you need to know. To make this model a good model of how quantum behaviors work, we're going to need to make an adjustment. For fast pictures, we're using a very powerful strobe, and the string is very very very light. Even the energy from the strobe is going to affect the string in unpredictable ways. Thus, you only get one good measurement. After that, the string is perturbed, and the measurements are now measuring some other modified waveform. Kind of a stretch for violin strings, but it's how it works when you string is the size of an electron!

So now we have an intuitive argument why you cannot know all the information about such waves, using discrete measurements. What is left is to explain why this is meaningful for particles. After all, particles aren't waves right?

Enter the double-slit experiments. They do something very important for this argument: they provide experimental evidence that electrons and photons do have wavelike behaviors -- their behavior is not really well modeled in these situations as pure particles. Electrons and photons are behaving differently than either the simple wave or the simple particle models suggest (go figure, they behave like electrons and protons ;-) ). They do have some wave like behaviors. And, with some handwaving of the mathematics, and some clever references to the results of the double-slit experiment, it becomes reasonable to suggest that position and momentum are coupled in a way remarkably similar to the amplitude and phase of our violin string above.

Beyond that, I tend to cheat and use an appeal to authority: if you disbelieve the results, you really should learn the mathematics needed to understand these results in an intellectual manner. You cannot disagree with the double-slit experiment, as much as you might like. It is experimental results, not theoretical ones. We have observed photons and electrons behaving in the way described.

I often treat this topic the same way I do relativity. I begin talking, and explaining. I watch their eyes glaze over, and get confused. Finally they will jump in with an expletive along the lines of, "bull----!" At that point I smile and say, "Excellent. Now we can really start the discussion."

  • $\begingroup$ I like this explanation up to the point where the measuring tool is messing with the wave's behavior. I think that is too misleading (and sadly, it's the easiest part to understand). $\endgroup$
    – Jon Coombs
    Jan 14, 2016 at 3:09
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    $\begingroup$ @JonCoombs Maybe its too easy to understand because its my crutch. Would you help me to see if this is the case or not? I was under the impression that one of the reasons you could not measure both the position and momentum of an electron using separate measurements is that the first measurement disrupts the waveform, "collapsing" if the measurement is classical, or merely perturbing it for weak measurements, so that when you take the second measurement, the position and momentum are no longer sufficiently correlated to the original position and momentum to gather any additional information. $\endgroup$
    – Cort Ammon
    Jan 14, 2016 at 4:57
  • $\begingroup$ I'm afraid I don't know enough to be of much help to you, but I think John Forkosh's answer might help: "imagine an instantaneous sound. Neither your ear, nor any apparatus, can say anything about its frequency at all -- you'd have to sense some finite portion of the waveform to analyze its shape/components, but "instantaneous" precludes that." I like that analogy because it has nothing to do with measurements affecting the sound wave. $\endgroup$
    – Jon Coombs
    Jan 14, 2016 at 21:06
  • $\begingroup$ @Jon Coombs Jon, there is no such thing as "instantaneous sound." There is a sound wave, and you can take an "instantaneous sample" from such a wave. If "instantaneous" is defined as 1 nanosecond, then you will get its amplitude and "curvature" - at that instant! So, measuring is involved. In addition, a portion of the sound energy went into (absorbed by) the measuring apparatus, so "measuring" does affect the object being measured. $\endgroup$
    – Guill
    Jan 20, 2016 at 23:53
  • $\begingroup$ @Guill Thanks, yes, that's exactly the point. I do realize that for us observers, measuring is always involved. But that really distracts from the most shocking/confusing/important claims that the HUP makes. That is, even if there were a way to measure it without affecting it, an "instantaneous sound" still would not have a pitch, so in that sense it "doesn't exist" (i.e. isn't really a "sound"). Right? $\endgroup$
    – Jon Coombs
    Jan 22, 2016 at 4:05

maybe not the kind of answer you are looking for, but from a theological perspective it is necessary so that electrons don't collapse into protons thus destroying the universe.

this is what would happen without it http://www.feynmanlectures.caltech.edu/II_01.html#Ch1-S1

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    $\begingroup$ Let's say "anthropic principle" and not "theology"... $\endgroup$
    – Martino
    Jan 14, 2016 at 11:44
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    $\begingroup$ why? to try to pretend it's not just pure theology by using "technical sounding" words?? $\endgroup$
    – Fattie
    Jan 15, 2016 at 20:49

The uncertainty principle is a mathematical effect related to Fourier duals. In normal maths everything disappears down to infinetesimals, so is (was) rarely mentioned. (IIRC its the point where Newton's difference between two points 'only just' disappeared)

Heisenberg identified that in QM, with it's fixed wave speed (radio, EM, Light, Gravity waves), there was a definite limit.

Ref: "A Friendly Guide to Wavelets", G.Kaiser, 1994, 0-8176-3711-7, p 52, footnote.

See also Ch9 regarding wave propagation and the wave-particle plurality.

  • $\begingroup$ This is a great insight. $\endgroup$
    – Fattie
    Jan 15, 2016 at 20:51

First of all: your last paragraph describes the Observer Effect and not the Heisenberg Uncertainty Principle. So that paragraph is absolutely out as any explanation.

There is an intuitve explanation for half of the phenomenon, and you already have this explanation beautifully written by user John Forkosh in his answer. In more technical language, his answer is an intuitive description of a property of the Fourier transform: that a distibution and its Fourier transform cannot both have compact support.

But the FT arises because the transformation between co-ordinates wherein observables corresponding to conjugate variables are respectively multiplication operators is needfully the Fourier transform by dint of the canonical commutation relationship (as illuminated by the Stone-von Neumann theorem).

So the question then becomes, why don't these conjugate observables commute? What is a physical explanation for the canonical commutation relationship? And the only answer I can come up with is that they simply don't. However, many if not most operations in the everyday world don't commute (the shoe and sock donning operators are an example I like to give). Most cooking recipes go terribly awry if you switch the order of the operations. So we shouldn't be too surprised if classical measurements and their commutativity don't always hold in physics.


There has been some disagreement above about the appropriate way to explain the HUP. I think that the more abstract explanation is the correct way to explain it, and that it can be illustrated with examples to make the abstraction clearer.

The classical way of thinking about the world runs something like this. There are particles and waves and fields and similar stuff. You can pick out a particular spot and say that the value of the field at that point is $F$, or you can say that a particle is at that place etc. In short, there is some set of measurable quantities that have a particular value at any particular place that can in principle be measured. And to measure some non-local quantity you would measure some number at one place, another number at some other place and then add them up, or whatever.

This is not true in quantum mechanics. Rather, it is in general the case that any particular measurable quantity does not have a single value. If you measure any particular quantity you will, in general, get a different value each time. Also, if you try to understand what is happening in an experiment such as an interference experiment, there is in general no explanation in terms of a system having a single measurable value of a particular quantity. For example, if you consider a single particle two-slit interference experiment, you will have to say that something is going through both of the slits. What you do to each slit can change the outcome of the experiment. But if you do measurements during the experiment the detector will only go off in one place at any given time. So the system does not have a single value of position.

Now, for at least some systems, you can prepare the system to have a value of some measurable quantity $X$ so that it has a probability arbitrarily close to one of having some particular value. What happens to other measurable quantities when you do this? At least some other measurable quantities change so that they have non-negligible probabilities of being in any one of a set of states.

For example, if you prepare an electron so that its position has variance $\delta x$, then the variance in its momentum $\delta p$ may increase. If you prepare a qubit so that $\sigma_z$ is sharp, then $\sigma_x$ will be unsharp.

If you want a crude way to explain what's going on then you could say that the state of the particle is like a blob of stuff where there is a limit to how small a volume it can occupy. The volume is not volume in physical space but rather a quantity defined in terms of probability distributions of some set of measurable quantities. If you squeeze the blob too tightly in one direction in this space, it will get fatter in another direction. This does not depend on whether you are poking the system or not, so to explain the HUP in terms of particles being disturbed by light shining on them is wrong.


Another way to explain it to lay persons is to first consider why we have effective laws of physics valid at the macroscopic scale in the first place. So, stripped of all its details, one should consider that there are mathematical laws that apply to some small microscopic scale. But this seems to preclude the possibility of there being simple mathematical laws that apply at a much larger scale due to the increasing complexity.

Now, lay people will be familiar with effective simple laws that are valid on large scales that are ultimately due to other laws valid at smaller scales. E.g. fluid dynamics can be described by simple effective laws while ultimately the fluid consists of molecules. If you zoom in so that the molecules are visible, there is no fluid visible that can be described by continuum dynamics.

So, what happens is that new laws emerge on larger scales, this is due to us being interested in describing what is observable in practice. As we increase the scale more and more, certain effects that in an exact mathematical description would be kept, become smaller and smaller. This then allows us to completely ignore such effects and replace the exact laws by effective laws where such effects are not present or only treated approximately.

Then usually the effective laws only become exactly true in some scaling limit where the system size or mass becomes infinitely large. One can then explain that according to quantum mechanics, momentum is defined by the wavelength of the wavefunction while to have a well define position, the wavefunction must have a finite width which precludes being able to define the wavelength.

However, if you are free to consider larger and larger scales, you can let the width of the wavefunction become larger but such that it doesn't scale as fast as your length scale, so it actually becomes smaller compared to your running length scale. But because in absolute terms the width does become larger, the wavelength and hence the momentum becomes better and better defined as well. In the infinite scaling limit, we then end up with both a well defined velocity and momentum.

This then allows us to build entire concepts that depend on particles having both a well defined velocity and momentum, which is strictly speaking impossible according to the exact laws of physics. But this should not be considered to be all that strange. We are used to dealing with analogies of this issue all the time. E.g., we don't have problems describing a lion chasing a zebra by saying that the lion is hungry and wants to eat, knowing full well that the lion is just a collection of molecules and all that is happening are interactions between these molecules.

There is no concept of hunger that can be rigorously defined at that the molecular level, this concept is an emergent phenomena that only arises when describing the system at a scale where the animal becomes visible.


Non-scientific joking answer may be like that:

The Product Release Uncertainty Principle says that you may know what your product will do or when it will be released - but not both things together.

Quick explanation: the company will never have enough "resources" to do the full testing in a certain amount of time. In one situation you can fix the release date, but your testing team won't tell what your developers have implemented correctly and what not. And in another situation you can give to the testers a full list of required product features, but you won't be able to tell your bosses the release date, because you don't know how long it will take to test all features.

Extreme case #1.

You have a product that no one has ever seen - and you say to release it right now. It's possible that you requested to develop Browser, while your developer team implemented Text Editor.

Extreme case #2.

You give to the tester complete freedom of actions - and they do test every possible combination and browsing scenario. Your product will never be released this way.


Imagine that the information that describes position and momentum is digital and of limited precision. There is a constant total precision for both of them, but you can slice it up differently. If you dedicate more bits to momentum, you get fewer bits for position and vice versa.


An intuitive explanation would require the situation to be translated to a non-quantum scale, away from the subatomic scale and into something that most people would understand.

Imagine a child is holding a balloon on a windy day. Suddenly, the wind pulls the balloon from their hand. The balloon is moved around unpredictably due to the wind blowing on it. You want to catch the balloon, but to do that, you need to know the speed (velocity) and where it is (position).

The problem is that a velocity is a measure of distance traveled over time, while a position is a measure of where the balloon is at a certain single point in time. Because of this, the more accurately you measure the position of the balloon, the less accurately you can measure the speed, because you don't have a time interval to work with. And the more accurately you measure the speed, the less accurately you can measure the position, because you don't have a single point in time to work with.

Now, imagine that the balloon is floating on a cord tied to the ground in a windless day. You should be able to measure both speed and position accurately, right? Well, no. The problem is that the balloon is still moving around in really slow motions because the sun is shining on it, and the light of the sun is slowly moving the balloon. In addition, tiny movements in the air that you cannot stop are also moving the balloon around.

The only way to avoid these 2 effects is by watching in a sealed room with no air and no light at all shining on it, not even the light that we can't see. However, if there is no light shining on the balloon, we can't see it. To see the balloon and measure where it is, we need to interact in some way, and we can't interact with the balloon without changing where it is or where it is going.

  • $\begingroup$ The problem with this answer is that it gives the impression that there are things you can do to the situation to improve on the total amount of information you can know about the balloon's velocity and position (by tying it down, reducing wind, etc.), which makes it sounds like the HUP is an issue of the quality of our measurement techniques and equipment, not the inherent property of the universe that it really is $\endgroup$
    – Kevin
    Oct 18, 2016 at 21:29
  • $\begingroup$ @KevinWells And AFAIK, that analogy still holds. There are things you can do to particles that make it so they move less (which is what tying a balloon down or reducing the wind does), like cooling them down to near absolute zero or using strong fields to keep them in place. But if they move less, that means that velocity measurements will be less accurate, while position calculations will be more accurate. $\endgroup$
    – Nzall
    Oct 18, 2016 at 22:11
  • $\begingroup$ If you hold a balloon still you can very accurately measure its position (where you are holding it) and its velocity (0 mph), so I don't see how the analogy holds $\endgroup$
    – Kevin
    Oct 19, 2016 at 15:52
  • $\begingroup$ @KevinWells and if you hold a subatomic particle still by freezing it to absolute zero, you can also very accurately measure its position and velocity in the same way. $\endgroup$
    – Nzall
    Oct 19, 2016 at 21:25

Depending on your level, the wave packet might be intuitive enough. But let me add a metaphysical justification that's intuitive on a very satisfying level: The uncertainty principle exists because the universe has a smallest scale.

If you zoom in on a digital image you get a grid. Reality doesn't have a regular grid in the same sense, but you can think about the grain in an analog medium like silver nitrade emulsion. Only the grains are not simply randomly scattered on the page, but appear centered over wherever you decide to take a peek.

  • $\begingroup$ you need to provide more to answer the OP. The relation between the 2 paragraphs is not obvious . This doesn't help to understand the uncertainty principle. However, I'm curious to understand your idea ... $\endgroup$
    – user46925
    Jan 18, 2016 at 11:53

Basic premise of uncertainty is simple,- in order to locate the particle so that you could say "hey particle is there at position $x,y,z$",- you have to interact with particle somehow,- by some field force, scattering other test particles from it, etc. But due to momentum conservation, you altogether pass some test particle/field momentum to target particle, hence in effect changing original momentum of target particle before interaction. The more you want to be sure where particle is,- the stronger interaction with a particle you have to make, the less you are sure how you affected it's kinetic energy and/or speed vector.

It's like when you capture a a fly with your hand,- so you can say "I know fly is located in my hand now". But at what speed it was flying before capture ? Can you measure that information exactly with your "hand capturing" event alone ? (Hint: capturing event will change fly initial speed too. And you have to align your hand more or less to a fly movement, which you are only able to do so because you can see a fly, because photons bumping-off a fly doesn't scatter fly off it's trajectory. But imagine that fly is microscopic, so that you don't see it, so you run and wave your hand like crazy across all room in a desperate need to find it).

Similarly, radar systems (radio, laser based, others) only works because detectable object is a lot more massive than energy of detecting wave. If for example plane or car size/mass would be comparable to the elementary particle,- any radar system will not work anymore in effect. You could only tell where car/plane is OR at what speed it was going on, but not BOTH. That's why uncertainty principle is not noticed at everyday situations,- because it is in effect at small object scales.


The Heisenberg uncertainty principle can be understood intuitively.

In order to make a complicated thing simple, you first have to get a good understanding about the accessory conditions:

-The principle of complementarity (momentum-position, energy-time etc.) with the delta-function

-The multidimensionality: Complex spaces may be doubling the number of dimensions, you may have additional parameters such as electromagnetic processes which may equally be shown as dimensions (such as the electromagnetic wave). And I even don't mention the infinite dimensions of a wave function…


Taking into account these principles of quantum mechanics, the uncertainty principle may receive a very simple visualization: Imagine at the place of a straight line a helicoid line which is following the straight line, or depending on the case some other helicoid form. The result: A point which is thought to be on the straight line is found somewhere very close to the straight line. Imagine the helicoid to be very thin, at the scale of Planck's constant, escaping to human observation. The result is, that the point has some deterministic place, but its direction with regard to the straight line is changing when the point follows the helicoid form, and also its distance may change (e.g. if the helicoid form is a 2-dimensional surface between the straight line and the helicoid line), and by consequence its position, even if it always on a clearly determined place - is considered to be random.

This model helps in many constellations to approach certain quantum phenomena. If it would be true (I have no idea!) the world would be deterministic, if not, it would be a deterministic model to help understand probabilistic quantum phenomena.

  • $\begingroup$ What? What has a helicoid to do with quantum mechanics? $\endgroup$
    – ACuriousMind
    Jan 13, 2016 at 16:10
  • 1
    $\begingroup$ A model is required for getting an intuitive concept of quantum physics. Quantum physics describes, but is not able to explain quantum physical phenomena. $\endgroup$
    – Moonraker
    Jan 13, 2016 at 16:26
  • $\begingroup$ If we're going to go freeform we may as well say the Uncertainty Principle is like driving a car. You can either look out the window to see where you are or look at the speedo to see how fast you're going. $\endgroup$ Oct 8, 2017 at 22:47

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