I agree with the fact that the principle points out to the inaccuracy in the measurement of the two quantities of the particles (momentum and position).

But measurements apart, does it explain anything about how nature works, in general? As in, I think the particle would have some exact value of momentum at that point in space (if not, please explain why).

So why not just tell that 'okay it does possess some momentum at that position, but I can't tell what that exact value is'?

Edit: I understood that the principle points out at nature as a whole, in general, and does not just point out at measurements

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Mar 6, 2020 at 10:11

11 Answers 11


Heisenberg uncertainty is not a measurement effect - it's a fundamental property of objects in the physical universe.

Historically, the uncertainty principle has been confused with a related effect in physics, called the observer effect, which notes that measurements of certain systems cannot be made without affecting the system, that is, without changing something in a system. Heisenberg utilized such an observer effect at the quantum level (see below) as a physical "explanation" of quantum uncertainty. It has since become clearer, however, that the uncertainty principle is inherent in the properties of all wave-like systems, and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects. Thus, the uncertainty principle actually states a fundamental property of quantum systems and is not a statement about the observational success of current technology.

(Emphasis mine)

Therefore you cannot say "it does possess some momentum, I just don't know what it is". If it did possess that, it would be a so-called hidden variable, most versions of which have been excluded by experiment.


Although I agree with Allure's answer (obviously, otherwise I would have to reject Bell's theorem!), at the risk of going off topic (in which case downvotes will let me know), I just wanted to make a comment on physics, models, observations and the operational approach, particularly about this statement in your question:

But measurements apart, does it explain anything about how nature works, in general?

Remember that the goal of physics is to make models which explain observations. No matter how hard you think about how the universe actually works, whatever that means, that is a question that's better left to philosophers. The only things we can talk about is stuff we can measure, physically, in a lab.

And to an extent, this is not a bad way to define "how the universe works": this is the operational approach. To put it in a harsh way, if I can't distinguish two models with experiments, they're the same, and none is better or closer to the truth than the other. In a sense, they're both the truth, as the true workings of the universe are ill defined unless we reference experiments. There is no a priori way to describe the universe as we are fundamentally observers.

Let's now get back to the uncertainty principle. People will tell you something along the lines of:

Position and momentum of a particle truly aren't defined simultaneously because $\hat x$ and $\hat p$ do not commute, and noncommuting observables share no eigenstates.

this is true, but when you hear this keep in mind that $\hat x$ and $\hat p$ are nothing but a model for our observations. At its core, the uncertainty principle is only about measurements! We can build a model in which the statistics of our measurements are calculated in a different way, using something called "hidden variables" that sit closer to our intuitive understanding of how the universe should work, but it turns out that these two models are distinguishable experimentally, and John Bell proved so. So people set out and did the experiment, and the uncertainty principle won. But keep in mind that hidden variables theories still talk about a hidden variables influencing measurement statistics.

In this light, what the uncertainty principle and Bell's theorem tell us is that experimentally we can never know both the position and momentum of a particle exactly at the same time, and there isn't anything that, if we could measure it, would help us gather this knowledge (a hidden variable).

Whether or not this means that the particle doesn't really have a position or a momentum, or even if the particle exists at all in any way that our human mind can conceive, is a question that according to the operationalist is outside the domain of physics.

  • $\begingroup$ Does the downvoter have an issue with this being a bit off topic or with the argument? I'm actually interested in hearing arguments against this view $\endgroup$ Mar 5, 2020 at 14:37
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    $\begingroup$ You've mentioned about trying to escape from understanding about how 'the universe actually works'. Please note that the word 'physics' comes from 'phusis' which is Greek for nature. And in my opinion, it should describe how nature works, in which case it can describe about the natural state of the particle, even without any observation. Hence, I just wanted to know if the moving electron possesses a definite momentum and a definite position in it's natural state. So, I do think that it should be included in the domain of physics. $\endgroup$ Mar 6, 2020 at 10:28
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    $\begingroup$ @SwaroopJoshi operationalism is a philosophical stance, so it's natural to disagree. My answer to your comment is that I've yet to see a satisfying definition of "natural state of a particle" or any property of a physical system that does not involve measurement, and I believe such a definition is impossible to give as all the knowledge we can gather about the universe, we gather through measurement. In addition, I don't believe the name a discipline was given thousands of years ago has any relevance to an argument about it, and I disagree that an operationalist view doesn't talk about nature. $\endgroup$ Mar 6, 2020 at 10:53
  • $\begingroup$ maybe you're defining nature by what you, through the tool experience. But I just believe that we may not be able to experience everything through the tool. It's your's or the choice of the person/group who is in charge on this planet to choose to exclude the actual reality from the domain of physics. I won't talk much about it, then. $\endgroup$ Mar 6, 2020 at 11:11
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    $\begingroup$ the first part of your comment is a good argument, but if there was some parts of the "true workings of the universe" that wasn't accessible through any experiments we humans could hypothetically conceive, what would be the point of building theories around it? We could never verify it, and besides, if we can't measure it, it means that it has no effect on the world that we can detect, and it's hence maybe an interesting philosophical point of discussion, but not science. $\endgroup$ Mar 6, 2020 at 11:30

In order to understand the Heisenberg Uncertainty Principle it is very helpful if you are able to first understand why it is that any sound having a finite duration in time cannot be a pure note, that is, a single frequency. This is a statement about classical physics, and it is not about any measurement precision. It is simply that when you do the Fourier analysis you find that anything with finite duration $\Delta t$ has a spread of frequency components $\Delta \omega$ satisfying $$ \Delta \omega \Delta t > 1. $$ This follows from the very definitions of time and frequency.

The new thing brought in by quantum theory is that a similar statement can be made about position and momentum. It tells us not about a limit to our ability to observe or measure, but simply that there is no such thing as a state having well-defined position and momentum together. Such a state would contradict what position and momentum are, says quantum mechanics.

One use of the Principle is that it gives a handy way to estimate ground state energies. Suppose there is a potential well having some form $V(x)$. The ground state will have some standard deviation $\Delta x$. The Uncertainty Principle then tells us that the momentum has standard deviation $\Delta p \ge \hbar / 2 \Delta x$. If the average momentum is zero (which it will be for the ground state of a static potential well) then $\langle p^2 \rangle = \Delta p^2$ and therefore the kinetic energy satisfies $$ \langle {\rm k.e.} \rangle \ge \frac{1}{2m} \left(\frac{\hbar}{2 \Delta x} \right)^2 $$ hence the total energy is at least $$ E \ge \frac{\hbar^2}{8 m} \frac{1}{\Delta x^2} + \langle V \rangle $$ where $\langle V \rangle$ can be estimated as the average potential energy over a region of width $\Delta x$. We then have a simple function of $\Delta x$ which can be minimised to find an estimate of the ground state energy.

Thus, Heisenberg Uncertainty offers the following general statement about ground states: Everything sinks to as close to the bottom of its potential well as it can, until the kinetic energy it then has to have (owing to Heisenberg Uncertainty) balances any further possible reductions of potential energy.

Another useful general observation is that as you cool things they spread into a wider position uncertainty. This is because cold means small kinetic energy which implies small absolute size of momentum, which in turn means the momentum must be well-defined, hence the position is not. Therefore cold things become more extended and wave-like.


I would like to add to Allure's answer, that once quantum mechanics was developed as a complete theory of the underlying framework of nature, i.e. has a mathematical expression, it became clear that the Heisenberg uncertainty principle is an envelope of the behavior of quantum mechanical operators.

By a mathematical expression I include the higher level quantum field theory, which allowed calculating interactions at the level of quantum particles. At the moment this physics theory is validated and is the mainstream physics one for particle physics.

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    $\begingroup$ I don't agree that QM is "a complete theory of the underlying framework of nature", except in the strict mathematical sense that it has a mathematically set of axioms. I would argue that the fact that there are things in nature, that cannot be explained by QM says that it isn't a complete theory. $\endgroup$
    – j4nd3r53n
    Mar 5, 2020 at 14:55
  • $\begingroup$ The accuracy of this answer really heavily depends on what you include in the umbrella term "quantum mechanics". For any definition of "quantum mechanics" that doesn't include quantum field theory, this answer is wrong. $\endgroup$ Mar 5, 2020 at 15:57
  • $\begingroup$ @j4nd3r53n I mean it is assumed to be by mainstream physics , at the level of the hbar dimensionsions, . Can you give an example of the things in nature that cannot be modeled by QM (except gravity which is in the research stage). $\endgroup$
    – anna v
    Mar 5, 2020 at 17:40
  • $\begingroup$ @probably_someone afaik QFT is included in what is meant by quantum mechanics . $\endgroup$
    – anna v
    Mar 5, 2020 at 17:43
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    $\begingroup$ @probably_someone I have added to remove the ambiguity $\endgroup$
    – anna v
    Mar 6, 2020 at 5:39

Yes, the uncertainty principle does tell us something about nature. It helps us understand the fact that a quantum state does not have a well defined value for an observable property unless the state is an eigenstate of the corresponding operator. In particular it says that, for any observables $\hat A$ and $\hat B$ $$\sigma_A^2 \sigma_B^2 \ge \left (\frac{\langle [\hat A, \hat B] \rangle}{2i} \right )^2$$ where $\sigma_A$ and $\sigma_B$ correspond to how precisely the values of $A$ and $B$ are defined, roughly the widths of the wavefunctions in the corresponding bases. From this it follows that $A$ and $B$ can only be simultaneously well defined when $\langle [\hat A, \hat B] \rangle = 0$. For the case of position and momentum, we can say that a particle in the state $| \psi \rangle$ only has a well defined position $x$ if $$\hat x | \psi \rangle = x | \psi \rangle$$ where $\hat x$ is the position operator. Since the position $\hat x$ and momentum $\hat p$ operators do not commute $$[\hat x, \hat p] = \hat x \hat p - \hat p \hat x = i \hbar \ne 0$$ there cannot be a simultaneous eigenstate of both operators, meaning that there cannot be a state that has both a well defined position and momentum. This particular fact is not a direct consequence of the uncertainty principle, but can still be useful in understanding how it applies here. For example, if we put a particle in a position eigenstate, then its position is well defined, so $\sigma_x = 0$. The uncertainty principle then tells us that $\sigma_p$ is infinite, which is a much stronger notion than just saying that $p$ is not well defined, as we said before.

In general, the key takeaway from the uncertainty principle is that it gives us a bound on how precisely two observable properties can be defined, not just how precisely they can be known.

  • $\begingroup$ It tells us that a quantum state does not have a well defined value for an observable property unless the state is an eigenstate of the corresponding operator This isn't an uncertainty principle. This is just true for quantum systems in general. $\endgroup$ Mar 5, 2020 at 0:50
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    $\begingroup$ @AaronStevens yes, I'm not saying that's what the uncertainty principle is, rather that it is how we make sense of the significance of the uncertainty principle. $\endgroup$
    – Sandejo
    Mar 5, 2020 at 3:22
  • $\begingroup$ That doesn't make sense. The part I quoted doesn't arise from uncertainty principles either. $\endgroup$ Mar 5, 2020 at 13:04
  • $\begingroup$ @AaronStevens It is not a direct consequence of the uncertainty principle, but it is a useful what way of understanding what the uncertainty principle physically represents, without getting into too much math. $\endgroup$
    – Sandejo
    Mar 5, 2020 at 13:51
  • $\begingroup$ It's still misleading though. The fact that quantum states are not in definite states until measurement has nothing to do with an uncertainty principle. Your statement deals with measurements of single observables, whereas uncertainty principles relates the spread of measurements of two observables and sometimes even depends on the specific quantum system. In any case, uncertainty principles do no tell us that a quantum state does not have a well defined value for an observable property unless the state is an eigenstate of the corresponding operator. It's just a postulate of QM. $\endgroup$ Mar 5, 2020 at 14:04

From vector algebra we know that $$ |\vec{u}|\,|\vec{v}| \geqslant \vec{u} \cdot \vec{v}$$ This can be generalized for inner products : $$ \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle \geqslant |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2} $$ That general principle is known as Cauchy–Schwarz inequality in mathematics.

Using that principle and wave functions for position and momentum $\psi(x), \varphi(p)$ - Heisenberg uncertainty principle can be deduced, by substituting them into Cauchy–Schwarz inequality : $$ \langle x\cdot \psi (x)\mid x\cdot \psi (x)\rangle \cdot \langle p\cdot \varphi (p)\mid p\cdot \varphi (p)\rangle \geq |\langle x\cdot \psi (x)\mid p\cdot \varphi (p)\rangle |^{2}~ $$

This reduces to :

$$ \sigma _{x}^{2}\sigma _{p}^{2} \geq |\langle x\cdot \psi (x)\mid p\cdot \varphi (p)\rangle |^{2}~ $$

Solving this inequality further produces famous Heisenberg uncertainty principle. Exact derivation of it can be checked here (Proof of the Kennard inequality using wave mechanics).

So root cause of Heisenberg uncertainty principle is that particles behaves like waves and in general you can't very well localize a wave. That's why point-like particle model fails in reality and QM was born (hence, the wave function is cornerstone of QM).

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    $\begingroup$ Why the down-vote ? $\endgroup$ Mar 5, 2020 at 9:04
  • $\begingroup$ Unclear. I see this as both 1) correct, and 2) a more technical version of Andrew Steane's mildly successful answer, which came later. It is a bit opaque to anyone without a knowledge of vector algebra, though, which probably includes the downvoters. That is, it could benefit from some elaboration at a more accessible level. $\endgroup$
    – hobbs
    Mar 6, 2020 at 6:27
  • $\begingroup$ @hobbs At this point I can only agree with Feynman and others that physics is just about "Shut up and calculate!". The more complex and more modern Physics framework is - the harder it will be to explain it in down-to-earth analogies. It is highly questionable if these analogies shows anything useful at all. In other cases they can pinpoint in wrong direction. All in all - to be able to truly grasp Physics - you need to grasp underlying math methods first. No other alternatives exist. $\endgroup$ Mar 6, 2020 at 10:01
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    $\begingroup$ @AgniusVasiliauskas fun fact, the one that said "shut up and calculate" was probably Mermin, not Feynman. After all, if you think about it, Feynman was always more about physical intuition than he was about trusting math. $\endgroup$ Mar 6, 2020 at 10:04
  • $\begingroup$ @user2723984 That's why I added "others" and I didn't say that this quote belongs to Feynman, did I ? Agree, physical intuition is important, but sorry - I haven't seen any Physicist who would have great physical intuition and at the same time bad understanding of mathematics. And btw, most greatest Physicists in history pushes boundaries in math too ! For example Newton invented calculus and Einstein pushed Tensor algebra and Riemannian geometry into Physics. So... $\endgroup$ Mar 6, 2020 at 10:16

The HUP is what tells us that the world is fundamentally QM in nature. All the elementary particles that you see defined in the SM are behaving in a way that will obey this simple rule, and it is not just about momentum and position (though that is a pair of observables or physical quantities that is commonly used in examples), but is true for any pair of observables.

In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities[1] asserting a fundamental limit to the precision with which the values for certain pairs of physical quantities of a particle, known as complementary variables or canonically conjugate variables such as position x and momentum p, can be predicted from initial conditions, or, depending on interpretation, to what extent such conjugate properties maintain their approximate meaning, as the mathematical framework of quantum physics does not support the notion of simultaneously well-defined conjugate properties expressed by a single value. The uncertainty principle implies that it is in general not possible to predict the value of a quantity with arbitrary certainty, even if all initial conditions are specified.


The HUP is important because it tells us that the error is not in our measuring devices, but that nature is weird (when looking at it with a classical view).

  • $\begingroup$ Why the downvote? $\endgroup$ Mar 6, 2020 at 4:51
  • $\begingroup$ no idea but I gave you an upvote to compensate the mean downvote :) $\endgroup$
    – undefined
    Mar 6, 2020 at 8:55
  • $\begingroup$ @undefined thank you so much $\endgroup$ Mar 6, 2020 at 22:54
  • $\begingroup$ Actually the quote you posted is misleading as the HUR is not limited in its application to conjugate variables. The derivation does not involve any assumption about the nature of the commutation relations either. Moreover, this has nothing to do with the ability to predict the value of a quantity, but rather is linked to the product of variances of joint measurements, i.e. necessarily involves the spread in outcomes of two variables measured by the same setup. $\endgroup$ Mar 8, 2020 at 5:58

The uncertainty principle is an unfortunate name. There are two things at play: a very certain relationship between the wavefunction in position space and the wavefunction in momentum space, and our inability to measure either well.

Unlike the classical world, position and momentum are one and the same for quantum particles; if we know how a particle is spread over space, we also know how it will be spread over space in a few moments (which is the entire point of the Schrödinger equation). Were we to know the exact position of a particle, that is to say, its amplitude for each point in space, then we would apply the Fourier transform to go from the position space into momentum space and we would also know the momenta exactly. There’s no uncertainty here. However, there’s a mathematical fact about the Fourier transform: if the original function has a sharp peak, the transformed function will be more evenly spread, and vice versa. So if a particle is localized, that is, its amplitude for being outside a certain tiny region of space is very small, then it will have a significant amplitude for a big chunk of momentum space.

At this point we get a “spread principle”, devoid of uncertainties, that merely states that the position and the momentum wavefunctions can’t both have peaks around which the functions quickly vanish; instead, one or both functions must be spread over space to a certain extent. One way to formalize this is the Kennard inequality.

However, we can’t get our macroscopic instruments to determine the entire wavefunction of a microscopic thing, it’s likely impossible. We can try to measure coordinates and momenta, and the numbers we get give us but a glimpse at the underlying reality, which is immensely more complex than two 3D vectors. The probabilistic relationship between what we measure and what the reality is follows some rules that we think we understand, but what that tells us about nature is an open question that the various interpretations of QM are trying to answer, using vastly different approaches.

Thus when we take the quantum relationship between position and momentum wavefunctions, which is exact, and apply our flawed measurement techniques to both, we get an inexact relationship between the measured values which we call the uncertainty principle.

  • $\begingroup$ position and momentum are one and the same for quantum particle is simply false. Position and momentum are NOT the same: for starters the quantities don’t have the same units. $\endgroup$ Mar 8, 2020 at 6:04
  • $\begingroup$ @ZeroTheHero When either can be derived from the other using an entirely deterministic transformation, it’s pretty safe to say the two are merely two different ways to look at the same thing. $\endgroup$ Mar 8, 2020 at 9:15

Look at a neutron star. The particles are under so much compression that all position locations will be occupied. Since we don't see matter more dense than this we assume that the position locations approach maximal definition.

This constraint means, according to the Heisenberg Uncertainty Principle that the momenta of the neutrons must be highly undefined.

Basically speaking, the denser the neutron matter becomes the more momentum space we get. As more mass is added the radius of the star decreases but the momentum space increases.

Once a critical mass is reached the radius of the matter in position decreases to its Schwarzschild radius and we can no longer speak about its position or momentum.

Nature's mystery is cloaked by an Event Horizon.

I like to give this example because it shows quantum effects on a stellar scale and challenges our intuition.

I don't think we can understand fully the Heisenberg Uncertainty Principle though because our brain is too large.

  • $\begingroup$ What does brain size correlate with the inability to understand something? Could you clarify your last statement? $\endgroup$ Mar 6, 2020 at 13:00
  • $\begingroup$ @Mindwin - I guess it sounds silly the way I wrote it. I mean something more along the lines of ... there are limits to our comprehension because of how the brain parts combine to work as an efficient unit. $\endgroup$
    – Pagoda
    Jun 18, 2020 at 18:24

There are a lot of large good answers here already, but something that is concerning to me in the existing answers is a lack of accessibility, due to introducing unnecessary complexity. The "Heisenberg Uncertainty Principle" is a property of ALL waves. Sinusoidal (plane) waves have a well defined wavelength $\lambda$, and thus wavenumber $k=\frac{2\pi}{\lambda}$. However, a sine wave cannot said to have any particular location in space, it has infinite spatial extent. A property of the Fourier Transform, which is what you use to go between the $x$ (spatial) version of the wave and the $k$ (frequency) version of the wave, is that the product of the extent of the wave in $x$ and in $k$ is guaranteed to be greater than or equal to some minimum value. If you measure extent using standard deviation (as is typical in quantum mechanics), then you have $\sigma_x \sigma_k \geq \frac{1}{2}$, where $\sigma_{\text{whatever}}$ represents the standard deviation.

This should make sense. If you hear a very quick beep from a speaker, it's difficult to place what frequency/pitch it was, while it has a pretty definite time of occurrence. However a long tone you can easily match, but it doesn't happen at a very specific time.

It just so happens that the sinusoidal waves which are solutions of the Schroedinger equation are defined to have a momentum $p=\hbar k$, and so $\sigma_x \sigma_p \geq \frac{\hbar}{2}$. The reason that it makes sense to define the momentum this way is that the speed of a sinusoidal solution is $v = \frac{\hbar k}{m}$.

However the Heisenberg Principle applies to classical (meaning not quantum) Electromagnetic waves, sound waves, and water waves as well. The difference is that the wavenumber no longer has this special association with momentum.

As far as what it says about nature, nature is full of waves everywhere you look, even when you know nothing about quantum mechanics. So you see or in some way experience the effects of HUP all the time whether you realize it or not. Especially electrical engineers have to deal with it a lot. You may have heard high bitrate internet connections referred to as high bandwidth, and it's exactly because of the HUP that these are one and the same.

I hope people who aren't versed in the meaning of operators, or the relevant mathematical theorems such as Cauchy-Schwarz, are able to gain something from this explanation. I am happy to answer any further questions about this. Cheers!


Increasingly, an information-theoretic interpretation of Nature is evolving among physicists. Nature may be saying that its small constituents contain a finite amount of information held by conjugate properties such as position and momentum. So the Heisenberg Indeterminacy Principle, (Heisenberg did not use the word uncertainty) may indicate the total information content of a system for those conjugate variables.

So if you finely interrogate one of those conjugate variables, there is less information available in the other because of the finite information'volume' overall.


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