I found these two examples in a books which demonstrate Heiseberg's uncertainty relation:


enter image description here

It shows that when we try to locate a moving electron,we transferred momentum via the photon that we send in order to measure the position or momentum of the electron.This gives rise to the position uncertainty Δx that is roughly equal to the width of the opening of the apparatus and the uncertainty in momentum ΔPx due to the transfer of momentum.


enter image description here

This time,we again have uncertainty Δx roughly equal to the width of the slit and uncertainty in momentum ΔPy due to the wave diffraction(or you can also say because of photons hitting the edges of the slit).

The funny thing about these two examples is that they show you how the uncertainties in the experiment arise from the interaction of the particle that we want to observe with its surroundings(either due to the measurement or just passing through a slit).
So this brings me to my question,is Heisenberg's uncertainty principle something that rises ONLY from the measurement process(interaction)?Because if we don't interact with a particle,then there is no change/transfer of momentum so the particle has a definite momentum AND position(but this contradicts the fact that many claim to be truth-that the particle does not have an exact location until it is measured) but when we try to interact with it we "mess up" the situation in ways that are described by the two experiments that i aforementioned.Did i got something wrong here?

And if Heisenberg's uncertainty principle isn't something that rises ONLY from the measurement process(interaction) and it is something much more fundamental(i.e. it's not the interaction that causes there uncertainties),then how would you define the uncertainty principle in order for me to understand it more specifically(and what does a particle do when it does not interact with something?-details about position and momentum)?

The important part of the question is:In order to fully understand and answer my question just follow the my thinking as i present it.Follow the 2 examples that i showed above.They imply that its the interaction/measurement that make it impossible to know both momentum and position because you mess it all up(transfer of momentum and stuff).Keep this in mind while also having in mind the statement that "a particle does not have a position(it isn't anywhere) or momentum until measured" and you can see that what confuses me is that with these two in mind,my conclusion is that without )interacting with the particle,it does in fact have a certain momentum and position(not as the above statement says).
To clarify a bit more: Position and momentum do not HAVE values until measured,but HUP rises from the interaction with a particle.It interacts with a particle at a certain definite position and it transfers some momentum to the definite momentum that it already has(if it did not have a position,how could they interact,and if it did not have a momentum then how can we even talk about transfer of momentum?).

Note:Bear in mind that i don't want an explanation that is purely mathematical(like just saying that momentum is just the fourier transform of position-which in my opinion is a RESULT of the principle and not the cause of it as someone might claim-again,correct me if i am wrong here).I am not saying that maths are not required in order to give me a complete explanation,just that i also want some kind of intuition and deeper understanding(because i think that most students take these fundamentals as granted and just proceed to solve exercises). Also, the pictures are from Eisberg and Resnick's book Quantum Physics.

  • $\begingroup$ An object for which you don't measure position doesn't have a position and if you don't measure its momentum, it doesn't have momentum (by the simple definition of science as rational explanations for observations). As soon as you measure either, the uncertainty relation kicks in. Having said that, the uncertainty of quantum states "exists" at all times in the math. We are not constructing measurable quantum states that don't obey the uncertainty relation. $\endgroup$
    – CuriousOne
    Commented Jun 23, 2015 at 0:29
  • $\begingroup$ I am not saying that. I am saying that when not measured it does have a position and momentum but we just do not know them and this comes in contradiction with the statement that a particle does not have a position(it isn't anywhere) or momentum until measured(not to be confused with "we do not know its momentum and position until measured") $\endgroup$ Commented Jun 23, 2015 at 0:34
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    $\begingroup$ You understood something completely different from what i am trying to say. I am not saying that we can measure something with 100% accuracy. I am talking about what the state of a particle is when it does not interact(or being measured) with something. Because its the interaction that gives rise to Hesenberg's uncertainty principle(i am not talking about uncertainties caused by the not perfect measurement tools),so without interaction something has a definite momentum and position but we just do not know what that momentum and position is simultaneously?That us what i am asking. $\endgroup$ Commented Jun 23, 2015 at 0:44
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    $\begingroup$ Your question is the naive imagination of a beginner who thinks that he can beat QM at its own game. We have all been there. Either you measure something and then that measurement is real or you don't and then there is no measurement. That's the definition of science, even in Newtonian mechanics (but you have probably never thought about these things in the case of e.g. Brownian motion which already invalidates your reasoning in a completely classical case). $\endgroup$
    – CuriousOne
    Commented Jun 23, 2015 at 1:07
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    $\begingroup$ Landos, I sort of see what you are saying. Can I suggest you put your last comment in your post and format it as clearly as you can, with the contradiction as your main point. I have seen your previous questions and they are similar to ones I would ask myself. $\endgroup$
    – user81619
    Commented Jun 23, 2015 at 1:11

4 Answers 4


I would boldly claim that this thought experiment (also known as the Heisenberg microscope) is simply the wrong picture to understand the origin of uncertainty principle. The reason why it is so is because it mixes up between uncertainty due to measurement and uncertainty due to quantum state; nonetheless it had made its way into numerous textbooks and confused numerous undergraduates (including me) by including quantum mechanical objects such as electrons and photons and giving some results that has the factor $\hbar$ in it.

I will try to explain this confusing business to the best of my abilities about your questions in three parts - firstly, what is Heisenberg uncertainty principle; secondly, why is it unique to quantum mechanics; and finally, why the Heisenberg microscope is a wrong way of understanding the uncertainty principle. I am sorry that I may have to include a bit of maths from time to time, but I hope you will follow (and I hope I am right about this - do comment if I made mistakes).

Firstly, what is uncertainty principle? The best way that I know of to understand it physically is the following scenario: imagine that you have prepared a huge quantity of identical quantum states, and you measured the position of half of these states and the momentum of the rest with perfect precision (see below). At the end of the day, you will obtain a list of positions and momenta, you will notice that these results do have uncertainties due to the probabilistic nature of quantum states.

Here is where the uncertainty principle kicks in: regardless of what quantum state you prepared in the first place, if you calculate the uncertainties of positions and momenta respectively by the data you obtained from that long list, it will always be the case the uncertainties calculated from the list obey the uncertainty principle $\Delta x \Delta p \geq \hbar/2$. A more interesting way of rephrasing it would be you can never prepare a quantum state of which the uncertainties calculated from the list $\Delta x \Delta p$ is smaller than $\hbar/2$.

Before moving on, it is worth discussing a few things in this imaginary scenario. First thing is obviously what do I mean by the phrase with perfect precision? I certainly do not mean that there is some 'position' and 'momentum' that the quantum state has prior to measurement, what I meant is that the measurement results are completely due to the quantum states themselves, and are subjected to no external disturbance by other physical systems. Well you may argue that it is physically impossible to do that for any experimental apparatus would introduce some perturbation of the system, but since we are living in the imaginary thought experiment well we get to decide what we can do and what we can't do.

And here's the point which is very important under the context of the problem: even in this ideal world we can obtain positions and momenta directly from the quantum states, the uncertainty principle still holds. Throw away apparatus like the microscope or any other fancy equipments, you still have uncertainty - and this property is fundamentally due to the nature of quantum states themselves.

Still need convincing why this is justified? Well here we enter the second part on uncertainty due to measurements. Look back to any experiments with classical systems - you can almost certainly find no experiments where there is 0% uncertainty as there are bound to be errors introduced by the environment; nonetheless it doesn't stop us from imagining a perfect experiment where the results are completely due to the physical system we are studying. Say you are measuring acceleration due to gravity in a lab - you can be certain that almost nobody will ever get $9.80665$ meter per second squared (unless you are a cheater) because of errors due to gravitational attraction to the surrounding objects, the grids on your ruler are not fine enough, etc. etc. But you have no problem convincing yourself that under the perfect and ideal condition you still will be able to get $9.80665$.

And the crux of the matter here is that the uncertainty due to environment (or errors) happens to all systems, be it classical or quantum. Nonetheless, the uncertainty principle only applies to quantum systems. In Newtonian mechanics, you can characterise the motion of a particle in one dimension by a pair of quantities $(x,p)$, or position and momentum, and you can make it such that following the experimental procedure we described in part 1, that by preparing a huge number of identical states and measure their positions and momenta, $\Delta x \Delta p \leq \hbar/2$. In fact, it is very easy: by preparing a bunch of particles having same position and momentum, $\Delta x = \Delta p = 0$. But in quantum mechanics, it simply cannot be done, because we are talking about an entirely different beast here: instead of $(x,p)$, you need to describe a quantum state with a wavefunction $|\psi\rangle$, and they must obey the uncertainty principle.

So here we are at the third part - why is the Heisenberg microscope the wrong picture to understand the origin of uncertainty principle. I suspect that you can now already answer that - the thought experiment basically attributes the origin of the uncertainty principle to error introduced in the experiment, but not the quantum state itself. In a perfect experiment, according to Heisenberg microscope, there will be no uncertainty; we can even try to perceive measuring the position and momentum of the electron using other methods - say shooting one electron off a gun and bouncing them off by a wall (maybe?) - that can give you uncertainties below the uncertainty principle according to the picture described by Heisenberg microscope. But this is simply not the case and you simply can't do that - because the state is described by a wavefunction $|\psi \rangle$, but not a pair of $(x,p)$, so it is simply wrong to use '$x$' or '$p$' to describe the electron.

This also leads to the complication about interactions, as you have mentioned in your question. The interaction between photon and electron cannot be simply described by 'momentum transfer' for this implicitly assumes that the physical state photons and electrons are characterised by some momenta. As stated before, the interaction can only be described in terms of $|\psi\rangle$; and to be absolutely strict the best way of understanding such interaction is from QED, rather than this semi-classical picture. Nonetheless, let me reiterate my statement again - remove the interaction (regardless of whether it is photon-electron interaction or whatever physical processes you use to probe the electron), you still have uncertainty principle, because it is a fundamental property of a quantum state.

Regardless, I suspect the reason why the Heisenberg microscope is so successful is because the way it mixes quantum mechanical interactions between electrons and photons and classical interpretations to give results involving the infamous $\hbar$ simply by manipulating the errors introduced in an experiment, which gives us the illusion that we can intuitively understand the uncertainty principle and it is simply not the case. I feel it's fitting to use this (mis)quote - certainly quantum mechanics has never allowed herself to be won; and at present every kind of intuition stands with sad and discouraged mien—IF, indeed, it stands at all! - but this is, I guess, why we love quantum mechanics so much :)


Others have explained why those textbook examples are bad (which they are, but are historically accurate I guess). Let me try to explain the problem itself by an analogy and without just saying that the quantum state character itself leads to ill-defined positions and momentums.

Suppose you are a farmer on a flat infinite Earth (pre-medieval times?), and you're obsessed by planting apple trees in long nice rows on your field, with exactly the same distance between the rows:

equidistant apple orchard

This practice of planting apples in long rows keep spreading, because apples are so tasty, so eventually the whole planet is filled with huge apple orchards.

At this point, an alien scout patrol flies by Earth and snaps a couple of close-up photos of the planet surface and their top alien scientists sit down and analyzes it. They start by writing down the positions in the photos of all the apple trees.

This is the position representation of the apple orchards.

Now in the alien spaceship they have a problem - they have almost filled up all their hard-drives, so they have to conserve data somehow, and one clever scientist comes up with the idea that since all the orchards in the photos look the same except perhaps for the row distances and row orientations, they can save space just by storing the number of rows per photo (all photos have the same size), orientation, and "phase" (if the first row in the photo is close to the border for example, or some distance in).

This number, phase and orientation is the momentum representation of the apple orchards.

Now remember, it's the same information. There is no point for the aliens to store both. Doing that would lead to arguments and general confusion between the alien scientists over if the position or momentum gangs have the "right" data.

So the aliens go home and report, and then their pointy-haired boss who is hungry for new kinds of foodsources comes and asks them "How many appletrees do the earthlings have in total?". The alien scientists go silent - they only stored the intervals of the orchard rows, took only photos of finite size and forgot to measure the size of flat Earth (we know it's infinite but they don't).

apple orchard from above

This shows that for a pure momentum state representation, where the row interval is known to infinite precision, there isn't and can't be any variation (or even limit) to the positions of the individual trees. They simply can't end after 1 km, because then the row intervals wouldn't be known as a single value (there would be a tight row interval for the first km, then the interval would go to infinity suddenly, etc). You could only know the row interval with infinite precision if your "photo" was infinitely large (encompassed all apple trees in the infinite apple orchard universe).

If there only had been a single row of trees on the entire planet when the aliens did their survey, and they had a method to deduce there were no more rows anywhere at all, they could assign an exact position of the trees, but they could never assign a row interval. This would have been a pure position state representation.

Interpolating between these two extreme situations give all kinds of intermediate representations where the position information is accurate to a certain degree and the momentum information is accurate to another degree, but they never meet in a 100% / 100% point. This relationship as discussed in quantum physics is referred to as the heisenberg uncertainty relation but it's neither fundamental nor strange. It's a fairly simple result of how an underlying physical reality (the orchards in this case) behave. Historically this was not the first conclusion, as the scientists in the 1920's came from a classical physics point of view, thus the mess in the old textbooks.

Going into the exact mapping of this to how quantum mechanics arises and behaves requires so much unraveling of common concepts that it's really out of scope of a single post here. I would summarize it as in the top paragraph: there is nothing fishy going on, and there is no problem of measurement as such. The issue is in trying to impose the classical notions of complementary positions and momentums to an object (a quantum mechanical state) that simply does not have both.

Consider this as well - you can run an accurate QM simulation using only one of the representations (or any other representation for that matter).

Disclaimer: this is not meant as an exact analogy, not the least due to the fact that electrons (probably) don't look like apples. Also in particular, the analogy does not stretch to discuss probabilities, decoherence, "disturbing the state by measurement" and all those things that are also referred to in the Question.

  • $\begingroup$ Thanks for the answer,although i know how a quantum system works,i know about the uncertainty.I just couldn't understand why do these two examples show that uncertainty in quantum systems comes from the interaction of the quantum system with something(slit or measuring device). $\endgroup$ Commented Jun 25, 2015 at 10:24
  • $\begingroup$ the experiments just illustrate the issue in a graphically simple way. also like I explained, the "uncertainty" arises because you try to impose redundant descriptions on the quantum state. there really is no such thing as a momentum of a particle (if you're working in the position basis) or vice versa, and it's not an artifact of measurement. to be honest, i think the last half of your question has some misunderstandings.. you can't take the classical concepts of position and momentum and particle and end up with modern QM/QFT without a fair amount of confusion (which is normal :). $\endgroup$
    – BjornW
    Commented Jun 25, 2015 at 19:59

A quantum particle can't simply be a classical particle with unknown but well-defined values for position and momentum. Quantum uncertainty really is a different sort of thing from simple lack of knowledge about the state of an object.

One can see this pretty clearly by considering the double-slit experiment. If you send electrons (or any kind of particle) through a wall with two slits in it, they form an interference pattern on the screen beyond. This happens even if you send the electrons through one by one, so it can't be that multiple electrons are interacting with each other. If an electron really did have a definite position and momentum at each time, then it would have a specific trajectory that it followed, and that trajectory would have to pass through just one of the slits. So how could it be affected by the presence of the other slit, and "know" to form the interference pattern?

It's clear that something must pass through both slits and interact with itself to generate the pattern, even if there's just one electron. Whatever that something is, it's what we describe by the wavefunction, and it's subject to the uncertainty principle.

Bell tests are another set of experiments that falsify the idea of quantum particles as having unknown but definite values for all their properties. These are hard to explain concisely as they involve a more complicated setup with a pair of entangled particles and some subtle reasoning about probabilities, but you can read the details on the web. The upshot is that the experiments have been done and so far, have definitively shown that for properties like spin/polarization as well as position and momentum, the Heisenberg principle applies and they really do not have any definite values (even unknown ones) before "measurement".

I don't consider such thought experiments like the ones you describe to be an explanation of the uncertainty principle. They're really just to answer the objection, "Can't I just measure both the position and the momentum? Who's going to stop me?" Well, the laws of physics will conspire to stop you. :D

By the way, since in your edit you mentioned interactions: particles can interact even without definite positions. If their lumps of probability density (as represented by their joint wavefunction) come to overlap in space, there is some amplitude that they interact and some amplitude that they don't, with both possibilities represented in the resulting wavefunction. Similarly, a range of possibilities for the amount of momentum transferred will also be represented in the wavefunction, but won't become definite until "measurement".


Here are a few things that you may need clarification on:

  1. What is uncertainty?
  2. What am I trying to achieve through a measurement?

Firstly, as you pointed out, we DO NOT need a measurement to be made for the uncertainty relation to hold good. Going by the statistical interpretation, uncertainty is defined as the standard deviation of the position/momentum/any other physically measurable quantity. What we mean by uncertainty is the non-zero nature of this standard deviation. For a quantum particle, there always exists an uncertainty in position/momentum, (unless in some extreme cases) and the HSU is just a mathematical statement that makes use of the complementary nature of the variables.

Secondly, a measurement of position/momentum causes a collapse of the wavefunction to an 'eigenstate' of the position/momentum.

Eigenstate of position -> wavefunction has a spike at one location only.

Eigenstate of momentum -> wavefunction becomes a pure sinusoid, with infinite spread and a definite wavelength, and hence momentum.

Since the posn. and momentum variables are complementary, AFTER MEASUREMENT TOO, given posn, we can't get a FIXED value of momentum with arbitrary precision and vice versa.

Hope this helped.


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