I have heard two descriptions of the uncertainty principle, and I am quite confused about the uncertainty principle now.

The first one is dependent on the wave function of the particle, and it says that if you have something like the sine wave, you are very sure about the momentum, but not the position, because according to the sine wave, it could be almost anywhere, and if you have something that just rises up and goes down once, you are very sure about the position, but not the momentum, since you only have one wavelength to measure.

The second description I heard of is that in order to observe a particle, you must shine light at it. If you shine a large amount of light, you will be very sure about the position, but then your confidence in momentum goes down because the energy transferred from the photons to the particle you're observing, and if you shine a low amount of light, you are able to observe the momentum very well, but not the position.

Which one is the correct one, or are both correct?

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    $\begingroup$ Neither of these is a definition in the technical sense. Where have you heard these? $\endgroup$ – ACuriousMind Mar 26 at 17:54
  • $\begingroup$ I see that this is collecting a few votes to put it on hold as opinion-based. I really don't think that close reason fits this question (definitely not in its current form, anyway). I'd be interested to understand it, if anyone who cast and/or agrees with those close votes would be willing to offer an explanation. $\endgroup$ – David Z Mar 31 at 4:03
  • $\begingroup$ actually the uncertainty principle springs up (mathematically) due to the non-commutative nature of position and momentum operators present in Hilbert Space. But there is a famous account of observing uncertainty principle from the pattern of spectral lines of hydrogen atom. $\endgroup$ – Ashwin Balaji Mar 31 at 6:20

The first one is correct, the second is not.

The second definition1 is actually describing the observer effect. Explanations written by non-experts often mix the two up. But one key difference is that the observer effect only applies to situations where some external "probe" (like a particle) is interacting with the system. The uncertainty principle, on the other hand, applies even to a system which is isolated and not interacting with anything external.

1A couple other people have pointed out that these are not really definitions of anything, but I'll use that word for consistency with your question.


While other answers say the first one is correct, there is something that should be pointed out. The issue is with the beginning of your statement:

The first one is dependent on the wave function of the particle...

The Heisenberg Uncertainty Principle is very useful because it actually doesn't depend on the particular wave function. In other words, $\Delta x\Delta p\geq\hbar/2$ is true for all wave functions, not just sine waves.

There are more general uncertainty principles that do depend on the wave function, but those aren't as famous.

Another thing to keep in mind is that neither of your two statements define the uncertainty principle. Your first statement is the closest to being correct, but even then it's more of an application of it, not a definition.

Also, the uncertainty principle isn't a statement of how "sure" or "confident" we are about the position and momentum of a particle, which seems to be a common idea in both of your statements.

  • $\begingroup$ Good points. I had interpreted "dependent on the wave function" colloquially, to mean something like "applies to the wave function". $\endgroup$ – David Z Mar 25 at 23:46
  • $\begingroup$ @DavidZ Yeah, that makes sense to me too. Best to cover all possible cases among our answers I guess :) $\endgroup$ – Aaron Stevens Mar 26 at 0:10

Before answering the question, I would first look at HUP from more technical standpoint:

The uncertainty principle is given by noncommutativity of measurement. When you have wave function $|\psi\rangle$ the measurement changes it to another wave function $|\alpha\rangle$ - this is the famous collapse of the wave function - and produces number $a$, for example component of momentum of the particle. The measurement can be represented as operator: $$\hat{O}_a: |\psi\rangle\rightarrow |\alpha\rangle,$$ where $|\alpha\rangle$ is now the state of the particle with definite value $a$. Before that, the particle could have been in superposition of states with several possible values of measured quantity, but once you measured it, you collapsed the wave function to that particular state. Because, now the particle is in the state of definite value $a$, succesive measurement will produce the same number $a$.

Now what if you decided to immediately after this measurement measure different quantity? Again you measure the value of $b$ and collapse the wave function to wave function of this particular state: $$\hat{O}_b: |\alpha\rangle\rightarrow |\beta\rangle.$$

The uncertainty principle follows from the fact, that measuring the quantity $a$ first and then the $b$ is not equivalent to doing it the other way around. That is: $$\hat{O}_b\hat{O}_a \neq \hat{O}_a\hat{O}_b $$

To show this would take some time, but intuitively this makes sense. If the particle could have definite value of quantity $a$ and $b$ at the same time, then measuring it should produce those two values. But since the values are already given, then it should not matter which you measure first. We know, however, it does and therefore the particle cannot be in state with definite value of $a$ and $b$ at the same time. These two values are simply incompatible. If the particle is in the state of definite value of $a$, then it must not be in state of definite value of $b$. The most notorious example of such quantities is position and momentum you wrote about.

This is however not really property of the wave function of the particle as such. It is property of the operators $\hat{O}_b$ and $\hat{O}_a$, i.e. property of the measurement itself. Every such operator/measurement has some wave functions associated with it, which are wave functions of definite values of the measured quantity. And this wave functions associated to the operators/measurement are simply incompatible.

Now to answer the question:

The first "definition" is taken from the point of view of wave function. It says that when you have wave function with definite value of position, then it is not function of definite value of momentum and vice versa.

The second "definition" is taken from the point of view of operators. It tells you that measurement of position changes the wave function in such a way, that it is now in the state of superposition of many momenta and there is no answer to which of the these momenta particle has and vice versa.

They are therefore equivalent. But note neither of your "definitions" is really a definition. They are more like different interpretations of HUP.

  • 2
    $\begingroup$ It tells you that measurement of position changes the wave function in such a way, that it is now in the state of superposition of many momenta and there is no answer to which of the these momenta particle has and vice versa. This is just a property of QM measurement. It isn't the uncertainty principle. You don't need to change the state of the system for the uncertainty principle to be true. $\endgroup$ – Aaron Stevens Mar 25 at 14:32
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    $\begingroup$ Yeah, I agree with what Aaron is saying. I think this answer misses the key point that the uncertainty principle is a statement about (among other things) quantum states which applies regardless of whether any physical change happens to the state. The mathematical interpretation in terms of commutators is correct as you describe it, but it sounds like you go from there to saying that the uncertainty principle is about how the state responds to sequential measurements, and that logical leap is not justified IMO. $\endgroup$ – David Z Mar 25 at 16:05
  • $\begingroup$ @DavidZ before I reply, what other things? $\endgroup$ – Umaxo Mar 26 at 5:56
  • $\begingroup$ I meant that, for example, in addition to being a statement about quantum states, the uncertainty principle can also be taken as a statement about waves of any kind. $\endgroup$ – David Z Mar 26 at 7:09
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    $\begingroup$ The point I was trying to make is that the HUP constrains the possible values for the spread of position measurements and spread of momentum measurements for similarly prepared states. The fact that a position measurement results in a states that can described as a superposition of momentum states is not the HUP. Sure, both ideas are linked together by the non-commutative nature of the operators, but that doesn't mean they are the same thing. $A\to B$ and $A\to C$ doesn't necessarily mean $B=C$. $\endgroup$ – Aaron Stevens Mar 26 at 18:02

Your first interpretation is correct as the second one merely implies that the observational capabilities are the only things which seem to effect our measurements on the momentum and position of particles. This is not the case. H.U.P is not limited by our technological abilities but is integral to the entirety of quantum mechanics i.e. it is just as good as a law. I've often found that the second one is used to explain the principle to those who initially deny a probabilistic universe and believe in absolute measurements (like me ;D)

The first interpretation uses the Fourier series (for a detailed and fabulous description-look here) and showcases a more advanced understanding in how eventually on combining (or rather-deconstructing for Fourier) waves can show us an uncertainty in position and velocity. It is more appropriate as it gives you insights in the mathematics and helps us to understand that HUP is the explanation to a phenomenon.

As a prof. once told me: A universe with H.U.P may lead to many fabulous things but a universe with the laws of quantum mechanics must begin from H.U.P.

Edit: Look into the observer effect as pointed out in David Z's answer. Missed that out here.

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    $\begingroup$ I've always seen the uncertainty principle following from postulates of QM, but I've never seen the uncertainty principle as a postulate itself. Where have you seen QM arise out of an uncertainty principle? $\endgroup$ – Aaron Stevens Mar 25 at 5:29
  • $\begingroup$ I've always considered HUP to be the basis of QM second only to the SWE. Also, if you're referring to the quote, I've only accepted this quote after being taught the various applications of HUP which I think would justify how QM arrives out of HUP. $\endgroup$ – user220704 Mar 25 at 5:56
  • $\begingroup$ I don't see how the HUP being applicable means that it is the basis of QM. $\endgroup$ – Aaron Stevens Mar 25 at 12:48
  • $\begingroup$ @AaronStevens in some of its applications, you could use HUP to explain the fine structure of hydrogen series by the energy-time relation of HUP. This was a drawback of the Bohr model and could be seen as the cornerstone of a new model. $\endgroup$ – Shreyas JV Mar 26 at 7:55
  • $\begingroup$ @ShreyasJV The energy-time relationship and the HUP are two different things. $\endgroup$ – Aaron Stevens Mar 26 at 11:30

They are in fact both correct, but the relationship is not obvious. Heisenberg explained his original argument for the uncertainty principle using a thought experiment called Heisenberg's microscope, which was essentially the second argument expressed a bit more thoroughly. Heisenberg took the argument further, showing that it leads to a reductio ad absurdum, meaning that position and momentum are not precise quantities, as assumed in classical mechanics, and can only be defined in terms of probability. His argument shows that this is a matter of principle, not one of technological limitation.

An observer seeks to measure the position and momentum of an electron of known momentum by bouncing a photon off it. If the photon has low energy, so that it will not disturb the momentum of the electron, then it has long wavelength and position cannot be discovered to any accuracy. If the photon has short wavelength, and correspondingly high energy, position can be measured precisely, but the photon scatters randomly, and an unknown momentum is transferred to the electron.

Heisenberg then noted that any attempt to measure the momentum transferred to the microscope necessarily results in a loss of knowledge of position of the microscope. Likewise, better measurement of the position of the microscope leads to loss of knowledge of its momentum. Attempts at more precise measurement of one property necessarily result in less precise determination of the other. This applies to the apparatus, and to any further apparatus used to measure the apparatus and so on ad infinitum.

Consequently, in the general case, the position and the momentum of an electron can only be stated in terms of probability distributions, where the more precise the probability of the one property is, the less precise is the other.

This argument can be used (as can others) to motivate the probabilistic structure of quantum mechanics, in which is based on the (mathematical) principle that probabilities can be expressed in terms of wave functions obeying the Born Rule. It is then possible to derive the relationship which is now called Heisenberg's Uncertainty principle:

  • uncertainty in position multiplied by uncertainty in momentum is greater than Planck’s constant divided by 4$\pi$

The derivation was not given by Heisenberg, but by Earle Hesse Kennard, in 1927.

  • $\begingroup$ The key feature here is that it is the experimental apparatus that has the effect. We have improved accuracies of measurement and intricacies of measurement, not by changing the 'particles' under investigation, but by changing the apparatus! Also, part of the uncertainty is a wave-particle duality regarding continuity in Fourier space (e.g. see "Friendly Guide to Wavelets" for theory). If the signal is continuous you can't have uncountable steps, or frequencies, hence countable 'quantum's. $\endgroup$ – Philip Oakley Mar 26 at 17:27

The basic premise of quantum mechanics is that every sub-atomic particle is (or is associated with) a wave. To be more precise a wave packet of finite size. If the wave is very long, the frequency and wavelength (but not the position) can be determined with high precision . If the packet is short, Fourier analysis says it can be described as a superposition of many long waves with a spread of frequencies (and energies).

  • $\begingroup$ Beware the Fourier analysis argument (confounding of countable and uncountable number of frequencies in a continuous wave). Try Wavelets for the appropriate properties of localisation and duality. $\endgroup$ – Philip Oakley Mar 26 at 17:31

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