Here's what I know. The Uncertainty Principle states that $$\sigma_x \cdot \sigma_p \geq {{\hbar} \over 2}$$

However, I also know that this principle refers to measurements performed over many identically prepared systems. So this should mean that you can know the exact position and momentum of a particle in a system, if you measure it. However, you can't predict the exact position and momentum of a future particle released into an identical system.

In addition, the uncertainty principle says nothing about repeated measurements on a particle within a system if the measurements are performed at different times. In other words, you can continuously measure a particle's position and momentum evolution within a system with respect to time. What you can't do is predict the evolution of another particle's momentum and position within an identically prepared system.

To sum things up, the uncertainty principle is a statement about limits on our predictive abilities of identically prepared systems rather than our measurement abilities of individual systems.

Is this correct at the theoretical level? If so, could you indeed "see" an individual particle's evolution within a particular system with respect to time?

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/24068/2451 , physics.stackexchange.com/q/54184/2451 , physics.stackexchange.com/q/114133/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 20, 2015 at 18:06
  • $\begingroup$ @Qmechanic All of these are asking about the uncertainty principle with the context of measurements of observables on the wave function. I'm not trying to find out what happens when there is average, but only a single trial experiment. Technically speaking, the uncertainty principle should have nothing to do with this, since there is no standard deviation, but there's insistence that there is. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 18:20

4 Answers 4


It is correct that the uncertainty principle is not a statement about experimental precision as such.

It is incorrect, however, to state that you can know position and momentum of a quantum system exactly, because it presupposes such a thing as "exact position" or "exact momentum" exists. It doesn't, and especially not simultaneously. Any two observables which have a non-trivial uncertainty relation do not commute - and if they do not commute, not every eigenstate of one is an eigenstate of the other. So if you measure an "exact position", you get a position eigenstate, which is not a momentum eigenstate - it has no such thing as "exact momentum". If you measure its momentum, it becomes a momentum eigenstate, but now this state hasn't any such thing as an "exact position".1

The uncertainty principle is also not really about predictive power - quantum mechanics is fully deterministic in the sense that if you have any quantum state, its time evolution is fully determined by the Schrödinger equation. How this squares with the process of measurement is the famous measurement problem of how collapse, decoherence, or whatever else you think happens there happens. This is not the content of the uncertainty principle, the uncertainty principle is a statement about what the standard deviations are if you repeat the measurement on identically prepared states. In natural language, it tells you how much the individual measurements will fluctuate around the expected value, i.e. how much of an error you make when you think of the quantum state as having a single, well-defined value for the observable instead of a probability distribution. (This is your main error, I think - you keep talking of "position of the particle" and "momentum of the particle" when there really is no such thing for the quantum state)

For the somewhat unrelated question of continued measurement, if you continually measure a quantum state, you might evoke the quantum Zeno effect where essentially no time evolution happens because the system is forced to decohere before it had time to evolve into a superposition.

1There is a further subtlety that there are no such things as momentum or position eigenstates inside the Hilbert space - the "eigenstates" are not permissible physical states, and you can only ever measure position and momentum inside an (however small) interval dictated by your experimental configuration. Formally, the probability to measure any exact momentum or position is zero since points are zero-measure subsets of the real line, and the wavefunction is a probability density that has to be integrated to yield a probability.

  • $\begingroup$ I understand that the observables don't commute, but I'm under the impression that this is only with respect to inserting the operator in $|\Psi|^2$. However, experimentally, $|\Psi|^2$ could be constructed from repeated measurements. In my case, I consider an individual experiment rather than the "ensemble" that gives Born's Interpretation. How does what you say apply to measurements conducted in one system? No averages, no ensembles, just a single trial observation. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 17:54
  • $\begingroup$ @Zach466920: Measurement destroys the state (by collapse/decoherence), forcing it into an eigenstate. Excepting certain techniques like weak measurement, one measurement per state is all you get. I'm not sure what your question about that one measurement is? $\endgroup$
    – ACuriousMind
    Aug 20, 2015 at 17:59
  • $\begingroup$ So you can't have two measurements, but you can have one? If you take this measurement, is there a limit on its precision? To answer your question, When I talk about single measurements I do away with averages and deviations. If there is only one measurement there can't be an average etc. there's just a value. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 18:08
  • $\begingroup$ @Zach466920: Yes, the one measurement will be one of the eigenvalues of the operator you measure with a probability given by the Born rule. The uncertainty principle has nothing to do with it, but speaking of its "precision" is ill-defined - there is no "true value" to compare to, because the state didn't have a definite value for the observable prior to measurement. $\endgroup$
    – ACuriousMind
    Aug 20, 2015 at 18:23
  • $\begingroup$ Uhm, so in a very crude analogy, if I measure an electron's position to be $[1,0.5,0.2]$ at time $t$ it was really at that position at time t, correct? So the only problem left is measuring the eigenvalues continuously. That's a different question that I'll look into. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 18:40

The uncertainty principle is about when you pick a state. When you pick a state you might pick one with high uncertainty in one observable or high uncertainty in the other or one with high uncertainty for both. See https://physics.stackexchange.com/a/169757 But the uncertainty is 100% not about an experimental difficulty or a lack of knowledge of some preexisting property.

There is a huge problem with most presentations about measurement. And the problem is that we call things measurements when they aren't.

Take the observable $\hat\sigma_z,$ when you do a "measurement" the result is something that always gives a consistent result when you "measure" with $\hat\sigma_z.$

But if you take the result of $\hat\sigma_z$ and then "measure" with $\hat\sigma_x$ you get something different than you started with you get something that gives a consistent result when you "measure" with $\hat\sigma_x,$ but gives result for a "measurement" of $\hat\sigma_z$ that give every result equally likely and that provably don't depend on the spin state.

So you used to have something that consistent gave a fixed result when you "measured " with $\hat\sigma_z$ so that was part of its nature that it did that. And then after you "measured" with $\hat\sigma_x$ you get something that no longer consistently gives that fixed result under "measurements" of $\hat\sigma_z.$

So these so called measurements definitely change the thing you supposedly measure.

And there is a trade off for certain measurements, giving fix repeatable results for one might not give a fixed repeatable result for the other.

So a state might be nice for one or nice for the other. But your choice of state might have a limit to how good it can be for both if they are the kind of things that mess each other up.

It would not sound deep if we hadn't picked words that make it sound like these interactions don't change things.


It seems your biggest issue is with the notion of measurement, so I'm going to give an answer that doesn't mention it at all.

Suppose a compass needle is pointing in some direction. We say it is in state 'A' if it's pointing north/south, and in state 'B' if it's pointing west/east. So if it points north it's definitely A and definitely not B, but if it points northeast it's some combination of A and B.

In such a situation, it is totally meaningless to ask "is northeast north/south, or is it east/west? Which one is it?" It is also totally meaningless to ask "would the situation improve if we kept looking at the needle? If we looked at the needle upside-down? If we only half looked at it?" It has nothing to do with looking at the needle, or how or when you look at it. It's just true; northeast is neither north/south or east/west, it's both at once.

Now say the needle is in state 'C' if it's pointing northwest/southeast, and in state 'D' if it's pointing northeast/southwest. Then in our previous example, the needle is definitely D and also definitely not C.

There exists no direction the needle can point that both has determinate A/B state, and determinate C/D state! For example, here the C/D state is totally determinate, but in exchange the A/B state is totally indeterminate.

Everything I've said is exactly analogous to position and momentum ("A/B" corresponds to the position basis, "C/D" corresponds to the momentum basis), and the uncertainty principle is true for the exact same reason. A well defined position and momentum just can't exist at the same time.

  • $\begingroup$ Yep, I've already been disabused of that misnomer. But it seems as though even just measuring A/B is worthless, see comments with top answer. Any comment? $\endgroup$
    – Zach466920
    Aug 21, 2015 at 1:14
  • $\begingroup$ There are similar but subtly different things going on here. $\endgroup$
    – knzhou
    Aug 21, 2015 at 1:29
  • $\begingroup$ First, if you accept only the usual postulate of quantum mechanics, but don't mention measurement, you already have a notion of "it can't be both definitely A/B and definitely C/D", where "being" "definitely" one state or another is defined in terms of the state space (i.e. here, A means north/south, the state of the compass is described by a single direction. in QM, you would have to accept the structure of position/momentum eigenstates and the fact that the whole state of a system is in its wavefunction. we can define "uncertainty" purely using this math.) $\endgroup$
    – knzhou
    Aug 21, 2015 at 1:30
  • $\begingroup$ Next, if you add in the Born rule, you can use the previous results to get statistical behavior about many measurements on identical systems. Here we can interpret our "uncertainty" from earlier as "the standard deviation of the numbers you get if you did the measurements a lot". $\endgroup$
    – knzhou
    Aug 21, 2015 at 1:31
  • $\begingroup$ However, this requires accepting the Born rule plus all the other postulates. So if you believe in the second type of uncertainty, you must believe in the first, so you can't do any better with individual systems. $\endgroup$
    – knzhou
    Aug 21, 2015 at 1:33

So this should mean that you can know the exact position and momentum of a particle in a system, if you measure it.

(1) A particle state with exact position and exact momentum doesn't exist.

(2) A particle state with exact position or exact momentum doesn't exist.

(3) One couldn't in principle measure exact position or exact momentum since our measurement apparatus is made from the same 'stuff' and thus, is subject to (1) and (2).

  • $\begingroup$ I think you missed the comment section, where I elaborate on what I intended by the use of the word "exact". I meant, that taking the measurement wouldn't be subject to the requirements of the uncertainty principle as you'd be measuring one system rather than an ensemble. If your stance is still the same after that, you directly contradict the above answer. Please address this. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 18:55
  • $\begingroup$ @Zach466920, nothing in my answer contradicts ACM's answer; there is necessarily an inexactness in one's measurement and it is related to the uncertainty principle. $\endgroup$ Aug 20, 2015 at 20:05
  • $\begingroup$ In the comments, its said "Yes, the one measurement will be one of the eigenvalues of the operator you measure with a probability given by the Born rule." sorry, but there is contradiction. You can't have a mean, deviation, etc. if you only make one measurement. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 20:13
  • $\begingroup$ @Zach466920, you're assuming that eigenkets of the position and momentum operators are physical states - they aren't. See (2) above. ACM mentions this in his answer. $\endgroup$ Aug 20, 2015 at 20:16
  • $\begingroup$ ACM says "Yes, the one measurement will be one of the eigenvalues of the operator you measure with a probability given by the Born rule. The uncertainty principle has nothing to do with it, but speaking of its "precision" is ill-defined - there is no "true value" to compare to, because the state didn't have a definite value for the observable prior to measurement." So after the measurement there is an exact value. If you want, you could tackle my next question below. $\endgroup$
    – Zach466920
    Aug 20, 2015 at 20:18

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