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Uncertainty of an operator $\hat{A}$ when observing a state $|\psi\rangle$ is defined as \begin{equation} \Delta A_{\psi} = |(\hat{A} - \langle\hat{A}\rangle)\psi| \end{equation} Now assume that there is some uncertainty for an operator $\hat{A}$ i.e., when I operate it on a state $|\psi\rangle$ there will be some error every time I measure it.(This is what I think physical interpretation of having uncertainty is)
How does this statement and the definition of uncertainty imply each other. i.e., how is this statement and the definition of uncertainty same?

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    $\begingroup$ Are you asking about uncertainty or uncertainty principle? The question says the former, but your accepted answer talks about the latter. $\endgroup$ Sep 22 at 11:41
  • $\begingroup$ @BioPhysicist My question is about uncertainty. I accepted the answer because he implied interpretation of uncertainty from uncertainty principle $\endgroup$ Sep 22 at 11:44
  • $\begingroup$ @BioPhysicist I am not sure if it will make much sense but how about the amount of uncertainty inside the uncertainty principle itself? Does averaging the very high number of measurements from the experiments eliminate it? $\endgroup$
    – Xfce4
    Sep 22 at 22:10
  • $\begingroup$ @Xfce4 Yes, there can be error in an empirically determined standard deviation. More measurements will reduce this as long as the errors are only due to imprecision of the measuring device around the correct value. $\endgroup$ Sep 22 at 22:39
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when I operate it on a state $|ψ⟩$ there will be some error every time I measure it

This is a common misconception. Performing a measurement on a system is not mathematically represented by applying the corresponding Hermitian operator to that state. The operator just tells us the possible measurement outcomes through its eigenvalues, and it allows us to determine the probability of measuring each of those values by expressing our state in the eigenbasis of the operator. $\hat A|\psi\rangle$ will not give you the measurement outcome.

How does this statement and the definition of uncertainty imply each other. i.e., how is this statement and the definition of uncertainty same?

Uncertainty and error are two different things in the context of Quantum Mechanics. For a given state we can compute the uncertainty of a measurement as you have described. What this means is that if we were to prepare a bunch of similar states and measure the observable in question, we would find the standard deviation of those measurements to (ideally) have a value of $\Delta A_\psi$. Error would come into play in the method of measurement, where the returned values have some error associated with them. Uncertainty is "baked into" Quantum Mechanics and is independent of the method of measurement, whereas error depends on how the measurement was done. A system can have no uncertainty with respect to an observable yet the measurements will still have non-zero error.


To address another confusion, uncertainty is not the same thing as the uncertainty principle, which relates the uncertainties of two observables. For a given state, we can determine the uncertainty of a single observable. Or, using the language from above, we can determine the standard deviation of measurements of an observable from similarly prepared states.

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There is always uncertainty associated with measurements,a nd which is analyzed using statistical methods. This uncertainty however can be made (in principle) vanishingly small with improving equipment. I say in principle, because there are obviously some limitations - e.g., our measurement cannot last longer than the lifetime of the universe, etc.

The uncertainty that one deals in quantum mechanics is however of different nature, resulting from the laws of nature rather than the imperfection of our measuring apparatus. Thus, this uncertainty can never be eliminated by simply improving our measurement device. Yet, when carrying out actual measurements it is calculated in the same way - in this respect one has to distinguish the mean and the sample average. See this answer for some additional details.

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    $\begingroup$ @BioPhysicist I agree but the reason we have such states who are not eigenstates of a given observable is precisely that we have incompatible observables. If we didn't have access to such observables who don't commute with the observable at hand then we'd not find such states in nature that are superpositions of different eigenstates of the given observable (leading to the uncertainty). This is roughly what happens when we have superselection sectors. $\endgroup$
    – Dvij D.C.
    Sep 22 at 13:19
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    $\begingroup$ @BioPhysicist The point is that we can have the measurements with zero uncertainty in quantum mechanics too. The distinction that I am trying to make is between the uncertainty due to the laws of nature and that due to imperfection of the measurement apparatus. $\endgroup$ Sep 22 at 13:24
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    $\begingroup$ @BioPhysicist No, absolutely, once we have a given state, one should be absolutely clear that you can talk about the uncertainty in a single observable, end of discussion. My comment pertains to the reason why we would have such a given state that has an uncertainty in a given observable in the first place (as you already recognize) ;-) $\endgroup$
    – Dvij D.C.
    Sep 22 at 13:29
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    $\begingroup$ @DvijD.C. As long as we are all certain on our understanding of uncertainty! $\endgroup$ Sep 22 at 13:30
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    $\begingroup$ @BioPhysicist I changed the sentence in the second paragraph, to reduce the risk of misunderstanding without getting too much into the details. $\endgroup$ Sep 22 at 13:39
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No, you should not think of the uncertainty principle as meaning that measurements necessarily have associated errors- it is more fundamental than that. What the uncertainty principle means is that there are certain combinations of properties- such as position and momentum- for which a particle can never have simultaneous exact values. It is not that the particle has exact values but we cannot measure them- it is that the particle does not actually have exact values.

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  • $\begingroup$ Right, so I can say that if there is some uncertainty, a wavefunction will collapse on different eigenstate. So how does the definition and this statement imply one another? $\endgroup$ Sep 22 at 11:37

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