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In reference to this elaborate answer by @DanielSank, I would like to pose the following question(s) in order to verify my understanding of the subject matter--in particular, that of the nature of uncertainty and measurements in quantum mechanics. For context, I would like to quote the most relevant paragraph (most relevant with respect to my question) from the linked answer:

The crucial thing here is that you never ever measure anything to infinite accuracy, so the wave functions resulting from your measurement are not exact eigen-states of what you think is the measurement operator. This is not just "experimental dirtiness". This is a fundamentally important aspect of QM which you should keep near your mental centre as you learn more.

Now, in my understanding, we never ever make a measurement that can produce an exact eigenstate of, say, the momentum operator because there is no normalizable (and thus, physically realizable) eigenstate of the momentum operator. And that, in turn, is the reason why we can never ever measure the momentum precisely--no matter how arbitrarily small we can make the error-bars nonetheless. Any attempt at measuring the momentum will never reduce the given state to a momentum eigenstate but only to a superposition of multiple momentum eigenstates--no matter how closely distributed (i.e. spiked near some particular eigenstate) those contributing eigenstates might be. Thus, in the measurement of the momentum, we will always be endowed with an uncertainty of the scale of the width of that spiked superposition of momentum eigenstates.

Now, all of that is fine and fundamental and not a result of the experimental dirtiness--but in the cases where there are physically realizable eigenstates, any inability of a ''measurement'' to produce a true eigenstate must be credited to the experimental dirtiness--I think. For example, for a Hermitian operator with discrete non-generate spectrum, there seems to be no way for a measurement to produce anything but a specific true eigenstate. In such a case, there would certainly be an indeterminacy in the outcome of the measurement when one starts with a generic state which can be a superposition of more than one eigenstates (and this would lead to an uncertainty in the measurement in the sense that when we are measuring the same observable over an ensemble of identically prepared states, we will not get the same value because different measurements would pick out different eigenstates--generically speaking) but this indeterminacy (and resulting uncertainty of the kind I described) is different from the uncertainty in the measurement of an observable corresponding to an operator whose eigenstates are simply not physical. In particular, once the measurement is made on this kind of an operator which admits discrete non-degenerate eigenstates, all the subsequent measurements are guaranteed to yield the exact same outcome with literally $100\%$ probability except for the experimental dirtiness. And this is the reason I think it is not quite right to say that "you never ever measure anything to infinite accuracy" in quantum mechanics (for example, we always measure the exact spin of an electron to my understanding).

Thus, my core question is that is it correct that it is not right to say that we never ever measure anything to infinite accuracy in quantum mechanics? Or at least can we assert that there are certain operators for which any uncertainty in the measurement must be credited to the experimental dirtiness?

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  • $\begingroup$ What if I measure something quantized? Like spin along an axis? $\endgroup$ – Ryan Thorngren Sep 7 '18 at 0:23
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    $\begingroup$ @RyanThorngren Yes, that is precisely what I am thinking that the measurement any operator with a discrete spectrum would give me an exact measurement (i.e. the measurement will reduce the state to an exact eigenstate and spit out the exact eigenvalue). And this should suffice to say that it is not right to say that we can never measure anything exactly in QM. $\endgroup$ – Dvij Mankad Sep 7 '18 at 0:48
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The crucial thing is that we never measure anything to infinite accuracy.

This is true but it has nothing to do with classical or quantum physics and has purely to do with the practicalities of measurement. We never measure anything exactly and there is always a degree of error.

Sp the wave functions resulting from your measurement are not exact eigenstates of what you think is your measurement operator.

However, this does not imply that nature is not exact. Classically speaking, nature is exact. Notice, here, that this is an epistemic statement as it should be. When we think about nature we think through a theory of nature - no matter how naive or sophisticated.

Likewise, in QM, the wave function to be measured is exact. But it's ontological status is still under discussion. Even as an amplitude whose square is probability, it is - theoretically speaking - exact. To be honest, the nomenclature to describe the various kinds of epistemic and ontological commitments here - despite a century of thinking over it - is all over the place. No wonder people are still confused about it.

For instance, people still say off the bat that QM is strange and bizarre. But personally, I find the notion of a purely deterministic nature strange and bizarre. This is hardly ever noted.

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  • $\begingroup$ Thanks for your answer! I follow you until the end of the second paragraph but I lost you in the discussion about the ontological and epistemological status of the wavefunction. I guess that you are pointing out that there are people who would assign the wavefunction an ontological status whereas others would just an epistemological one and I think I have nothing more but to agree that there is that disagreement. [...] $\endgroup$ – Dvij Mankad Sep 8 '18 at 1:12
  • $\begingroup$ [...] But nevertheless, my question still stands I guess--whether the wavefunction is ontological or epistemological, whether or not it is exact in a certain situation or is supposed to be exact in a certain situation stands, right? Now, regarding your last comment, I am really curious what makes you think so. Is that an epistemic claim you are making that it is not possible to know enough and compute enough to have a ''deterministic knowledge''? [...] $\endgroup$ – Dvij Mankad Sep 8 '18 at 1:15
  • $\begingroup$ [...] If so then certainly everyone agrees I guess. But it is the ontological reality of indeterminacy claimed by QM that seems strange and bizarre to most--and rightly so I would say! $\endgroup$ – Dvij Mankad Sep 8 '18 at 1:15
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Conceptually it seems there is a difference here between conjugate observable pairs (non commuting ones) such as momentum and position, and other observables with no conjugate, such as spin.

One cannot in principle give an exact value to an observable belonging to the first class, to which the uncertainty principle (a dispersion relation between both ends of a Fourier transform) apply. This translates experimentally in limitations such as the amount of energy required for a position measurement, which increases with the wanted precision. On the other hand one can measure a spin component exactly (e.g. Stern-Gerlach).

Conjugate pairs of observables directly come from Hamiltonian mechanics and reflect the symplectic structure of phase space. They are intimately associated with phase space being continuous. I am no expert in that domain but I would guess that this is what breaks down in QM, and that in fact there is no actual physical continuity (which comes anyway with its own problems in terms of infinities, as pointed out by The_Sympathizer answer), but a discrete structure instead, at least when measurements are involved. This is measurement problem territory though, and still an open question as far as I know.

One intriguing point is that conjugate variables are reciprocal derivatives of the action. The action is also driving quantum interference in a very general way as seen in the path integral formulation. I am not clear on the link beetween these aspects of action, but it seems to point towards the idea that the phase space of a quantum system in-between measurements (measurements here considered as boundary conditions in the path integral view), although mathematically useful and well-defined, is in fact not a really helpful concept in understanding intuitively the nature of the system evolution.

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  • $\begingroup$ The phase space of a quantum system is a perfectly valid concept. What's not valid is assuming that the system occupies a single point in it, cf. Phase space formulation, Wigner functions,... $\endgroup$ – ACuriousMind Sep 7 '18 at 8:26
  • $\begingroup$ @ACuriousMind. Yes, that is what I mean by "not a 'physically' valid concept". I just mean we cannot picture a system "there", so to say. Maybe I should rewrite that part? $\endgroup$ – Stéphane Rollandin Sep 7 '18 at 8:27
  • $\begingroup$ But we can - we just need to picture it as a smeared out blob (the quasiprobability distribution), not a point. $\endgroup$ – ACuriousMind Sep 7 '18 at 8:30
  • $\begingroup$ @ACuriousMind. Yes but I am talking about the physical intuition here. A blob represents a superposition, and we do not know how to (again, intuitively) relate the classical view of a system with a superposition of states. My last sentence is about the "nature" of the system evolution. I do not think that the picture of a blob in phase space tells us much about the nature of what is happening. But granted, this is about interpretation of QM, not formalism. $\endgroup$ – Stéphane Rollandin Sep 7 '18 at 8:34
  • $\begingroup$ @ACuriousMind. I have rewritten the last part, hopefully making its meaning and intent more clear. $\endgroup$ – Stéphane Rollandin Sep 7 '18 at 8:52
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As has been pointed out here the inability to measure exactly is more due to the physical limitations of measurement. At the very least, an infinitely precise measurement of an arbitrary real-valued physical quantity, as can be had in both classical and quantum mechanics, generically requires with probability one the storage of an infinite amount of information - in particular, if I were to measure an exact and arbitrary real number, I need to physically retain a literally infinite number of digits. It would be like knowing ALL of the digits of $\pi$ - all $\aleph_0$ of them, and put that onto a physical diskette. As far as we can tell, Nature abhors an actual infinity, at least within a finite volume of space and time. (GR's singularities create conflicts with our other theories and so cannot be trusted to actually exist, and every time we've gotten to observe what should be an "actual" singularity according to some theory, that theory shows itself incomplete.)

This would apply even if particles were classical - we'd still need an infinite storage device to store that much information coming out of the measurement.

However, QM differs from CM in that once you get something to an "eigenstate" of these parameters hypothetically, the theory soon quits making sense. In particular, suppose you measure position of a particle exactly. Now we have an "eigenstate" of position. The momentum distribution for this, however, is not normalized. Suppose now we measure the magnitude of momentum afterward. What happens now is you effectively draw a random real number from $\mathbb{R}^+$ with a uniform distribution. The expected value of this draw is $\infty$, meaning it should typically be an infinitely big real number. (Or alternatively, that the number $\infty$ belongs to $\mathbb{R}$.) This is a logical contradiction within the theory if you assume every state can be subjected to measurement, which means that in a way such states are theoretically unsound, when they are not in classical mechanics. Technically it is then a trivial theory (as inconsistent, the principle of explosion tells you you can deduce everything, including literally that the President of the US is a giant tomato), and you must exclude these as states a particle can actually attain, or that some measurements are undefined and not merely probabilistic (thus making the theory puke hard if you tried to imagine describing an apparatus subjected to one of them), or formulate in paraconsistent logic, to restore sanity.

Whereas CM, on the other hand, has no problem with such parameters as exact - they always are, and the theory makes sense with that.

That said, CM is itself subject to other pathologies that may be harder to remove. There are CM systems in which that the evolution of the system is nondeterministic in a more severe sense than QM, in particular where there is literally no law that can choose how they will evolve. One such example is "Norton's dome", where we imagine a particle at the peak of a "dome" whose height is given by the continuous but not everywhere-differentiable function

$$h(r) = \frac{2}{3g} r^{3/2}$$

The peak ($r = 0$) is a cusp. If the particle is placed there, the CM equations predict that there are consistent solutions to $r(t)$ which are constant for an arbitrarily long time before motion commences, meaning that the future behavior is not determined by the present behavior, and moreover there is not even a probabilistic law to choose which one will actualize.

Of course this is an unphysical situation just as an infinitely precise measurement is, but that is irrelevant to thinking of the extent of theoretical logical consistency and determinism of the theories themselves.

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    $\begingroup$ The argument at the beginning of this answer doesn't really work: You don't need infinite bits to store, for instance, representations of the values $0,1,\sqrt{2},\pi,...$, evidenced by the fact that I managed to type them into this comment box. The rest of your answer works perfectly fine for variables with continuous spectrum like position and momentum, but it doesn't seems to address the situation of "Hermitian operator with discrete non-[de]generate spectrum" the question explicitly asks about. $\endgroup$ – ACuriousMind Sep 7 '18 at 8:32
  • $\begingroup$ @ACuriousMind : There is in most circumstances a probability of zero that an arbitrary dynamical quantity, with real-valued physical variables, will be in one of those easily-representable quantities. Moreover, even if it is one of them, I believe that finding a suitable finite representation that can handle full arbitrariness would be a non-computable problem. $\endgroup$ – The_Sympathizer Sep 7 '18 at 9:33

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