3
$\begingroup$

What would be the meaning of the uncertainty principle in the case of hidden-variable interpretations of quantum mechanics, especially contextual hidden-variable theories?


Edit: Another way to put the question would be whether it is possible to circumvent the uncertainty principle within the framework of contextual QM. If not, why is there still irreducible uncertainty in a deterministic theory?

Doesn't this mean that the observational part of the wave function (Quantum Information-al!) is responsible for the uncertainty?

Even more, do we need to inevitably resort to the Observer Effect interpretation (aka noise-disturbance) of the Uncertainty Principle to explain such an irreducible uncertainty within the framework of a deterministic theory?

$\endgroup$
4
  • $\begingroup$ uncertainty product of what observable? $\endgroup$ Jun 23 at 22:32
  • $\begingroup$ Any two conjugate coordinate operators. Should not make any difference in the contextual interpretations of QM. $\endgroup$ Jun 23 at 23:50
  • $\begingroup$ Any hidden variable theory to be considered as an interpretation, would have to reproduce the QM successes in fitting data, including the HUP, imo. $\endgroup$
    – anna v
    Jun 24 at 4:35
  • 1
    $\begingroup$ @annav I guess there's no way you could have understood the question. I'm sorry. $\endgroup$ Jun 24 at 8:08

3 Answers 3

3
$\begingroup$

I think you are asking what is the hidden-variable interpretation/description of unsharply defined observables when, in the standard interpretation of QM, these observables are incompatible and the quantum state is not eigenstate of both variables.

For systems quantistically defined in finite dimensional Hilbert spaces, existence of non commuting elementary observables already implies contextuality in realistic hidden variable interpretations of QM, in view of the Kochen Specker theorem.

That theorem assumes hypotheses on the (non-quantum) valuation functions $v_\lambda(A)$ only for pairs of compatible observables $A,B$ and proves that the theory must be contextual. That is the powerfulness of the theorem.

How the quantum phenomenology of uncompatible observables is described in hidden variable theories is matter of each concrete hidden variable theory (independently from the fact that all these theories must be contextual).

Contextuality alone is by no means able to provide an intepretation of the quantum phenomenology of incompatibke variables, it is only a necessary constraint.

Presumably, a description should be related to the epistemic (vs ontic) stochasticity of the hidden-variable state in a hidden variable theory. In other words an explanation should be given in terms of the probability distribution which describes our knowledge of the hidden variable $\lambda$.

A problem is the the (standard version of the) KS theorem applies to finite dimensional Hilbert spaces and bounded observables. Therefore it cannot be directly applied to, e.g., a particle on the real line (where the Heisenberg principle applies!) without stronger hypoteses on the set of observables and on the valuation function. However, you can focus on hidden variabe theories explicitely constructed to deal with this case. I know only one of them: the Bohm theory. It is contextual and it includes an explanation of the Hesinberg principle

$\endgroup$
8
  • $\begingroup$ I did not know that Kochen-Specker sharply depends on finite dimensionality of the Hilbert Space. Can you introduce some reference that points to this aspect particularly please? $\endgroup$ Jun 24 at 10:58
  • 1
    $\begingroup$ The proof depends on the nature of the observables and the valuation function. The standard hypoteses are that there is a non-zero map $v: B(H)_{sa} \to H$ such that it is linear and preserves the product of compatible observables (the selfadjoint elements of $B(H)$). This map cannot exist if $H$ is finite dimensional. If $H$ infinite dimensional the physically relvant observables are unbounded and issues with domains arise. On the other hand the contraddiction arises from the Gleason theorem which needs that the valuation map is strongly continuous. In finite dimension it is automatic. $\endgroup$ Jun 24 at 11:05
  • 1
    $\begingroup$ I am not saying that the KS theorem does not hold in infinite dim spaces, I am just saying that it needs more specific hypotheses. $\endgroup$ Jun 24 at 11:06
  • $\begingroup$ (See e.g. chapter 5 of my book link.springer.com/book/10.1007/978-3-030-18346-2#aboutBook) $\endgroup$ Jun 24 at 11:09
  • $\begingroup$ Nice answer. For the general account of the Quantum Phenomenology, can I assume that incompatibility simply stands for the fact the measurement device(aka commutative Ring of operators) will change the value of the measured observable that is incompatible with the operators of the same ring? $\endgroup$ Jun 24 at 11:25
1
$\begingroup$

The uncertainty principle has the exact same meaning with hidden-variable interpretations as with any other interpretation of QM. It says that in order to reproduce the predictions of QM, the probability distribution over the hidden variables must be such that the product of the statistical uncertainties of the measured value of certain observables satisfy certain lower bounds.

There might be some philosophical sense in which the quantum state isn’t “truly” uncertain for hidden-variable theories, but the uncertainty of what result will be measured simply reflects our classical lack of knowledge of the exact state of the hidden variables. But the practical relevance is the same - the mathematical mapping between the state of the hidden variables and the outcome of measuring observables must be such that certain inequities are satisfied.

$\endgroup$
9
  • $\begingroup$ About the second part, are you saying that the uncertainty principle reduces to the so-called classical observer effect? The fact that measuring a system will definitely changes the system. I actually want to know the root cause of the uncertainty in a deterministic theory? Why should uncertainty still survive there without taking into account the role of "measurement"? $\endgroup$ Jun 24 at 11:03
  • $\begingroup$ @BastamTajik In my opinion, the HUP is largely independent of the fact that measurement changes the state of the system. You could certainly imagine a logically coherent alternative theory to QM in which the outcomes of measurements still follow the Born statistics, but measurements leave the state of the system unchanged. Such a theory would still have the HUP but wouldn't have an "observer effect". $\endgroup$
    – tparker
    Jun 24 at 11:20
  • $\begingroup$ Then please let me know your interpretation of incompatibility within HV scenarios? $\endgroup$ Jun 24 at 11:44
  • $\begingroup$ Hidden-variable models of quantum mechanics certainly must have an observer effect in which making a measurement changes the state of the system, because that is what we observe empirically. This effect explains incompatibility. I'm just saying that the HUP is a logically independent result that also happens to follow from the axioms of QM. You could have the HUP even if there was no observer effect, as it simply describes the product of uncertainties in future measurements that have not yet been made. $\endgroup$
    – tparker
    Jun 24 at 12:37
  • $\begingroup$ You are probably imaging a hidden-variable theory in which repeated measurements in the same basis always result in the same outcome. I am imaging a hidden-variable theory in which the hidden variables stochastically reset in between measurements, so that there is no "wave function" collapse at all, and repeated measurements even in the same basis remain stochastic. $\endgroup$
    – tparker
    Jun 24 at 12:39
0
$\begingroup$

The Hidden Variable Interpretation does not rule out probabilities. It just states that QM is not truly random and there is a parameter we do not know about. And since right now we have found no hidden parameter, we still don't know where the particle would be. That means probability density is still probability density, and uncertainties are still uncertainties. It's just that in this interpretation, probabilities and uncertainties are subjective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.