5
$\begingroup$

Kurt Jacobs (Quantum Measurement Theory and its applications, 2014) writing about the Measurement Problem says (italics mine)

..there is no collapse. Any set of mutually orthogonal subspaces defines a set of mutually exclusive possibilities but there is no physical mechanism by which one of these is chosen as the one we experience as reality. Measurements do not exist.
So if measurements are unreal, how do we decide when it is appropriate to pretend that a measurement has occured, and select one of the orthogonal subspaces as the outcome?

He then goes on to the standard explanation about the statistical irreversibility of measurement processes to justify this pretending.

Thus it is the emergent irreversible behaviour that leads to thermodynamics that provides the irreversibility necessary for the emergence of an apparent measurement process.

My question: if measurements are only pretend and apparent and any measurement process can be understood within the unitary evolution of standard quantum mechanics (something that many writers on Physics SE seem to accept), is this not the same thing as saying that quantum mechanics contains within it a much larger universe than the part we can experience through measurement i.e. we are a part of an emergent multiverse, to use David Wallace's term for the Everett "interpretation" (something that only a few writers on Physics SE accept)?

I fear that those who argue that quantum mechanics deep down is just deterministic unitary dynamics, but that our observable macroscopic universe is the only real universe want to have their cake and eat it. What am I missing?

$\endgroup$
  • $\begingroup$ Physics is based on observation. What is an observation if not a set of measurements? I find it hard for a physicist to defend the position that measurements do not exist. $\endgroup$ – Stéphane Rollandin Oct 14 '16 at 8:24
  • $\begingroup$ @StephaneRollandin I have unfairly picked out 3 sentences from a 500 page book, then you have picked out one of those sentences. We both know Kurt Jacobs doesn't literally mean that. $\endgroup$ – Bruce Greetham Oct 14 '16 at 9:10
  • $\begingroup$ Maybe, but at times I really wonder what people think when they wash away the notion of measurement in such radical terms. Unitary evolution is given a theoretical primordial standpoint, which I can understand, but on the experimental side it is evidently measurements that are primordial. No theory can afford to do without them without running the risk of versing into solipsism or similar radical philosophical views. $\endgroup$ – Stéphane Rollandin Oct 14 '16 at 9:40
4
$\begingroup$

Quantum mechanics is a theory of (noncommutative) probability, that is in my opinion best understood in terms of Bayesian probability. The information on a system is encoded in the state (probability) and in its unitary dynamical evolution (analogously, one can think of the evolution only affecting observables i.e. random variables, but that's not important for the question at hand).

The act of observing is in some sense "outside" of any probability theory (for the latter simply describes the expected outcomes of such observations, but not the act of observing itself). But let us put that aside for a moment, and focus on measurements.

Measurements by means of instruments and measurement processes can be understood pretty well within the probability theory of quantum mechanics. But one needs to enlarge the system considered to include the instrument. The complex of "system+instrument" evolves unitarily, but the effective dynamics on the system alone (tracing out the degrees of freedom of the instrument) is not unitary and thus irreversible. This is the von Neumann scheme of measurement.

What happens to the noncommutative probability in this measurement scheme is the following. Let us suppose that the starting state of the system is a pure state (a state with maximal bayesian information), described by the orthogonal projection $P_{\psi}$ on the span of the vector $H\ni \psi=\sum_{n\in\mathbb{N}}a_n \varphi_n$ (where the latter is the decomposition in eigenvectors of the random variable $T$ we are measuring corresponding to $\lambda_n$). The instrument is in the state $P_\Psi$, $\Psi\in K$. So the resulting initial state of the complex "system+instrument" is a tensor product state $P_{\psi\otimes \Psi}$ on $H\otimes K$. The measurement process is given by a unitary evolution $U(t)$ on $H\otimes K$ that mixes $\psi$ and $\Psi$: $U(t)(\psi\otimes\Psi)$ is no more a tensor product.

At time $t^*$ when the measurement is finished, we trace out the degrees of freedom of the instrument to obtain the effective dynamics of the system after the measurement. This process is "irreversible", and yields a mixed state for the system. The resulting mixed state is $$\sum_{n\in\mathbb{N}}\lvert a_n\rvert^2 P_{\varphi_n}\; .$$

Such state is then evolved by means of the unitary dynamics $u(t+t^*)$ of the system when isolated (it is no more interacting with the instrument), and is therefore $\sum_{n\in\mathbb{N}}\lvert a_n\rvert^2 P_{u(t+t^*)\varphi_n}$.

It is, in my opinion, misleading to say that we cannot know when a measurement has occurred. The maximal information on the system includes that you know if and when/how many times you put your system in interaction with an instrument (or another environment for what is worth). That essentially amounts to the knowledge of the unitary evolution of the complex "system+instrument(s)(+environment(s))"; and it is not different from the supposed maximal knowledge that we can have in classical mechanics (that in fact is the commutative probability theory that emerges from quantum mechanics when we neglect the noncommutative effects).

Of course, as it happens in any probability theory, if you have done $M$ measurements, the correct way of calculating the probability for the outcome $(x_1,\dotsc,x_M)$ of the measurements is using conditional probability. Now suppose that the time evolution preserves the projectors $P_{\varphi_n}$ above, i.e. $$\sum_{n\in\mathbb{N}}\lvert a_n\rvert^2 P_{u(t+t^*)\varphi_n}=\sum_{n\in\mathbb{N}}\lvert a_n\rvert^2 P_{\varphi_n}\; .$$ Then suppose you do $M$ identical measurements as the one described above. Of course the resulting probability of measuring the outcomes $(\lambda_1,\dotsc,\lambda_M)$ is zero unless $\lambda_1=\dotsc=\lambda_M=\lambda_n$ that has associated probability $\lvert a_n\rvert^2$.

This is a pretty convincing scheme in my opinion, if interpreted (as it should be) as a probability theory. In addition, it does not need to assume any "emergent multiverse" or the alike, simply you assume to have the probabilistic knowledge of the system and surrounding environment, essentially up to how many times the two are interacting/coupled, and how.

References. The existence of measurement processes behaving as described above for any QM observable (with both discrete and continuous spectra) has been proved by Ozawa. This has recently been extended to all (interesting) observables in QFT as well by Okamura and Ozawa. An introduction to von Neumann's measurement scheme can be found in his book.

$\endgroup$
  • $\begingroup$ Yes, I accept it may make sense to treat QM as a Bayesian probability theory and you are satisfied with the mathematical coherence of this, so I take your word on that. Can you clarify how you are answering my question: I take it that you are saying that even with pure state knowledge of system and measurement apparatus QM is still indeterministic about the outcome of individual measurements (due to non-commuting observables) and you just accept this indeterminism as a feature of physics (possibly to be resolved by a deeper theory). Therefore you reject the premise of my argument. $\endgroup$ – Bruce Greetham Oct 14 '16 at 10:55
  • 1
    $\begingroup$ @BruceGreetham I am saying that as every probability theory, QM only predicts the probabilities of possible outcomes for observables. This is true for both classical and noncommutative probabilities. If one is willing to accept QM as a meaningful description of nature (and there are plenty of observations that support this) then you have to accept an intrinsic probabilistic interpretation. Within that picture, measurement instruments and processes fit well. So I don't see the need for introducing multiverses and the alike $\endgroup$ – yuggib Oct 14 '16 at 11:45
  • $\begingroup$ That's fine - but my current question is "does a deterministic interpretation of QM lead to multiverses". There are arguments on SE which say not. You may wish to stay silent on this debate. $\endgroup$ – Bruce Greetham Oct 14 '16 at 11:57
  • $\begingroup$ @BruceGreetham If I understand correctly, the "deterministic interpretation" of QM is that its evolution is unitary. If it is so, the answer to your question is "not necessarily", as explained by the purely unitary theory of measurements explained above. If what you mean is something else, then I don't have a definite opinion. $\endgroup$ – yuggib Oct 14 '16 at 12:04
  • $\begingroup$ Please see physics.stackexchange.com/questions/269587/… by Arnold Neumaier as an example of a deterministic non-multiverse interpretation of QM which motivated this question. I'm trying to reconcile with your ideas. $\endgroup$ – Bruce Greetham Oct 17 '16 at 19:30

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.