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In quantum Bayesianism (QBsim) interpretation, the wave function $| \psi \rangle$, or density operator $\hat{\rho} = | \psi \rangle \langle \psi |$, is not objective. It is instead interpreted as the coding tool of statistical knowledge of each individual agent (or: observer). This is so far my understanding of the QBsim. In addition I also assume that, the QBism wouldn't provide new rules for predicting quantum phenomena. Assuming the above understanding is accurate and I haven't missed important points of the QBsim, I have a question about it.

Using the famous example in Feynman's book, for the double-slit experiment of an electron, suppose that there are simultaneously two independent agents:

1) the first agent sets up the measurement device near two slits, detecting which path the electron goes by (assuming the device does not destroy the electron, instead disturbs the electron inevitably).

2) the second agent sets up an array of "clicking" device on the screen of far-field, which records the electron's interference pattern.

Then the two agents independently predict whether there is an interference pattern on the screen or not. Let the two agents be independent and not communicate about the wave-function information. How can predictions of the two agents be consistent with the subjective nature of wave functions? Or put it in another way:

1) if the first agent predicts correctly, which appears to be obvious, and

2) the second agent predicts wrongly, which appears to be due to the uninformed "state" knowledge of this agent,

would the contradiction recover the "objectivity" of two agent's wave functions? More specifically, the first agent's "state" knowledge "collapsed" to the correct (or: objective) wave function, while the second agent was ignorant and possessed the wrong wave function. I believe I could write all above reasoning in the usual quantum formalism. So I am wondering if I misunderstood the QBism, or otherwise this example reveals a fundamental problem in the QBism interpretation.

EDIT

The motivation for my question came from my reading Englert's paper:

http://www.physics.nus.edu.sg/~phyebg/arXiv.1308.5290v2.pdf

where on page 8 (section 6) it is talking about the state reduction, which is interpreted as the observer's knowledge update. I just realized that, my original question is not appropriate concerning of what to predict for the two agents. But the point should still be valid:

1) from the first agent experience, the state reduced to a mixture of two paths:

$(|path_1, A_1\rangle \langle path_1, A_1| + |path_2, A_2\rangle \langle path_2, A_2|) / 2$ (1)

2) while from the second agent experience, the state evolved to an entanglement of two paths (before the electron hits the screen).

$(|path_1, A_1\rangle + |path_2, A_2\rangle) / \sqrt{2}$ (2)

where $A_i$ (i = 1, 2) refers to the state of observer 1's detector.

those two states are two different mathematical objects, which are considered "subjective" for each agent. Now it is true that both agents make the same prediction about "no-interference" on the screen. However, would it possible for the two agents to make contradictory predictions about other experiments - since the states in eq. (1) and (2) are two different mathematical objects?

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  • $\begingroup$ The second observer has not properly understood the physical process they are trying to model: it is not surprising that they are wrong. Over time, with sufficient measurements, the second observer would be able to perform process tomography on the system (double slit experiment + mischievous first observer) and make the correct prediction about the future behaviour using their "subjective" state. This subjective state would differ from the first observer's but still yield the same predictions for observables accessible to the second observer. No fundamental problems here. $\endgroup$ – Mark Mitchison Jan 12 '15 at 8:16
  • $\begingroup$ @MarkMitchison My question is not really about why the second observer's prediction was wrong. My question is concerning about the "objectivity" of the observer's wave function. Please see my response to Hindsight below for more detailed discussion. Besides, in order not to make the question more complicated, I would like to focus on one-step prediction, instead of continuing to collect data and update the prediction. $\endgroup$ – G. Xu Jan 12 '15 at 17:26
  • $\begingroup$ Okay. My point is simply that second observer should try to assign the correct state to predict the future based on his own subjective experience. Clearly, this should not be the same state as that of the first observer, whose experience is completely different. Note that there is no such thing as "one-step" prediction in science. You always feed forward experience of the past into your model for predicting the future. $\endgroup$ – Mark Mitchison Jan 12 '15 at 17:31
  • $\begingroup$ @MarkMitchison Sorry for not making my point clear on "one-step" prediction. What I really meant is - let's look at the prediction of the first step. Hope this clears out the confusion. $\endgroup$ – G. Xu Jan 12 '15 at 18:26
  • $\begingroup$ @MarkMitchison I have edited and updated my question based on the correct state for the second observer. $\endgroup$ – G. Xu Jan 12 '15 at 19:16
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QBism is much more than that: the second observer's subjective experience contains the two slits, the measuring device installed by the first observer, the first observer himself and the rest of the world. Only by taking into account all this data can he estimate probabilities properly.

But even if their experiences disagree, there is no way they could know that. Because each one thinks of himself as supreme and the other one being just quantum matter.

I would like to quote David Mermin here:

I would say that each user of quantum mechanics builds a personal representation of reality based entirely on his experience. Since what you tell me about your experience is part of my experience, we can expect considerable overlap in our different models of reality.

Note that "considerable overlap" does not mean that everything we feel should be the same.

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  • $\begingroup$ @MarkMitchison My question is concerning about the "objectivity" of the observer's wave function. The way saying that "the second observer's subjective experience contains the two slits + the measuring device installed by the first observer + the rest" is actually suggesting that, his/her wave-function is not completely subjective. More specifically, the second observer's wave-function is not fully equivalent to his/her knowledge (or: information-only) state. $\endgroup$ – G. Xu Jan 12 '15 at 17:55
  • $\begingroup$ @MarkMitchison This indeed makes more sense to me - event if the second observer is not informed of the first observer's device installation, the wave-function for him/her should still include that device's state for making a correct "objective" model. $\endgroup$ – G. Xu Jan 12 '15 at 17:56
  • $\begingroup$ @G.Xu In QBism there is only the subjective wavefunction. There is nothing "objective". $\endgroup$ – Mark Mitchison Jan 12 '15 at 17:57
  • $\begingroup$ @MarkMitchison Ok, I should not use the word "objective" for the correct model. But my main concern is, would sill some testable contradictory predictions for two agents exist? Why are we so sure there would never exist such experimental tests? $\endgroup$ – G. Xu Jan 12 '15 at 19:29
  • $\begingroup$ @G.Xu in QBism, there is an objective world of yet unknown structure. Quantum mechanics is a tool used by agents to build personal realities based on their experience and this objective world. Wave functions are measures of their expectations of the world having some properties. $\endgroup$ – Prof. Legolasov Jan 13 '15 at 1:47
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This is an answer to the second version of the question, namely

would it possible for the two agents to make contradictory predictions about other experiments

The answer is no, assuming that your original distinction between the two observers holds. To be precise, this distinction is that observer 1 has access to the state of the detector after the measurement, while observer 2 does not. This means that 2 can trace over the states of the detector, yielding the state $$\rho_2 = \mathrm{Tr}_A\; |\psi_2\rangle\langle \psi_2| = \frac{1}{2}(|path_1\rangle\langle path_1| + |path_2\rangle\langle path_2|).$$

It is easy to verify that observer 1 obtains the same state when tracing over the detector. Therefore, both observers agree on all measurements that could be performed on the $path$ variables only, including whether or not an interference pattern can be found on the screen. In other words, no measurement on the $path$ variables alone can distinguish the two states.

It is only possible to distinguish the two states if you measure some global property of the particle+detector system, i.e. look at correlations between path measurements and detector measurements. But by assumption, observer 2 does not have access to the results of observer 1's measurements.

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  • $\begingroup$ Yes, this is clearly right. This is the usual no-go proof with mixed states. I guess I am trying to seek a generalized statement like this - for a given quantum system, with a set of measurements $M$ and a set of observers, each of whom are given the access to a certain subsets of $M$, a necessary condition for the QBism interpretation to be valid is that there should exist no contradictory predictions among all observers. $\endgroup$ – G. Xu Jan 13 '15 at 0:00
  • $\begingroup$ @G.Xu Yeah, but the point (which I think Englert was also trying to make) is that QBism, like all interpretations, ultimately makes predictions in exactly the same way as every other interpretation, i.e. by assigning a density matrix consistent with all the prior information available, and then applying the Born rule to measurements. So QBism doesn't predict contradictions because QM doesn't. The only thing that sets QBism apart (philosophically) is its abject denial of realism. $\endgroup$ – Mark Mitchison Jan 13 '15 at 0:12
  • $\begingroup$ I disagree that, an interpretation of QM would only differ philosophically from the QM itself. I know this is sort of subtle. My simple reasoning is as follows: any interpretation would have to add additional assumptions beyond the QM itself, or substitute a few assumptions by new ones. For an example, Von Neumann's wave-function reduction is an extra assumption beyond the QM theory. Due to additional assumptions, there is no reason to sufficiently infer that an interpretation always predicts the same as the QM for every possible experiment. But finding out an experiment is not easy. $\endgroup$ – G. Xu Jan 13 '15 at 2:23
  • $\begingroup$ @G.Xu Maybe I don't understand what you mean, but von Neumann's projection postulate is not an additional assumption beyond QM theory. It is part of the standard theory of quantum measurement. In fact, interpretations --- in the usual sense of the terminology --- differ only philosophically by definition. $\endgroup$ – Mark Mitchison Jan 13 '15 at 2:34
  • $\begingroup$ It might be true that von Neumann's projection postulate is part of the standard theory of quantum measurement. But the theory of quantum measurement cannot be solely derived from the standard quantum theory. Besides, the projection postulate has never been experimentally tested. Or equivalently, so far, there is no experiment whose result exclusively depends on the prediction by using von Neumann's projection postulate. In this sense I mean the projection postulate is an additional assumption. $\endgroup$ – G. Xu Jan 13 '15 at 5:13
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You might like https://arxiv.org/abs/hep-th/0110253 and the references here, especially the discussion of the simple interferometer on page 11...

http://physics.bu.edu/~youssef/quantum/quantum_refs.html

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    $\begingroup$ Answers with references only are not encouraged. Please summarize the main points that would answer the question and put the references in addition. $\endgroup$ – flaudemus Mar 2 at 22:26

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