Does quantum mechanics need a conscious observer?

1. Jan 6, 2014

atyy

Textbooks like Landau and Lifshitz avoid the observer having to be conscious, by saying a measurement is the interaction of a quantum system with classical apparatus to produce a classical result.

However, there is an analogy that is often drawn that wave function collapse is like statistical updating or Bayesian conditioning. If one accepts this analogy, doesn't the observer have to be conscious, since it is the observer's knowledge that is being updated?

2. Jan 6, 2014

StevieTNZ

Some people (including physicists) advocate that view that it is indeed a conscious observer that performs the measurement. I personally subscribe to that interpretation.

However, its very hard to tell which interpretation is correct with the lack of experimental means of testing which interpretation is correct over others. It all depends whether standard QM is correct after all, and doesn't need modifying (i.e. conforming to GRW theory, which includes collapse as part of its equations; effectively a physical collapse of the wave function).

3. Jan 6, 2014

Staff: Mentor

The question is - does QM NEED a conscious observer.

The obvious answer is no - since many many interpretations such as my ignorance ensemble interpretation do not require it.

Remember this idea came out of Von-Neumann's famous textbook when he showed the Von-Neumann cut can be placed anywhere. But when you trace it back the only place different is human consciousness - so that's where he placed it. But nowadays with decoherence we know a place that is different - just after decoherence. Place it there and all this silly conscious observer created rubbish goes out the door.

The other high priest of conscious causes collapse, Wigner, when he heard of some early work by Zurek on decoherence, did a total 180% about face and totally rejected it.

So the answer is - why bother - simply place it just after decoherence. I think SOME (not all, but some) people that hold to it have a penchant for deliberately wanting the world to be weirder than it actually is. Also some new age touchy feely types use it as justification for whacko silly stuff like What The Bleep Do We Know Anyway that would undoubtedly make a confirmed rationalist like Von-Neumann wince.

BTW - I hasten to add - its a valid interpretation some people hold to even today - but they are well in the minority.

Thanks
Bill

Last edited: Jan 6, 2014
4. Jan 6, 2014

atyy

I should have said, "Does QM need a conscious observer if wave function collapse is interpreted as a kind of statistical updating" If it is statistical updating of knowledge, doesn't it need a conscious observer, since only such observers have knowledge?

5. Jan 6, 2014

Staff: Mentor

Hmmm.

Expressed that way it's a more interesting question.

The same could be said of probability theory in general.

Personally I think no. Remember it's a model - and models are something only understood by rational entities. Its part of the model that this updating occurs - its purely a theoretical construct.

Thanks
Bill

6. Jan 7, 2014

.Scott

If you require a "conscious observer", you would then need to define what exactly is a "conscious observer".

7. Jan 7, 2014

Jano L.

As Bhobba says, it is the same as with probability in general. But the answer is no. I am sure that there are cases where computers take data as they come in and use them to calculate corresponding probabilities. One could program computer that monitors results of sequential Stern-Gerlach experiment and calculate the same things physicist would do.

8. Jan 7, 2014

Jano L.

Good point! I guess most people would say machine is conscious if it were capable to respond in an impressive way to inputs they provide. Perhaps machine capable of doing calculations with $\psi$ and predicting results of atomic experiments would qualify as conscious at least in some technical sense:-)

9. Jan 7, 2014

StevieTNZ

I define conscious the easiest way possible: what I perceive, hear, touch, in my mind. Without getting too philosophical on what those terms mean (especially 'mind'). What I'm seeing right now is in my mind, for example, is the easiest approach for me.

10. Jan 7, 2014

atyy

How about if I asked whether the rational entity must be in the model itself? Say if I distinguish two layers of model: (1) reality and (2) the model of reality. In a model of such as Newtonian mechanics, it does seem that I must put a rational agent like myself in (1), but not necessarily in (2). So could I refine the question to ask, that while all models require a rational agent in (1), do Bayesian conditioning models of the wave function collapse require that the rational agent also be in (2)?

As you say, it is really a question about probability. Based on a Bayesian interpretation, I believe Scott Aaronson gives one definition of a "rational person" http://www.scottaaronson.com/blog/?p=822: "As for what “rational” means, all we’ll need to know is that a rational person can never assign a probability of 0 to something that will actually happen.'

Does a frequentist interpretation of probability (eg. Durr, Goldstein, Zanghi's version of dBB http://arxiv.org/abs/quant-ph/0308039) avoid a rational agent in (2)? Actually, is there even such a thing as frequentist updating? Or is frequentist updating necessarily irrational?

Last edited: Jan 7, 2014
11. Jan 7, 2014

strangerep

QM is not only about the association between a (modified) notion of probability and squared matrix elements.

It's also about spectra of the observables.

Consider angular momentum. If the generators are not represented as Hermitian operators on a Hilbert space, then you can't derive the well-known half-integer spectrum.

So one might as well ask: "is an electron still an electron if there's no conscious observer?"

Or: does the hydrogen atom collapse (electron falling into the nucleus) if there's no conscious observer"?

(Duh)

12. Jan 7, 2014

Jano L.

That is a strange definition. There were events that had predicted probabilities 0 and still happened. One king won an island in the game of dice, because he managed to throw 7. The probability of such event is rationally 0, but nevertheless it happened (the die split into two pieces while rolling and these ended up on 3 and 4). Still, the prediction that ' "the die will end up showing 7" has probability 0' was rational.

The same thing applies in any application of probability theory - events that we deem irrelevant have probability 0, but no theorizing can prevent them from happening.

I'm sorry, but I do not understand your two layers. I do not see how introduction of rational entity into theoretical scheme helps in any way.

13. Jan 7, 2014

atyy

@Strangerep and @Jano L, the question is in the context of the many attempts to understand collapse as a form of Bayesian conditioning. It's a very old idea. bhobba often mentions it casually - it was really his common use of the analogy between collapse and statistical updating, as well as his saying that the wave function doesn't need a conscious observer ("rational" might have been a better word choice) that lay behind my question, since they weren't obviously compatible to me. The analogy is hinted at in Cohen-Tannoudji, Diu and Laloe's text, and it is mentioned in the text of Wiseman and Milburn.

Some recent attempts include:

http://arxiv.org/abs/quant-ph/0106133
Quantum probabilities as Bayesian probabilities
Carlton M. Caves, Christopher A. Fuchs, Ruediger Schack
Phys. Rev. A 65, 022305 (2002)

http://arxiv.org/abs/1107.5849
Towards a Formulation of Quantum Theory as a Causally Neutral Theory of Bayesian Inference
M. S. Leifer, R. W. Spekkens
Phys. Rev. A 88, 052130 (2013)

Bayesian coherence is a formal theory of rationality, and one of the requirements is that the prior must assign non-zero probability to the truth, which is why I believe Aaronson used that as his definition of a rational agent - one is using a Bayesian framework.

A similar line of query is what the nature of hidden variables behind quantum mechanics can be. Depending on the relationship between the hidden variables, the wave function can be interpreted as "psi-ontic" or "psi-epistemic", where I think the latter means something like "belief". One popular set of definitions is in:

http://arxiv.org/abs/0706.2661
Einstein, incompleteness, and the epistemic view of quantum states
Nicholas Harrigan, Robert W. Spekkens
Found. Phys. 40, 125 (2010)

Although maximally psi-epistemic interpretations seem to have been ruled out http://arxiv.org/abs/1208.5132 , the existence of psi-epistemic interpretations is said to have been proved in:

http://arxiv.org/abs/1201.6554
Distinct Quantum States Can Be Compatible with a Single State of Reality
Peter G. Lewis, David Jennings, Jonathan Barrett, Terry Rudolph
Phys. Rev. Lett. 109, 150404 (2012)

http://arxiv.org/abs/1303.2834
Psi-Epistemic Theories: The Role of Symmetry
Scott Aaronson, Adam Bouland, Lynn Chua, George Lowther
Phys. Rev. A 88, 032111 (2013)

I don't know whether a psi-epistemic interpretation necessarily requires a Bayesian interpretation, but I believe the search for the possibility of a Bayesian interpretation is a motivation for psi-epsitemic interpretations.

Interestingly, there is a form of many-worlds in which Bayesian probability plays a role:

http://arxiv.org/abs/quant-ph/0312157
Quantum Probability from Subjective Likelihood: improving on Deutsch's proof of the probability rule
David Wallace

Last edited: Jan 7, 2014
14. Jan 7, 2014

strangerep

The notion of collapse is not necessary to derive the correct angular momentum spectrum.

But... I guess you already know that I consider the whole wave-function-collapse-to-eigenstate-after-measurement thing to be a load of nonsense. :yuck:

15. Jan 7, 2014

atyy

Assuming your view about collapse is more or less in line with Ballentine's book, let's agree to disagree about that specific issue in this thread.

16. Jan 8, 2014

martinbn

Why only conscious observers can have knowledge, and not any classical or classical-like object? A classical measuring device records the result of the measurement and that is its knowledge.

17. Jan 8, 2014

atyy

If I understand .Scott and Jano L correctly, a conscious observer is probably a classical object. So in Landau and Lifshitz, it is the irreversible reading on the classical measuring apparatus that is intuitively specified. In the Bayesian analogy, it is the rational agent that is intuitively specified (let's use "rational" instead of "conscious", since that seems closer to standard Bayesian terminology). So the question then is, what would be a more precise specification of "rational" agent? Presumably it would do "Bayesian" updating, and have a "rational" prior. In the framework of Bayesian coherence, I think it can be shown that any prior which assigns non-zero probability to the truth will converge to the truth, and so is rational (at least that's how I understand Scott Aaronson's definition of a rational agent http://www.scottaaronson.com/blog/?p=822).

Last edited: Jan 8, 2014
18. Jan 8, 2014

StevieTNZ

Everything is quantum, though (in principle). So in reality there is no classical apparatus.

I could go further to state that the fundamental equation of QM, the Schrodinger equation, only predicts probabilities, and that you never end up with a definite state of knowledge if there exists Classical devices - if a quantum system interacts with a Classical device.

19. Jan 8, 2014

atyy

But if everything is quantum, eg. many-worlds, then there is no wave function collapse. However, apparent collapse must still hold for an observer in many-worlds. In some forms of many-worlds that observer uses decision-theoretic principles or Bayesian inference. So in that context we can talk about the question: if collapse is analogous to Bayesian updating, must quantum mechanics contain a "Bayes-rational" agent?

http://arxiv.org/abs/0906.2718
A formal proof of the Born rule from decision-theoretic assumptions
David Wallace

20. Jan 8, 2014

StevieTNZ

Exactly. But it doesn't stop us from seeing a definite reality.

So maybe we need to re-think what 'wave function collapse' actually means if everything is quantum. Is there a wave function collapse per say that causes a definite reality to arise, or does the Schrodinger equation still hold when we see a definite reality - but we're just seeing one component of the superposition, where the superposition principle still holds true?