Insights 9 Reasons Quantum Mechanics is Incomplete - Comments

  • #51
Lüders rule, as I understand it (see below), seems to require an observer outside the system being measured. This is not the case in QFT, where observers and everything else are comprised of excitations of the quantum fields.

http://philsci-archive.pitt.edu/4111/1/Lueders_rule_BuschLahti.pdf
 
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  • #52
Avodyne said:
Lüders rule, as I understand it (see below), seems to require an observer outside the system being measured. This is not the case in QFT, where observers and everything else are comprised of excitations of the quantum fields.

http://philsci-archive.pitt.edu/4111/1/Lueders_rule_BuschLahti.pdf
Well, for quantum field theory you may need the thermal interpretation, where observers can be part of the system and collapse happens as an approximate effect due to the ignored environment.
 
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  • #53
Avodyne said:
Lüders rule, as I understand it (see below), seems to require an observer outside the system being measured. This is not the case in QFT, where observers and everything else are comprised of excitations of the quantum fields.

http://philsci-archive.pitt.edu/4111/1/Lueders_rule_BuschLahti.pdf
How does this affect an epistemic view? Even if you include the observer themselves within the quantum description there is no need for the quantum state to be a physical/ontic object and thus no need for a dynamical or mathematically detailed description of where the other branches "go".

For example in a dice roll there are probabilities for each roll ##p(1)##, ##p(2)## and so on. I could also include the observer in this:
$$p(1, Observer\hspace{2pt} sees \hspace{2pt}1)$$
$$p(2, Observer\hspace{2pt} sees\hspace{2pt} 2)$$

However there is no need to explain where ##p(2, Observer\hspace{2pt} sees\hspace{2pt} 2)## "goes" upon the outcome ##1##.

Since observers can be included within the system even in QM, this isn't particular to QFT.
 
  • #54
DarMM said:
How does this affect an epistemic view?
Because you still need rules to decide when to modify the state by chopping off branches. If you do it under the wrong circumstances, you will destroy observable quantum coherence effects.

Psi-epistemic interpretations have all sorts of other problems as well:

https://arxiv.org/abs/1303.2834
 
  • #55
Avodyne said:
Because you still need rules to decide when to modify the state by chopping off branches. If you do it under the wrong circumstances, you will destroy observable quantum coherence effects.
You remove the branches related only to unobtained outcomes for the systems you actually observe. This is not in contradiction with a superobserver retaining a superposed description of your entire lab. See for example Richard Healey's "The quantum revolution in philosophy" Chapter 11 for a good summary of this.

Avodyne said:
Psi-epistemic interpretations have all sorts of other problems as well:

https://arxiv.org/abs/1303.2834
That paper concerns psi-epistemic views where there is an underlying ontic space obeying the ontological framework axioms, just like the PBR theorem. It says nothing about epistemic theories in general such as for example acausal views. It's not really an issue for any of the epistemic views actually being worked on. Especially ones where there is no underlying ontic hidden variable space.
 
  • #56
I consider orthodox quantum mechanics to be incomplete (or maybe inconsistent) and I don't think it really makes any difference whether you consider the wave function to be epistemic or ontological. Wigner's friend or Frauchiger and Renner shows this incompleteness.

Rather than going through the F&R argument again, let's just take something much simpler. Alice and Bob prepare an electron so that it is a superposition of spin-up and spin-down. Alice measures the spin. According to orthodox quantum mechanics, she either gets spin-up with such-and-such probability, or she gets spin-down with such-and-such probability.

But Bob, considering the composite system of Alice + electron, will predict that that composite system will evolve into a superposition of "the electron is spin-up and Alice measured spin-up" and "the electron is spin-down and Alice measured spin-down". From Bob's point of view, Alice doesn't get a definite result, but somehow turns herself into a superposition of both results.

So it seems to me that either you're led to (1) the "fantasy" that is many-worlds, or (2) quantum mechanics is inconsistent, or (3) somehow Alice being in a superposition of states doesn't preclude her being in a definite state.

The latter possibility is realized by the Bohmian interpretation, but I would say that if you take that way out, then you are admitting that standard quantum mechanics is incomplete, because the Bohmian interpretation has additional variables (the actual positions of particles) that are not present in standard QM.

Calling the wave function "epistemic" rather than "ontological" doesn't actually accomplish anything, in my opinion. So you want to interpret Alice in her superposed state as being epistemological? Does that mean that she's ACTUALLY in some definite state, and we just don't know which? The problem with that is that there is actually a difference between "Alice is in this state or the other, we just don't know which" and "Alice is in a superposition of this state and that state". I don't think you can consistently treat the latter as epistemological. In any case, if you say that the superposition reflects your lack of information about the "true" state of Alice, that to me means again that quantum mechanics is incomplete, since it presupposes a "true" state.
 
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  • #57
Parsing your description of the classic thought experiment, StevenDaryl

Alice and Bob prepare an electron so that it is a superposition of spin-up and spin-down. [\quote]
fine so far.
Alice measures the spin. According to orthodox quantum mechanics, she either gets spin-up with such-and-such probability, or she gets spin-down with such-and-such probability.
Yes and so you should then taking into account that she made this measurement and recorded her result in a classical copiable memory by describing Alice + electron via a density operator:
\rho =\rho_A\otimes \left( \begin{array}{cc} 0.5&0\\0&0.5\end{array}\right)
But Bob, considering the composite system of Alice + electron, will predict that that composite system will evolve into a superposition of "the electron is spin-up and Alice measured spin-up" and "the electron is spin-down and Alice measured spin-down".
If Bob does this he is making a grave mistake. Alice is not at zero entropy and so cannot be described sharply. He, if he wishes to describe Alice plus electron before the physical and dynamic act of measurement Alice will make occurs would presumably use a density operator to include the entropy dump Alice necessarily must utilize to amplify the electron's spin component into a macroscopic record.
From Bob's point of view, Alice doesn't get a definite result, but somehow turns herself into a superposition of both results.
From Bob's point of view, if he has correctly represented the measurement process of Alice, if he correctly represents the dynamic that makes her recording of the electron spin possible, he will end up with a density operator of the form I mentioned above. In short Alice as an observer necessarily must interact with the system entropically and decoher it. At this point Bob is not describing this Alice but any such Alice electron pair given the same or equivalent initial conditions. Then Bob classically measures the record of Alice and collapses the classical probability distribution down to a single classical sharp state just the same way that the LOTTO commission does each time they bounce their ping pong balls and collapse the value distribution over the tickets to sharp "win" or "lose" states.
 
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  • #58
Well first of all after going through the paper more I don't think FR adds anything beyond Wigner's friend.

Secondly an epistemic account will by nature be taking the view that QM is incomplete as it views a central object in it as non-representational. What the epistemic approach is supposed to add is not a complete and closed account of the ontology of nature (explicitly because they are epistemic). Rather they seek to provide an explanation of several of the properties of quantum states by saying that they are epistemic objects. So for example teleportation, local indistinguishability, no cloning, remote steering, interference, non-commutativity are viewed as having a more natural explanation if you understand ##\psi## as epistemic.

However saying that these properties of ##\psi## have a more natural explanation if ##\psi## is a sort of generalized probability distribution, does not mean you're claiming to know the fundamental ontology of nature. In fact you're explicitly not, since you are providing an epistemic account.

The two main classes of epistemic views then differ on what the "completion" of QM will be like. Roughly speaking ##\psi##-statistical views will see the underlying theory as having some strange ontology like retro or acausality. ##\psi##-epistemic/doxastic views like QBism, Bub or Copenhagen view a completion as impossible for reasons that differ between them.

However to reiterate, by their nature they deny that QM is a complete representational account of the world. The simply seek to explain the properties of ##\psi##.

stevendaryl said:
I don't think you can consistently treat the latter as epistemological
Spekken's toy model gives an epistemic set of states that have interference at the observer + system level, while having a definite outcome at the system level. Superpostion of the total system isn't incompatible with definite outcomes for subsystems. If it was epistemic views would be finished.
To explain briefly in Spekkens toy model superposition arises from the structure of the space of probability distributions under the presence of an epistemic limit. When considered over "system + environment" that limit has a different form than over just "system", so we have a superposition for the former, but a definite outcome for the latter.
Actual epistemic views are more complicated than this, but it gives a good example of how superpositions are compatible with definite subsystem outcomes.
 
  • #59
jambaugh said:
Yes and so you should then taking into account that she made this measurement and recorded her result in a classical copiable memory by describing Alice + electron via a density operator:

No, you shouldn't. That's introducing an ad hoc element to quantum mechanics. Through unitary evolution a pure state never turns into a mixed state. It becomes a mixed state through a modeling decision on OUR part. We can trace over unobservable degrees of freedom and produce a mixed state, but as I said, that's something WE do. It's not a physical effect.
 
  • #60
Let me add again, the collapse is not mysterious for classical probability distributions because they too are fundamentally (in the frequentist's intepretation of probability) phenomenological descriptions. One is not describing the state of a six sided die, one is describing the behavior of that die when tossed. The phenomena of its relative frequencies of outcomes when tossed repeatedly.

I make this distinction in my Prob Stat classes by beginning the semester having the students estimate the probability that at least two class members have the same birthday. I give them two guesses which is a major hint, and write my two guesses down on a piece of paper. We then go through and verify if it is the case (highly probable for my classes of 30 or more students). I then ask them again, what's the probability as I turn over my guesses "Either it's P=1 or P=0". The point here is you can calculate the probability that an arbitrary set of 30 people will have a common birthday but that is not a description of the reality of our set of students, it is a description of the class of such sets for which ours is an instance. Saying our specific group has a certain probability distribution is shorthand for saying that group belongs to a class of such groups with the given probability distribution.
 
  • #61
stevendaryl said:
No, you shouldn't. That's introducing an ad hoc element to quantum mechanics. Through unitary evolution a pure state never turns into a mixed state. It becomes a mixed state through a modeling decision on OUR part. We can trace over unobservable degrees of freedom and produce a mixed state, but as I said, that's something WE do. It's not a physical effect.
Right, but to give a valid pure state description of a measuring device requires you break the device by measuring it so exactly its entropy becomes zero. This precludes it being a measuring device. That's the fallacy of Wigner's friend arguments.
When you include the measuring device prior to measurement you have automatically started with a "mixed state" because to measure it must carry with it an entropy dump. From Bob's perspective the system evolves unitarily and by virtue of the measurment having occurred the description will necessarily involve entanglement between measuring device and system measured. The entropy of the composite doesn't change but the sum of the sub-entropies does go up due to this entanglement. The entropy dump and measurement record by themselves have higher entropy, the measured system by itself has completely decohered because it has been maximally measured i.e. maximally entangled with the measuring device.
 
  • #62
jambaugh said:
Right, but to give a valid pure state description of a measuring device requires you break the device by measuring it so exactly its entropy becomes zero. This precludes it being a measuring device. That's the fallacy of Wigner's friend arguments.

I consider that completely bogus. You don't have to know the initial state precisely in order to do reasoning about it. You do conditional reasoning: If it is in state x, then it will evolve into state y.

From Bob's perspective the system evolves unitarily and by virtue of the measurment having occurred the description will necessarily involve entanglement between measuring device and system measured.

From Bob's perspective, it's false that Alice got spin up or spin down. So what, exactly is the content of the claim that a measurement produces one of the eigenvalues with such-and-such a probability?
 
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  • #63
stevendaryl said:
I consider that completely bogus. You don't have to know the initial state precisely in order to do reasoning about it. You do conditional reasoning: If it is in state x, then it will evolve into state y.
You're thinking in terms of classical states. You disagree with getting a mixed state after the fact. I point out that your problem is starting with as sharp state and claiming part of the system is able to make a measurement.

Ok. Let me try it without that, basically a deferred measurement scenario let "Alice" be a simple enough system to be sharply described in a pragmatic sense. To say "Alice measured the electron" Alice and the electron become maximally entangled. Bob's proper description of the Alice-electron composite necessarily represents this and so Bob cannot describe it in terms of Alice's state and the electron's state. To speak of Alice alone he must trace over the electron's density operator (while the composite is still sharp) yielding a mixed state of Alice with 50-50 percent probabilities of having a "up" vs a "down" record of the electron state. Likewise to describe the electron alone Bob must trace over Alice's density matrix yielding again a mixed state of the electron in that 50-50 classical superposition of outcomes. And we're back to playing Lotto.

What I'm doing here is basically letting Alice borrow Bob's entropy dump. It's not Wigner's friend but an indirect measurement. I would not describe Alice as having made a measurement here because she's too damned cold. But it shows again that Yes the composite system as a whole evolves unitarily but the factor system does not since it becomes entangled with the other factor system.

The "measuring system" must have enough spare degrees of freedom to make the states of nearly maximal entanglement much more likely than a spontaneous return to un-entangled. But it is this inevitable entanglement that is the 2nd law of thermodynamics in action. It is the fact that though Me plus the photon that bounced off my head and is heading toward Alpha Centauri may be evolving unitarily, I cannot access the photon any more since I am now outside its future light cone. I must discard any hope of recovering its part of our entangled information and trace it out. I alone now have a "mixed state".

( It grates on me to use this "my state" language because it is the very source of confusion by pretending one is making ontic descriptions using QM. But it is the language of the intended audience.)
 
  • #64
jambaugh said:
You're thinking in terms of classical states.[/QUOTES]

That's not true. Schrodinger's equation tells us how to evolve a quantum state from one time to another.

I point out that your problem is starting with as sharp state and claiming part of the system is able to make a measurement.

Why don't you restate what you're claiming without mentioning the word "measurement"? A measurement is nothing more than interaction that causes a macroscopic variable (such as whether there is a spot on a photographic plate at a particular location) to become correlated with a microscopic variable, such as the z-component of the electron's spin. You can (presumably) derive from physics the fact that if a spin-up electron passes through the magnetic field then it will be deflected in one direction and will produce a spot over here. If a spin-down electron passes through the magnetic field then it will be deflected in the opposite direction and will produce a spot over there.

But the linearity of quantum mechanics implies that a superposition of spin-up and spin-down will result in a superposition of the dot being here and the dot being there. (Okay, since the macroscopic device interacts with the rest of the universe, the prediction would be that the whole universe ends up in a superposition of states. Thus Many-Worlds, fantasy or not, is a consequence of quantum mechanics. It really is. Unless you want to add hidden variables that say that only one of the possibilities is "real").

Ok. Let me try it without that, basically a deferred measurement scenario let "Alice" be a simple enough system to be sharply described in a pragmatic sense. To say "Alice measured the electron" Alice and the electron become maximally entangled. Bob's proper description of the Alice-electron composite necessarily represents this and so Bob cannot describe it in terms of Alice's state and the electron's state. To speak of Alice alone he must trace over the electron's density operator (while the composite is still sharp) yielding a mixed state of Alice with 50-50 percent probabilities of having a "up" vs a "down" record of the electron state. Likewise to describe the electron alone Bob must trace over Alice's density matrix yielding again a mixed state of the electron in that 50-50 classical superposition of outcomes. And we're back to playing Lotto.

As I said earlier, this business of tracing over degrees of freedom is something WE do. It isn't a physical interaction. It's a modeling decision. We model the situation in which you have a system entangled with the rest of the universe as if there was a collapse to one of a number of macroscopically distinguishable states, and the mixed state represents the lack of information about exactly which state it is. But that's really bogus. The mixed state resulted from our decision to draw the boundary of the system we are interested in. There can't be physical consequences of a modeling choice.
 
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  • #65
jambaugh said:
[..]
I cannot access the photon any more since I am now outside its future light cone. I must discard any hope of recovering its part of our entangled information and trace it out. I alone now have a "mixed state".

( It grates on me to use this "my state" language because it is the very source of confusion by pretending one is making ontic descriptions using QM. But it is the language of the intended audience.)
I like what you've written because tracing out 'Me' has physical justification. But it cannot be a physical process because the entanglement is not broken. So it seems the tracing is to remove information that is inaccesible and not because of a change of any state.

I happen to believe what you say ( in so far as I understand it) and also that superposition does no physics and is only an accounting procedure. So I'd like to think that tracing out models actuality correctly.
 
  • #66
DarMM said:
Spekken's toy model gives an epistemic set of states that have interference at the observer + system level, while having a definite outcome at the system level. Superpostion of the total system isn't incompatible with definite outcomes for subsystems. If it was epistemic views would be finished.
To explain briefly in Spekkens toy model superposition arises from the structure of the space of probability distributions under the presence of an epistemic limit. When considered over "system + environment" that limit has a different form than over just "system", so we have a superposition for the former, but a definite outcome for the latter.
Actual epistemic views are more complicated than this, but it gives a good example of how superpositions are compatible with definite subsystem outcomes.

Spekkens's toy model assumes quantum mechanics is incomplete, ie. that there are hidden variables. Spekkens's work is motivated by the measurement problem, and aims to build on things like Bohmian Mechanics, and ask whether any other types of completions of quantum mechanics are possible. So I don't think Spekkens's use of "epistemic" is the same as @jambaugh's, since @jambaugh doesn't seem to be a fan of ontic states.
 
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  • #67
Let me make my scenario even simpler. Imagine a world where there is basically nothing in the world except Alice, her measuring equipment, and her electron. That's the entire universe. This presumably could be described by an enormously complicated wave function. We can reason that if the electron starts off in a spin-up state, then the whole system will evolve into a state in which Alice is in the "having measured spin-up" state. If the electron starts off in the spin-down state, then the whole system will evolve into a state in which Alice is in the "having measured spin-down" state. So the linearity of quantum mechanics implies that if the electron starts in the state which is a superposition of spin-up and spin-down, with equal amplitudes, then the whole system will evolve into a state in which Alice is in a superposition of "having measured spin-up" and "having measured spin-down". That seems almost unquestionable to me.

(The reason I say "almost" is because measurement involves irreversible changes, and I'm not sure how to model irreversible changes using pure quantum mechanics.)

So if you want to insist that Alice's measurement of the electron's spin (along the z-axis, say) always produces the result of Alice having measured spin-up or the result of Alice having measured spin-down, then that means that we have to be able to interpret a superposition of macroscopically different states as being either in one state or the other.

But it's inconsistent to interpret it that way. A superposition of a spin-up electron and a spin-down electron is not either one or the other, it's an electron in a definite state (spin in some direction other than ##\pm \hat{z}##).
 
  • #68
atyy said:
Spekkens's toy model assumes quantum mechanics is incomplete, ie. that there are hidden variables. Spekkens's work is motivated by the measurement problem, and aims to build on things like Bohmian Mechanics, and ask whether any other types of completions of quantum mechanics are possible. So I don't think Spekkens's use of "epistemic" is the same as @jambaugh's, since @jambaugh doesn't seem to be a fan of ontic states.
Certainly, as I mentioned above there are two classes of epistemic views ##\psi##-statistical and ##\psi##-doxastic. Spekkens toy model is more in the tradition of the former kind.

QBism for example, a ##\psi##-doxastic view, would also say QM is incomplete, but not because there are hidden variables. Incomplete is not quite the same as "missing hidden variables".
 
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  • #69
stevendaryl said:
So if you want to insist that Alice's measurement of the electron's spin (along the z-axis, say) always produces the result of Alice having measured spin-up or the result of Alice having measured spin-down, then that means that we have to be able to interpret a superposition of macroscopically different states as being either in one state or the other.
Okay let's look at Spekkens toy model. We have a system with an electron, a measuring device and then the lab environment. We'll model each with a qubit.

In the model a qubit has four ontic states ##1, 2, 3, 4##. However there is an epistemic limit so you can only resolve the ontic space to half the maximum limit. This means there are six maximum knowledge epistemic states:
$$
|\uparrow\rangle = \{1,2\} \\
|\downarrow\rangle = \{3,4\} \\
|+\rangle = \{1,3\} \\
|-\rangle = \{2,4\} \\
|i\rangle = \{2,3\} \\
|-i\rangle = \{1,4\} \\
$$
So for instance ##|\uparrow\rangle## is an epistemic state indicating the ontic state is ##1## or ##2##. Superposition of two epistemic states like
$$|\uparrow\rangle + |\downarrow\rangle = |+\rangle = \{1,3\}$$
can then be seen to not be "or" from Kolmogorov probability but a bilinear mapping from maximum knowledge epistemic states to maximum knowledge epistemic states where one ontic state from each is present in the pair of the superposition.

We also have states of non-maximal knowledge, mixed states, like:
$$\frac{1}{2}\mathbb{I} = \{1,2,3,4\}$$

Now let us consider an entangled state:
$$|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle = \{1.1, 2.2, 3.3, 4.4\}$$
where ##a.b## means the first particle is in ontic state ##a## and the second is in ontic state ##b##.

As you can see this means that the first particle is in one of the states:
$$\{1,2,3,4\}$$
which is a mixed state.

By gaining more knowledge of the entire system, i.e. that the two particles are always in the same ontic state, I know less about a single particle. I don't what state a single particle is in at all and thus have non-maximum knowledge of it, a mixed state.

Now consider the three particle state:
$$|\uparrow\uparrow\uparrow\rangle + |\downarrow\downarrow\downarrow\rangle = \{1.1.1, 1.2.2, 2.1.2, 2.2.1, 3.3.3, 3.4.4, 4.3.4, 4.4.3\}$$

Take the first particle to be the atomic system, the second to be the device and the third to be the lab environment. A superobserver outside the lab might use the above state. An observer within the lab, who doesn't track his lab environment (it's impossible otherwise he'd be a superobserver) might see his equipment read "up" in which case he knows that "atomic system + device" is in the state:
$$|\uparrow\uparrow\rangle = \{1.1, 1.2, 2.1, 2.2\}$$
which can easily be seen to be compatible with the above use of a superposition by the superobserver.

This is because states of maximum knowledge of the three particle system are consistent with states of maximum knowledge of subsystems that remove one of its "branches". This in turn is because one is reducing the systems tracked and thus one can increase knowledge of the subsystems.

This is essentially what happens in all epistemic models, but it's easier to see here in the toy model.

See Chapter 11 of Richard Healey's "The Quantum Revolution in philosophy" for a longer exposition on it in the ##\psi##-doxastic case.
 
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  • #70
To follow on from the above, I recommend anybody to have a read of Spekkens paper:
https://arxiv.org/abs/quant-ph/0401052
It shows a bunch of "quantum" effects show up in a purely local classical theory with an epistemic limit, e.g.
  1. Superposition
  2. Non-commutativity of operators
  3. Uncertainty principle
  4. Interference
  5. Wigner's Friend ambiguities
  6. No cloning
  7. Superdense coding
  8. Indistinguishability of unknown states
  9. Collapse
  10. Entanglement Monogamy
  11. Teleportation
  12. Non-locality without entanglement (from https://arxiv.org/abs/quant-ph/9804053)
So it allows you to see what are "pure" quantum effects, i.e. Contextuality, Quantum Computing speedups and Non-classical correlations (e.g. CHSH violating correlations).
 
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  • #71
stevendaryl said:
(The reason I say "almost" is because measurement involves irreversible changes, and I'm not sure how to model irreversible changes using pure quantum mechanics.)
This needs interactions with the environment, which (for tractability) is usually taken to be a heat bath consisting of infinitely many harmonic oscillators with a continuous, unbounded frequency spectrum, and eliminating the bath degrees of freedom. The result is (after invoking the Markov approximation) a Lindblad-type equation for the reduced density operator. Pure states do not work because dissipation usually destroys pureness of the initial state.

If purity is to be preserved, one can use the Schrödinger equation with an optical potential, i.e., an imaginary contribution to the Hamiltonian. This is less accurate than Lindblad models but often adequate.
 
  • #72
DarMM said:
Now let us consider an entangled state:
$$
|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle = \{1.1, 2.2, 3.3, 4.4\}
$$

This doesn't look right. It seems like ##|\uparrow\uparrow\rangle## should be ##\{1.1, 1.2, 2.1, 2.2\}##, and similarly for ##|\downarrow\downarrow\rangle## with ontic states ##3## and ##4##.
 
  • #73
PeterDonis said:
This doesn't look right. It seems like ##|\uparrow\uparrow\rangle## should be ##\{1.1, 1.2, 2.1, 2.2\}##, and similarly for ##|\downarrow\downarrow\rangle## with ontic states ##3## and ##4##.
That is the correct form for ##|\uparrow\uparrow\rangle##, but I gave the entangled state not ##|\uparrow\uparrow\rangle##.
 
  • #74
DarMM said:
That is the correct form for ##|\uparrow\uparrow\rangle##, but I gave the entangled state not ##|\uparrow\uparrow\rangle##.

The entangled state is just a superposition of ##|\uparrow\uparrow\rangle## and ##|\downarrow\downarrow\rangle##, which just means you include all the ontic possibilities from both one-qubit states in the set of ontic possibilities for the two-qubit states. So I still don't see how ontic states like ##1.2## or ##2.1## (or ##3.4## or ##4.3##) are excluded.
 
  • #75
PeterDonis said:
The entangled state is just a superposition of ##|\uparrow\uparrow\rangle## and ##|\downarrow\downarrow\rangle##, which just means you include all the ontic possibilities from both one-qubit states in the set of ontic possibilities for the two-qubit states.
No it doesn't as that would lead to a non-maximal knowledge state. The ##+## of superposition is a mapping between maximal knowledge (in standard terminology "pure") states. See for example the case given at the start of my post, we have:
$$
|\uparrow\rangle = \{1,2\} \\
|\downarrow\rangle = \{3,4\}
$$
However it is not the case that ##|+\rangle = |\uparrow\rangle + |\downarrow\rangle## is given by ##\{1,2,3,4\}## as that is a mixed state, instead it is given by ##\{1,3\}##
 
  • #76
Can you explain your notation? How can a ket be the same as a bunch of numbers in curly brackets?
 
  • #77
It's in my post #69 above. You have an ontic space with four elements or states ##1,2,3,4##. Restricted epistemic states over such a space replicate many, though not all, features of quantum mechanics. That is they provide a local classical model that replicates what I listed in post #70. Spekkens paper I linked to goes into more detail.
 
  • #78
DarMM said:
The ##+## of superposition is a mapping between maximal knowledge (in standard terminology "pure") states.

I understand that for the standard notation using kets; I'm just not seeing how that standard notation maps to the ontic states for more than one qubit. I'll look at the Spekkens paper.
 
  • #79
PeterDonis said:
I understand that for the standard notation using kets; I'm just not seeing how that standard notation maps to the ontic states for more than one qubit. I'll look at the Spekkens paper.
It's very "formal" just to tell you not transparently obvious, i.e. it's just a bilinear map with the right properties.
 
  • #80
Well, it's highly misleading to label the toy model with the same symbols as the quantum (state) kets. The toy model obviously cannot reproduce all established quantum mechanical facts; for sure not the violation of Bell's inequality. I've a look at Spekken's paper later...
 
  • #81
DarMM said:
it's just a bilinear map with the right properties.

There's a bilinear map for single qubits, yes; you give it explicitly in your post. But I don't see how the bilinear map for a two-qubit system is constructed.
 
  • #82
PeterDonis said:
There's a bilinear map for single qubits, yes; you give it explicitly in your post. But I don't see how the bilinear map for a two-qubit system is constructed.
The details are given in Spekkens paper section III.A

You might prefer the exposition in https://arxiv.org/abs/1103.5037.
 
  • #83
vanhees71 said:
Well, it's highly misleading to label the toy model with the same symbols as the quantum (state) kets. The toy model obviously cannot reproduce all established quantum mechanical facts; for sure not the violation of Bell's inequality. I've a look at Spekken's paper later...
Spekkens does it, I'm not going to deviate from his notation when quoting him. However in his paper it's obvious what he's doing.

And the model doesn't replicate the Bell Inequalities as I said, that's the point. It provides a local classical model of some features of QM in order to show that those features aren't specifically "quantum". Then the features it doesn't replicate are what is especially quantum. It's being used to pinpoint uniquely quantum features. See post #70

The reason I brought it up was because Wigner's friend shows up in this model rather simply.
 
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  • #84
People might be interested to know that it's specifically because of models like Spekkens that people were led to trying the Frauchiger-Renner and Brukner Objectivity theorems.

(Brukner discussed here: https://www.physicsforums.com/threa...-basic-wigners-friend-type-experiment.968181/)

Initially there seems to be some contradiction between QM in the Wigner's friend scenario and the fact of definite outcomes for subsystems. However Spekkens toy model (and similar) shows that your not going to find such an issue because Wigner's friend can occur in purely local classical theories.

Thus the attempt to find a Wigner's friend that incorporates uniquely quantum features not replicated by these classical models like the CHSH inequalities or Hardy's paradox. Frauchiger-Renner is "Wigner + Hardy" and Brukner's objectivity theorem is "Wigner + CHSH".

However it seems neither do the job. There is no contradiction between a superobserver using a superposition and definite outcomes for subsystems, because the superposition refers to the statistics of superobservables of the friend's entire lab.
 
  • #85
I'm trying to reconcile the following statements:
jambaugh said:
There is one reason that quantum mechanics is a complete physical theory.
DarMM said:
If you look at @jambaugh 's post he's clearly taking an epistemic view of the wavefunction where collapse is just (generalized) Bayesian conditioning.
DarMM said:
Secondly an epistemic account will by nature be taking the view that QM is incomplete as it views a central object in it as non-representational.
Question: is there a non-many-worlds interpretation of QM for which QM is "complete"?
 
  • #86
Avodyne said:
Question: is there a non-many-worlds interpretation of QM for which QM is "complete"?
The thermal interpretation has this property.
 
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  • #87
Avodyne said:
I'm trying to reconcile the following statements:
Basically @jambaugh is (I think) saying QM is complete because there is no further you can go, i.e. you cannot obtain the deeper explanation or any deeper theory will disagree with it. I'm saying epistemic views are not complete in the sense @stevendaryl gives of not providing the explanation for certain predictions.

Or to be brief I'm saying it's not complete and @jambaugh is saying it's as complete as it is going to get.

Avodyne said:
Question: is there a non-many-worlds interpretation of QM for which QM is "complete"?
Bohmian Mechanics. I would say there is no fully worked out complete interpretation for QFT.
 
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  • #88
DarMM said:
Bohmian Mechanics. I would say there is no fully worked out complete interpretation for QFT.
Bohmian mechanics is a ''completion'' of ordinary QM by hidden variables. It is not QM itself but a different theory which contains QM as a subtheory.
 
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  • #89
A. Neumaier said:
Bohmian mechanics is a ''completion'' of ordinary QM by hidden variables. It is not QM itself but a different theory which contains QM as a subtheory.
True, but most people include it in the notion of "complete" interpretations, however out of equilibrium it is more general than QM as you say. Strictly speaking it's not an interpretation, but then few of the interpretations of QM are when you look at them closely.

I'd say your Thermal Interpretation is the clearest realist and complete view of QFT I've seen.
 
  • #90
stevendaryl said:
Thus Many-Worlds, fantasy or not, is a consequence of quantum mechanics. It really is. Unless you want to add hidden variables that say that only one of the possibilities is "real"
That depends on a very particular view of what a superposition is. How can Many-Worlds for example cope with the fact that any system confined to a finite volume is always in a mixed state?
 
  • #91
DarMM said:
The details are given in Spekkens paper section III.A

This section doesn't say anything about two-qubit systems; the paper only starts discussing those in Section IV. The first time in that section that I see a state that looks like ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle## is in Section IV.C, p. 15, just before equation 78. There the state ##[ (1\ \text{V}\ 2) \cdot (1\ \text{V}\ 2)] \ \text{V}\ [(3\ \text{V}\ 4) \cdot (3\ \text{V}\ 4)]## is given, which looks to me like the equivalent in the paper's notation of ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle##--but the paper says this state is a state of non-maximal knowledge, whereas you're saying the state you mean by ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle## is a state of maximal knowledge.

So either I'm misunderstanding the paper's notation and how it relates to standard ket notation, or you're using standard ket notation to mean something other than what it obviously seems to map to in the paper's notation.
 
  • #92
DarMM said:
You might prefer the exposition in https://arxiv.org/abs/1103.5037.

This isn't really helping me because I'm not familiar with qubit stabilizer notation to begin with, so I would have to learn two unfamiliar things instead of one.
 
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  • #93
PeterDonis said:
This section doesn't say anything about two-qubit systems; the paper only starts discussing those in Section IV.
Sorry how the mapping is defined there can just be extended to two qubit systems to give the results I have. The same method of construction goes through. I can show how based on the single qubit case the two qubit case works if you wish.

PeterDonis said:
The first time in that section that I see a state that looks like ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle## is in Section IV.C, p. 15, just before equation 78. There the state ##[ (1\ \text{V}\ 2) \cdot (1\ \text{V}\ 2)] \ \text{V}\ [(3\ \text{V}\ 4) \cdot (3\ \text{V}\ 4)]## is given, which looks to me like the equivalent in the paper's notation of ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle##--but the paper says this state is a state of non-maximal knowledge
No that's the papers version of the mixed state:
$$\frac{1}{2}|\uparrow\uparrow\rangle \langle\uparrow\uparrow\rangle + \frac{1}{2}|\downarrow\downarrow\rangle \langle\downarrow\downarrow|$$
the state ##| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle## would be:
$$(1.1) \lor (2.2) \lor (3.3) \lor (4.4)$$
I gave this as:
$$\{1.1, 2.2, 3.3, 4.4\}$$
Which was actually a bit stupid of me as I should stick to Spekkens notation.
 
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  • #94
DarMM said:
I can show how based on the single qubit case the two qubit case works if you wish.

That would be helpful. In particular, I'm still not seeing how this

$$
| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle
$$

matches up to this

$$
(1.1) \lor (2.2) \lor (3.3) \lor (4.4)
$$

I understand fine what the latter means in terms of the ontic states of the model. I just don't understand how to get from the former to the latter.

DarMM said:
the mixed state:
$$
\frac{1}{2}|\uparrow\uparrow\rangle \langle\uparrow\uparrow\rangle + \frac{1}{2}|\downarrow\downarrow\rangle \langle\downarrow\downarrow\rangle
$$

I think you mis-formatted this; it looks to me like it should be

$$
\frac{1}{2}\vert\uparrow\uparrow\rangle \langle\uparrow\uparrow\vert + \frac{1}{2}\vert\downarrow\downarrow\rangle \langle\downarrow\downarrow\vert
$$
 
  • #95
PeterDonis said:
That would be helpful. In particular, I'm still not seeing how this

$$
| \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle
$$

matches up to this

$$
(1.1) \lor (2.2) \lor (3.3) \lor (4.4)
$$

I understand fine what the latter means in terms of the ontic states of the model. I just don't understand how to get from the former to the latter.
Perfect, I'll put it up tomorrow.

PeterDonis said:
I think you mis-formatted this; it looks to me like it should be

$$
\frac{1}{2}\vert\uparrow\uparrow\rangle \langle\uparrow\uparrow\vert + \frac{1}{2}\vert\downarrow\downarrow\rangle \langle\downarrow\downarrow\vert
$$
Yes indeed, I was even staring there at my post trying to see what was wrong! Took me a while to see the little bracket. Corrected now.
 
  • #97
Avodyne said:
IMO, many-worlds with the Vaidman interpretation of the Born rule is complete:

http://philsci-archive.pitt.edu/14590/
If I'm reading him right, he's saying that the measure of the subset of uncountably infinite worlds with outcome ##A## is given by the square of the coefficient of ##|A\rangle## in the expansion of the universal wavefunction in the (quasi-)classical worlds basis?
 
  • #98
Yes. But the starting point is counting branches in simpler situations. See also Carroll & Sebens, who elaborate basically the same idea, with slightly different language:

https://arxiv.org/abs/1405.7907
 
  • #99
DarMM said:
That depends on a very particular view of what a superposition is. How can Many-Worlds for example cope with the fact that any system confined to a finite volume is always in a mixed state?

What do you mean by that? What theorem is that?
 
  • #100
stevendaryl said:
What do you mean by that? What theorem is that?
In QFT for a variety of subregions of spacetime you can prove the algebra of observables is a type ##III## C*-algebra factor and these have no pure states.

The meaning of this for Copenhagen-like views and the Thermal Interpretation is clear enough for me, but what it means for MWI is not clear.
 

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