Does the EPR experiment imply QM is incomplete?

[stevendaryl]

I'm not sure I understand.
Do not assume there is any decoherence. The state that you would use for predicting outcome of interference measurement between $|d\rangle |D\rangle$ and $|u\rangle |U\rangle$ then should be $\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle$. Exactly the same as in the second case because they are the same case. And your Born probabilities remain hidden variables that you can't observe because you performed interference measurement.

I don't know what you mean by "remain hidden".

If you use the postulate that a measurement always results in an eigenvalue of the thing that is measured, then there would be no interference between the two alternatives. If you use Schrodinger's equation to compute the probabilities, then there would be interference. So the two axioms are contradictory. They predict different probabilities for winding up in the state $|final\rangle$.

[edit]This assumes that measurement is a physical process by which a microscopic variable is amplified to make a macroscopic difference in the measuring device. If you define measurement to mean "a conscious observer becomes aware of the result" then nothing that devices do can be considered a measurement.

 

[zonde]

If you use the postulate that a measurement always results in an eigenvalue of the thing that is measured, then there would be no interference between the two alternatives.

Can you explain why do you claim that? Because I can imagine different reasons why do you think so, and I would like to avoid guessing.

 

[vanhees71]

Not having followed this thread in detail, I think the problem between your mutual understanding is the usual one not to distinguish clearly between measurements of observables on a system and the preparation of the system in some state. Let's keep the story as simple as possible and assume a non-degenerate observable $A$ to be measured. This means that the eigenspaces for each eigenvalue of the corresponding representing self-adjoint operator $\hat{A}$ are all one-dimensional. Let's denote the corresponding normalized eigenvectors as $|u_a \rangle$, where $a$ denotes the eigenvalue. Also let's suppose these are true eigenvectors, normalizable to 1 and the spectrum of possible eigenvalues $a$ thus discrete.

Let's furter assume the system is prepared in a pure state, represented by the statistical operator $|\psi \rangle \langle \psi|$ with $|\psi \rangle$ a normalized vector. The only meaning, within the minimal interpretation, of this state is that the probability to find the value $a$ when the observable $A$ is measured is given according to Born's rule by
$$W_{\psi}(a)=|\langle u_a|\psi \rangle|^2.$$
This implies, for this most simple case, that the system is prepared with a determined value $a$ of the observable $A$ if and only if $|\psi \rangle \langle \psi|=|u_a \rangle \langle u_a|$, and then $W_{\psi}(a)=1$ as it must be.

Otherwise
$$|\psi \rangle=\sum_{a} \psi_a |u_a \rangle$$
is in a superposition of eigenvectors of $\hat{A}$ and thus
$$W_{\psi}(a)=|\psi_a|^2.$$
For such an ideal measurement, each outcome at an individual system is always one of the possible values $a$ of $A$, which are the eigenvalues of $\hat{A}$, but since the observable $A$ doesn't take a determined value there are only the given probabilities and nothing else due to the preparation of the system in the state $|\psi \rangle \langle \psi|$.

What's usually meant when one talks about the "measurement problem" is this very assumption that there's always a well-defined outcome when $A$ is measured, no matter whether the value of $A$ is determined or note due to the preparation of the system. This is, however, only a metaphysical problem. From a physical, i.e., operational point of view of preparations (defining the quantum state) and measurements, there's no such problem as long as all observations agree with this postulate of Born's rule.

 

[zonde]

What's usually meant when one talks about the "measurement problem" is this very assumption that there's always a well-defined outcome when $A$ is measured, no matter whether the value of $A$ is determined or note due to the preparation of the system. This is, however, only a metaphysical problem. From a physical, i.e., operational point of view of preparations (defining the quantum state) and measurements, there's no such problem as long as all observations agree with this postulate of Born's rule.

The question is not about "measurement problem" itself but rather about inconsistency that this "measurement problem" creates (or does not create) in the model.
In simple words the question is where do the relative phase factors go when measurement is performed. Schrodinger's equation says that relative phase factors live on, while Born rule leaves no place for them to live on. @stevendaryl says there is inconsistency because relative phase factors are fine according to Schrodinger's equation, but they die according to Born rule. I say that relative phase factors could be just fine even after we apply Born rule.

 

[stevendaryl]

What's usually meant when one talks about the "measurement problem" is this very assumption that there's always a well-defined outcome when $A$ is measured, no matter whether the value of $A$ is determined or note due to the preparation of the system. This is, however, only a metaphysical problem. From a physical, i.e., operational point of view of preparations (defining the quantum state) and measurements, there's no such problem as long as all observations agree with this postulate of Born's rule.

My point is that the assumption that there is a well-defined (single) outcome is contradicted by unitary evolution, if you consider the measurement device itself to be a quantum system.

 

[vanhees71]

The question is not about "measurement problem" itself but rather about inconsistency that this "measurement problem" creates (or does not create) in the model.
In simple words the question is where do the relative phase factors go when measurement is performed. Schrodinger's equation says that relative phase factors live on, while Born rule leaves no place for them to live on. @stevendaryl says there is inconsistency because relative phase factors are fine according to Schrodinger's equation, but they die according to Born rule. I say that relative phase factors could be just fine even after we apply Born rule.

This I don't understand. The relative phase factors are the very point that "matter waves" have been introduced and modern QT was discovered in the first place. The relative phases are crucial for, e.g., the result of the double-slit experiment, showing interference fringes in the probability distribution (later demonstrated to be correct by Davisson and Germer with electrons for the first time).

 

[vanhees71]

My point is that the assumption that there is a well-defined (single) outcome is contradicted by unitary evolution, if you consider the measurement device itself to be a quantum system.

This I never understood either. Measurement devices as macroscopic objects are never described by unitary time evolution but by "master equations" of "open quantum systems". That's the whole point of the decoherence program. There's no contradiction between unitary time evolution for closed systems and effective descriptions of macroscopic, i.e., heavily coarse-grained, observables.

 

[stevendaryl]

This I never understood either. Measurement devices as macroscopic objects are never described by unitary time evolution but by "master equations" of "open quantum systems".

Sure. But I'm talking about an idealized situation in which you have an isolated composite system that consists of a measurement device plus the system that it's measuring. You can describe the composite system quantum mechanically.

In the real world, of course, there are interactions between any macroscopic measuring device and the rest of the universe: It's interacting gravitationally and electromagnetically and so forth. But in principle, we can consider an isolated system that contains a measuring device. If the theory is inconsistent in that case, it shows that the theory is inconsistent, period.

 

[vanhees71]

Measurements are done with real-world macroscopic apparati. I let these problems happily to the philosophers to have some food of thought for writing papers with more footnotes than main text...

 

[stevendaryl]

Measurements are done with real-world macroscopic apparati. I let these problems happily to the philosophers to have some food of thought for writing papers with more footnotes than main text...

That's why I said that possibly only philosophers care whether our theories are consistent. But regardless of your attitude toward it, if a theory is inconsistent, then it can't be actually correct. So that gets back to the claim that quantum mechanics is incomplete, or at least, our understanding of it is incomplete.

 

[vanhees71]

QT seems not to be very inconsistent but very successful in describing the observed and quantitatively measured world. It's obvious that our contemporary theoretical understanding of nature is incomplete. It's quite arrogant to expect something else!

 

[zonde]

My point is that the assumption that there is a well-defined (single) outcome is contradicted by unitary evolution, if you consider the measurement device itself to be a quantum system.

But I'm talking about an idealized situation in which you have an isolated composite system that consists of a measurement device plus the system that it's measuring. You can describe the composite system quantum mechanically.

My objection is that quantum system can't be considered isolated at the moment of collapse. Consider one subsystem of entangled state. When it undergoes pre-measurement the state of other subsystem becomes determined (assuming detection as rather passive process in respect to measured property).
So unitary evolution after collapse can be there, but only if you consider larger system.

 

[stevendaryl]

My objection is that quantum system can't be considered isolated at the moment of collapse.

In an hypothetical world in which there is nothing but a measuring device and a particle that it measures, then what happens? Does the measuring device get a result that is an eigenvalue, or does it become a superposition?

 

[zonde]

In an hypothetical world in which there is nothing but a measuring device and a particle that it measures, then what happens? Does the measuring device get a result that is an eigenvalue, or does it become a superposition?

Hypothetically in such a hypothetical world particle does not interact with measurement device at all.

 

[stevendaryl]

Hypothetically in such a hypothetical world particle does not interact with measurement device at all.

Why would you say that? Two subsystems can't interact if they are the only things in the universe?

 

[zonde]

Why would you say that? Two subsystems can't interact if they are the only things in the universe?

For a particle to end up in new quantum state this state has to be available. If there are no quantum states to which particle can change it stays in the quantum state in which it is already.

 

[stevendaryl]

For a particle to end up in new quantum state this state has to be available. If there are no quantum states to which particle can change it stays in the quantum state in which it is already.

Who said that there were no other quantum states available?

 

[zonde]

Who said that there were no other quantum states available?

You proposed two options and none is valid. Particle can't change to other state (eigenvalue of measurement operator) if the system together can't satisfy Schrodinger equation. And particle can't become delocalized superposition because particles are localized.

 

[Lord Jestocost]

@vanhees71
In comment #143 you describe what Schlosshauer terms “measurement-as-axiom”. But this is not of help if one starts to think about “measurement-as-interaction”: How does individual measurement outcomes come about dynamically?

 

[stevendaryl]

You proposed two options and none is valid. Particle can't change to other state (eigenvalue of measurement operator) if the system together can't satisfy Schrodinger equation. And particle can't become delocalized superposition because particles are localized.

Sorry, I have no idea what you are talking about.

 

[Mentz114]

In an hypothetical world in which there is nothing but a measuring device and a particle that it measures, then what happens? Does the measuring device get a result that is an eigenvalue, or does it become a superposition?

It depends on the Hamiltonian and the initial states of the system and apparatus.

My point being that it is possible for unitary evolution to result in an eigenstate.

 

[stevendaryl]

It depends on the Hamiltonian and the initial states of the system and apparatus.

My point being that it is possible for unitary evolution to result in an eigenstate.

But there are certainly cases where you can prove that that can't happen.

For example, if the initial state has reflection symmetry about some point, but each of the "pointer states" lacks this symmetry, then you can show that unitary evolution cannot result in a definite pointer state.

 

[Mentz114]

But there are certainly cases where you can prove that that can't happen.

For example, if the initial state has reflection symmetry about some point, but each of the "pointer states" lacks this symmetry, then you can show that unitary evolution cannot result in a definite pointer state.

True. But that would be a poorly designed instrument. From what I've read it is important that the apparatus has an observable that has the same eigenstates (or close to) as the system being tested.

 

[stevendaryl]

True. But that would be a poorly designed instrument. From what I've read it is important that the apparatus has an observable that has the same eigenstates (or close to) as the system being tested.

That's what I'm talking about. Take the example of a spin measurement: If the particle is spin-up, then some pointer points to the left. If the particle is spin-down, then the pointer points to the right. The initial state of the pointer is a neutral position that is left-right symmetric.

So you prepare an initial state for the particle that is an equal superposition of spin-up and spin-down. The initial state is left-right symmetric. The final state (if it gives a definite result) is not.

 

[PeterDonis]

The initial state is left-right symmetric. The final state (if it gives a definite result) is not.

More to the point of your previous comment, unitary evolution results in a superposition of "spin-up and pointer pointing to the left" and "spin-down and pointer pointing to the right", i.e., an entangled state which is left-right symmetric, but which does not have a definite state for the pointer. To get a definite state for the pointer from a starting state that is left-right symmetric, you would need some non-unitary process somewhere.

 

[zonde]

That's what I'm talking about. Take the example of a spin measurement: If the particle is spin-up, then some pointer points to the left. If the particle is spin-down, then the pointer points to the right. The initial state of the pointer is a neutral position that is left-right symmetric.

So you prepare an initial state for the particle that is an equal superposition of spin-up and spin-down. The initial state is left-right symmetric. The final state (if it gives a definite result) is not.

Your reasoning is valid (as far as I can tell), but it does not mean that contradiction with unitary evolution is unavoidable.
We can say that yes there is sudden change in the state of the system, if there is another sudden change in another system somewhere (nearby). And both systems taken together satisfy Schrodinger equation. It's like entangled pair of particles, taken separately there is sudden change in particle state. Taken together both particles add to the same combined state even after this sudden change.

 

[PeterDonis]

We can say that yes there is sudden change in the state of the system, if there is another sudden change in another system somewhere (nearby). And both systems taken together satisfy Schrodinger equation.

Where are you getting this from? Do you have a peer-reviewed reference that proposes a model like this?

 

[zonde]

Where are you getting this from? Do you have a peer-reviewed reference that proposes a model like this?

No, I have references for experiments that observe phenomena like this:
http://arxiv.org/abs/1508.05949
http://arxiv.org/abs/1511.03189
http://arxiv.org/abs/1511.03190

 

[PeterDonis]

I have references for experiments that observe phenomena like this

These are all recent experiments testing for violations of the Bell inequalities with more loopholes closed. That has nothing to do with what we're discussing. Measuring the spins of a pair of entangled particles that start out in a left-right symmetric state still cannot produce a state that is not left-right symmetric by unitary evolution. Read my post #165; the extension of what I said there to the case of a pair of spin measurements on entangled particles is straightforward and doesn't change my conclusion.

 

[stevendaryl]

Your reasoning is valid (as far as I can tell), but it does not mean that contradiction with unitary evolution is unavoidable.
We can say that yes there is sudden change in the state of the system, if there is another sudden change in another system somewhere (nearby). And both systems taken together satisfy Schrodinger equation. It's like entangled pair of particles, taken separately there is sudden change in particle state. Taken together both particles add to the same combined state even after this sudden change.

I'm not sure if I understand what you're suggesting, but something similar happens in Many-Worlds. If one pointer points to the left in one world, it points to the right in another, so the Many-Worlds model remains left-right symmetric.

 

[zonde]

Read my post #165; the extension of what I said there to the case of a pair of spin measurements on entangled particles is straightforward and doesn't change my conclusion.

I am not questioning your conclusion. I completely agree with it. And if am not mistaken @stevendaryl agrees with it as well.

 

[zonde]

I'm not sure if I understand what you're suggesting, but something similar happens in Many-Worlds. If one pointer points to the left in one world, it points to the right in another, so the Many-Worlds model remains left-right symmetric.

Yes, that's very similar to MWI, only it is restricted to single world.

 

[Mentz114]

[]
So you prepare an initial state for the particle that is an equal superposition of spin-up and spin-down. The initial state is left-right symmetric. The final state (if it gives a definite result) is not.

I don't understand why left/right and up/down are significant. Is the initial state not also 'up/down' symmetric. The final state of what ?
I don't know what point you are making.

The superposition of states in the X basis after the spin has been prepared as Z+ (say) cannot be a physical state, because at that moment the is no angular momentum in any direction but Z. So the superposition refers to non-existent values or at best two zeros.

The only physically consistent interpretation is that a mixture is prepared.

 

[PeterDonis]

I am not questioning your conclusion. I completely agree with it.

No, you don't. You said:

it does not mean that contradiction with unitary evolution is unavoidable.

This is incorrect; the contradiction between unitary evolution and definite pointer states is unavoidable. That is true even if you measure a pair of entangled systems instead of a single system; under unitary evolution, each individual pointer device still ends up entangled, not in a definite state.

 

[PeterDonis]

I don't understand why left/right and up/down are significant.

Because that's how the spin measurement device was oriented for that particular measurement. You could indeed orient the device in any direction and still apply the same reasoning. But any particular spin measurement has to be done along a particular direction.

I don't know what point you are making.

The point he is making is that, if having a pointer that is not in a definite state after measurement means the measurement is poorly designed, then unitary evolution predicts that all measurements are poorly designed, since unitary evolution will never give you a pointer that ends up in a definite state after measurement.