Does the EPR experiment imply QM is incomplete?

[stevendaryl]

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.

That the assumption that a measurement always gives an eigenvalue is contradictory with the assumption that evolution is unitary.

The initial state of the lab plus particle is left-right symmetric. The final state of the lab is not.

I don't understand your difficulty. I'm imagining a measuring device with a literal pointer. It measures the z-component of the spin of an electron, and the pointer swings right if the result is spin-up, and swings left if the result is spin-down.

 

[zonde]

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.

Ok, you are identifying definite state with definite (pure) quantum state. I don't.
Hmm, may this is exactly the same problem in my discussion with @stevendaryl.

 

[stevendaryl]

Ok, you are identifying definite state with definite (pure) quantum state. I don't.
Hmm, may this is exactly the same problem in my discussion with @stevendaryl.

Well, the difference between a macroscopic object, such as a measuring device, and a microscopic object, such as an electron, is that for any given macroscopic state (what we would intuitively, pre-quantum mechanics, think of a state, such as "the readout shows the number 32" or "the pointer points to the left" or "the left light is on") there are many, many microscopic states that correspond to it.

I don't have the mathematical sophistication to accurately describe the situation using quantum mechanics, but perhaps it's something like the following:

The complete system perhapse can be described by three variables: $|s, S, j\rangle$, where $s$ is the observable corresponding to the system being measured (an electron's spin, maybe), $S$ is the corresponding value of the "pointer variable", and $j$ represents all the other degrees of freedom.

To make it both simple and definite, we will assume that there are two possible values for $s$:$u$ and $d$, and three possible values for $S$: $0, U, D$. There are many (possibly infinitely many) values for the other degrees of freedom, $j$.

To say that the pointer accurately measures the z-component of spin is to say something like the following:

• If you start in the state $|u, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{ujk} |u, U, k\rangle$
• If you start in the state $|d, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{djk} |d, D, k\rangle$
• It follows from the linearity of the evolution operator that if you start in a superposition of the form $\alpha |u, 0, j\rangle + \beta |d, 0, j\rangle$, then you will end up in a superposition of the form $\alpha \sum_k c_{ujk} |u, U, k\rangle + \beta \sum_k c_{djk} |d, D, k\rangle$

By "end up", I mean applying the evolution operator $e^{-iHt}$.

Now, what I'm a little hazy about is how to deal with irreversibility in quantum mechanics. A measurement is irreversible. I don't know whether the irreversibility is completely explained by the fact that the final state is massively degenerate, compared to the initial state. That's the classical explanation. I don't know whether anything we have to say hinges on the interpretation of irreversibility.

Anyway, given the above assumptions about the evolution, we can always add classical uncertainty, by letting the initial state be an incoherent mixture of

$\alpha |u, 0, j\rangle + \beta |d, 0, j\rangle$

for different values of $j$.

 

[PeterDonis]

you are identifying definite state with definite (pure) quantum state. I don't.

Then what definition are you using? And why do you think it's relevant?

 

[PeterDonis]

what I'm a little hazy about is how to deal with irreversibility in quantum mechanics. A measurement is irreversible.

And unitary evolution is reversible, so right there you have stated the key inconsistency between unitary evolution and measurements.

 

[stevendaryl]

And unitary evolution is reversible, so right there you have stated the key inconsistency between unitary evolution and measurements.

Not necessarily. Shattering a mirror is irreversible, but that doesn't mean that it can't be described by classical mechanics, which is reversible. The classical resolution to the contradiction is that shattering a mirror isn't really irreversible, but the reverse process requires unattainable initial conditions. There could be a similar explanation in quantum mechanics.

 

[stevendaryl]

Then what definition are you using? And why do you think it's relevant?

He's probably thinking of the state being described by an incoherent mixture, rather than a pure state. My feeling is that that doesn't change anything.

 

[zonde]

Now, what I'm a little hazy about is how to deal with irreversibility in quantum mechanics. A measurement is irreversible. I don't know whether the irreversibility is completely explained by the fact that the final state is massively degenerate, compared to the initial state. That's the classical explanation. I don't know whether anything we have to say hinges on the interpretation of irreversibility.

Measurement record is supposed to be rather permanent. So it is supposed to be stable i.e. to be end state of irreversible process.
But I don't think that you can model irreversible process using QM (I can be wrong of course).

 

[zonde]

Then what definition are you using? And why do you think it's relevant?

I am not using definition. I am understanding it as a basic term. I probably would describe it as some stable state from which we can make consistent measurement records.

 

[PeterDonis]

I am not using definition. I am understanding it as a basic term.

I have no idea what you mean by this. If you don't know how it's defined, how can you understand it "as a basic term"? (Or any other way, for that matter?)

 

[Mentz114]

That the assumption that a measurement always gives an eigenvalue is contradictory with the assumption that evolution is unitary.
[..]

I did not assert this because it is obviously not true. What I did say was that it is not impossible. You might not agree with that either.

I'm reading ( again) a fascinating paper called Interaction of superpositions of coherent states of light with two-level atoms published in the Journal of Modern Optics (1992, vol 39, No 7 pp. 1441) where the unitary evolution of superpositions and mixed states shows the atom oscillating rapidly between $|ground\rangle$ and $|excited\rangle$ states which is a kind of physical superposition.

Not necessarily. Shattering a mirror is irreversible, but that doesn't mean that it can't be described by classical mechanics, which is reversible. The classical resolution to the contradiction is that shattering a mirror isn't really irreversible, but the reverse process requires unattainable initial conditions. There could be a similar explanation in quantum mechanics.

Irreversiblily means no recurrences ever. But a recurrence time of 10⁶ years is nearly as good and can be achieved by unitary evolution.

(I'm sorry my last post sounded like a whinge.)

 

[vanhees71]

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.

You forget "the environment". For a measurement you need to fix the outcome, and this involves to read off the pointer state, leading to disentanglement through decoherence.

 

[vanhees71]

Not necessarily. Shattering a mirror is irreversible, but that doesn't mean that it can't be described by classical mechanics, which is reversible. The classical resolution to the contradiction is that shattering a mirror isn't really irreversible, but the reverse process requires unattainable initial conditions. There could be a similar explanation in quantum mechanics.

Of course. Macroscopic objects are hard to prepare in all details of its microscopic constituents. That's why we use macroscopic observables to describe its behavior, and this "coarse grained view" often leads to "classical behavior" looking irreversible.

 

[PeterDonis]

For a measurement you need to fix the outcome, and this involves to read off the pointer state, leading to disentanglement through decoherence.

This doesn't get rid of the superposition produced by unitary evolution. It just explains why there are no "cross terms" like, for example, "spin up but measuring device measured spin down".

 

[vanhees71]

But it explains, why there's only one outcome. It's forced by looking at a coarse grained macroscopic observable that is not described by a pure entangled state.

 

[stevendaryl]

But it explains, why there's only one outcome. It's forced by looking at a coarse grained macroscopic observable that is not described by a pure entangled state.

Coarse-graining can't turn a symmetric state into an asymmetric one. If the initial state and the Hamiltonian are left-right symmetric, but each of the results violates left-right symmetry, then you cannot explain getting a single result just using coarse-graining. Or if you are positive that you can, I'd like to see an example showing this effect.

Consider a particle in the following potential well. Initially, the particle has a wave function that is even under the operation $x \rightarrow -x$ and is well-localized at $x=0$ (for example, $\psi(x) = e^{-\lambda x^2}$).

Later, you measure the position of the particle. Most likely, you'll either find the particle in the left well or the right well. Each of these results violates the left-right symmetry of the initial conditions.

How does coarse-graining explain this loss of symmetry?

 

[vanhees71]

It's not clear to me what you are after. I thought we discuss the Stern-Gerlach apparatus, where you get an entanglement between position and the component of the spin in direction of the magnetic field. A beam of particles (ensemble) splits into two partial beams whose relative strength is given by the probabilities of QT. Where is there a problem with parity?

The irreversable element in this case is the registration of the particle at the screen (or nowadays some electronic CCD-like device). This gives always one spot (which is why Schrödinger's original interpretation of a particle as a "smeared charge distribution" doesn't work and why Born discovered the probabilistic meaning of Schrödinger's wave function), and thanks to the entanglement of position with spin a unique outcome for the spin component. The coarse grained macroscopic observable is the point on the screen.

 

[stevendaryl]

It's not clear to me what you are after.

I'm not "after" anything. I'm just showing the inconsistency between smooth evolution under the Schrodinger equation and the prediction of the Born rule for measurements.

I thought we discuss the Stern-Gerlach apparatus, where you get an entanglement between position and the component of the spin in direction of the magnetic field. A beam of particles (ensemble) splits into two partial beams whose relative strength is given by the probabilities of QT. Where is there a problem with parity?

The initial state is left-right symmetric. The final state has a dot on the left photographic plate, or a dot on the right photographic plate. So the final state is not left-right symmetric.

Coarse-graining does not explain why you get a unique result. @PeterDonis is right. Entanglement explains why you don't see interference between macroscopically different alternatives, but it does not explain why there is only one result produced.

 

[vanhees71]

There is no contradiction between quantum mechanics and Born's rule. Born's rule is used to compare the predictions of the formalism with experiment and obviously with great success. Nature seems to prefer the probabilistic behavior described by QT and not some philosophical prejudice of some scientists.

I still don't know, what you mean by "left-right symmetric" in context of the SG experiment, but suppose we have prepared the system in a state such that there's a 50% probability for the particle to show spin up and 50% probability to show spin down (in the measured direction determined by the magnetic field), then of course the probability distribution will be "left-right symmetric", i.e., supposed you have a well-designed SG experiment, you'll find the particles with 50% in the region of space referring to spin up and 50% for spin down. That's the content of this symmetry, i.e., it refers to the probabilistic meaning of the quantum state according to Born's rule. The outcome for each individual measurement cannot be predicted, only the probabilities.

If there is no interference between macroscopically different alternatives (here to register an individual particle as a spot on a screen), then there's one and only one result of the measurement, and this is in accordance with all observations so far.

 

[stevendaryl]

There is no contradiction between quantum mechanics and Born's rule.

I think that's wrong. The Born rule predicts that the measurement device will either be in one macroscopic state or another. The pointer will either point to the left or to the right. The dot will either be on the left photographic plate or the right photographic plate. Smooth evolution predicts that it (together with the rest of the universe) will be in a superposition of those states. Those are different states. They have different properties. They in principle lead to different probabilities for future states.

In practice the differences are not observable because the interference effects are too tiny to measure. That, once again, is why I call it a "soft contradiction". But whether it is a soft contradiction or a hard contradiction, if a theory is inconsistent, then it can't be correct.

 

[vanhees71]

QT in its minimal interpretation is consistent. I also don't see what you mean by "different probabilities for future states". Of course you have to analyse for each individual measurement device what happens to the measured system, the measurement device etc. For the standard SG experiment, the particles hit the screen and are lost. Then Stern and Gerlach proudly send the photographs, which btw. worked out so well only because of cheap cigars they smoked in the lab (i.e., today the wouldn't succeed within the Goethe University anymore ;-)) of the particles around to colleagues.

 

[stevendaryl]

QT in its minimal interpretation is consistent.

I know you believe that, but it's not true.

Okay, I want to distinguish between the minimal "recipe" and the minimal "ontology". The minimal recipe doesn't attempt to apply quantum mechanics to macroscopic devices. You apply quantum mechanics only to microscopic systems.

But a minimal ontology would imply that quantum mechanics applies to macroscopic systems as well as microscopic. If you assume that, then the minimal interpretation (Schrodinger's equation + Born rule) becomes inconsistent.

 

[stevendaryl]

The minimal recipe, which is consistent, even if ad hoc, does not assume that QM applies to arbitrarily large systems, does not assume that measurement is a physical process, does not assume that the measurement act respects speed-of-light limitations, does not assume that there is only one possible world, does not assume that collapse is or is not physical. It simply takes no position on any of those.

The minimal ontology, though, does take a position on all of those, and altogether, they are inconsistent.

 

[PeterDonis]

it explains, why there's only one outcome

Only if you add a collapse postulate. In the MWI there isn't only one outcome; all outcomes occur. And the MWI is perfectly consistent with decoherence--in fact, I believe decoherence was originally developed in order to explain why the different branches in the MWI don't interfere.

 

[vanhees71]

I know you believe that, but it's not true.

Okay, I want to distinguish between the minimal "recipe" and the minimal "ontology". The minimal recipe doesn't attempt to apply quantum mechanics to macroscopic devices. You apply quantum mechanics only to microscopic systems.

But a minimal ontology would imply that quantum mechanics applies to macroscopic systems as well as microscopic. If you assume that, then the minimal interpretation (Schrodinger's equation + Born rule) becomes inconsistent.

Quantum mechanics in its minimal interpretation (usually physicists working on physics and not philosophy follow the "shutup and calculation interpretation", which is just the minimal interpretation adding ignorance of unnecessary philosophical quibbles irrelevant to physics) is applied with great success to all kinds of mesoscopic and macroscopic systems. Condensed-matter physics is all about this. There is no contradiction between QT and experiment nor is there any inconsistency within QT in its application to macroscopic systems. To the contrary, condensed-matter physics is one of the success stories of QT with evergrowing numbers of scientists involved in it and leading even to applications in everyday life like the laptop I'm hacking this posting right now.

 

[vanhees71]

Only if you add a collapse postulate. In the MWI there isn't only one outcome; all outcomes occur. And the MWI is perfectly consistent with decoherence--in fact, I believe decoherence was originally developed in order to explain why the different branches in the MWI don't interfere.

There is no collapse. It's decoherence, which is very hard to avoid to the dismay of quantum-computer engineers.

 

[stevendaryl]

Quantum mechanics in its minimal interpretation (usually physicists working on physics and not philosophy follow the "shutup and calculation interpretation", which is just the minimal interpretation adding ignorance of unnecessary philosophical quibbles irrelevant to physics) is applied with great success to all kinds of mesoscopic and macroscopic systems. Condensed-matter physics is all about this. There is no contradiction between QT and experiment nor is there any inconsistency within QT in its application to macroscopic systems. To the contrary, condensed-matter physics is one of the success stories of QT with evergrowing numbers of scientists involved in it and leading even to applications in everyday life like the laptop I'm hacking this posting right now.

I don't have any disagreement with any of that. The issue is not its usefulness but its completeness and consistency. Yes, ad hoc procedures can deal with inconsistencies.

The inconsistency is that the claim that a measurement device always returns an eigenvalue of the observable being measured contradicts the claim that that measurement device obeys quantum mechanics. The resolution is to have rules of thumb for when you treat the measurement device as a measuring device and when you treat it as a quantum-mechanical system. You have rules of thumb for resolving the contradictions.

 

[vanhees71]

Where are, in your opinion, inconsistencies?

 

[stevendaryl]

Where are, in your opinion, inconsistencies?

The Born rule predicts that after a measurement, a measuring device is in a definite pointer state. Unitary evolution predicts that it's not.

 

[stevendaryl]

The Born rule predicts that after a measurement, a measuring device is in a definite pointer state. Unitary evolution predicts that it's not.

In the particularly simple case of measuring the spin of a Fermion, let's assume that the state of the device + environment + whatever can be described using a basis: $|P,j\rangle$, where $P$ is the "pointer state" variable, which can take on values "U" (measured spin-up) or "D" (measured spin-down) or "0" (no measurement yet), and where $j$ describes all other variables that we're not interested in.

If we assume that in the initial state, the Fermion was in the superposition $\alpha|u\rangle |u\rangle + \beta |d\rangle$, and that the device had pointer state $0$, then at a later time, the state of the composite system will be described by a density matrix of the form:

$\rho_{jkk'uu} |U, k\rangle \langle U, k'\rangle + \rho_{jkk'ud} |U, k\rangle \langle D, k'\rangle + \rho_{jkk'du} |D, k\rangle \langle U, k'\rangle + \rho_{jkk'ud} |U, k\rangle \langle D, k'\rangle$

In contrast, the Born rule would predict that the coefficients for the "cross" terms, $\rho_{jkk'ud}$ and $\rho_{jkk'du}$, would be zero.

Those are different predictions. It's not a matter of interpretation. Those are different states.

 

[PeterDonis]

There is no collapse.

Then there isn't one outcome. So your statement in post #190, which I quoted...

it explains, why there's only one outcome.

...doesn't make sense if there is no collapse. So which is it?

 

[zonde]

Well, the difference between a macroscopic object, such as a measuring device, and a microscopic object, such as an electron, is that for any given macroscopic state (what we would intuitively, pre-quantum mechanics, think of a state, such as "the readout shows the number 32" or "the pointer points to the left" or "the left light is on") there are many, many microscopic states that correspond to it.

I don't have the mathematical sophistication to accurately describe the situation using quantum mechanics, but perhaps it's something like the following:

The complete system perhapse can be described by three variables: $|s, S, j\rangle$, where $s$ is the observable corresponding to the system being measured (an electron's spin, maybe), $S$ is the corresponding value of the "pointer variable", and $j$ represents all the other degrees of freedom.

To make it both simple and definite, we will assume that there are two possible values for $s$:$u$ and $d$, and three possible values for $S$: $0, U, D$. There are many (possibly infinitely many) values for the other degrees of freedom, $j$.

To say that the pointer accurately measures the z-component of spin is to say something like the following:

• If you start in the state $|u, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{ujk} |u, U, k\rangle$
• If you start in the state $|d, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{djk} |d, D, k\rangle$
• It follows from the linearity of the evolution operator that if you start in a superposition of the form $\alpha |u, 0, j\rangle + \beta |d, 0, j\rangle$, then you will end up in a superposition of the form $\alpha \sum_k c_{ujk} |u, U, k\rangle + \beta \sum_k c_{djk} |d, D, k\rangle$

By "end up", I mean applying the evolution operator $e^{-iHt}$.

You say: "It follows from the linearity of the evolution operator". But how can I check that evolution operator is indeed linear? I suppose that for evolution operator to be linear Hamiltonian should be linear as well, right? But is this the case when we have configuration that can evolve by avalanche type process? Total potential energy for system of several particles interacting by coulomb potentials would have cross terms. Is this not a problem if we assume that Hamiltonian is linear?

 

[stevendaryl]

You say: "It follows from the linearity of the evolution operator". But how can I check that evolution operator is indeed linear?

That's what QM says about it.

I suppose that for evolution operator to be linear Hamiltonian should be linear as well, right? But is this the case when we have configuration that can evolve by avalanche type process? Total potential energy for system of several particles interacting by coulomb potentials would have cross terms. Is this not a problem if we assume that Hamiltonian is linear?

Saying that evolution is linear means that if $\psi_1(x,t)$ and $\psi_2(x,t)$ are solutions to Schrodinger's equation, then so is a linear combination:

$\alpha \psi_1(x,t) + \beta \psi_2(x,t)$

Quantum mechanics assumes that evolution is linear.

An example of a nonlinear theory might be something like this:

$\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2} + V(x) \Psi + (\Psi)^3 = i \hbar \frac{d\Psi}{dt}$

The presence of second or higher powers of $\Psi$ would make it nonlinear.

Actually, field theory does consider differential equations that look like that, but in those cases, $\Psi$ is not the wave function, it's a field operator. There is something corresponding to the wave function in quantum field theory, and its evolution is still linear.

 

[vanhees71]

Then there isn't one outcome. So your statement in post #190, which I quoted...
...doesn't make sense if there is no collapse. So which is it?

There's one outcome also in the minimal interpretation. That there is a unique outcome of a measurement is part of the definition of the word "measurement" here. If the measurement doesn't deliver one outcome (including a systematic and statistical error estimate) it's not a valid measurement by definition, and any experimentalist's paper sending such a result to a serious science journal will be rejected by peer review.

 

[Mentz114]

There's one outcome also in the minimal interpretation. That there is a unique outcome of a measurement is part of the definition of the word "measurement" here. If the measurement doesn't deliver one outcome (including a systematic and statistical error estimate) it's not a valid measurement by definition, and any experimentalist's paper sending such a result to a serious science journal will be rejected by peer review.

Agree. The definition of 'measurement' is particularly important. We could rewrite the Born rule as 'a successful measurement is one where the apparatus indicates correctly the eigenstate of the system'. It would save some confusion in my opinion.