Does the EPR experiment imply QM is incomplete?

In summary: This is a difficult question. It's difficult to say because it seems like it would be a pretty big discovery if it existed.
  • #106
bhobba said:
just know its something every interpretation has and is what on this forum (and in all textbooks I am aware of) we call the Born Rule.
Except for my thermal interpretation! The latter says that as a property of measurements, Born's rule is only an approximation, valid only in very simple, idealized systems.
See the thread https://www.physicsforums.com/threads/many-measurements-are-not-covered-by-borns-rule.894095/ where this is argued from well-known cases in the literature.
 
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  • #107
I actually do think that EPR shows that we don't completely understand QM. Whether that means it's incomplete or not is a matter of interpretation.

I don't think that we completely understand what a "measurement" is in QM. There are two competing ways of understanding measurements, and they are at odds. And both come into play in the EPR experiment.

One conception of measurement is that it's passive, at least in principle: When we measure some property of a system, we're just discovering a fact about that system. That's the classical notion of measurement.

A second conception of measurement is that it's active: The result of measurement is not a pre-existing quantity that you just become aware of, but is brought into existence through the participation and interaction of both the system being measured and the measuring device. The act of measuring disturbs the system being measured.

Both conceptions come into play in EPR. When it comes to measuring the polarization of a single photon, we can actually prove (at least in a one-world ontology) that the measurement disturbs the state of the photon being measured. The way that we measure a photon's polarization is by using a polarizing filter: If the photon passes through the filter, then its polarization is aligned with the orientation of the filter. If the photon is absorbed, then its polarization is perpendicular to the filter's orientation. To see that the filter is actually changing the state of the photon (as opposed to just filtering out photons of the wrong polarization), you can do the following:

  • Send photons through one filter. The photons that come through will have a specific polarization.
  • Have those that pass through go through a second filter, oriented at angle ##\theta## relative to the first.
  • Have those that pass through the second filter go through a third.
  • Etc.
With enough intermediate filters (actually, just three is enough, with a relative angle of 45o between the orientations of successive filters), you can get photons coming through oriented 90o away from the orientation they had after the first filter. It's obvious in this case that the filters have twisted the photons' polarizations, rather than simply passing those that had the wrong polarization. So the act of measurement of a photon's polarization actually does something to the photon. The fact that the photon coming out of the filter is polarization in a particular direction doesn't say anything definite about what it's polarization was before it went through.

In EPR, we have two correlated photons measured by Alice and Bob. Alice measures the polarization of one photon, and whatever answer she gets, she immediately knows the polarization of Bob's photon. Unless you allow for faster-than-light influences, Alice's measurement can't disturb Bob's photon. So if you disbelieve in FTL influences, you have to think of Alice's measurement as simply updating her information about Bob's photon, rather than changing Bob's photon.

But how can both be the case? When Alice measure's her own photon's polarization, her photon is interacting with her filter, and the state of her photon is affected by the filter. The fact that her photon comes through the filter polarized along a particular direction does not imply that the photon had the polarization prior to its passing through. But Alice can use the polarization of her photon after passing through the filter to deduce Bob's photon's polarization. How can that be possible? The disturbance model of measurement is justified in the case of Alice measuring her own photon, while the passive update model of measurement is justified in the case of Alice deducing something about Bob's photon.
 
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  • #108
stevendaryl said:
I actually do think that EPR shows that we don't completely understand QM. Whether that means it's incomplete or not is a matter of interpretation.[].
About entangled photons QT says the the probability of a coincidence between A's and B's results is ##\cos(\alpha-\beta)^2## and no-one misunderstands that.

You are concerned about the things that QT does not tell us. The fact that it says so little could mean it is incomplete, but who cares anyway except philosophers
 
  • #109
Nice way of looking at EPR stevendaryl, I never thought of it like that.

Just a small thing, since the EPR correlations can be simulated by a local classical model, would maybe a Bell situation be better? I think your main point carries over without change.

There you would have the frustration between the active and passive view, but without the easy "out" that QM is incomplete due to the existence of a local classical model to simulate the correlations.
 
  • #110
stevendaryl said:
When Alice measure's her own photon's polarization, her photon is interacting with her filter, and the state of her photon is affected by the filter. The fact that her photon comes through the filter polarized along a particular direction does not imply that the photon had the polarization prior to its passing through. But Alice can use the polarization of her photon after passing through the filter to deduce Bob's photon's polarization. How can that be possible?
Here's how this experiment is interpreted in Bohmian Mechanics: The photons detected by Alice and Bob are particles that each possesses just one (hidden) property: their actual position in 3D space at the point in time they are measured. All other quantum properties, such as polarization, are properties of the pilot wave with which these particles are entangled. When Alice measures the polarization of her photon, the wave function of her measuring device becomes entangled with the wave function of the photon. It is the pilot wave of the entangled photon/measuring device system that produces the measured outcome. So long as Bob's photon remains entangled with Alice's photon, its polarization will be correlated with the outcome of Alice's measurement.
 
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  • #111
Lish Lash said:
Here's how this experiment is interpreted in Bohmian Mechanics: The photons detected by Alice and Bob are particles that each possesses just one (hidden) property: their actual position in 3D space at the point in time they are measured. All other quantum properties, such as polarization, are properties of the pilot wave with which these particles are entangled. When Alice measures the polarization of her photon, the wave function of her measuring device becomes entangled with the wave function of the photon. It is the pilot wave of the entangled photon/measuring device system that produces the measured outcome. So long as Bob's photon remains entangled with Alice's photon, its polarization will be correlated with the outcome of Alice's measurement.
You are explaining entanglement with a help of entanglement. It's circular explanation, don't you see?
 
  • #112
Mentz114 said:
About entangled photons QT says the the probability of a coincidence between A's and B's results is ##\cos(\alpha-\beta)^2## and no-one misunderstands that.

You are concerned about the things that QT does not tell us. The fact that it says so little could mean it is incomplete, but who cares anyway except philosophers

I actually don't think that it "says so little". I think that what it does (quantum mechanics in the usual rule of thumb) is actually inconsistent, but it's a "soft" inconsistency. I've said before the reason I think that:

  • You have the Born rule, which says that when a measuring device interacts with a system to measure some property of that system, then you get a result that is an eigenvalue of the corresponding operator with probabilities given by the squares of the relevant amplitudes.
  • Getting a particular result means that the measuring device is in some definite configuration.
  • However, if you treat the measuring device as a quantum system, then the interaction of the measuring device with the system it's measuring doesn't produce a single result, but produces a superposition of all possible results.
So on the one hand, you have a prediction from quantum mechanics (Schrodinger's equation) that the measurement device will be described by one state, a superposition of various possible macroscopic results. This prediction is deterministic. On the other hand, you have a prediction from quantum mechanics (Born's rule) that the measurement device will be described by one or another state, where each has a definite macroscopic result. Those are two different, and contradictory, predictions, both made by quantum mechanics.

I call that a "soft" contradiction, because nobody actually computes the state of a macroscopic device by treating it as a quantum-mechanical system and applying Schrodinger's equation. It's too complex to do that. They treat the device classically, or semi-classically, and only consider the Born prediction. But I think that it's actually inconsistent, logically. Maybe only philosophers care about inconsistencies, if there is a work-around.
 
  • #113
Lish Lash said:
Here's how this experiment is interpreted in Bohmian Mechanics: The photons detected by Alice and Bob are particles that each possesses just one (hidden) property: their actual position in 3D space at the point in time they are measured. All other quantum properties, such as polarization, are properties of the pilot wave with which these particles are entangled. When Alice measures the polarization of her photon, the wave function of her measuring device becomes entangled with the wave function of the photon. It is the pilot wave of the entangled photon/measuring device system that produces the measured outcome. So long as Bob's photon remains entangled with Alice's photon, its polarization will be correlated with the outcome of Alice's measurement.

Yes, Bohmian mechanics doesn't actually treat measurements according to the Born rule, but instead, only uses the Born rule to give probabilities of various locations in configuration space (the position space for all particles involved). So the two aspects of measurement don't directly come into play.
 
  • #114
stevendaryl said:
[..]
So on the one hand, you have a prediction from quantum mechanics (Schrodinger's equation) that the measurement device will be described by one state, a superposition of various possible macroscopic results.
[..]
We've been here. I don't think macroscopic superpositions are possible. The SE does not assert that. People interpreting the SE say that.
 
  • #115
Mentz114 said:
We've been here. I don't think macroscopic superpositions are possible. The SE does not assert that.

I don't agree. Quantum mechanics certainly has no criterion for when a system is too large to be described by the Schrodinger equation. And if it is really the case that sufficiently large systems are no longer described by the Schrodinger equation, then that would definitely mean that quantum mechanics is incomplete. (It would actually mean that it is false, but only an approximation good for small systems.)
 
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  • #116
stevendaryl said:
I don't agree. Quantum mechanics certainly has no criterion for when a system is too large to be described by the Schrodinger equation. And if it is really the case that sufficiently large systems are no longer described by the Schrodinger equation, then that would definitely mean that quantum mechanics is incomplete. (It would actually mean that it is false, but only an approximation good for small systems.)
I am not saying that the SE cannot describe a macroscopic object. I'm objecting to the assumption of superpositions of solutions.
They are mathematically valid but so what. Not every solution to a set of constraints necessarily has a real counterpart.
 
  • #117
stevendaryl said:
So on the one hand, you have a prediction from quantum mechanics (Schrodinger's equation) that the measurement device will be described by one state, a superposition of various possible macroscopic results. This prediction is deterministic. On the other hand, you have a prediction from quantum mechanics (Born's rule) that the measurement device will be described by one or another state, where each has a definite macroscopic result. Those are two different, and contradictory, predictions, both made by quantum mechanics.
These two predictions are contradictory only if you consider QM complete.
If QM is incomplete there is no contradiction, because the state that is associated with different macroscopic configurations describes only one property of these configurations. That property is the same while other properties are different.
 
  • #118
Mentz114 said:
I am not saying that the SE cannot describe a macroscopic object. I'm objecting to the assumption of superpositions of solutions.

These two statements are inconsistent. The SE gives you "superpositions of solutions" just by time evolution. You can't arbitrarily exclude "superpositions of solutions" and still use the SE at all.
 
  • #119
PeterDonis said:
These two statements are inconsistent. The SE gives you "superpositions of solutions" just by time evolution. You can't arbitrarily exclude "superpositions of solutions" and still use the SE at all.
I could be wrong but I thought

##\hat{H}=\sum c_n|E_n\rangle\langle E_n|##, so that ## \hat{H}E_n\rangle =c_n |E_n \rangle##

The evolution is driven by ##e^{\hat{H}t}## but ##\hat{H}## is idempotent so I think my point below stands.

i.e. an eigenvalue cannot evolve to a superposition ? And a superposition can only evolve to a superposition.
 
  • #120
Mentz114 said:
an eigenvalue cannot evolve to a superposition ?

An eigenstate of the Hamiltonian will stay an eigenstate of the Hamiltonian, yes. But an eigenstate of the Hamiltonian has no interactions whatever--nothing ever happens to it. So no real object is ever in an eigenstate of the Hamiltonian. Any state that is a reasonable candidate to describe a real object will change under time evolution; and any state that, at some instant of time, happens to look like a reasonable classical state of a classical object, will not stay that way; it will evolve into a "Schrodinger's Cat" type state that does not describe anything like a classical state of a classical object.
 
  • #121
stevendaryl said:
So on the one hand, you have a prediction from quantum mechanics (Schrodinger's equation) that the measurement device will be described by one state, a superposition of various possible macroscopic results.
On the other hand, you have a prediction from quantum mechanics (Born's rule) that the measurement device will be described by one or another state, where each has a definite macroscopic result.

Please can you express these alternatives as equations. I cannot make sense of the words.

[edit]

OK, the first one is a superposition and the second is not. Sorry for the dyslexia ( or something).
 
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  • #122
PeterDonis said:
These two statements are inconsistent. The SE gives you "superpositions of solutions" just by time evolution. You can't arbitrarily exclude "superpositions of solutions" and still use the SE at all.
PeterDonis said:
An eigenstate of the Hamiltonian will stay an eigenstate of the Hamiltonian, yes. But an eigenstate of the Hamiltonian has no interactions whatever--nothing ever happens to it. So no real object is ever in an eigenstate of the Hamiltonian. Any state that is a reasonable candidate to describe a real object will change under time evolution; and any state that, at some instant of time, happens to look like a reasonable classical state of a classical object, will not stay that way; it will evolve into a "Schrodinger's Cat" type state that does not describe anything like a classical state of a classical object.

I think it is implicit in the statement of @stevendaryl's 'soft paradox' that there is an operator that acts on the macroscopic wave function (in the appropriate basis) . The superposition must be between one or more of its eigenvalues. So it must have started as a preparation of a superposition. If these preparations are not allowed, then obviously we can't evolve into a superposition.

It all comes down to preparation again.
 
  • #123
Mentz114 said:
So it must have started as a preparation of a superposition. If these preparations are not allowed, then obviously we can't evolve into a superposition.

That's not correct as you state it. An eigenstate of any operator other than the Hamiltonian will not stay an eigenstate of that operator under time evolution, unless that operator commutes with the Hamiltonian. It will evolve into a superposition of eigenstates.

Some useful operators, such as total momentum and total angular momentum, commute with the Hamiltonian. But I don't think any operator that might represent a realistic observable for a macroscopic system being measured--something like center of mass position, for example--will commute with the Hamiltonian.
 
  • #124
PeterDonis said:
That's not correct as you state it. An eigenstate of any operator other than the Hamiltonian will not stay an eigenstate of that operator under time evolution, unless that operator commutes with the Hamiltonian. It will evolve into a superposition of eigenstates.

Some useful operators, such as total momentum and total angular momentum, commute with the Hamiltonian. But I don't think any operator that might represent a realistic observable for a macroscopic system being measured--something like center of mass position, for example--will commute with the Hamiltonian.
Thank you. That is interesting. Two questions about the macroscopic apparatus come to mind,
If we are measuring a quantum property, would the apparatus have the same eigenvalues as the microscopic property ?
If the observable is an angle, would the operator commute with ##\hat{H}## ?
 
  • #125
Mentz114 said:
If we are measuring a quantum property, would the apparatus have the same eigenvalues as the microscopic property ?

Eigenvalues of what operator?

When we model quantum measurements as applying an operator to the system, that operator is an operator on the Hilbert space describing the measured system, not the measured system + measuring device. The eigenvalues of the operator are therefore eigenvalues applying to the measured system only. If you think about it, this is already an admission that such a model is incomplete.

When we try to construct a more complete model, where we include the measuring device and the interaction between it and the measured system, then we no longer model measurement as applying an operator to the system; we model it as just time evolving the system using the Schrodinger Equation with the Hamiltonian including the interaction between the measuring device and the measured system. This time evolution then puts the whole system (measured system + measuring device) into an entangled state. This state is not an eigenstate of any operator (or at least, not of any operator that anyone is writing down and using in the analysis), so there aren't any useful eigenvalues that apply to it.

In such a more complete model, "measurement results" are really encoded in the labels that are put on the terms in the entangled state. For example, say we put a qubit through a spin measurement. The resulting state will look something like (leaving out normalization factors): ##\vert + \rangle \vert M_+ \rangle + \vert - \rangle \vert M_- \rangle##. Here the measurement result "measured spin up" is encoded in the kets in the first term, and the measurement result "measured spin down" is encoded in the kets in the second term. It is not encoded in eigenvalues of any operator.

Mentz114 said:
If the observable is an angle, would the operator commute with ##\hat{H}## ?

I don't think so. Measuring an angle is not the same as measuring angular momentum.
 
  • #126
Mentz114 said:
Please can you express these alternatives as equations. I cannot make sense of the words.

Let's make it simple, and suppose that we have some measurement device that measures the spin of a particle along the z-axis. For the particle, ##|u\rangle## is the state that is spin-up in the z-direction, and ##|d\rangle is the state that is spin-down. Let's suppose that ##|0\rangle## is the initial "ready" state of the device, and let ##|U\rangle## mean "measured spin-up" and ##|D\rangle## mean "measured spin down". Those are sometimes called "pointer" states.

So the assumption that the device actually works as a measuring device is that:
  1. ##|u\rangle |0\rangle \Rightarrow |u\rangle |U\rangle##
  2. ##|d\rangle |0\rangle \Rightarrow |d\rangle |D\rangle##
(where ##\Rightarrow## means "evolves into, taking into account the Schrodinger equation")

By linearity of the Schrodinger equation, it follows that:

##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow \alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##

So if the particle starts off in a superposition of states, then the measuring device (and the rest of the universe, eventually, but we're not modeling that) ends up in a superposition of different "pointer" states.

So that's the prediction of the Schrodinger equation. The Born rule says, instead, that:
  1. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |u\rangle |U\rangle## with probability ##|\alpha|^2##
  2. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |d\rangle |D\rangle## with probability ##|\beta|^2##
The Born gives a probabilistic transition rule leading to a definite pointer state, while the Schrodinger equation gives a deterministic transition rule leading to a superposition of pointer states. Those are different, and contradictory, predictions.
 
  • #127
Mentz114 said:
I think it is implicit in the statement of @stevendaryl's 'soft paradox' that there is an operator that acts on the macroscopic wave function (in the appropriate basis) . The superposition must be between one or more of its eigenvalues. So it must have started as a preparation of a superposition. If these preparations are not allowed, then obviously we can't evolve into a superposition.

Think about an actual measurement. You pass an electron through a Stern-Gerlach device. The electron is either diverted to the right, if it's spin-up, or it's diverted to the left, if it's spin-down. If the electron is diverted left, it makes a visible dark spot on the left side of a photographic plate. If it's diverted to the right, it makes a visible dark spot on the right side of a photographic plate.

So connecting this to the mathematics above, ##|U\rangle## is the Stern-Gerlach system plus photographic plate, in which there is a dark spot on the right. ##|D\rangle## is the state where the dark spot is on the left.
 
  • #128
stevendaryl said:
By linearity of the Schrodinger equation, it follows that:

##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow \alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##

So if the particle starts off in a superposition of states, then the measuring device (and the rest of the universe, eventually, but we're not modeling that) ends up in a superposition of different "pointer" states.

So that's the prediction of the Schrodinger equation. The Born rule says, instead, that:
  1. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |u\rangle |U\rangle## with probability ##|\alpha|^2##
  2. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |d\rangle |D\rangle## with probability ##|\beta|^2##
The Born gives a probabilistic transition rule leading to a definite pointer state, while the Schrodinger equation gives a deterministic transition rule leading to a superposition of pointer states. Those are different, and contradictory, predictions.
They are not contradictory if they describe different aspects of reality.
One describes phase difference between different outcomes (HVs).
Other prediction describes actual outcome (when HV is revealed).

In case of microscopic systems you can't perform both measurements at the same time. In case of macroscopic systems we don't know how to measure phase relationship (perform interference measurement) between different outcomes, but hypothetically if we would know how to perform that interference measurement we can say that we would not be able to learn what was actual outcome in particular case.
 
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  • #129
stevendaryl said:
Let's make it simple, and suppose that we have some measurement device that measures the spin of a particle along the z-axis. For the particle, ##|u\rangle## is the state that is spin-up in the z-direction, and ##|d\rangle is the state that is spin-down. Let's suppose that ##|0\rangle## is the initial "ready" state of the device, and let ##|U\rangle## mean "measured spin-up" and ##|D\rangle## mean "measured spin down". Those are sometimes called "pointer" states.

So the assumption that the device actually works as a measuring device is that:
  1. ##|u\rangle |0\rangle \Rightarrow |u\rangle |U\rangle##
  2. ##|d\rangle |0\rangle \Rightarrow |d\rangle |D\rangle##
(where ##\Rightarrow## means "evolves into, taking into account the Schrodinger equation")

By linearity of the Schrodinger equation, it follows that:

##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow \alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##

So if the particle starts off in a superposition of states, then the measuring device (and the rest of the universe, eventually, but we're not modeling that) ends up in a superposition of different "pointer" states.

So that's the prediction of the Schrodinger equation. The Born rule says, instead, that:
  1. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |u\rangle |U\rangle## with probability ##|\alpha|^2##
  2. ##(\alpha |u\rangle + \beta |d\rangle)|0\rangle \Rightarrow |d\rangle |D\rangle## with probability ##|\beta|^2##
The Born gives a probabilistic transition rule leading to a definite pointer state, while the Schrodinger equation gives a deterministic transition rule leading to a superposition of pointer states. Those are different, and contradictory, predictions.
I don't see any ##e^{\hat{H}t}## in there. Unless you have explicitly different evolutions I can't see any contradiction. The state of the macroscopic apparatus needs to be there also because the point in question is final state. You need to give the assumptions concerning preparation of the microscopic state and the apparatus.
 
  • #130
Mentz114 said:
I don't see any ##e^{\hat{H}t}## in there.

Yeah, well that's the stumbling block for any discussion of the application of quantum mechanics to large systems is that it's difficult to do it rigorously.

Unless you have explicitly different evolutions I can't see any contradiction.

That's why I call it a "soft contradiction". The computations are intractable to actually derive the state from first principles.

However, we know that an electron hitting a photographic plate does cause a dark spot on the plate. To believe that this is not described by quantum mechanics seems to be equivalent to believing that quantum mechanics is incorrect or incomplete.

I spent a certain amount of time studying AI and one of the systems I looked at was the CYC project (I don't know whether that's been abandoned, or not). CYC had a notion of "microtheories" which described a particular small domain. For a physics example, statics. It knew how to reason within a microtheory. Then there were heuristics on top of the microtheories to decide which microtheory to use in what circumstances.

It was suspected that the microtheories were actually inconsistent, but it was a "soft" inconsistency, because you never applied more than one microtheory at a time.

Quantum mechanics is similarly composed of two microtheories for describing systems: For small systems, you use the Schrodinger equation. For large systems, you use the Born rule.
 
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  • #131
PeterDonis said:
Eigenvalues of what operator?

[cut for brevity]
Thanks for trying to unconfuse me. I have to keep asking stuff that should be included in the description of the problem and I have asked stevendaryl for clarification.
There may be a contradiction in the requirement that the apprartus state should be highly correlated with the incoming system unless the apparatus has some similarities with the incoming state - viz. common eigenvalues which implies an operator (?). The states being evolved must include the apparatus and the outcome depends only on the Hamiltonian and the initial state. This does not always result in a superposition but I don't know if that is relevant because the 'contradiction' is not unambiguously defined.
 
  • #132
zonde said:
They are not contradictory if they describe different aspects of reality.
One describes phase difference between different outcomes (HVs).
Other prediction describes actual outcome (when HV is revealed).

In case of microscopic systems you can't perform both measurements at the same time. In case of macroscopic systems we don't know how to measure phase relationship (perform interference measurement) between different outcomes, but hypothetically if we would know how to perform that interference measurement we can say that we would not be able to learn what was actual outcome in particular case.
Yes, that is true. In 'collapse' terms the the first case is uncollapsed, but the second is after measurement. That is pretty important.
 
  • #133
zonde said:
They are not contradictory if they describe different aspects of reality.

They predict different future probabilities, so they are contradictory predictions. To see this, suppose there is a final state ##|final\rangle##. Let ##\psi_{U}## be the probability of making a transition from ##|u\rangle |U\rangle## to the state ##|final\rangle##. Let ##\psi_{D}## be the probability of making a transition from ##|d\rangle |D\rangle## to the state ##|final\rangle##.

Then, the probability of later observing the system in the state ##|final\rangle## will be

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2##

under the assumption that the composite system nondeterministically transitioned to ##|u\rangle |U\rangle##, with probability ##|\alpha|^2## or to ##|d\rangle |D\rangle##, with probability ##|\beta|^2##.

In contrast, if the composite system is in the superposition ##\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##, then the probability of ending up in state ##|final\rangle## will be given by:

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2 + \alpha^* \psi_{U}^* \beta \psi_{D} +\beta^* \psi_{D}^* \alpha \psi_{U} ##

Those are different, and inconsistent, predictions.
 
  • #134
stevendaryl said:
They predict different future probabilities, so they are contradictory predictions. To see this, suppose there is a final state ##|final\rangle##. Let ##\psi_{U}## be the probability of making a transition from ##|u\rangle |U\rangle## to the state ##|final\rangle##. Let ##\psi_{D}## be the probability of making a transition from ##|d\rangle |D\rangle## to the state ##|final\rangle##.

Then, the probability of later observing the system in the state ##|final\rangle## will be

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2##

under the assumption that the composite system nondeterministically transitioned to ##|u\rangle |U\rangle##, with probability ##|\alpha|^2## or to ##|d\rangle |D\rangle##, with probability ##|\beta|^2##.

In contrast, if the composite system is in the superposition ##\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##, then the probability of ending up in state ##|final\rangle## will be given by:

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2 + \alpha^* \psi_{U}^* \beta \psi_{D} +\beta^* \psi_{D}^* \alpha \psi_{U} ##

Those are different, and inconsistent, predictions.
That is clearer. Coherence is the difference as Zonde pointed out. Is this not the problem that the decoherence theory addresses ?

In the case where ##\psi_{U}## and ##\psi_D## are orthogonal do the cross-terms disappear ?
 
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  • #135
stevendaryl said:
They predict different future probabilities, so they are contradictory predictions. To see this, suppose there is a final state ##|final\rangle##. Let ##\psi_{U}## be the probability of making a transition from ##|u\rangle |U\rangle## to the state ##|final\rangle##. Let ##\psi_{D}## be the probability of making a transition from ##|d\rangle |D\rangle## to the state ##|final\rangle##.

Then, the probability of later observing the system in the state ##|final\rangle## will be

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2##

under the assumption that the composite system nondeterministically transitioned to ##|u\rangle |U\rangle##, with probability ##|\alpha|^2## or to ##|d\rangle |D\rangle##, with probability ##|\beta|^2##.

In contrast, if the composite system is in the superposition ##\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle##, then the probability of ending up in state ##|final\rangle## will be given by:

##|\psi_{U}|^2 |\alpha|^2 + |\psi_{D}|^2 |\beta|^2 + \alpha^* \psi_{U}^* \beta \psi_{D} +\beta^* \psi_{D}^* \alpha \psi_{U} ##

Those are different, and inconsistent, predictions.
You are considering setup where ##|d\rangle |D\rangle## and ##|u\rangle |U\rangle## can end up in the same state ##|final\rangle##. This is clearly interference measurement. But you ignore phase relationship between two initial states, so you are assuming that the two initial states have decohered.
On the other hand for the state ##\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle## you assume no decoherence.
Can you explain your reasoning vs coherence/decoherence?
To me it seems like you are using two different microtheories for the two cases. In one microtheory each individual measurement outcome is completely independent from the rest of the world and/or from any other measurement in ensemble and so there can be no physical basis to consider any relative phase relationship between outcomes.
In the other microtheory two measurement outcomes sort of exist in two parallel but interacting worlds at the same time so there are means how to consider relative phase relationship between two outcomes.

Well, I'm not sure that inconsistency will persist if you would consider both cases in the same microtheory.
 
  • #136
zonde said:
You are considering setup where ##|d\rangle |D\rangle## and ##|u\rangle |U\rangle## can end up in the same state ##|final\rangle##. This is clearly interference measurement. But you ignore phase relationship between two initial states, so you are assuming that the two initial states have decohered.

Yes, that's the reason I call it a "soft" contradiction. You can't actually calculate phases accurately enough to calculate interference effects between macroscopic objects. But if we assumed unlimited computation power, we could in principle see the difference between the two evolution equations.

Decoherence is a matter of computational ability. Two systems have decohered if it is in practice impossible to accurately predict the phase relationship between them. I actually do not think that decoherence has any role in understanding the foundations of quantum mechanics, although it does explain why in practice we don't see interference effects for macroscopic objects.

Well, I'm not sure that inconsistency will persist if you would consider both cases in the same microtheory.

That's my point: the two microtheories are: (1) smooth evolution according to Schrodinger's equation, and (2) getting definite results of measurements according to the Born rule. It's the inconsistency between these two that I'm pointing out.
 
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  • #137
Mentz114 said:
That is clearer. Coherence is the difference as Zonde pointed out. Is this not the problem that the decoherence theory addresses ?

In the case where ##\psi_{U}## and ##\psi_D## are orthogonal do the cross-terms disappear ?

They don't disappear. They become negligible. That's why I call it a "soft" contradiction. The differences between the predictions are in practice impossible to observe. But they are different predictions.

Just for clarification, ##\psi_U## and ##\psi_D## are just complex numbers, not functions.

##\psi_U = \langle final |e^{-iHt}|u,U\rangle##
##\psi_D = \langle final | e^{-iHt}|d,D\rangle##
 
  • #138
stevendaryl said:
Decoherence is a matter of computational ability. Two systems have decohered if it is in practice impossible to accurately predict the phase relationship between them.

Actually, there are two things going on decoherence: One, as I said, is just the practical impossibility of computing phase relationships between states of a large system. The second is that if the system becomes entangled with yet other systems, then interference effects are not possible between states of the subsystem, only between different states of the larger composite system. And rapidly, that larger system becomes the entire universe (or the near-by part of it).
 
  • #139
stevendaryl said:
You can't actually calculate phases accurately enough to calculate interference effects between macroscopic objects. But if we assumed unlimited computation power, we could in principle see the difference between the two evolution equations.
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.

On the other hand if you assume decoherence in both cases then the state ##\alpha |u\rangle |U\rangle + \beta |d\rangle |D\rangle## becomes unphysical idealization. Again there is no inconsistency.

I guess you have your viewpoint without HVs and you are saying it is inconsistent. And my suggestion that bringing HVs into the picture makes it consistent is not satisfactory for you.
 
  • #140
I guess one of the questions about decoherence is why one basis is preferred over another. I would say that position basis is preferred because particles form bond states when they are near each other. That determines that macro world "lives" in position basis.
 

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