Von Neumann QM Rules Equivalent to Bohm?

[stevendaryl]

After Alice finds V for her photon, she knows that Bob finds H for his for sure,

Well, we're mixing up two different types of EPR here. I'm assuming the case where the two photons have the same polarization, but it doesn't matter.

and there is no hidden variable. Why should there be one?

Once again, X is the statemement "Bob will measure his photon to have horizontal polarization". I think we agree that after Alice's measurement. statement X is true. Yes or no? Was it true before Alice's measurement? Yes or no?

After A's measurement, finding V, for sure B will measure H. That's due to the entangled initial state. This follows directly from the usual assumptions of QT.

It doesn't matter what it is due to. The issue is just the truth value of the statement X = "Bob will measure his photon to have horizontal polarization". Is it true after Alice's measurement? Was it true before Alice's measurement?

 

[stevendaryl]

I don't know what you mean by "effects" under 2.

If there is a fact about Bob's situation that was not true at one time and becomes true at another time, I consider that to be a change in Bob's situation. I would say that Alice's actions have an effect on Bob if they cause a change to Bob's situation.

I suppose the disagreement is whether "Bob will measure result H" is a fact about Bob's situation, or not.

 

[atyy]

I suppose the disagreement is whether "Bob will measure result H" is a fact about Bob's situation, or not.

Especially in the Bell tests, since Bob chooses his measurement randomly at the last moment.

 

[stevendaryl]

Especially in the Bell tests, since Bob chooses his measurement randomly at the last moment.

In the particular exchange with vanhees71, I was talking about the situation in which Alice and Bob agree ahead of time to use the same filter orientation. For Bell's purposes, he didn't want to do this, but if we do stipulate that Alice and Bob use the same filter setting, and that it's the type of twin-pair photon generation that produces two photons with the same polarization, then we can reason as follows: (For definiteness, let's assume that it's asymmetric; Alice is much closer to the source of the photons than Bob is, so she gets her photon before Bob gets his.)

1. Immediately before Alice measure her photon's polarization, the most complete statement that can be made about Bob's measurement result is that it has a 50/50 chance of H polarization or V polarization.
2. Immediately after Alice measures her photon's polarization, and finds that it has polarization H, then she can claim, with 100% certainty: "Bob will measure polarization H for his photon"

So the issue is, what is the nature of the claim/prediction X = "Bob will measure polarization H for his photon"? Is is a fact about Bob's situation (photon + filter + detector)?

To me, it seems like it's as good a candidate for being a fact as anything else in quantum mechanics.

 

[TrickyDicky]

A clarification about my #145, I'm using atyy's distinction regarding necessity of collapse between sequential measurements and measurements to which the deferred measurement principle can be applied when I say that R(collapse) conditions effective perturbative QFT S-matrix time-ordered product, that must include microcausality(operators commuting at spacelike separation) to ensure coordinate independence(being spacetime fields).

 

[morrobay]

After Alice finds V for her photon, she knows that Bob finds H for his for sure, and there is no hidden variable. Why should there be one?
After A's measurement, finding V, for sure B will measure H. That's due to the entangled initial state. This follows directly from the usual assumptions of QT.

In your next post # 148, you refer to " non - local " correlations as described by entangled states. Could you elaborate on this in the context of whether Bob (above) would have measured H before Alice measured V
I see two options : 1. They were created at source , Alice V and Bob H (predetermined )
2. A superposition Alice 50/50 V / H and when Alice finds V Bob finds H ( non-local interaction or non local
correlation ) ?

 

[vanhees71]

Both. 1. and 2. are obviously wrong! Again: The biphoton was created by a local interaction of a laser beam with the non-linear birefringent crystal by parametric down conversion (LOCALITY of interactions). The polarization state is after some standard-optics manipulations given by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle - |VH \rangle).$$
One of the photons is registered at A's place and one at B's place. Both are using polarization filters in the same $x$-direction.

Now the states of the single photons are given by tracing out the other photon respectively, i.e., they are in mixed states:
$$\hat{\rho}_A=\mathrm{Tr}_{B} |\Psi \rangle \langle \Psi |=\frac{1}{2} \mathbb{1}.$$
Also Bob finds
$$\hat{\rho}_B=\mathrm{Tr}_{A} |\Psi \rangle \langle \Psi |=\frac{1}{2} \mathbb{1}.$$
Both have maximally unpolarized single photons! The polarization state of each of the single photons is maximally indetermined.

Now, if A measures $H$ for her photon, she updates her state to
$$|\Psi_{A|H} \rangle = \left (|H \rangle \langle H| \otimes \mathbb{1} \right )|\Psi \rangle=\frac{1}{\sqrt{2}} |HV \rangle.$$
This happens with the probability
$$P_A(H)=\|\Psi_{A|H} \|^2=\frac{1}{2}.$$
Bob still doesn't know about this, and he still uses $|\Psi \rangle$. So he only knows that he'll find with 50% probability H and with 50% probability V, while Alice knows that he must find V.

In this minimal interpretation, nothing happened to Bob's photon, and it cannot happen anything to it before Bob measures it, if the registration events for the photons at A's and B's place are space-like separated (according to Einstein causality). This is a consistent (in my opinion the only consistent) interpretation of the formalism given by relativistic local QFTs.

The entanglement enables correlations between far-distant registration events for the photons without predetermination of the measured observables, as is demonstrated clearly by this example. In this sense there are NON-LOCAL correlations in a theory where you have only LOCAL interactions, and the latter is the case by construction of QED as a local microcausal relativistic QFT.

 

[atyy]

Now, if A measures $H$ for her photon, she updates her state to
$$|\Psi_{A|H} \rangle = \left (|H \rangle \langle H| \otimes \mathbb{1} \right )|\Psi \rangle=\frac{1}{\sqrt{2}} |HV \rangle.$$
This happens with the probability
$$P_A(H)=\|\Psi_{A|H} \|^2=\frac{1}{2}.$$
Bob still doesn't know about this, and he still uses $|\Psi \rangle$. So he only knows that he'll find with 50% probability H and with 50% probability V, while Alice knows that he must find V.

Alice's knowledge is frame dependent.

 

[vanhees71]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

 

[TrickyDicky]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

For the calculation of a specific outcome of a measurement perturbative QED is not Lorentz invariant, both by the nonlinearity of the perturbative operations and per a theorem by Haag that I've seen you call "a quibble", so in strict terms it is the other way around, the presence of something like collapse is granted mathematically.

 

[vanhees71]

Well, mathematically QED is not proven to exist. Nevertheless the S-matrix elements as calculated in renormalized perturbation theory and which are compared to observations are Lorentz invariant (and gauge invariant). So the theory I'm talking about is this theory and not some fictitious "exact solution" of QED which might exist or not. It's of course a somewhant unsatisfying state that one of the most successful theories (nowadays one can say the entire standard model including QFD and QCD is the most successful theory ever) is shaky in its mathematical foundations.

 

[atyy]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

Well, the minimal interpretation is consistent, and has predictions that are Lorentz invariant. But it has collapse, which may or may not be real. What I don't think is right about what you write is that relativistic quantum field theory preserves Einstein causality. Einstein causality, as considered by EPR, is formalized by Bell as separability. Einstein causality is either empty if the wave function is not real, or it is violated if the wave function is real.

 

[TrickyDicky]

Well, mathematically QED is not proven to exist. Nevertheless the S-matrix elements as calculated in renormalized perturbation theory and which are compared to observations are Lorentz invariant (and gauge invariant). So the theory I'm talking about is this theory and not some fictitious "exact solution" of QED which might exist or not. It's of course a somewhant unsatisfying state that one of the most successful theories (nowadays one can say the entire standard model including QFD and QCD is the most successful theory ever) is shaky in its mathematical foundations.

We must be talking about different things. AFAIK there is a QED that is perfectly Lorentz invariant but divergent, and a QED that manages to obtainn finite and very accurate results out of the calculations that uses regularization (the S-matrix elements you mention), this operation is mathematically very bad behaved, and Lorentz invariance is not rigorously recovered. So I'm not sure what other case you are referring to.

 

[vanhees71]

Now I don't understand what you are talking about. The renormalized n-point functions and thus the S-matrix elements, cross sections, etc. calculated from them are manifestly covariant.

 

[vanhees71]

Well, the minimal interpretation is consistent, and has predictions that are Lorentz invariant. But it has collapse, which may or may not be real. What I don't think is right about what you write is that relativistic quantum field theory preserves Einstein causality. Einstein causality, as considered by EPR, is formalized by Bell as separability. Einstein causality is either empty if the wave function is not real, or it is violated if the wave function is real.

I think, we won't ever come to a conclusion about this, because we seem not to talk the same language. Again, I don't know, what you mean by "collapse" here. For me it's the update of the description of the state due to new information gained by a measurement, it's not a physical process, and I don't need to call it collapse, because nothing collapses here.

I also don't see, where Einstein causality is violated at any place here. The only thing what can obey or violate causality are observable facts, and nowhere in the above description is Einstein causality in this sense violated, and it's not empty. If you can realize a faster-than light signal propagation, then it's violated. So far no such thing has been observed (not even any clear case of falsification of the Standard Model has been found, although it's vigorously searched for one).

 

[atyy]

I think, we won't ever come to a conclusion about this, because we seem not to talk the same language. Again, I don't know, what you mean by "collapse" here. For me it's the update of the description of the state due to new information gained by a measurement, it's not a physical process, and I don't need to call it collapse, because nothing collapses here.

I also don't see, where Einstein causality is violated at any place here. The only thing what can obey or violate causality are observable facts, and nowhere in the above description is Einstein causality in this sense violated, and it's not empty. If you can realize a faster-than light signal propagation, then it's violated. So far no such thing has been observed (not even any clear case of falsification of the Standard Model has been found, although it's vigorously searched for one).

That's fine - it is very unusual to call "on faster than light propagation of classical information" "Einstein causality". I still think you are wrong, because you keep referring to "local interactions" - as in the interaction between the measurement apparatus and the photon. This is the usual meaning of Einstein causality - that interactions are local. It is not the same as "no faster than light transmission of classical information". So I don't think you are being consistent: what do you mean by "Einstein causality"?

(1) if Einstein causality is local interactions - no, quantum theory does have non-local interactions. If you have the apparatus interact with the photon physically, then there is physical collapse, and Einstein causality is violated.

(2) if Einstein causality is no faster than light transmission of classical information, then yes, quantum field theory preserves this.

But notions (1) and (2) are distinct.

 

[vanhees71]

Quantum field theory as we know it is by assumption only using local interactions and thus cannot have non-local ones as you claimed. A QFT is called local when it's action is defined by a Lagrange density expressed as polynomials of fields and its derivatives that transform in a local way under Poincare transformations, i.e., like the classical fields. Further it's assumed to be microcausal, i.e., local observables commute at space-like distances. This guarantees the linked-cluster principle and thus there's no faster-than light propagation of observable signals and thus also no faster-than-light transmission of information (see Weinberg, QT of Fields, Vol. I).

I don't know, what you want to emphasize by the term "classical information". How do you distinguish classical information from just information?

 

[atyy]

Quantum field theory as we know it is by assumption only using local interactions and thus cannot have non-local ones as you claimed. A QFT is called local when it's action is defined by a Lagrange density expressed as polynomials of fields and its derivatives that transform in a local way under Poincare transformations, i.e., like the classical fields. Further it's assumed to be microcausal, i.e., local observables commute at space-like distances. This guarantees the linked-cluster principle and thus there's no faster-than light propagation of observable signals and thus also no faster-than-light transmission of information (see Weinberg, QT of Fields, Vol. I).

Yes, but the interactions you talk about are not real, just like wave function collapse. The interactions are essentially the Hamiltonian, which is not real. However, you talk about the photon interacting with the measurement apparatus as if it is real. That is misleading. Only the events and the probabilities of the events are real.

I don't know, what you want to emphasize by the term "classical information". How do you distinguish classical information from just information?

That is something quantum theory assumes we know.

 

[vanhees71]

Well, the interactions between photons and the charged particles in the measurement apparatus are described by QED, by what else?

 

[atyy]

Well, the interactions between photons and the charged particles in the measurement apparatus are described by QED, by what else?

At some stage in the minimal interpretation, we cannot include the whole universe in the wave function, which means there is also no Hamiltonian for the universe. Let's acll whatever is outside the wave function the measurement apparatus. The wave function is not real, and since the Hamiltonian describes the wave function evolution, the Hamiltonian is also not real. So all the local interactions you describe are not real. In contrast, the measurement apparatus is real. So the Hamiltonian does not describe the measurement apparatus.

 

[vanhees71]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

 

[atyy]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

The measurement apparatus cannot be a "coarse grained" version of something unreal like the Hamiltonian or the wave function, unless coarse graining can produce reality from non-reality.

 

[atyy]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

I'm not sure this is right, but let me see if this argument can persuade you that what you say is at odds with the Bell theorem.

The locality of the interactions in the Lagrangian ultimately becomes the locality in the Hamiltonian. The Hamiltonian in the Heisenberg picture evolves according to classical local equations of motion and deterministically. The Bell theorem forbids local deterministic explanations of the nonlocal correlations. So the local deterministic Hamiltonian cannot explain the nonlocal quantum correlations - it needs other things like the quantum state, the nonlocal observables, and the Born rule - but by the time we calculate that, it is hardly clear that the calculation is local.

 

[TrickyDicky]

Now I don't understand what you are talking about. The renormalized n-point functions and thus the S-matrix elements, cross sections, etc. calculated from them are manifestly covariant.

But what I'm saying is that you are using the free fields in the interaction picture for that, and that is precisely what Haag's theorem proves not to exist as a mathematical construction. So it doesn't really matter what regularization procedure you use to perform the calculations , the symmetries have already been lost.

 

[Haelfix]

I'm not sure this is right, but let me see if this argument can persuade you that what you say is at odds with the Bell theorem.

The locality of the interactions in the Lagrangian ultimately becomes the locality in the Hamiltonian. The Hamiltonian in the Heisenberg picture evolves according to classical local equations of motion and deterministically. The Bell theorem forbids local deterministic explanations of the nonlocal correlations. So the local deterministic Hamiltonian cannot explain the nonlocal quantum correlations - it needs other things like the quantum state, the nonlocal observables, and the Born rule - but by the time we calculate that, it is hardly clear that the calculation is local.

I'm not sure what you think the problem is. Suppose I took a pair of socks, one red the other blue. Put one in a box, and send it to Alpha Centauri. If I waited some time and opened the box, I would instantly know the state of the other pair. This is completely classical, and shows that there is nothing wrong with nonlocal correlations, indeed almost every correlation is nonlocal. There would only be a problem if I could wiggle a sock on Alpha Centauri, such that it would instantenously cause a wiggle on the earth. Then I would have acausal observable physics.

The difference between quantum mechanics and this classical example lies encoded in the fact that when we do GHZ or EPR experiments, we can infer that the original state was NOT either red or blue, but rather something new eg a state that is in a weird mixture, like half red and half blue. That further the dynamics must allow interference of states and the noncommutativity of operators. However all these things stay manifestly local, in the sense that the lagrangian stays written in a certain form and that operators at spacelike separation don't commute.

 

[Ilja]

Now I don't understand what you are talking about. The renormalized n-point functions and thus the S-matrix elements, cross sections, etc. calculated from them are manifestly covariant.

The point is that, if we take Haag's theorem seriously, the only really consistent theories are similar to lattice theories (at least in their main property, being not relativistically covariant). If we have a lattice theory as a fundamental theory, the large distance limit is a continuous theory, which can have relativistic covariance. In a simple case, the lattice defined by atoms of a crystal gives a wave equation with constant speed of sound, and the symmetry of this simplest wave equation is, of course, the Poincare group. But this can be, by construction (once obtained from a lattice theory) only an approximate symmetry, the whole continuous theory is only an approximation.

 

[TrickyDicky]

The point is that, if we take Haag's theorem seriously,

We should. For some reason people often dismiss it using as argument the accuracy of the results obtained in QED when all the theorem says is those results are actually obtained in a way that is not exactly the way the idealized theory says. It should be looked as a real good hint instead of trying to ignore it.

the only really consistent theories are similar to lattice theories (at least in their main property, being not relativistically covariant).

This I disagree that is the only really consistent modification. It is too strong an assumption as departure point.

 

[TrickyDicky]

Nevertheless the S-matrix elements as calculated in renormalized perturbation theory and which are compared to observations are Lorentz invariant (and gauge invariant).

Say you take a regularization method like DR, you can say that since it's physical limit is d→4, it is in that sense trivially Lorentz invariant as the limit is 4-spacetime. But then you still have that as ε→0 the chiral(axial) and conformal symmetries are broken in general.

 

[stevendaryl]

I'm not sure what you think the problem is. Suppose I took a pair of socks, one red the other blue. Put one in a box, and send it to Alpha Centauri. If I waited some time and opened the box, I would instantly know the state of the other pair. This is completely classical, and shows that there is nothing wrong with nonlocal correlations, indeed almost every correlation is nonlocal.

But classically, a nonlocal correlation can always be explained in terms of local correlations plus ignorance of the physical state. So in this case, the two possible physical states for the pair socks are:

1. A = red sock to Alpha Centauri, blue sock to Earth
2. B = blue sock to Alpha Centauri, red sock to Earth

The "mixed state" of state A with probability 1/2, state B with probability 1/2 reflects, not anything nonlocal about the world, but merely reflects human ignorance.

Classically--and I should say, quantum-mechanically as well, in any situation not involving entanglement--a nonlocal correlation is always evidence of local correlations involving as yet unknown parameters.

 

[Haelfix]

I completely agree, again the only difference with quantum mechanics is that in so far as the analogy goes (say in the Bell thought experiment experiment), the classical mixed state is promoted to a pure entangled state. You still require preparation of the initial state within a local setting.

So what we have is a fundamentally new object (an entangled state) where we can have maximal knowledge about both of the subsystems by themselves but know nothing about the composite system, or viceversa. That's the difference with classical physics!

All that Bell says is that you can either make a big issue about locality, or you can accept the very pedestrian notion that there is nothing wrong with locality but rather that quantum mechanics requires a new object that is not describable in terms of classical physics.. I think that's all the minimal intepretation actually says, and there is really nothing wrong or mysterious with that point of view.