I Clarification of the postulates of QM

[stevendaryl]

The final sentence doesn't make sense to me. What do you mean by "local" here. Of course, in relativistic QFT by construction all interactions are local in space and time (you write down a Lagrangian with field operators multiplied at the same space-time point only).

And that has nothing to do with the reason that people suspect that QM is nonlocal. So it's a distraction to bring it up.

What's in some sense "non-local" in QT is related to our debate and entanglement, but that I'd not call "non-local" but long-range correlations between parts of a quantum system. One must not misunderstand long-range correlations with non-local interactions at a distance!

I would say that because QT has nonlocal correlations that do not reduce to local interactions on local variables, QT is inherently a nonlocal theory.

 

[vanhees71]

But the interesting point is that there is no problem with collapse in relativistic QFT.

If you think there is a problem, then you are using the wrong definition of causality.

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

 

[stevendaryl]

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

Well, I don't see any difference between a literal and nonliteral collapse unless you have a clear idea of the separation between what is real and what is subjective. To say that there are no nonlocal influences is to say that no change of a physical quantity here can affect a physical quantity at a spacelike separation. But what properties are physical, in quantum theory? Copenhagen says that only macroscopic properties are real, or only observed properties are real. But that requires a distinction between macroscopic/microscopic or between observer/observed which isn't made clear in the theory.

 

[atyy]

No, there's no problem in relativistic QFT, as I stress all the time. It's only a problem if you assume a litteral collapse, where instaneously Bob's particle is affected by Alice's measurement. In the "minimal interpretation" there's no such assumption and thus no such problem.

1) But if there is no difference in predictions between a literal and a non-literal collapse, why would one who holds to a minimal interpretation object to a literal collapse?

2) Why do you object to the collapse in Cohen-Tannoudji, Diu and Laloe's book? As far as I can tell, they are agnostic as to whether collapse is literal. (In fact, I have never heard of a collapse as literal, except from people who object to it.)

 

[vanhees71]

1) Well, if you take the state as a physical entity and claim that when A measures $\sigma_z$ instantaneously the state collapses, this is a real instaneous effect in the entire universe. This contradicts the very construction of local microcausal QFT. I don't see, why I should buy a self-contradictory postulate, which in fact I never need to describe observations using Q(F)T.

2) I'm a bit surprised, how inaccurate these authors (Nobel laureat included) state the fundamental postulates. I guess, they are pretty uninterested in "interpretation" and rather present the applications of the theory to observable phenomena, and this they do very well. So I don't say that it's a bad book, but, e.g., the formulation that if the system is prepared in state $|\psi \rangle$ the probability to find the system in state $|\phi \rangle$ is $|\langle \phi|\psi \rangle|^2$ is misleading. It made me crazy when I learned QT from another book (I don't remember which one it was), because I couldn't get how in this formulation anything can be independent of the picture of time evolution chosen, and that should be true, because how you choose the picture is quite arbitrary. The resolution is, of course, easy if you put it in the right way: If the system is prepared in the state $|\psi \rangle$ (more precisely the state is reprsented by $|\psi \rangle \langle \psi|$ or equivalently by the corresponding ray, which is another glitch in the chapter on the postulates), then the probability to measure the value $a$ of an observable $A$ is given by $P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2$, where $|a,\beta \rangle$ is the orthonormal basis of the subspace $\mathrm{Eig}(\hat{A},a)$ (modulo the possibility of continuous spectral values, where you have an integral instead of the sum).

 

[atyy]

1) Well, if you take the state as a physical entity and claim that when A measures $\sigma_z$ instantaneously the state collapses, this is a real instaneous effect in the entire universe. This contradicts the very construction of local microcausal QFT. I don't see, why I should buy a self-contradictory postulate, which in fact I never need to describe observations using Q(F)T.

The local microcausal construction is used for the Hamiltonian, which determines the unitary evolution between measurements. The collapse occurs at a measurement. There is no contradiction.

2) I'm a bit surprised, how inaccurate these authors (Nobel laureat included) state the fundamental postulates. I guess, they are pretty uninterested in "interpretation" and rather present the applications of the theory to observable phenomena, and this they do very well. So I don't say that it's a bad book, but, e.g., the formulation that if the system is prepared in state $|\psi \rangle$ the probability to find the system in state $|\phi \rangle$ is $|\langle \phi|\psi \rangle|^2$ is misleading. It made me crazy when I learned QT from another book (I don't remember which one it was), because I couldn't get how in this formulation anything can be independent of the picture of time evolution chosen, and that should be true, because how you choose the picture is quite arbitrary. The resolution is, of course, easy if you put it in the right way: If the system is prepared in the state $|\psi \rangle$ (more precisely the state is reprsented by $|\psi \rangle \langle \psi|$ or equivalently by the corresponding ray, which is another glitch in the chapter on the postulates), then the probability to measure the value $a$ of an observable $A$ is given by $P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2$, where $|a,\beta \rangle$ is the orthonormal basis of the subspace $\mathrm{Eig}(\hat{A},a)$ (modulo the possibility of continuous spectral values, where you have an integral instead of the sum).

But the Cohen-Tannoudji formulation on their p220 of volume 1 looks exactly the same as what you wrote (ie. one has to specify the measurement observable), except that they add that after the measurement, the state of the system is different from before the measurement.

And yes, if they only care about applications of the theory to observable phenomena, then they are agnostic as to whether the wave function and collapse are real or not. Isn't that the minimal interpretation?

 

[vanhees71]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either! On the one hand we have to use quantum theory, driven by observations that tell us that the classical theory is only an approximation. So to claim a measurement doesn't follow the laws of QT is not very satisfactory, and I don't see, why one should use this assumption nowadays, where we have understood much better the emergence of classical behavior of macroscopic systems from quantum theory than the "founding fathers" of QT could have known in the beginning. The interaction of a particle with a detector follows the rules of quantum theory and thus is described as a local interaction between the measured system.

I don't want to bash the textbook by Cohen-Tanoudji et al, but I think you should get the postulates as precise as possible, because this helps tremendously to study the subject.

 

[ShayanJ]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either! On the one hand we have to use quantum theory, driven by observations that tell us that the classical theory is only an approximation. So to claim a measurement doesn't follow the laws of QT is not very satisfactory, and I don't see, why one should use this assumption nowadays, where we have understood much better the emergence of classical behavior of macroscopic systems from quantum theory than the "founding fathers" of QT could have known in the beginning. The interaction of a particle with a detector follows the rules of quantum theory and thus is described as a local interaction between the measured system.

If you're talking about quantum decoherence, then what you've described in this thread till now, is in contradiction with it.
You say that a measurement doesn't change the wave-function at all, i.e. there is no collapse.
But when collapse is assumed, its assumed as a blackbox. No one says it has to come from somewhere else than the Schrodinger equation, its just assumed and the possibility of explaining it is left open. And decoherence has been able to explain it partially. So decoherence has been able to explain something(partially), that you've always denied. How can you consider it as a support for your arguments? Because it actually rules out what you suggest!

 

[vanhees71]

No, I don't say that measurements change the wave function (or better the state, because we discuss relativistic QT here, and there is no consistent descriptions of it by wave functions a la Schrödinger). I only say that the change of the state is due to quantum dynamics and not an instantaneous action at a distance leading to some collapse thing that is somehow outside of the general dynamical laws of QT. The emergence of classical behavior of macroscopic systems, among them measurement apparati, is quite well understood nowadys within quantum many-body theory, but this is just relativistic local microcausal QFT. So there are only local interactions, no actions at a distance by construction. So this cannot rule out what I suggest!

 

[AlexCaledin]

In the relativistic QFT, instead of "instantaneous action at a distance",

"it is impossible to keep a particle from traveling faster than light"

https://en.wikipedia.org/wiki/Path_integral_formulation#Quantum_field_theory

 

[bhobba]

"it is impossible to keep a particle from traveling faster than light"

That one must use such in the path integral does not imply any particles are actually traveling FTL.

Thanks
Bill

 

[AlexCaledin]

That ... does not imply any particles are actually traveling F[aster ]T[han ]L[ight]...

Quite right - because actual traveling always consists of acts of observing the particle (observable drops or bubbles or something else)...

 

[vanhees71]

How do you come to this conclusion? Massive particles are traveling with speeds less than the speed of light.

 

[stevendaryl]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either!

I agree, that's why the standard Copenhagen interpretation of QM is unsatisfying to me. It seems to rely on either a classical/quantum split, or a macroscopic/microscopic split, or an observer/observed split. In any of those types of split, you're distinguishing systems that should (in my opinion) be treated the same--they're just quantum systems.

But to me, the Born rule is the source of this, not just collapse. You can't formulate the Born rule without mentioning one of these splits.

 

[vanhees71]

Why do you need one of the splits to formulate Born's rule? It just gives the physical meaning to quantum states. You just prepare your system and measure the observable you like to observe, and then the quantum state you've prepared tells you the probabilities to find values of this observable. With a large enough ensemble you can check whether the prediction is correct. That's it. I don't need more to make sense out of quantum theory. Where do you need the split?

 

[stevendaryl]

Why do you need one of the splits to formulate Born's rule?

Well, the formulation used in standard QM is: When you measure an observable, you get an eigenvalue of the corresponding Hermitian operator, with probabilities given by the square of the projection of the wave function onto the subspace with that value of the operator. So that formulation uses the concept of "when you measure..." What makes an interaction into a measurement of an observable? I don't really think that the density matrix formulation is any different, conceptually: "The expectation value of an observable corresponding to an operator \hat{O} is the trace of \rho \hat{O}". How can you make sense of "expectation value of an observable" without making "measuring an observable" into something separate from other interactions?

Either "measurement" and "observation" are primitive concepts, which bakes the distinction into the formalism, or else they are derived concepts. As a derived concept, you might say something like "An interaction counts as a measurement of an observable if afterward there is a persistent macroscopic record of the value". But that involves the macroscopic/microscopic split.

 

[stevendaryl]

You just prepare your system and measure the observable you like to observe, and then the quantum state you've prepared tells you the probabilities to find values of this observable.

I consider that sentence to already have the split. "The probabilities to find values of this observable" already makes the distinction between observations and other interactions.

 

[vanhees71]

Well, the formulation used in standard QM is: When you measure an observable, you get an eigenvalue of the corresponding Hermitian operator, with probabilities given by the square of the projection of the wave function onto the subspace with that value of the operator. So that formulation uses the concept of "when you measure..." What makes an interaction into a measurement of an observable? I don't really think that the density matrix formulation is any different, conceptually: "The expectation value of an observable corresponding to an operator \hat{O} is the trace of \rho \hat{O}". How can you make sense of "expectation value of an observable" without making "measuring an observable" into something separate from other interactions?

Either "measurement" and "observation" are primitive concepts, which bakes the distinction into the formalism, or else they are derived concepts. As a derived concept, you might say something like "An interaction counts as a measurement of an observable if afterward there is a persistent macroscopic record of the value". But that involves the macroscopic/microscopic split.

You measure an observable by using the adequate measurement apparatus. How else? What you quote are just the postulates of the formalism, and a state of a system is not a self-adjoint trace-class operator in some Hilbert space or a measurement some projection operator to an eigenspace of a self-adjoint operator representing an observable but real-world preparation procedures and real-world measurement apparati, defining the quantities operationally. Of course at a certain point you must assume that the measurement apparatus measures what you want to measure, but that's so with observables within classical physics either.

 

[stevendaryl]

You measure an observable by using the adequate measurement apparatus. How else?

And what makes a measurement apparatus adequate? I think there is no way to formulate that in a non-circular way without making the kind of split I'm talking about.

 

[stevendaryl]

Of course at a certain point you must assume that the measurement apparatus measures what you want to measure, but that's so with observables within classical physics either.

No, that's not true. In classical physics, it's derivable. And also, in classical physics, the laws don't refer to observables at all. The distinction isn't baked into classical physics, so there is no need for the split.

 

[atyy]

The measurement is due to interaction of the measured system with a measurement apparatus. I also never bought the argument that this is something different than the interactions described by quantum theory. This doesn't make sense either! On the one hand we have to use quantum theory, driven by observations that tell us that the classical theory is only an approximation. So to claim a measurement doesn't follow the laws of QT is not very satisfactory, and I don't see, why one should use this assumption nowadays, where we have understood much better the emergence of classical behavior of macroscopic systems from quantum theory than the "founding fathers" of QT could have known in the beginning. The interaction of a particle with a detector follows the rules of quantum theory and thus is described as a local interaction between the measured system.

I don't want to bash the textbook by Cohen-Tanoudji et al, but I think you should get the postulates as precise as possible, because this helps tremendously to study the subject.

But if we follow your reasoning, and quantum theory applies to everything, then there should be a wave function of the universe, and it should make physical sense without many-worlds or Bohmian mechanics.

 

[vanhees71]

Here, I have a problem of course, because a probabilistic interpretation makes it necessary to be able to prepare an ensemble of systems. For the universe as a whole that's impossible. On the other hand, it's anyway fictitious since we can never observe the universe as a whole ;-).

 

[A. Neumaier]

we can never observe the universe as a whole ;-).

?

We can evaluate lots of observables of the universe as a whole by measuring its local fields and currents at particular positions of interest.

 

[vanhees71]

Even with an infinite lifetime we can only observe a part of the universe:

https://en.wikipedia.org/wiki/List_of_cosmological_horizons

 

[A. Neumaier]

But if [...] quantum theory applies to everything, then there should be a wave function of the universe, and it should make physical sense without many-worlds or Bohmian mechanics.

There must be a state of the universe, but not necessarily a wave function -- it could be a mixed state. It does indeed make physical sense:

According to my http://arnold-neumaier.at/physfaq/cei/ , we need a Hilbert space carrying a representation of the standard model plus some (not yet decided) form of gravity, unitary dynamics for operators, density operators for Heisenberg states, the definition of $\langle A\rangle:=\mbox{tr}~\rho A$ as mathematical framework, and for its interpretation a single rule:

Upon measuring a Hermitian operator $A$, the measured result will be approximately $\bar A=\langle A\rangle$, with an uncertainty
at least of the order of $\sigma_A=\sqrt{\langle (A-\bar A)^2\rangle}$. If the measurement can be sufficiently often repeated (on an object with the same or sufficiently similar state) then $\sigma_A$ will be a lower bound on the standard deviation of the measurement results.

Everyone doing quantum mechanics uses these rules (even those adhering to the shut-up-and-calculate mode of working), and they apply universally. No probabilistic interpretation beyond that is needed, so it applies also to the single universe we live in. Everything deduced in quantum field theory about macroscopic properties follows, and one has a completely self-consistent setting. The transition to classicality is automatic and needs no deep investigations - the classical situation is simply the limit of a huge number of particles. Whereas on the microscopic level, uncertainties of single events are large, so that state determination must be based by the statistics of multiple events with a similar preparation.

We cannot expect to measure all the observables of the whole universe, and perhaps never determine its precise state. But measuring all observables or finding its exact state is already out of the question for a small quantum system such as a shaken bottle of water. What matters for a successful physics of the universe is only that we can model (and then predict) the observables that are accessible to measurement.

 

[A. Neumaier]

Even with an infinite lifetime we can only observe a part of the universe:

sure, but this only means that we cannot expect to measure all the observables of the whole universe, and perhaps never determine its precise state.

But measuring all observables or finding its exact state is already out of the question for a small quantum system such as a shaken bottle of water, though nobody deduces from this impossibility that its state is fictitious.

 

[atyy]

Here, I have a problem of course, because a probabilistic interpretation makes it necessary to be able to prepare an ensemble of systems. For the universe as a whole that's impossible. On the other hand, it's anyway fictitious since we can never observe the universe as a whole ;-).

But don't you think Dirac, Landau & Lifshitz, Cohen-Tanoudji etc may have taken this into account when they state collapse as a postulate?

ie. you are not able to really defend quantum mechanics as applying to the whole universe.

 

[vanhees71]

sure, but this only means that we cannot expect to measure all the observables of the whole universe, and perhaps never determine its precise state.

Yes, and that's why "the state of the entire universe" is an empty phrase since, if I claim to know the "state of the entire universe" and give it to you in terms of some Stat. Op. you can never empirically check my claim. So I can claim whatever I like. Some theorists love such ideas since they cannot not disproven by observation, but than that's (perhaps interesting) math but no physics!

 

[vanhees71]

But don't you think Dirac, Landau & Lifshitz, Cohen-Tanoudji etc may have taken this into account when they state collapse as a postulate?

ie. you are not able to really defend quantum mechanics as applying to the whole universe.

Yes, but I don't want to defend any theory as applying to the whole universe since it's not observable.

Of course, it's an assumption made in cosmology that there's no preferred place or direction in the universe (at least in the large-scale coarse grained picture) and that all physical laws are thus the same at any point and time in this universe (cosmological principle), and so far we have not seen any contradiction to this assumption, which mathematically boils down to the statement that the large-scale coarse grained spacetime is described by a Friedmann-Lemaitre-Robertson-Walker metric, by the "local" observations we are able to make today, and you can thus keep the cosmological principle as hypothesis about the "state" of the entire universe, but you'll never be able to finally check it completely, because there are regions in spacetime we can never observe (given the observational fact that we live in an "accelerating" universe there's a "future horizon").

 

[A. Neumaier]

Yes, and that's why "the state of the entire universe" is an empty phrase

By the same reasoning, the state of a piece of metal would also be an empty phrase, since we never know it exactly, and we can probe only a few of its observables. But probing these variables is sufficient; it is the conventional test for adequacy of a proposed state of the metal. Otherwise no solid state physics would be possible.

Exactly the same holds for the universe as a whole.

if I claim to know the "state of the entire universe" and give it to you in terms of some Stat. Op. you can never empirically check my claim. So I can claim whatever I like.

One can claim whatever one likes about any system, but the claim is no physics if it is easily falsified. In case the system is the whole universe, you can claim whatever you like but unless your claim is very informed it can be easily falsified by computing from the state the expectation values of the electromagnetic field at points where we can measure it. It is very difficult to come up with a state that cannot be falsified in this or similar ways. For this would be a state that is compatible with everything we have ever empirically observed in the universe! Thus knowing this state amounts to knowing all physics accessible to us.

Thus, as for a metal, one must be content with describing this state approximately, but this is not impossible.