Wave function collapse in QED?

[QuasiParticle]

Does the measurement problem ("wave function collapse") or something similar somehow manifest itself in QED and other quantum field theories? Is it somehow built-in into the propagators etc. "away from sight"? If so, how does it affect the theories and is this a problem, which needs to be solved?

A related question: Is there a "barrier" between the classical world and the quantum world in the standard model, as there is in the old (Copenhagen) quantum mechanics?

 

[DrDu]

I don't think there is a principal difference between QFT and ordinary quantum mechanics wrt the measurement theory.
The separation between the quantum and classical world is already built in the definition of the vacuum state.
There is some nice discussion in the book "Local quantum physics" by R. Haag.

 

[QuasiParticle]

Thanks, I will take a look at the book. Yes, my understanding of the situation is also that QFT does not "solve the problem". But it is almost never discussed in the context of QFT.

One reason I am asking the question is that Lee Smolin in his book "The Trouble with Physics" stated that solving the measurement problem is likely an important piece in constructing the quantum theory of gravitation.

 

[atyy]

Yes, wave function collapse is just the same in QFT and QM (if one uses an interpretation with collapse). For example:
http://arxiv.org/abs/quant-ph/0102043
http://arxiv.org/abs/hep-th/0110205

Smolin's remarks about quantum gravity are speculative. His colleague Rovelli speculated differently, eg. " Pick your favorite interpretation of quantum mechanics, and use it for interpreting the quantum aspects of the theory." http://relativity.livingreviews.org/Articles/lrr-2008-5/fulltext.html (section 5.4).

 

[bhobba]

Thanks, I will take a look at the book. Yes, my understanding of the situation is also that QFT does not "solve the problem". But it is almost never discussed in the context of QFT.

QFT does not solve the measuement problem, its still there but in a slightly different form, but in many ways makes a lot of the mystery of QM easier to understand.

Check out the following:
https://www.amazon.com/dp/0473179768/?tag=pfamazon01-20

If you are just starting out in QM its a very good book, and the Kindle version is dirt cheap.

BTW don't get too caught up in the quantum gravity thing. Its not quite the issue some poularisations make it out to be:
http://arxiv.org/pdf/1209.3511v1.pdf
'Effective field theory shows that general relativity and quantum mechanics work together perfectly normally over a range of scales and curvatures, including those relevant for the world that we see around us. However, effective field theories are only valid over some range of scales. General relativity certainly does have problematic issues at extreme scales. There are important problems which the effective field theory does not solve because they are beyond its range of validity. However, this means that the issue of quantum gravity is not what we thought it to be. Rather than a fundamental incompatibility of quantum mechanics and gravity, we are in the more familiar situation of needing a more complete theory beyond the range of their combined applicability. The usual marriage of general relativity and quantum mechanics is fine at ordinary energies, but we now seek to uncover the modifications that must be present in more extreme conditions. This is the modern view of the problem of quantum gravity, and it represents progress over the outdated view of the past.'

Thanks
Bill

 

[bhobba]

A related question: Is there a "barrier" between the classical world and the quantum world in the standard model, as there is in the old (Copenhagen) quantum mechanics?

Copenhagen is bit outdated these days - but its a bit subtle:
http://motls.blogspot.com.au/2011/05/copenhagen-interpretation-of-quantum.html

Its not wrong - but modern interpretations fix up an issue it has. Copenhagen assumes a classical common-sense world observations appear in. But how can a theory that assumes such a world explain that world? That's the conundrum.

Nowadays everything is assumed to be quantum and how the classical world emerges is explained by decoherence:
http://www.ipod.org.uk/reality/reality_decoherence.asp

However some issues do remain, although research is of course ongoing, and progress is being made.

Thanks
Bill

 

[DrDu]

Personally, I can't see any difference between "decoherence" and the Copenhagen interpretation.
This explanation seems to have been detailed already by von Neumann long ago.
The point is that for decoherence really to explain the measurement, you have to exclude the possibility of recurrences, i.e. restoration of the original wavefunction after a long time. Mathematically, this requires coupling to an infinite system (although already for systems consisting of a dozen or so particles recurrence times can be practially infinite) which allows for classical operators.

 

[Demystifier]

Personally, I can't see any difference between "decoherence" and the Copenhagen interpretation. This explanation seems to have been detailed already by von Neumann long ago.

It is true that some essential ideas of decoherence have already been implicit in the von Neumann (1932) theory of quantum measurement. However, it does not mean that there is no difference between Copenhagen interpretation and decoherence. The main difference is that decoherence is not an interpretation. Decoherence more-or-less explains how the total wave function of the whole system (including the environment and even the observer) splits into separate branches. But this does not explain why only one of the branches becomes effective. To resolve that problem, von Neumann's interpretation was that a wave function collapse happens due to the observer's cosciousness. Not all interpretations agree on the relevance of consciousness, even when they do agree that decoherence plays an important role.

 

[DrDu]

Decoherence more-or-less explains how the total wave function of the whole system (including the environment and even the observer) splits into separate branches. But this does not explain why only one of the branches becomes effective.

I think that axiomatic QFT made really a contribution here.
Namely, a QFT contains infinite degrees of freedom and the theorem of von Neumann on the equivalence of all representations of the Heisenberg algebra breaks down.
I was struggling with this in the theory of superconductivity. There, you usually consider representations with unsharp particle number, which is not meaningfull for finite systems. For infinite systems, it can be shown that this description is equivalent to a description with fixed particle number, the former being irreducible, the latter being irreducible.
For infinite systems, there is no way to distinguish the two.
See the following article by R. Haag:
http://link.springer.com/article/10.1007/BF02731446
In the same spirit, we have no means to tell apart whether the statistical mixture obtained after a measurement process is in reality a coherent superposition of observers obtaining different outcomes or a projection to one definite outcome.

 

[bhobba]

Personally, I can't see any difference between "decoherence" and the Copenhagen interpretation.

Decoherence is not an interpretation, its implicit in the formulation itself.

It simply clarified a slight blemish Copenhagen had in exactly what counts as an observation without assuming a classical world. You simply assume its when decoherence occurs which is entirely quantum. It also explains apparent collapse - not actual collapse - but still it was a step forward.

It also led to interpretations where it takes central stage such as the ignorance ensemble interpretation and consistent histories.

Thanks
Bill

 

[bhobba]

To resolve that problem, von Neumann's interpretation was that a wave function collapse happens due to the observer's cosciousness. Not all interpretations agree on the relevance of consciousness, even when they do agree that decoherence plays an important role.

The models of how decoherence occurred in practice were not fully worked out eg tracing over the environment - at least as far as I know anyway - there is nothing along those lines in his mathematical foundations tome that I can recall. Even now they have not been fully worked out in all generality eg we don't even know for sure if its really an artefact of dividing the system into environment and system being observed.

Personally I am with Wigner who was in the Von Neumann camp. He realized this conciousness causes collapse wasn't required from some early work by Zurek. Just after decoherence is the obvious place to put the Von Neumann cut, for me anyway, and is exactly where I put it. When that's done tons of issues go out the door.

Thanks
Bill

 

[Demystifier]

In the same spirit, we have no means to tell apart whether the statistical mixture obtained after a measurement process is in reality a coherent superposition of observers obtaining different outcomes or a projection to one definite outcome.

When you say that "we have no means to tell apart", do you mean "tell by EXPERIMENT", or "tell by THEORY"? If you mean "tell by experiment", then I agree. But if you mean "tell by theory", then I disagree. By using theory I can definitely tell whether it is a coherent superposition or a projection. For example, by using Schrodinger equation (and nothing else) I can tell that it is a coherent superposition and not a projection.

Of course, you may say that a part of theory which does not have experimental consequences is irrelevant. My objection is that relevance or irrelevance of something is a subjective category. For example, it may be irrelevant to a practical engineer, but relevant to a philosopher. Physicists are somewhere in between, so it may be irrelevant to some physicists but relevant to other physicists.

So when you say that you "can't see any difference between decoherence and the Copenhagen interpretation", it probably means that you find the theoretical difference (which cannot be seen experimentally) irrelevant. And if I am right, then it probably means that by "Copenhagen" interpretation you mean the interpretation number 2:
https://www.physicsforums.com/showthread.php?t=332269

 

[DrDu]

When you say that "we have no means to tell apart", do you mean "tell by EXPERIMENT", or "tell by THEORY"? If you mean "tell by experiment", then I agree. But if you mean "tell by theory", then I disagree. By using theory I can definitely tell whether it is a coherent superposition or a projection. For example, by using Schrodinger equation (and nothing else) I can tell that it is a coherent superposition and not a projection.

So, so you think you can tell Heaven from Hell, blue skies from pain.

This problem is even more urgent in QFT than in QM. In QFT we use much more abstract constructs which are far from being - even theoretically - observable. For example creation and anihilation operators, vector potentials and not even to mention the unrenormalized quantities.

 

[bhobba]

In the same spirit, we have no means to tell apart whether the statistical mixture obtained after a measurement process is in reality a coherent superposition of observers obtaining different outcomes or a projection to one definite outcome.

So, so you think you can tell Heaven from Hell, blue skies from pain.

Mate I am with you on that one.

Observationally there is no difference between an improper mixture and a proper one - easiest solution - assume its a proper one - measurement problem a non issue. Seems simple to me.

You will find however convincing others of it is not quite that simple.

Thanks
Bill

 

[George Jones]

Just after decoherence is the obvious place to put the Von Neumann cut, for me anyway, and is exactly where I put it. When that's done tons of issues go out the door.

Not according to Steven Weinberg.

 

[bhobba]

Not according to Steven Weinberg.

Cant quite follow that one.

Von Neumann showed the cut can be placed anywhere - so why not just after decoherence? That is you consider the improper mixture as a proper one which amounts to the same thing.

Or did you have something else in mind?

Thanks
Bill

 

[Ilja]

The main difference is that decoherence is not an interpretation. Decoherence more-or-less explains how the total wave function of the whole system (including the environment and even the observer) splits into separate branches.

No, it doesn't explain anything. Because it presupposes some other fundamental structure, a subdivision of the world into systems.

Of course, if one names one thing "system", and the other one "observer" or "environment", this subdivision is hidden from the reader in a slightly better way that if it is named "classical part" and "quantum part". "System" and "observer" sounds quite objective, while "classical part" and "quantum part" is clearly artificial, a subdivision which can be made only by a scientist who decides that for one part the classical approximation is sufficient, clearly not an objective feature of this device. But the subdivision into "system" and "environment" is arbitrary too. So there remains an arbitrary subdivision in decoherence-based interpretations.

The subdivision is required, without subdivision no decoherence.

See http://arxiv.org/abs/0901.3262 why some additional structure is necessary, also http://arxiv.org/abs/0903.4657.

 

[stevendaryl]

Mate I am with you on that one.

Observationally there is no difference between an improper mixture and a proper one - easiest solution - assume its a proper one - measurement problem a non issue. Seems simple to me.

You will find however convincing others of it is not quite that simple.

Thanks
Bill

Yes, for practical purposes, that's the solution. It's not very satisfactory from a philosophical perspective, because we know that's it not a proper mixture, because proper mixtures do not evolve from pure states. So we can pretend that it's a proper mixture, to get on with our lives, but in the back of our minds, we know that it's only pretending.

(Actually, I think that Hawking claims, or has claimed in the past, that near a black hole, pure states can evolve into mixed states. I'm not sure if I understand that claim, but I think it is related to the claim that Hawking radiation is incoherent and thermal.)

 

[stevendaryl]

Thanks, I will take a look at the book. Yes, my understanding of the situation is also that QFT does not "solve the problem". But it is almost never discussed in the context of QFT.

One reason I am asking the question is that Lee Smolin in his book "The Trouble with Physics" stated that solving the measurement problem is likely an important piece in constructing the quantum theory of gravitation.

I agree that QFT doesn't actually solve the problem.

However, there is something that's interesting about QFT compared with QM. In the usual treatments of QM, the wave function is where the action is, and the wave function is a weird beast. People are misled by single-particle quantum mechanics into thinking of the wave function as a kind of field propagating through space, in the same way that the electromagnetic field does. That's not correct, and it becomes obvious when you consider multiparticle wave functions. The wave function does not exist in ordinary physical 3D space, it exists in 3N dimensional configuration space, where N is the number of particles. So when people talk about the wave function being local or nonlocal, I think that that's a little jumbled, because usually people mean "local in physical 3D space", while the wave function, if it is local in any sense, is only local in configuration space. It doesn't really make sense to talk about it being local in physical space.

The contrast with QFT is this: where the action is in quantum field theory is not the wave function, it's the field operators. The field operators ARE fields that propagate in ordinary space, they're just operator-valued fields, rather than real-number-valued fields. They obey a perfectly local evolution equation.

Of course, QFT has something in addition to the field operators, which is the state. The state doesn't get much discussion in QFT, because nobody really ever does much with the full state. Instead, what people usually deal with is the vacuum state, the state with no particles, or fields or anything. The quantities of interest are almost always rewritten in terms of matrix elements of operators sandwiched between vacuum states. So the formalism of QFT tends to de-emphasize the parts that are spooky and nonlocal.

The Heisenberg formulation of ordinary quantum mechanics does the same sort of thing--it puts all the dynamics into operators, and then the state is just something that is constant.

 

[DrDu]

Yes, for practical purposes, that's the solution. It's not very satisfactory from a philosophical perspective, because we know that's it not a proper mixture, because proper mixtures do not evolve from pure states.

I am not sure if we really know this. While it holds for simple QM systems it may not hold for QFT with infinite number of dimensions.
If we leave aside gravitation, which may or may not offer another solution, but probably an even more difficult one, space is flat and infinite. For an infinite volume, no wavefunction exists, vulgo also no unitary evolution of all space.

 

[Demystifier]

The contrast with QFT is this: where the action is in quantum field theory is not the wave function, it's the field operators. The field operators ARE fields that propagate in ordinary space, they're just operator-valued fields, rather than real-number-valued fields. They obey a perfectly local evolution equation.

This is wrong, there is no really such difference between QFT and QM. The real difference is between Schrodinger and Heisenberg picture, each of which can be used in both QFT and QM.

In Schrodinger picture, the thing which evolves with time is the state in the Hilbert space, both in QFT and QM.

In Heisenberg picture, the thing which evolves with time is the field operaror (for QFT) or position operator (for QM).

The misconception comes from the fact that in practical calculations, one usually uses Schrodinger picture for QM and Heisenberg picture for (free) QFT.

 

[DrDu]

The misconception comes from the fact that in practical calculations, one usually uses Schrodinger picture for QM and Heisenberg picture for (free) QFT.

...and especially in QFT the interaction picture in which both the states and the operators evolve.

 

[stevendaryl]

This is wrong, there is no really such difference between QFT and QM. The real difference is between Schrodinger and Heisenberg picture, each of which can be used in both QFT and QM.

In Schrodinger picture, the thing which evolves with time is the state in the Hilbert space, both in QFT and QM.

In Heisenberg picture, the thing which evolves with time is the field operaror (for QFT) or position operator (for QM).

The misconception comes from the fact that in practical calculations, one usually uses Schrodinger picture for QM and Heisenberg picture for (free) QFT.

You're basically repeating what I said, after saying I was wrong. The difference is in the way that the two subjects are typically done. QM could be done in the QFT way, but it typically isn't.

 

[Demystifier]

You're basically repeating what I said, after saying I was wrong. The difference is in the way that the two subjects are typically done. QM could be done in the QFT way, but it typically isn't.

Then I misunderstood you, sorry! :thumbs:

 

[stevendaryl]

Then I misunderstood you, sorry! :thumbs:

My main point was that there are proofs that QFT is causal, which are proofs about the field operators (something along the lines of field operators at spacelike separations commute, or anti-commute), so that might lead people to think that QFT lacks the "spooky action at a distance" that seems to present in non-relativistic QM. But that's misleading--it's comparing apples and oranges. If you do NRQM in the Heisenberg way, in terms of operators, then you won't see any nonlocality either. The nonlocality is in the state, not the operators.

 

[Demystifier]

My main point was that there are proofs that QFT is causal, which are proofs about the field operators (something along the lines of field operators at spacelike separations commute, or anti-commute), so that might lead people to think that QFT lacks the "spooky action at a distance" that seems to present in non-relativistic QM. But that's misleading--it's comparing apples and oranges. If you do NRQM in the Heisenberg way, in terms of operators, then you won't see any nonlocality either. The nonlocality is in the state, not the operators.

That is a good point, but there is also a way to describe nonlocality in QFT in terms of operators. For example, if psi(x) is a local operator, then the product of operators psi(x)psi(y) (which itself is an operator) is not a local operator. Hence, not all operators in QFT are local. And more to the point, when you act with a sum of such nonlocal product operators on the vacuum, you get an entangled many-particle state responsible for the nonlocality.

 

[atyy]

@Dr Du and @bhobba, decoherence does not solve the measurement problem because it doesn't make sense for the subsystem to give a definite outcome but the universe still be in a superposition. And yes, the wave function of "the universe" does matter, since without it there cannot be decoherence. Decoherence is only indistinguishable from collapse, provided collapse is assumed.

If there is a problem to be solved, decoherence does not solve it.

One can take an attitude of practically indistinguishable. But in that case there is already no problem to be solved, without decoherence, ie. Copenhagen works fine without decoherence.

 

[bhobba]

@Dr Du and @bhobba, decoherence does not solve the measurement problem because it doesn't make sense for the subsystem to give a definite outcome but the universe still be in a superposition.

Atyy. Von Neumann showed the Von Neumann cut can be placed anywhere. Placing it just after decoherence is not in anyway problematical for QM or for the 'wave-function' of the universe - although in that context state is more appropriate IMHO because you really can't observe the position of the universe so expanding it in terms of position eigenstates doesn't really make any sense. But with the view of a state as a preparation procedure the state of the universe is itself rather problematical - but still a valid concept IMHO with appropriate care.

I, and I think Dr Du would agree, have specifically stated it does not solve the measurement problem. I will state it again in bold letters - DECOHERENCE DOES NOT SOLVE THE MEASUREMENT PROBLEM. You can't get plainer than that. What it does however is allow interpretative assumptions that do. There are a few of those eg Consistent Histories and MWI. My interpretive assumption is taking the improper mixture as a proper one ie placing the Von Neumann cut just after decoherence. The formalism allows that - its completely valid.

If there is a problem to be solved, decoherence does not solve it.

If there even is a problem is very interpretation dependant - as what you say implies.

I have said it before, and will say it again, the measurement problem etc etc is not the central mystery in QM. Pick any issue that worries you and there is an interpretation where its non existent or its solution is trivial. What we do not have is an interpretation that solves all of them, and even worse no way to experimentally separate them. IMHO that's the central mystery.

Thanks
Bill

 

[DrDu]

@Dr Du and @bhobba, decoherence does not solve the measurement problem because it doesn't make sense for the subsystem to give a definite outcome but the universe still be in a superposition.

In my oppinion, it is not possible to define a wavefunction for an infinite universe. So it makes no sense to talk of the universe being in a superposition. That means that at least asymptotically, we can get a mixture instead of a superposition.

 

[Demystifier]

In my oppinion, it is not possible to define a wavefunction for an infinite universe.

How about finite universe?

 

[DrDu]

How about finite universe?

Although there are finite flat possibilities, like a toroidal space, I don't consider them very plausible without any evidence. Other kinds of finite universes are curved and require taking gravitation into account. As I said already, gravitation may offer alternative explanations, however, I completely ignore this field.

 

[stevendaryl]

In my oppinion, it is not possible to define a wavefunction for an infinite universe. So it makes no sense to talk of the universe being in a superposition. That means that at least asymptotically, we can get a mixture instead of a superposition.

Well, one day when (or if) there is a fully quantum theory of gravity, there would have to be something like a wave function for the universe, since the universe as a whole would be evolving according to quantum mechanics. Hawking actually wrote down a candidate wave function for the universe years ago, although I'm not sure how seriously to take it.

On the other hand, it could be that combining quantum mechanics and gravity might mean drastically changing quantum mechanics from its present form.

 

[atyy]

In my oppinion, it is not possible to define a wavefunction for an infinite universe. So it makes no sense to talk of the universe being in a superposition. That means that at least asymptotically, we can get a mixture instead of a superposition.

If there is no wave function of the universe, then the mixture cannot be improper. However, the universe cannot be in a proper mixture, since that also assumes that the universe has a wave function. So I don't see how anything is solved by saying the universe is in a mixture.

It would make sense to say that there is no wave function of the universe, and quantum mechanics only applies to subsystems of the universe. But if this is the case, there is no problem for decoherence to solve

 

[DrDu]

So I don't see how anything is solved by saying the universe is in a mixture.

Did I say so? Primarily, we want to explain how a subsystem coupled to the rest of the universe can end up in a mixed state, not the universe as such.

 

[atyy]

Did I say so? Primarily, we want to explain how a subsystem coupled to the rest of the universe can end up in a mixed state, not the universe as such.

But how can the subsystem be in an improper mixed state if the whole universe is not in a pure state?

 

[Vanadium 50]

These answers are pitched way, way over the head of the OP, who was reading a popularization.

 

[DrDu]

But how can the subsystem be in an improper mixed state if the whole universe is not in a pure state?

What I have in mind is some description of the measurement problem in terms of algebraic QFT the latter being presented in the book by Haag I already cited.
I just googled a thesis who tries to do so and looks interesting:
http://pub.uni-bielefeld.de/download/2303208/2303211I think the point with the impossibility to write down a wavefunction for the whole universe is as in the following example (which lacks completely mathematical rigor):
Consider an infinite lattice of spins $\sigma_i$. If $\xi_i$ is the wavefunction for spin i, we could formally introduce the wavefunction of the whole universe as $\prod_i \xi_i$, however, these functions are not well defined as we have no means as to define the convergence of the infinite product.
Assume now that all the $\xi_i$ are the same.
We could create a new wavefunction if we apply the same unitary transformation U to each of the $\xi_i$
$\xi'_i=U\xi_i$. Now, $|\langle \prod \xi_i|\prod \xi'_i \rangle|=\prod |\langle \xi_i|\xi'_i\rangle|=0$ as $|\langle \xi|\xi\rangle|<1$. Hence even the infinitesimally transformed wavefunctions would be orthogonal to each other and in fact no local operation acting on only a finite number of spins can transform one wavefunction into the other.
The ill defined states wouldn't even live in the same Hilbert space.
We now introduce the following operator: $S=\mathrm{lim}_{V\to \infty} 1/V \sum_{i \in V} \sigma_i$. This operator will commute with all operators constructed from a finite number of the $\sigma_i$ which form an algebra. The operator S is therefore a classical observable distinguishing the different universes.
S can be represented by a pure number according to Schur's lemma iff the representation of the algebra is irreducible however, there are also reducible representations possible. This is a shadow of the superposition principle.
Hence the notion of (ir-)reducible representations of the algebra has replaced the ill defined notion of superpositions of whole universes.
However, we have no means to decide whether we live in a reducible or an irreducible representation.

We could now imagine that decoherence will transform asymptotically in the limit $t \to \infty$ an initial superposition $\psi+\psi'$ into a reducible representation i.e. a mixture of different possible classical outcomes.
The convergence of the limit $t \to \infty$ is typically very rapid, i.e. already in a very short time, we can't distinguish a superposition from a statistical mixture of measurement outcomes as our measurements have but finite precision.

 

[QuasiParticle]

Thanks to everyone for the replies and for the reading suggestions, I will definitely look into them. Even though the discussion seems to have derailed slightly, it is nevertheless interesting.

I'm not new to QM, I just like to read some popular physics books for fun Every now and then there are some interesting analogies etc., especially in the fields of physics I'm not so familiar with. But anyway, QFT isn't really my field of true expertise and all the QFT courses I have taken have been "technical" in the sense that they only teach how to calculate and have not concerned themselves with things like interpretations or the measurement problem. And the measurement problem is rarely discussed in the context of QFT, therefore my question.

Perhaps the answer is in the references provided, but I will just quickly ask (if there happens to be a quick answer): how does the discontinuity of a measurement appear in QFT? In the Copenhagen QM the non-Schrödinger evolution of the system due to a measurement is quite apparent. DrDu mentioned that the separation between classical and quantum worlds is already built in the definition of the vacuum state. Even though this sounds sensible, I don't quite see it.

 

[atyy]

What I have in mind is some description of the measurement problem in terms of algebraic QFT the latter being presented in the book by Haag I already cited.
I just googled a thesis who tries to do so and looks interesting:
http://pub.uni-bielefeld.de/download/2303208/2303211I think the point with the impossibility to write down a wavefunction for the whole universe is as in the following example (which lacks completely mathematical rigor):
Consider an infinite lattice of spins $\sigma_i$. If $\xi_i$ is the wavefunction for spin i, we could formally introduce the wavefunction of the whole universe as $\prod_i \xi_i$, however, these functions are not well defined as we have no means as to define the convergence of the infinite product.
Assume now that all the $\xi_i$ are the same.
We could create a new wavefunction if we apply the same unitary transformation U to each of the $\xi_i$
$\xi'_i=U\xi_i$. Now, $|\langle \prod \xi_i|\prod \xi'_i \rangle|=\prod |\langle \xi_i|\xi'_i\rangle|=0$ as $|\langle \xi|\xi\rangle|<1$. Hence even the infinitesimally transformed wavefunctions would be orthogonal to each other and in fact no local operation acting on only a finite number of spins can transform one wavefunction into the other.
The ill defined states wouldn't even live in the same Hilbert space.
We now introduce the following operator: $S=\mathrm{lim}_{V\to \infty} 1/V \sum_{i \in V} \sigma_i$. This operator will commute with all operators constructed from a finite number of the $\sigma_i$ which form an algebra. The operator S is therefore a classical observable distinguishing the different universes.
S can be represented by a pure number according to Schur's lemma iff the representation of the algebra is irreducible however, there are also reducible representations possible. This is a shadow of the superposition principle.
Hence the notion of (ir-)reducible representations of the algebra has replaced the ill defined notion of superpositions of whole universes.
However, we have no means to decide whether we live in a reducible or an irreducible representation.

We could now imagine that decoherence will transform asymptotically in the limit $t \to \infty$ an initial superposition $\psi+\psi'$ into a reducible representation i.e. a mixture of different possible classical outcomes.
The convergence of the limit $t \to \infty$ is typically very rapid, i.e. already in a very short time, we can't distinguish a superposition from a statistical mixture of measurement outcomes as our measurements have but finite precision.

But is the limit $t \to \infty$ really "typically very rapid"? If I understand you correctly, these two papers discussed something related.

Hepp, Helevtiva Physica Acta, 1972
Quantum Theory of Measurement and Macroscopic Observables
http://retro.seals.ch/digbib/view?pid=hpa-001:1972:45::1204
The generation of probabilities from probability amplitudes in a quantum mechanical measurement process is discussed in the framework of infinite quantum systems. In several explicitly soluble models, the measurement leads to macroscopically different 'pointer positions' and to a rigorous 'reduction of the wave packet' with respect to all local observables.

Bell, Helevtiva Physica Acta, 1975
On wave packet reduction in the Coleman-Hepp model
http://retro.seals.ch/digbib/view?pid=hpa-001:1975:48::824
The quantum mechanical measurement problem is considered in a model due to Hepp and Coleman. Whereas Hepp emphasized a 'rigorous "reduction of the wave packet"', in a certain mathematical limit, it is emphasized here that no such reduction ever actually occurs. Some general remarks are made on the advantages of the Heisenberg picture for such considerations, especially in connection with extension to relativistic theories. The non-reduction of the wave packet is directly related to the deterministic character of Heisenberg equations of motion.

 

[bhobba]

how does the discontinuity of a measurement appear in QFT?

It appears when the quantum field suddenly becomes localised. The book Fields of Colour gives the lay explanation. In QFT the field is a superposition of the number states of particles. When that superposition changes to an actual particle or particles being observed an observation has occurred. How does that happen - decoherence is the modern take - but I will leave it to you to investigate that one. But just to give an all too brief overview, interaction with the environment changes the superposition to an improper mixed state which looks like collapse has occurred - ie the superposition of their being 1 particle, 2 particles etc etc is changed to being one particle, 2 particles etc etc with an actual probability for each on observation. If 'localisation' actually had occurred it would be a proper mixed state, the difference being for proper states the system was one particle, two particles etc randomly presented for observation, but observationally they are the same which leads to the view apparent 'localisation' has occurred. However the issue is controversial, not in the sense of the math, that's rock solid, but what can be read into it. You can do a search on Physics Forums to see the details of the debate. Personally I don't like to get drawn into such discussions because I know from bitter experience they go nowhere - hence my previous comment - 'You will find however convincing others of it is not quite that simple.'

Thanks
Bill

 

[atyy]

Thanks to everyone for the replies and for the reading suggestions, I will definitely look into them. Even though the discussion seems to have derailed slightly, it is nevertheless interesting.

I'm not new to QM, I just like to read some popular physics books for fun Every now and then there are some interesting analogies etc., especially in the fields of physics I'm not so familiar with. But anyway, QFT isn't really my field of true expertise and all the QFT courses I have taken have been "technical" in the sense that they only teach how to calculate and have not concerned themselves with things like interpretations or the measurement problem. And the measurement problem is rarely discussed in the context of QFT, therefore my question.

Perhaps the answer is in the references provided, but I will just quickly ask (if there happens to be a quick answer): how does the discontinuity of a measurement appear in QFT? In the Copenhagen QM the non-Schrödinger evolution of the system due to a measurement is quite apparent. DrDu mentioned that the separation between classical and quantum worlds is already built in the definition of the vacuum state. Even though this sounds sensible, I don't quite see it.

One way to argue informally that QFT is just QM is that QFT is used in condensed matter physics as well as high energy physics. In condensed matter physics, the systems are typically described by the non-relativistic Schroedinger's equation for many identical particles. This can be rewritten using a language called "second quantization", so that it becomes a quantum field theory. So non-relativistic QM of many identical particles is QFT.

http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-2-qftc3.pdf
http://www.its.caltech.edu/~yehgroup/NTU_2007 Summer Lectures/NTU2007_Part_II_1&2&3.pdf

 

[DrDu]

One way to argue informally that QFT is just QM is that QFT is used in condensed matter physics as well as high energy physics. In condensed matter physics, the systems are typically described by the non-relativistic Schroedinger's equation for many identical particles.

That's true. However, one of the decisive new features in condensed matter physics is the appearance of phase transitions due to the (almost) infinite extent of the phases. In my oppinion, the measurement process shares many similarities with phase transitions.

 

[atyy]

That's true. However, one of the decisive new features in condensed matter physics is the appearance of phase transitions due to the (almost) infinite extent of the phases. In my oppinion, the measurement process shares many similarities with phase transitions.

I agree that phase transitions and collapse of the wave function can be made to look mathematically similar.

In the case of phase transitions, we know that they do not really occur, since there are only finite numbers of particles in real life. There is no conceptual problem with saying that phase transitions do not really occur.

In the case of wave function collapse, we do believe real life is only at finite N and t, but we do have a conceptual problem if we say that there are not really definite outcomes. So for quantum measurement, somehow we must insist that the "phase transition to collapse" really occurs in spite of finite N and t, which is essentially the Copenhagen interpretation with its division of the universe into classical and quantum worlds, and the assumption of collapse.

 

[stevendaryl]

What I have in mind is some description of the measurement problem in terms of algebraic QFT the latter being presented in the book by Haag I already cited.
I just googled a thesis who tries to do so and looks interesting:
http://pub.uni-bielefeld.de/download/2303208/2303211


I think the point with the impossibility to write down a wavefunction for the whole universe is as in the following example (which lacks completely mathematical rigor):
Consider an infinite lattice of spins $\sigma_i$. If $\xi_i$ is the wavefunction for spin i, we could formally introduce the wavefunction of the whole universe as $\prod_i \xi_i$, however, these functions are not well defined as we have no means as to define the convergence of the infinite product.
Assume now that all the $\xi_i$ are the same.
We could create a new wavefunction if we apply the same unitary transformation U to each of the $\xi_i$
$\xi'_i=U\xi_i$. Now, $|\langle \prod \xi_i|\prod \xi'_i \rangle|=\prod |\langle \xi_i|\xi'_i\rangle|=0$ as $|\langle \xi|\xi\rangle|<1$. Hence even the infinitesimally transformed wavefunctions would be orthogonal to each other and in fact no local operation acting on only a finite number of spins can transform one wavefunction into the other.
The ill defined states wouldn't even live in the same Hilbert space.

Okay, you're really talking about an infinite number of particles, rather than the universe being infinite. In ordinary quantum mechanics, the universe is assumed to be infinite, but it is assumed that there are only finitely many particles. In QFT, the number of particles is allowed to be unbounded, but in the usual ways of computing things, people consider particles as a perturbation of a vacuum state with no particles.

I'm not sure whether people know how to handle the case of a truly infinite number of particles.

 

[DrDu]

Okay, you're really talking about an infinite number of particles, rather than the universe being infinite. In ordinary quantum mechanics, the universe is assumed to be infinite, but it is assumed that there are only finitely many particles. In QFT, the number of particles is allowed to be unbounded, but in the usual ways of computing things, people consider particles as a perturbation of a vacuum state with no particles.

I'm not sure whether people know how to handle the case of a truly infinite number of particles.

No, I am not talking about an infinite number of particles but of field operators. The sigmas only serve as an example.
In qft, we have creation and anihilation operators $a^+(x)$ and a(x) with a continuous index x instead of the discrete i from the example.