Wave function collapse in QED?

[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.