Undergrad Is quantum collapse an interpretation?

[tom.stoer]

The unitary time evolution occurs after the collapse. So neither collapse nor unitary time evolution are real.

I don't agree.

For an instrumentalist (or positivist) neither time evolution nor collapse describe something that is "happening out there in the real world as described". But for a realist something in our mathematical model does indeed describe what is "really happening...". Because unitary time evolution and collapse are contradictory they cannot be real "in th for same sense". So you have to make a choice!

The choice of Everett's supporters is to interpret the unitary time evolution as a realistic description of a process happening out there in the real world (read Deutsch, as an example) and to reject the collapse.

I haven't seen any interpretation doing it the other way round, i.e. to reject unitary time evolution as being "real" but chose the collapse:-)

 

[zonde]

There is no collapse here. Collapse is needed when you make a sequence of measurements (ie. more than one measurement per setup). However, you have only described two different setups, each with a single measurement.

Would you agree if I change the wording a bit? Say, collapse is needed when we make a sequence of projections i.e. more than one projection per setup?

 

[zonde]

The collapse happens at no specific point in space.

But decoherence is physical process that happens at specific limited span of time, right? So can you point out the "point" in time when decoherence happens? Does it happen when photon got past the polarizer or when it reaches detector?

 

[zonde]

If you measure an eigenvalue then you project the full state to the corresponding subspace. So the collapse happens in your mind when you readjust the information you have.

If physical situation out there changes with time and you reflect that change in time within your model using projection then projection represents some real dynamics. And while projection is done by you the changes of physical situation that are represented by projection are not done by you. Map is not the territory.

 

[tom.stoer]

But decoherence is physical process that happens at specific limited span of time, right? So can you point out the "point" in time when decoherence happens? Does it happen when photon got past the polarizer or when it reaches detector?

Decoherence happens whenever quantum system and measurement device get entangled with environmental degrees of freedom.

If physical situation out there changes with time and you reflect that change in time within your model using projection then projection represents some real dynamics.

As I explained the non-unitary collaps cannot represent real dynamics b/c it mathematically contradicts dynamics represented by unitary time evolution.

From a realist perspective: the interaction of a quantum system with a measurement device (its atoms, electrons etc.) is a quantum mechanical process and therefore the dynamics must be compliant with unitary time evolution; b/c the collapse is non-unitary it violates the fundamental rule for any quantum mechanical process and cannot be "real".

The way out is that the collapse is only an apparent one, i.e. that the overall dynamics is unitary, compliant with quantum mechanics, but that due to decoherence it appears as collapse.

 

[zonde]

Decoherence happens whenever quantum system and measurement device get entangled with environmental degrees of freedom.

So you don't know where to place decoherence. Because it works neither way.

 

[martinbn]

As I explained the non-unitary collapse cannot represent real dynamics b/c it mathematically contradicts dynamics represented by unitary time evolution.

This is something I don't understand. Why is it a contradiction? It seems that it is a contradiction only if you assume that there can be only one type of evolution. Why is it not possible for the system to evolve unitary for some time, then non-unitary and so on? I can see this would raise more questions about the specifics, but why is it a contradiction by itself?

 

[vanhees71]

So you think that you cannot have an interpretation without projection and at the same time you are convinced that you can have one without collapse. Aren't these two the same?

No, because it's just an effective description, neglecting the irrelevant microscopic details of absorption. The absorption is a local process, i.e., the photon hitting the polarizer and getting absorbed happens at the place where the polarizer is located, i.e., the largest relevant spatial extend is the size of the polarizer, but "collapse" (in this case the absorption of the photon filtering it out as an unwanted polarization state) means it's something happening instantaneously in the entire space, and that's violating causality and also the very construction of QED as a local relativistic field theory obeying the linked-cluster principle.

 

[martinbn]

No, because it's just an effective description, neglecting the irrelevant microscopic details of absorption. The absorption is a local process, i.e., the photon hitting the polarizer and getting absorbed happens at the place where the polarizer is located, i.e., the largest relevant spatial extend is the size of the polarizer, but "collapse" (in this case the absorption of the photon filtering it out as an unwanted polarization state) means it's something happening instantaneously in the entire space, and that's violating causality and also the very construction of QED as a local relativistic field theory obeying the linked-cluster principle.

Consider this, a photon is prepared in a state $\alpha|V\rangle+\beta|H\rangle$, later it is in the state $|H\rangle$. How did the state evolve unitarily? It is not possible. How do you explain that?

 

[vanhees71]

To get a unitary description, you'd have to consider an entire closed system, consisting of the photon and the polarizer. According to Q(F)T the time evolution of a closed system is unitary with the self-adjoing Hamiltonian of the entire system as the "generator". For the photon alone, it's of course not unitary since you "integrate out" a huge system, namely the macroscopic polarization filter!

 

[martinbn]

To get a unitary description, you'd have to consider an entire closed system, consisting of the photon and the polarizer. According to Q(F)T the time evolution of a closed system is unitary with the self-adjoing Hamiltonian of the entire system as the "generator". For the photon alone, it's of course not unitary since you "integrate out" a huge system, namely the macroscopic polarization filter!

Correct me if i am wrong but if you do that you end up with a mixed state either |V> or |H>, not just |H>. So you have a measurement problem, how do you get away with only unitary evolution?

 

[zonde]

To get a unitary description, you'd have to consider an entire closed system, consisting of the photon and the polarizer. According to Q(F)T the time evolution of a closed system is unitary with the self-adjoing Hamiltonian of the entire system as the "generator". For the photon alone, it's of course not unitary since you "integrate out" a huge system, namely the macroscopic polarization filter!

Well, but if you are not specifically after unitary description but rather after description of photon alone? Say you want to do some other things with photon and consider everything that happened with photon before, including polarizer, a state preparation.
You would consider that there is a photon with certain polarization, right?

 

[vanhees71]

Correct me if i am wrong but if you do that you end up with a mixed state either |V> or |H>, not just |H>. So you have a measurement problem, how do you get away with only unitary evolution?

But at the end you only consider the photons that are going through the polarizer, and these have a definite polarization!

 

[tom.stoer]

Why is it a contradiction? It seems that it is a contradiction only if you assume that there can be only one type of evolution. Why is it not possible for the system to evolve unitary for some time, then non-unitary and so on?

a) unitary time evolution:

$$|\psi(t\rangle) \to |\psi(t^\prime)\rangle = U(t^\prime,t) \, |\psi(t\rangle)$$

b) non-unitary collapse to some eigenstate:

$$|\psi(t\rangle) \to \hat{P}_\lambda |\psi(t)\rangle \sim |\lambda(t)\rangle$$

U is invertible, P is not. U is deterministic, P is not.

When i.e. for which time shall the last evolution apply? which axioms or rules tell you when to use (a) and when to use (b)? what is the difference between an interaction with a measuremt device (b) and "something else" (a) ? what selects the target eigenstate to which the state shall collapse? ...?

 

[martinbn]

When i.e. for which time shall the last evolution apply? which axioms or rules tell you when to use (a) and when to use (b)? what is the difference between an interaction with a measuremt device (b) and "something else" (a) ? what selects the target eigenstate to which the state shall collapse? ...?

I agree, I even said that it poses many questions. But you say that it leads to a contradiction. My question is why? Is it obvious, not to me, or do you mean that it would be problematic as it poses more questions than it solves?

 

[martinbn]

But at the end you only consider the photons that are going through the polarizer, and these have a definite polarization!

We are going in circles. My questions is how do you explain that with only unitary evolution, given that the map from the fist state to the second is not unitary?

 

[vanhees71]

I can only repeat what I said before. Only the full time evolution of the entire closed system is unitary. If you project out parts of the system, their evolution is not unitary, because of that projection. I don't know, how I can reformulate this in other words. It's about evolution equations for open quantum systems. Maybe, it helps to read about the Lindblad equation:

https://ocw.mit.edu/courses/nuclear...s-fall-2012/lecture-notes/MIT22_51F12_Ch8.pdf

A more simple example, which entirely works with unitary time evolution is the Stern-Gerlach apparatus. After the particles are running through the inhomogeneous magnetic field you get a state, where the position of the particles is entangled with the spin state. Provided you have a well designed magnetic field you split thus a particle beam into sufficiently well separated partial beams, each of which contains particles with a certain value of the spin component in the direction of the magnetic field (usually this is taken as the $z$ direction, i.e., in each partial beam the spin state of the particles is a pure state $|\sigma_z \rangle$). If you just consider this partial beam (you can block all other partial beams by just putting an appropriate absorber in front), and what you have effectively is the projection of an arbitrary (spin state) $\hat{\rho}$ to the pure state $|\sigma_{z} \rangle \langle \sigma_{z}$. Of course you have a beam with less particles since you absorbed all the "unwanted" ones. At this point of course you have again an effective non-unitary description due to the absorption, where you don't consider the full dynamics of the particles + absorbing material.

The separation into partial beams with determined spin components, however was completely unitary, and you can as well do without any blocking of the unwanted particles, but only experimenting with particles out of one of the partial beams by just using only those at the corresponding location. Then you have a preparation with a unitary time evolution.

 

[tom.stoer]

The original question seems to be

Consider this, a photon is prepared in a state $\alpha|V\rangle+\beta|H\rangle$, later it is in the state $|H\rangle$. How did the state evolve unitarily? It is not possible. How do you explain that?

Sorry to say that, this is trivial.

If both $\alpha|V\rangle+\beta|H\rangle$ and $|H\rangle$ are normalized this is simply a unitary rotation. The unitary operator $U(t)$ which is nothing else but a $U(2) = U(1) \cdot SU(2)$ matrix can be constructed explicitly.

The easiest way to see this is setting

$$\beta(t) = \sqrt{1 - |\alpha(t)|^2} $$

and for some later time

$$\alpha(t_0) = 0 \;\Rightarrow\; \beta(t_0) = 1 $$But that is not really what decoherence and Everett is all about. In this formalism after a measurement of an observable the overall system is not in an eigenstate of this observable!

 

[Demystifier]

I can only repeat what I said before. Only the full time evolution of the entire closed system is unitary. If you project out parts of the system, their evolution is not unitary, because of that projection.

You are failing to distinguish projection from tracing out.

When the full system evolves unitarily, the subsystem evolves non-unitarily due to tracing out of the degrees of freedom that do not belong to the subsystem. This is related to decoherence, but a priori it has nothing to do with projection.

Projection, on the other hand, is related to update of information, or in a slightly different interpretation, with collapse. But projection cannot be explained by decoherence or tracing out of unobserved degrees.

Of course, both operations introduce non-unitarity in a description of the system. Yet those operations are different mathematically, physically and philosophically.

 

[vanhees71]

Well what @martinbn had in mind is the projection rather than the unitary "rotation". I.e., according to the collapse postulate you have a transition
$$|\psi \rangle \mapsto |H \rangle \langle H| \psi \rangle, \qquad (1)$$
and if $|\psi \rangle$ and $|H \rangle$ are both normalized in general the projection is of course not normalized anymore.

But it's not said that this is really the result of the interaction of the photon with the measurement apparatus (that's the case for ideal filter measurements), but all that's really said by the formalism and what's accepted in the minimal interpretation exclusively is that in such a case the probability that one finds a photon to be H-polarized, given it's prepared in the polarization state $|\psi \rangle$ is given by $|\langle H|\psi \rangle|^2$. There is no evolution as in one, except in the sense of an effective description of the interaction between the photon and the polarizer, and since this is a description looking at the photon only in the sense of an open quantum system this "evolution" doesn't need to be unitary (as it must be for a closed quantum system, where the dynamics is due to a self-adjoint Hamiltonian).

 

[vanhees71]

You are failing to distinguish projection from tracing out.

When the full system evolves unitarily, the subsystem evolves non-unitarily due to tracing out of the degrees of freedom that do not belong to the subsystem. This is related to decoherence, but a priori it has nothing to do with projection.

Projection, on the other hand, is related to update of information, or in a slightly different interpretation, with collapse. But projection cannot be explained by decoherence or tracing out of unobserved degrees.

Of course, both operations introduce non-unitarity in a description of the system. Yet those operations are different mathematically, physically and philosophically.

Well, I think I was a bit sloppy here. In case of an ideal filter measurement (or better preparation) the tracing out, however, should indeed lead to a projection, or are you saying that any ideal filter measurement disproves quantum theory since it cannot be understood by unitary time evolution and tracing out the irrelevant degrees of freedom?

 

[Mentz114]

a) unitary time evolution:

$$|\psi(t\rangle) \to |\psi(t^\prime)\rangle = U(t^\prime,t) \, |\psi(t\rangle)$$

b) non-unitary collapse to some eigenstate:

$$|\psi(t\rangle) \to \hat{P}_\lambda |\psi(t)\rangle \sim |\lambda(t)\rangle$$

U is invertible, P is not. U is deterministic, P is not.

When i.e. for which time shall the last evolution apply? which axioms or rules tell you when to use (a) and when to use (b)? what is the difference between an interaction with a measuremt device (b) and "something else" (a) ? what selects the target eigenstate to which the state shall collapse? ...?

Have you seen this paper by L S Schulman ? I don't want to pay the price ...

Model apparatus for quantum measurements
Abstract

We present a model system that behaves as a measurement apparatus for quantum systems should. The device is macroscopic, it interacts with the microscopic system to be measured, and the results of that interaction affect the macroscopic device in a macroscopic, irreversible way. Everything is treated quantum mechanically: the apparatus is defined in terms of its (many) coordinates, the Hamiltonian is given, and time evolution follows Schrödinger's equation. It is proposed that this model be itself used as a laboratory for testing ideas on the measurement process.

https://link.springer.com/article/10.1007/BF01026572

 

[Demystifier]

Well, I think I was a bit sloppy here. In case of an ideal filter measurement (or better preparation) the tracing out, however, should indeed lead to a projection. I guess, there are simple models where such a thing is calculated in the literature.

If that was true, you would be right that there is no problem of measurement and there would be no need for various non-minimal interpretations of QM. But unfortunately it is not true, and this is exactly the origin of the problem of measurement. There is no model (simple or complicated), based merely on standard QM, in which tracing out leads to projection.

 

[vanhees71]

Hm, that's indeed an obstacle, because blocking out particles from a partial beam is almost trivial in the lab. You just put a beam dump.

 

[Demystifier]

are you saying that any ideal filter measurement disproves quantum theory since it cannot be understood by unitary time evolution and tracing out the irrelevant degrees of freedom?

I am saying that any ideal filter measurement disproves some of your claims on quantum theory.

 

[vanhees71]

Do you have a reference, where such a no-go theorem is proven?

 

[Demystifier]

Hm, that's indeed an obstacle, because blocking out particles from a partial beam is almost trivial in the lab. You just put a beam dump.

Now you are failing to distinguish blocking from projection.To paraphrase A. Peres, blocking happens in the laboratory, while projection happens in the Hilbert space.

 

[Demystifier]

Do you have a reference, where such a no-go theorem is proven?

See e.g. the Schlosshauer's book on decoherence, especially the section on the problem of outcomes.

 

[vanhees71]

Well, if I filter out all partial beams from a SG apparatus and keep only the one I want, it's pretty much what I'd (perhaps too naively) describe by a projection operator in the formalism. It gives me the correspondingly less intense beam (with the intensity drop given by the probabilities according to Born's rule). So you say that there is a difference between the description of blocking in the formalism and projection?

 

[Demystifier]

Well, if I filter out all partial beams from a SG apparatus and keep only the one I want, it's pretty much what I'd (perhaps too naively) describe by a projection operator in the formalism. It gives me the correspondingly less intense beam (with the intensity drop given by the probabilities according to Born's rule). So you say that there is a difference between the description of blocking in the formalism and projection?

Ah, we were talking about two different types of projection.

The projection above which you are talking about is deterministic. The non-unitary evolution due to tracing out is also deterministic.

I was talking about non-deterministic projection, when you don't know in advance in which state the system will end up. It is only this latter non-deterministic projection that lies at the origin of the problem of measurement in QM.