Is the collapse indispensable?

[A. Neumaier]

it should be very clear that the state has a meaning for an individual system

This is not very clear - on the contrary, it directly contradicts the minimal statistical interpretation. It is neither minimal nor statistical, but the source of all troubles, including atyy's claim that the statistical interpretation needs a collapse. Every interpretation that attaches a state to the individual system needs the collapse, whereas if an individual system has no associated state there isn't even a way to say what the collapse should mean.

it's a (measurable!) difference whether you prepare N times a polarization entangled two-photon Fock state or some state with 2N photons.

Indeed, the two sources create two essentially different systems. In the first case, each individual system contains exactly 2 photons, while in the second case, each individual system contains exactly 2N photons.

In both cases, one can verify what the source produces by making experiments on a large number of these systems, and in this way distinguish the two.

And in both cases, the individual system has no state, it has only the general characteristics of a quantum system that are independent of its state (in this case, the fixed number of photons they contain).

 

[zonde]

I would like to say something on the topic of this thread.
I think that collapse can't be simply removed for the following reasons:
1. QM with collapse is standard. So if we say that QM predictions are experimentally verified we mean QM with collapse. Present attempts at removing collapse are making additional assumptions (about how to model measurement). Adding assumptions is justified only if we arrive at new predictions so that we can verify these new assumptions. Otherwise we have to stick to less specific model.
2. Attempts at removing collapse assume that macroscopic (classical) objects (experiment equipment) are very complex quantum systems that can be approximated as simple quantum systems. I think this is assumption is false. And the reason is following. In case of simple quantum systems we take into consideration any particle that participates in interactions within the system, that way we make a clear cut between the system under consideration and the rest of the world. However in case of large classical systems we ignore environment even so the system is constantly interacting with environment. So basically the cut is impossible to make, at one moment photon is part of environment (does not yet belong to the system) at next moment it is absorbed (belongs to the system) and yet a moment later it (another one) is heading away from the system (does not belong to the system any more). We can compare it with living being who is breathing the air. We can not place a clear cut between living thing and it's environment while we can do that for a piece of rock.

 

[A. Neumaier]

QM with collapse is standard.

No. Collapse occurs only in some of the standard interpretations. Collapse is certainly not part of shut-up-and-calculate, the part on which everyone agrees.

Only the Born rule (which doesn't say anything about the state after measurement) is standard, when applied in the appropriate context (e.g., in scattering experiments).

In case of simple quantum systems we take into consideration any particle that participates in interactions within the system.

No. You simply ignore most things. The whole detector participates in the interaction with the system, otherwise it couldn't detect anything. Thus it would have to be taken into consideration, according to your claim. Simply ignoring this and replacing the interaction by collapse is obviously an approximation. In addition, there are losses due to contact with the transmission medium and its boundary. Again these consist of lots of particles interacting with the system (otherwise the system could not lose anything to them). You ignore this, too. The typical Bell-type experiments ignores all these issues and replaces them by approximate reasoning about efficiency.

 

[zonde]

No. Collapse occurs only in some of the standard interpretations. Collapse is certainly not part of shut-up-and-calculate, the part on which everyone agrees.

It's not so much a question about the part on which everyone agrees but rather about the approach that is most universally used to get clear unequivocal predictions for real experimental setups. But it would be better to her a confirmation from some experimentalist.

In addition, there are losses due to contact with the transmission medium and its boundary. Again these consist of lots of particles interacting with the system (otherwise the system could not lose anything to them). You ignore this, too. The typical Bell-type experiments ignores all these issues and replaces them by approximate reasoning about efficiency.

Yes, here you are right.

 

[A. Neumaier]

that is most universally used to get clear unequivocal predictions

The predictions depend only on the shut-up-and-calculate part.

 

[stevendaryl]

I think we (kith, you , rubi, and I) have agreed many times that this uses the deferred measurement principle, and is a way of calculating the same probabilities without using collapse. However, what we have is a simultaneous measurement at single late time, not two measurements in sequence. This is the same as avoiding nonlocality in quantum mechanics by saying that there is no reality to the distant observer, since the distant observer does not need to be real until she meets Bob. So yes, collapse can be avoided, just like nonlocality. However, one has to place some non-standard restriction on what one considers real (sequential measurements or distant observers).

Okay, but isn't that more of a matter of qualms about the interpretation of QM? If you can get the same numbers (in theory) without collapse, doesn't that show that collapse isn't "indispensable"?

 

[vanhees71]

Indeed, the two sources create two essentially different systems. In the first case, each individual system contains exactly 2 photons, while in the second case, each individual system contains exactly 2N photons.

In both cases, one can verify what the source produces by making experiments on a large number of these systems, and in this way distinguish the two.

And in both cases, the individual system has no state, it has only the general characteristics of a quantum system that are independent of its state (in this case, the fixed number of photons they contain).

Now you yourself admit that the state (an equivalence class of a preparation procedure of a single (!) system) has a meaning for the individual system. You can't have ensembles if you can't prepare individual systems. I also agree that you can test the assumption whether you really have prepared some specific state can only be done on the ensemble since probabilistic statements are meaningless for an individual system. That's the minimal interpretation: Specifying the state of an individual system has only very limited meaning concerning the observable facts about this system. The only statement you can make is that, if you have prepared the system in some state, where some observable is determined, i.e., if measured on the individual system you get a predetermined value, which is the eigenvalue of the self-adjoint operator representing that observable. The statistical operator describing the state then must be of the form
$$\hat{\rho}_A(a)=\sum_{j} p_j |a,j \rangle \langle a,j|, \quad \sum_j p_j=1,\quad p_j \geq 0.$$
The $|a,j \rangle$ span the eigenspace of $\hat{A}$ of eigenvalue $a$.

Your last paragraph simply describes an unprepared system. Then you cannot even associate a state to an ensemble or better said you don't even have an ensemble, because it's not said how to specify it.

 

[A. Neumaier]

You can't have ensembles if you can't prepare individual systems

One prepares individual systems, according to the statistical interpretation, but these individual systems have no state, since the state is a property of the ensemble only, not of the individual systems.

you yourself admit that the state (an equivalence class of a preparation procedure of a single (!) system) has a meaning for the individual system.

No. Only measurable properties that are definite in the state of the ensemble have a meaning for the individual; in the present case the number of particles specifying the individual system - since this is common to all individual systems by definition of the ensemble. But the individual system has no state - it only has the definite properties common to all individual systems in the preparation.

Your last paragraph simply describes an unprepared system.

No. It describes the individual systems in a preparation whose state has definite particle number but otherwise only statistical properties that depend on what is prepared and measured. For example, in an electron accelerator one prepares an electron beam whose individual systems are known to be electrons but whose other properties are undetermined and depend on the measurement performed on them, according to the momentum distribution determined by the state (the detailed preparation).

In fact, strictly speaking, the measurement results are not even properties of the individual system but properties of the detector in contact with the particle field determined by the preparation. One can completely avoid mentioning the individual microscopic systems. Indeed, what one measures in a collision experiment are ionization tracks and tracks of deposited energy - properties of the detection fluid or wires. Quantum mechanics predicts how the statistics of the tracks in the detector is related to the state of the source, both macroscopically determined stuff.

The particles themselves remain invisible and their properties may even be regarded as completely hypothetical. That we say we measured the track of a particle is already an interpretation of the measurement results, even a problematic one: In the most orthodox setting where only properties can be measured that correspond to commuting operators, a quantum particle should not have a track, since a track implies fairly definite position and momentum simultaneously!

 

[naima]

It shows that every quantum theory that requires collapse can be converted into one that evolves purely unitarily and makes the same predictions. Here is the recipe:
We start with a Hilbert space $\mathcal H$, a unitary time evolution $U(t)$ and a set of (possibly non-commuting) observables $(X_i)_{i=1}^n$. We define the Hilbert space $\hat{\mathcal H} = \mathcal H\otimes\underbrace{\mathcal H \otimes \cdots \mathcal H}_{n \,\text{times}}$. We define the time evolution $\hat U(t) \psi\otimes\phi_1\otimes\cdots\otimes\phi_n = (U(t)\psi)\otimes\phi_1\otimes\cdots\otimes\phi_n$ and the pointer observables $\hat X_i \psi\otimes\phi_1\otimes\cdots\otimes\phi_n = \psi\otimes\phi_1\otimes\cdots\otimes (X_i\phi_i)\otimes\cdots\otimes\phi_n$. First, we note that $\left[\hat X_i,\hat X_j\right]=0$, so we can apply the previous result. Now, for every observable $X_i$ with $X_i\xi_{i k} = \lambda_{i k}\xi_{i k}$ (I assume discrete spectrum here, so I don't have to dive into direct integrals), we introduce the unitary von Neumann measurements $U_i \left(\sum_k c_k\xi_{i k}\right)\otimes\phi_1\otimes\cdots\otimes\phi_n = \sum_k c_k \xi_{i k} \otimes\phi_1\otimes\cdots\otimes \xi_{i k} \otimes\cdots\otimes\phi_n$. Whenever a measurement of an observable $X_i$ is performed, we apply the corresponding unitary operator $U_i$ to the state. Thus, all time evolutions are given by unitary operators (either $\hat U(t)$ or $U_i$) and thus the whole system evolves unitarily. Moreover, all predictions of QM with collapse, including joint and conditional probabilities, are reproduced exactly, without ever having to use the collapse postulate.

Of course, this is the least realistic model of measurement devices possible, but one can always put more effort in better models.

In this paper Pati shows that when an unknown bit becomes a part of two entangled bits there is a unitary process such that the inital state of the bit is not erased but copied to a third bit.

 

[vanhees71]

One prepares individual systems, according to the statistical interpretation, but these individual systems have no state, since the state is a property of the ensemble only, not of the individual systems.

No. Only measurable properties that are definite in the state of the ensemble have a meaning for the individual; in the present case the number of particles specifying the individual system - since this is common to all individual systems by definition of the ensemble. But the individual system has no state - it only has the definite properties common to all individual systems in the preparation.

No. It describes the individual systems in a preparation whose state has definite particle number but otherwise only statistical properties that depend on what is prepared and measured. For example, in an electron accelerator one prepares an electron beam whose individual systems are known to be electrons but whose other properties are undetermined and depend on the measurement performed on them, according to the momentum distribution determined by the state (the detailed preparation).

In fact, strictly speaking, the measurement results are not even properties of the individual system but properties of the detector in contact with the particle field determined by the preparation. One can completely avoid mentioning the individual microscopic systems. Indeed, what one measures in a collision experiment are ionization tracks and tracks of deposited energy - properties of the detection fluid or wires. Quantum mechanics predicts how the statistics of the tracks in the detector is related to the state of the source, both macroscopically determined stuff.

The particles themselves remain invisible and their properties may even be regarded as completely hypothetical. That we say we measured the track of a particle is already an interpretation of the measurement results, even a problematic one: In the most orthodox setting where only properties can be measured that correspond to commuting operators, a quantum particle should not have a track, since a track implies fairly definite position and momentum simultaneously!

I can fully agree with that formulation. The association of the state with the individual systems forming the ensemble is the common equivalence class of preparation procedures. For the single system only that observables have a definite value that have been prepared. About everything else you have only probabilistic information, which has a meaning only for the ensemble.

The question about the "tracks" of single particles in, e.g., a cloud chamber has been fully understood by Mott as early as 1929

N. F. Mott, The Wave Mechanics of $\alpha$-ray tracks, Procs. Roy. Soc. A 126, 79 (1929)
http://rspa.royalsocietypublishing.org/content/126/800/79

It's, of course, due to the interaction of the particle with the matter in the detector which makes the particle appear as if moving on a "track". Of course, that's a pretty coarse-grained picture of the partice.

 

[Nugatory]

This thread is closed at the starter's request. An interesting and important followon conversation has been moved to the new thread https://www.physicsforums.com/threads/measurement-and-preparation.856391/