B Nature of collapse / does collapse exist?

entropy1

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If we look at the combined system: "measured system - environment" as being a closed system, and we would measure a pure or proper mixed state, would coherence be conserved?
 
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Sorry, I thought I said something really wrong so I deleted my latest reply. Now I don't remember the exact words I said but this new reply should be clearer. It might also answer your new question "would coherence be preserved?"

I thought that any kind of measurement that yields a macroscopic result is the result of decoherence in the measuring device?
If the system is actually already in some pure state that is an eigenstate of the observable I measure, then I will just get that state. No decoherence is needed.

But to always interpret a mixed state to be such a correct proper mixed state is unfounded.
 

entropy1

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Thank you @Truecrimson Do you also know the answer to my question #26? :smile:
 
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What do you mean by "would coherence be conserved"? The joint "(original) measured system + environment" would just become my new "measured system" and we go back to the same discussion.
 

entropy1

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What do you mean by "would coherence be conserved"? The joint "(original) measured system + environment" would just become my new "measured system" and we go back to the same discussion.
So a measurement is always prone to decoherence (heat bath)?

What I mean is this:
Wikipedia on decoherence said:
[..](although the combined system plus environment evolves in a unitary fashion).
 
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It's probably more correct to say that there is no measurement without decoherence. Decoherence is what provides the measurement record.

Now you might wonder what about the proper mixed states that I talked about. My point is just that, proper mixed states are introduced to be conceptually the same as classical probability distribution, and classically we don't need decoherence. But QM is different.

If we look at the combined system: "measured system - environment" as being a closed system, and we would measure a pure or proper mixed state, would coherence be conserved?
If you make a measurement on the combined system, it's not closed anymore.
 

entropy1

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So making a measurement can't be done in a closed system, and that is why decoherence comes in play, right?

Or is it rather that we can't say for sure whether we are measuring a proper mixed state (or pure state) or an improper one?

And what is the difference between a proper mixed state and a pure state anyway?
 
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Sorry for coming into the discussion with far less knowledge than all of you have. As I see it there are 2 kinds of interactions, some which produce information which has the potential to influence the future and others which do not produce such information. In the classical double slit experiment, both results (generating which-path information and not generating it) are observations, both are measurements. And yet, the setup where we do not attempt to generate which-path information is not considered as a "wavefunction collapse". We have observed the system, undoubtedly so, we have a definite persistent record of the result of the observation (an interference pattern) and yet this observation did not cause the electron's probability wave to collapse, it just maintained its superposition of 50% going through slit A and 50% going through slit B, causing the interference result.

So, observation or measurement 'per se' does not cause collapse. It is only when the process generates irreversible information which could potentially alter the future that we get collapse.

The fundamental issue is consistency. Take the quantum Zeno experiment by Itano et al in 1990. They put some 5000 beryllium atoms in a magnetic trap. By applying a shower of microwaves they could bring the beryllium atoms to an excited energy state. Then by shining a laser beam they could tell how many atoms where in their ground state or in the excited state (because non-excited atoms absorbed some of the laser energy while excited ones did not).

Long story short (if anyone wants the details I will post them), the beryllium atoms were clearly inferred to be in a true superposition of both states non-excited and excited when not observed. But if we shine the laser to generate information about the beryllium atoms' state, their superposition is destroyed. Simply because of consistency, we can not have information about the atoms' state in the laser beam and a superposition of states in the atoms. As simple as that. No definite state information existing = all the possible states consistent with the information coexist in superposition. Definite state information exists = the physical system is in that state, "collapsed". It's just consistency.

What makes the difference between processes which do not require generating definite information, such as our double slit experiment without which-path detectors, and those which demand information to be generated, such as our double slit with which-path detectors switched on? Simply the physical arrangement of the system. It is not necessary that we observe any results, we may as well send them to outer space without looking at them, it is just the physical arrangement of the system which defines if definite state information must be generated or not.
 
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So making a measurement can't be done in a closed system, and that is why decoherence comes in play, right?
It's the other way around, decoherence is what provides the measurement record, and since decoherence requires an open system, measurement requires an open system. (Beware that the notion of a "closed system" here is different from that in classical physics, which usually means that the system doesn't give or take mass with other system. "Closed system" here means that it doesn't give or take information in a sense that can be made precise. Then it's a tautology to say that I can't measure a closed system, because I can't get any information out of it!)

The notion of a proper mixed state, IMO, is introduced only to highlight the conceptual tension between them and mixed states in QM and prevent the interpretation of collapse as a redistribution of probabilities (unless some additional assumption is made e.g. hidden variables).
 

vanhees71

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Dear Entropy1,

The mathematical formalism adopted by Schroedinger leaves clear that in the instant the location of the particle is made, all probabilities disappear. Strangely, since the formulation to this day, numerous discussions about the significance of this disappearance occur, maintaining that there is something mysterious in it. Nevertheless, when we have a dice in hand, before we throw it the possibility of each face falling upside is one to six. In the moment it falls upon the table and immobilize, it's clear one can no more speak in probabilities, as one of the faces was defined. It is no more probable, it's certain. Its obvious, there is nothing mysterious in it, as even Einstein and Niels Bohr concurred.
No! Even if you believe in collapse and take wave mechanics (i.e., non-relativistic (and only non-relativistic!) QT) of a single particle and even if you believe in collapse, in no way a particle can take a fully determined exact position. If so the probability distribution whould be a Dirac-##\delta## distribution, i.e., $$|\psi(\vec{x})|^2=\delta^{(3)}(\vec{x}-\vec{x}_0).$$
Since you cannot take a square root of the ##\delta## distribution, this ##\psi## doesn't exist.

On the other hand you may insist that there is a eigenstate of the position operator. This becomes more clear in the momentum representation, in which
$$\hat{\vec{x}}=\mathrm{i} \vec{\nabla}_{p},$$
and the putative eigenstates are
$$\tilde{u}_{\vec{x}_0}(\vec{p})=\frac{1}{(2 \pi)^{3/2}} \exp[-\mathrm{i}(\vec{x}-\vec{x}_0)],$$
but that's not a square-integrable function and thus cannot describe a state. Fourier-transforming to the position representation you get
$$u_{\vec{x}_0}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}_0),$$
which confirms this result, i.e., you cannot even square the ##\delta## distribution!

Here you have generalized eigenvectors of unbound operators. This always happens if you have operators with a continuous spectrum (or part of whose spectrum is continuous). Then you have generalized eigenfunctions (in the sense of distributions of functional analysis) rather than Hilbert-space eigenfunctions, and these distributions can never describe states of the particle.

In other words a true pure state of a particle within Schrödinger's formulation of non-relativistic QT of a single particle is always represented by a square-integrable wave function, which may very narrowly peak around some value ##\vec{x}_0## of the position of the particle, but it's always of finite width and thus the standard deviation of the position is never 0. This is also underlined by Heisenberg's uncertainty relation ##\Delta x \Delta p_x \geq 1/2##, which shows that neither position nor momentum can ever be exactly determined, and that the more you specify the position the less well specified is the momentum of the particle.
 

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