I Is wave function collapse equivalent to entanglement with the observer?

[dns]

TL;DR Summary

Is wave function collapse equivalent to entanglement with the observer/detector?

While electron is propagating as quantum wave, we don’t know its, position only possible distribution, so we are (and our detector) definitely are not correlated with its position.

When that electron arrives at the position detector, it becomes entangled with it (detector registering position must correlate with electrons arrival position, otherwise its subpar detector)

So “quantum wave collapse” is “entanglement with the detector”.

It seems obvious simplification (one less concept), but I do not find this formulation anywhere, so I must be wrong.

Where am I wrong? (Pointer to experiment that contradicts such view would be much appreciated)

Thanks in advance for enlightenment.

 

[PeterDonis]

Moderator's note: Thread moved to the QM interpretations subforum, and thread level changed to "I".

 

[PeterDonis]

Is wave function collapse equivalent to entanglement with the observer/detector?

That depends on what interpretation of QM you adopt. In an interpretation like the MWI, what appears to us to be "wave function collapse" is indeed just entanglement of the measured system with the measuring device, followed by decoherence. But in that interpretation, there are no single outcomes of measurements; all possible outcomes happen, each in its own branch of the wave function.

But on an interpretation in which wave function collapse actually physically changes the wave function, so that only a single branch corresponding to the measured outcome survives, no, wave function collapse is not equivalent to entanglement with the measuring device. The state after measurement in such an interpretation is not entangled.

 

[Demystifier]

TL;DR Summary: Is wave function collapse equivalent to entanglement with the observer/detector?

Where am I wrong? (Pointer to experiment that contradicts such view would be much appreciated)

Perhaps the simplest example to understand it is the Schrodinger cat thought experiment. The cat is entangled with the atom, which means that the cat (with atom) is in the superposition of dead and alive. More generally, entanglement is always a superposition. On the other hand, when the observation/detection happens, then there is no longer a superposition, the cat is either dead or alive. After the detection, the cat is correlated with something. Before the detection it is entangled with something. Entanglement precedes correlation. Entanglement is a cause of correlation. But entanglement is not equivalent to correlation. In classical physics there is correlation, but there is no entanglement.

 

[PeroK]

TL;DR Summary: Is wave function collapse equivalent to entanglement with the observer/detector?

While electron is propagating as quantum wave, we don’t know its, position only possible distribution, so we are (and our detector) definitely are not correlated with its position.

When that electron arrives at the position detector, it becomes entangled with it (detector registering position must correlate with electrons arrival position, otherwise its subpar detector)

So “quantum wave collapse” is “entanglement with the detector”.

If we take the detection of an electron by a detector, then the electron is absorbed by the apparatus and no longer exists as an independent particle. The detector and electron are now part of the same system.

That's not what's usually meant by entanglement or wave-function collapse. Also, the detector is not correlated with the electron, but (if anything) with the source of the electron.

It's good to think about these things, but you should try to use terminology as precisely as possible.

It's interesting at this point to think about what a detector does. We have a microscopic particle, whose impact with a macroscopic device should hardly register. But, by clever design, the detector amplifies the tiny energy of the interaction in order to make a light flash and/or make a digital display and/or record some data on a computer disk. The direct interaction of the electron with the detector is only the first part of this process.

 

[javisot20]

Quantum objects only interact with quantum objects, otherwise the cat would be "entangled" to the atom in superposition being able to see live and dead cats, but we don't see that happening like that. (In MWI yes we can)

If a quantum object interacts with a macroscopic object, the macroscopic object absorbs the quantum object (disappearing), or the quantum object loses its quantum properties. We don't see the macroscopic object acquiring quantum properties. Quantum particles are not point balls that follow a certain trajectory, they are extensive objects in space-time that follow an integral of paths, since their evolution is conditioned by the uncertainty principle.

Entanglement is a stronger correlation than a classical correlation, typical in quantum objects, while collapse is a redistribution of the probabilities associated with a quantum system. They are different things, but their common link is quantum objects.

 

[PeterDonis]

Quantum objects only interact with quantum objects

Everything is a quantum object in quantum mechanics.

If a quantum object interacts with a macroscopic object, the macroscopic object absorbs the quantum object (disappearing), or the quantum object loses its quantum properties.

This is not what quantum mechanics says. QM only says that in "macroscopic" objects, what you call "quantum properties" are almost always undetectable because they get averaged out over the huge number of degrees of freedom in the system.

 

[javisot20]

I was simply referring to what PeroK comments here, which is true,

If we take the detection of an electron by a detector, then the electron is absorbed by the apparatus and no longer exists as an independent particle. The detector and electron are now part of the same system.

Everything is a quantum object in quantum mechanics.


This is not what quantum mechanics says. QM only says that in "macroscopic" objects, what you call "quantum properties" are almost always undetectable because they get averaged out over the huge number of degrees of freedom in the system.

That's why what I said is correct, quantum objects are not entangled with macroscopic objects (or do you claim that QM doesn't says that?)

 

[PeterDonis]

I was simply referring to what PeroK comments here, which is true,

What @PeroK stated is true, but it doesn't mean what you think it means.

That's why what I said is correct

No, it isn't.

quantum objects are not entangled with macroscopic objects (or do you claim that QM doesn't says that?)

That's exactly what I'm claiming, yes--that QM doesn't say that. QM doesn't even say there is a sharp distinction between "quantum objects" and "macroscopic objects". QM says that in principle everything is a "quantum object", as I posted before.

There are some interpretations of QM, such as at least some versions of Copenhagen, that have to claim some sort of special status for "measuring devices" and "observers"--but those are interpretations, not "QM". They don't change the basic math of QM, and the basic math of QM is perfectly capable of representing any object, including a "macroscopic" one, as a "quantum object".

I strongly suggest that you think very carefully before posting further in this thread.

 

[Fra]

TL;DR Summary: Is wave function collapse equivalent to entanglement with the observer/detector?
...
So “quantum wave collapse” is “entanglement with the detector”.

I think of it as two views of the "same thing". But they two views can not be said to be "equivalent" in a meaningful way as they do not even contain the same amount of information.

The hilbert space of the system + detector, is obviously bigger than just the system, so they kind of correspond to two inequivalent "observing systems"; one bigger than the other one. In there I think lies also the issue of to what extent "observer equivalence" really means; and how it is defined, without using yet more external concepts.

/Fredrik

 

[Demystifier]

Everything is a quantum object in quantum mechanics.

What about time?

 

[martinbn]

What about time?

He means every object is a quantum object.

 

[javisot20]

That's exactly what I'm claiming, yes--that QM doesn't say that. QM doesn't even say there is a sharp distinction between "quantum objects" and "macroscopic objects". QM says that in principle everything is a "quantum object", as I posted before.

QM says that all objects are quantum objects, then it holds that in QM only quantum objects are entangled. I don't understand why you deny it, can you clarify it? please

 

[PeterDonis]

QM says that all objects are quantum objects, then it holds that in QM only quantum objects are entangled. I don't understand why you deny it

I haven't denied that at all. If all objects are quantum objects, then "only quantum objects can be entangled" is the same as "all objects can be entangled".

 

[Nugatory]

QM says that all objects are quantum objects, then it holds that in QM only quantum objects are entangled. I don't understand why you deny it, can you clarify it? please

Any macroscopic object is made up of an enormous number of particles, each of which individually behaves quantum mechanically. We can show mathematically that these quantum effects tend to average out so that the object as a whole behaves classically even though everything it is made of is behaving quantum mechanically (google for "quantum decoherence" for more about this). Thus we can treat the interaction between an electron and a macroscopic electron detector as you describe in post #6 (quantum electron absorbed by non-quantum detector) or in pure quantum mechanical terms (quantum electron becomes entangled with all the quantum particles that collectively present as a non-quantum detector). Different interpretations will choose differently here.

 

[dns]

try to use terminology as precisely as possible.

Will do.
I apreciate any corrections of my mistakes.
(I do undrstand importance of precise terminology, but I am not related with formal physics education, so I am bound to make such mistakes :(, I can only ask of patience and please don't take any my followup questions the wrong way, I just want to understand)

then the electron is absorbed by the apparatus

I imagined counter at the slit in double slit experiment. Is that possible?

entanglement is always a superposition

Can we say that in QM everything, every paticle is in superposition technicaly all the time* ?
* except for that point in time of the observation, which is instant and has time measure of 0s.

followed by decoherence

I need help with this term.
Can I interpret decoherance as a "restart" of the wave function evolution? (New wave beeing started at the point of measurement)

 

[PeterDonis]

I need help with this term.

It's an important one, but you won't find much about it in many introductory QM textbooks because the theory of decoherence only started being developed in, IIRC, the late 1970s and early 1980s, and it still has not made its way fully into textbooks.

The Wikipedia page on decoherence is rather long, but I think it has some good references:

https://en.wikipedia.org/wiki/Quantum_decoherence

Can I interpret decoherance as a "restart" of the wave function evolution? (New wave beeing started at the point of measurement)

Not really, no.

First, exactly how to interpret decoherence depends on which QM interpretation you adopt.

Second, decoherence by itself does not account for why measurements have single outcomes. (Note that not all QM interpretations even agree that measurements have single outcomes: for example, in the MWI, all possible outcomes of a measurement happen, each one in its own branch of the wave function. In the MWI, decoherence explains why the branches don't interfere with each other.) So if you want to "restart" the wave function based on a measured outcome, decoherence by itself doesn't justify why you can do that.

 

[PeterDonis]

entanglement is always a superposition.

No, this is not correct. See below.

Can we say that in QM everything, every paticle is in superposition technicaly all the time* ?

No. "Superposition" is basis dependent. For any quantum state, including an entangled one, you can find a basis in which that state is not a superposition: just pick that state as one of the basis vectors.

Entanglement itself is basis independent; whether or not a state is entangled remains the same no matter what basis you choose. So entanglement is a better thing to focus on than "superposition".

 

[Demystifier]

No, this is not correct. See below.


No. "Superposition" is basis dependent. For any quantum state, including an entangled one, you can find a basis in which that state is not a superposition: just pick that state as one of the basis vectors.

Entanglement itself is basis independent; whether or not a state is entangled remains the same no matter what basis you choose. So entanglement is a better thing to focus on than "superposition".

Basis vectors of what? To be able to speak of entanglement, we must first decompose the Hilbert space into a tensor product of two or more Hilbert subspaces. Such a decomposition is not unique. The notion of entanglement depends on this decomposition. Once we fix the decomposition, we have a freedom to choose the basis of each Hilbert subspace, the entanglement does not depend on this basis. But if we change the basis such that we take an entangled state as a basis vector, then this is a vector in the whole Hilbert space, not in a subspace. This is a very strange thing to do given the decomposition of the Hilbert space. Such a change of basis is akin to changing the decomposition, and entanglement is not invariant under the change of decomposition.

 

[Morbert]

Basis vectors of what? To be able to speak of entanglement, we must first decompose the Hilbert space into a tensor product of two or more Hilbert subspaces. Such a decomposition is not unique. The notion of entanglement depends on this decomposition.

Wouldn't the decomposition be specified in the posing of the question? E.g. If I ask you "Are systems A and B entangled?" then we consider the tensor product of Hilbert spaces for A and for B.

Once we fix the decomposition, we have a freedom to choose the basis of each Hilbert subspace, the entanglement does not depend on this basis. But if we change the basis such that we take an entangled state as a basis vector, then this is a vector in the whole Hilbert space, not in a subspace. This is a very strange thing to do given the decomposition of the Hilbert space. Such a change of basis is akin to changing the decomposition, and entanglement is not invariant under the change of decomposition.

Once we have a Hilbert space decomposition in mind, we can always expand a separable state in a basis of vectors that that span the entire Hilbert space, but the condition of a separable state is merely that we can express it as a product of states of the attended systems.

 

[pines-demon]

Entanglement itself is basis independent; whether or not a state is entangled remains the same no matter what basis you choose. So entanglement is a better thing to focus on than "superposition".

Minor point for precision, it depends on what you mean by basis, can't you just choose somekind of Bell basis and then there is no explicit non-separability? For precision what you cannot do is to choose new independent "single particle" bases to remove entanglement.

Edit: sorry I did not see Demystifier post, but probably it is the same issue I am pointing here.
Edit2: non-separability.

 

[Morbert]

Minor point for precision, it depends on what you mean by basis, can't you just choose somekind of Bell basis and then there is no explicit entanglement? For precision what you cannot do is to choose new independent "single particle" bases to remove entanglement.

Edit: sorry I did not see Demystifier post, but probably it is the same issue I am pointing here.

Two particle 1 and 2 in the state $\ket{0}_1\otimes\ket{0}_2 = (\ket{\Phi^+}_{12} + \ket{\Phi^-}_{12})/\sqrt{2}$ are not entangled even though we have written the state as an expansion of Bell states, which span the entire Hilbert space, because the state can be written as a product of states of 1 and of 2.

Similarly, two particles 1 and 2 in the state $|\Phi^+\rangle_{12}$ are entangled even if we write it as $|\Phi^+\rangle_{12}$ because the state can't be written as a product of states of 1 and of 2.

 

[pines-demon]

Two particle 1 and 2 in the state $\ket{0}_1\otimes\ket{0}_2 = (\ket{\Phi^+}_{12} + \ket{\Phi^-}_{12})/\sqrt{2}$ are not entangled even though we have written the state as an expansion of Bell states, which span the entire Hilbert space, because the state can be written as a product of states of 1 and of 2.

Similarly, two particles 1 and 2 in the state $|\Phi^+\rangle_{12}$ are entangled even if we write it as $|\Phi^+\rangle_{12}$ because the state can't be written as a product of states of 1 and of 2.

Exactly, it is when you decide what are the particle Hilbert spaces and the allowed basis transformations that you can speak about entanglement. Again it was just a minor remark.

 

[PeterDonis]

To be able to speak of entanglement, we must first decompose the Hilbert space into a tensor product of two or more Hilbert subspaces. Such a decomposition is not unique.

While this is true, what decomposition we choose is not arbitrary. It is based on the physical composition of the system. For example, if we have a system of two electrons that were produced from a common source and are now spatially separated, the only decomposition that makes sense is the one that has each electron as a subsystem, where spatial position is what defines "which subsystem" and each subsystem is a spin-1/2 qubit. Then entanglement of the spins is independent of which basis we choose for each spin.

if we change the basis such that we take an entangled state as a basis vector, then this is a vector in the whole Hilbert space, not in a subspace.

Ah, I see what you mean; once we choose a decomposition, we are also choosing to express any state in the system's Hilbert space in terms of products of states of the subsystems. Then it is obvious from the definition of an entangled state that any such state must be written as a sum of multiple such products; if it could be written as a single such product, it would be separable, not entangled.

I agree with all that; but I still don't think the term "superposition" is a very good one to describe it, precisely because it is still basis independent, as you say. At the very least, I think it would be a good idea if this crucial issue were better explained in the literature.

 

[bhobba]

Interesting thread.

What was it John Wheeler said, 'No phenomenon is a real phenomenon until it is an observed phenomenon.'

Lenny Susskind explains the connection to decoherence in his theoretical minimum on Quantum Mechanics (IMHO, THE book for beginners, even though it involves a bit of calculus). It is, however, an area where much research has been done, and Decoherence: and the Quantum-To-Classical Transition by Maximilian Schlosshauer is more complete (but more advanced).

Decoherence does not generate actual wave-function collapse. It only provides a framework for apparent wave-function collapse, as the components of a quantum system entangle with other quantum systems within the same environment, and you get what is called a mixed state. Here, our ignorance is described by classical probability. However, it does not explain why QM is probabilistic to begin with. Gleason's theorem suggests it is related to the possible outcomes are the eigenvalues of an observable operator. Why the potential outcomes are encoded that way is a profound mystery. It is easy to think of the possible outcomes as the diagonal of some matrix; the observable is this matrix in diagonal form. But why is this the correct way to mathematically describe the possible outcomes? What was it Newton said; hypotheses non fingo (Latin for "I frame no hypotheses"). Well, some attempt to - they are called interpretations. For gravity, Einstein formed a hypothesis called GR. Do we need another Einstein?

Thanks
Bill