I Do macro objects get entangled?

[Lord Jestocost]

Isn't "collapse" physical as well? I think QT just doesn't tell us.

From an operational point of view, the formalism of quantum mechanics amounts to nothing but a calculational recipe, set up in the last resort to predict the probabilities of various directly observed macroscopic outcomes. Born's rule is the link and the only link that connects our perceptions - what we think to be “observations of a physical reality” - with the mathematics of quantum mechanics. If you don't have this link, the mathematical formalism of quantum mechanics has nothing to do with “reality”, viz. the symbols occurring in the quantum mechanical formalism, such as the probability amplitudes, correspond to nothing in the ”real world”. The wave function is merely an intellectual tool and there is nothing which has to “collapse” physically.

 

[entropy1]

Not only does unitary evolution "come with" the MWI, but that is a big piece of its appeal - MWI uses just unitary evolution and ...

I think that decoherence involves unitary time evolution, right? So does decoherence also come with MWI?

 

[entropy1]

How can MWI accommodate a split of the worldthread in the case of entangled particles, if the probabilities of the threads are unknown? For example: a thread split could be: 0.25|A> + 0.75|B>. What if the coefficients of the threads cannot be known as in the case of entanglement?

 

[PeterDonis]

How can MWI accommodate a split of the worldthread in the case of entangled particles, if the probabilities of the threads are unknown?

Nature always knows the amplitudes. It's only us humans who often don't know them.

 

[entropy1]

I mistaked probabilities and amplitudes: should be amplitudes indeed. (really?? )

 

[entropy1]

If a state $|U\rangle(|A\rangle+|B\rangle)$ decoheres into a state $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$ at t₀, and the outcome of the measurement is $|A\rangle$, is the history (before t₀) of this outcome $|A\rangle$ possibly different than the history (before t₀) if the outcome had been $|B\rangle$?

I wonder, because with decoherence the WF of the whole universe is of significance, and if this is the case, then this could include the whole history as well as the whole future of the universe, I figured.

 

[PeterDonis]

If a state $|U\rangle(|A\rangle+|B\rangle)$ decoheres into a state $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$

That's not decoherence, that's a measurement interaction. Decoherence is just the statement that, once you have the state $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$, the two terms in the state will never interfere with each other again.

 

[entropy1]

That's not decoherence, that's a measurement interaction. Decoherence is just the statement that, once you have the state $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$, the two terms in the state will never interfere with each other again.

Ok, but if we make a measurement, we don't measure $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$, but rather either $|U_A\rangle|A\rangle$ or $|U_B\rangle|B\rangle$, right?

So, could the history of $|U_A\rangle$ be different than the history of $|U_B\rangle$?

 

[PeterDonis]

if we make a measurement, we don't measure $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$, but rather either $|U_A\rangle|A\rangle$ or $|U_B\rangle|B\rangle$, right?

Yes.

could the history of $|U_A\rangle$ be different than the history of $|U_B\rangle$?

I don't know what you mean. You already wrote down the "history"--the state started out as $|U\rangle \left(|A\rangle + |B\rangle \right)$. You can't go back and change that based on a measurement result. It is what it is.

Your question is like looking at a tree with a trunk and two branches, and asking if one branch has a different trunk than the other.

 

[entropy1]

Perhaps I would write $(|U_A\rangle+|U_B\rangle)(|A\rangle+|B\rangle)$. And then let this evolve into $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$. Would this be a proper approach?

A yes/no answer would be helpful too.

 

[PeterDonis]

Perhaps I would write $(|U_A\rangle+|U_B\rangle)(|A\rangle+|B\rangle)$. And then let this evolve into $|U_A\rangle|A\rangle+|U_B\rangle|B\rangle$. Would this be a proper approach?

No. That would not be a unitary evolution.

 

[entropy1]

Could you say that $"U_A"$ is a condition for the measurement outcome to be possible to be $A$?

 

[PeterDonis]

Could you say that "$U_A$" is a condition for the measurement outcome to be possible to be $A$?

I don't know what you mean by that. I think you are making this much harder than it needs to be.

You have a measuring device that starts out in state $|U\rangle$. If the measured system is in state $|A\rangle$, the measuring device transitions to state $|U_A\rangle$, which is the state described as "outcome $A$ has been measured". If the measured system is in state $|B\rangle$, the measuring device transitions to state $|U_B\rangle$, which is the state described as "outcome $B$ has been measured". So if the state of the system is a superposition of $|A\rangle$ and $|B\rangle$, the interaction between the measuring device and the system will transition the overall state to $|U_A\rangle |A\rangle + |U_B\rangle |B\rangle$.

The above is what QM says. I don't see the point of trying to ask all these other questions about "conditions" or "history" or anything else.

 

[entropy1]

Ok.

P.S.: I can't tell you why I ask these questions for that reason probably wouldn't be understood by most people on this forum.

BTW, thanks for answering It still is the case that the people here know what they are talking about and I am still searching.

I didn't realize you could and probably should see the measuring apparatus as part of $U$. Further, the kind of distributiveness of $|U\rangle$ puzzles me a little.

 

[PeterDonis]

I didn't realize you could and probably should see the measuring apparatus as part of $U$.

The measuring apparatus is not "part of" $U$, it is $U$, at least for the cases we've been discussing. What else would it be?

the kind of distributiveness of $|U\rangle$ puzzles me a little.

What do you mean by "distributiveness"?

 

[entropy1]

The measuring apparatus is not "part of" $U$, it is $U$, at least for the cases we've been discussing. What else would it be?

I thought: "The entire universe" (of which the measuring apparatus is a part I guess).

What do you mean by "distributiveness"?

Mathematical distributiveness, as in $|U\rangle(|A\rangle+|B\rangle)$ evolves into $|U_1\rangle|A\rangle+|U_2\rangle|B\rangle$.

 

[PeterDonis]

I thought: "The entire universe" (of which the measuring apparatus is a part I guess)

It can't be the entire universe, because you separated out the system which can be in either state $A$ or state $B$, or some superposition of them.

If you mean "the entire universe except for that system", then you're leaving out an awful lot of information, since the rest of the universe contains a lot more than just the measuring device (which of course it must contain). And since (a) you're not interested in all that other information, and (b) in general you don't know what it is anyway, it seems much better to keep your model limited and just consider $U$ to be the measuring device, since its states are the states you're interested in, and admit that the rest of the universe is simply left out of your model. Otherwise you're going to confuse yourself; for example, see below.

Mathematical distributiveness, as in $|U\rangle(|A\rangle+|B\rangle)$ evolves into $|U_1\rangle|A\rangle+|U_2\rangle|B\rangle$.

Ok, got it. This is just the definition of a measurement interaction, assuming that $U$ is the measuring device. What's puzzling about it?

I can see how it might be puzzling if you are thinking of $U$ as the entire rest of the universe, instead of just the measuring device. But anyone else who writes down a state transition like the one above is just thinking of $U$ as the measuring device, for the reasons I gave above. So if you find the transition puzzling, my advice would be to stop thinking of $U$ as the entire rest of the universe and realize that it's only the measuring device.

 

[entropy1]

my advice would be to stop thinking of $U$ as the entire rest of the universe and realize that it's only the measuring device.

Well, I guess it makes a lot of sense like that. But it seems like saying: "If the outcome is A, the measuring device will show outcome A", or: "If the measuring device shows outcome A, the outcome is A", which, if that are the implications, you couldn't be sure of, because you couldn't know if the outcome has been A if the device measured A, right? You could only be sure if one required the other. Just to define it that way seems like a tautology.

Sorry to bother you some further

 

[PeterDonis]

you couldn't know if the outcome has been A if the device measured A, right?

Um, what? I have no idea where you're getting this from. If the device measured A, that means the outcome is A. That's what "measurement" means.

 

[entropy1]

Um, what? I have no idea where you're getting this from. If the device measured A, that means the outcome is A. That's what "measurement" means.

Is that an assumption or does it follow from some? For instance, can you calculate that?

 

[Mentz114]

Is that an assumption or does it follow from some? For instance, can you calculate that?

According to Von Neumann’s phenomenological picture of the measurement process, the coupling of a microsystem, S, to a measuring instrument, $L$ leads to the following two effects.
(I) It converts a pure state of S, as given by a linear combination $\sum_r c_r u_r$ of its orthonormal energy eigenstates $u_r$, into a statistical mixture of these states for which $|c_r|^2$
is the probability of finding this system in the state $u_r$. This effect is often termed the
‘reduction of the wave packet’.
(II) It sends a certain set of classical, i.e. intercommuting, macroscopic variables M of $L$
to values, indicated by pointers, that specify which of the states $u_r$ of S is realized.

see for instance Sewell in https://arxiv.org/abs/0710.3315v1 (it is above B level though)

 

[entropy1]

see for instance Sewell in https://arxiv.org/abs/0710.3315v1 (it is above B level though)

Thank you. That is something I will try to read.

 

[Mentz114]

On first sight, this seems pretty revolutionary, for it seems to solve the measurment problem if I am correct.

It does not do the first, I think.

Initial Conditions.
We assume that the the systems S and I are coupled together
at time t = 0 following independent preparation of S in a pure state and I in a mixed one,
as represented by a normalised vector ψ and a density matrix Φ, respectively.

 

[PeterDonis]

Is that an assumption

It's a definition. We have to have some definition of what symbols like $|U_A\rangle$ mean. That's it.

 

[entropy1]

So, if the state of the measuring apparatus is $|U_A\rangle$, we will have measured outcome $|A\rangle$. But why does unitary evolution require that we get a superposition of states of $|U_x\rangle|x\rangle$?

 

[PeterDonis]

So, if the state of the measuring apparatus is $|U_A\rangle$, we will have measured outcome $|A\rangle$.

Yes.

why does unitary evolution require that we get a superposition of states of $|U_x\rangle|x\rangle$?

Because the state of the measured system didn't start out as $|A\rangle$. It started out in a superposition. That means the measurement interaction, which entangles the states of the measured system and the measuring device, will put $U$ in a superposition as well. (More precisely, it will put the total quantum system, which consists of the measured system and the measuring device together, into a superposition.)

 

[entropy1]

Ah, that is illuminating. Thanks!

 

[Mentz114]

[]
Because the state of the measured system didn't start out as $|A\rangle$. It started out in a superposition. That means the measurement interaction, which entangles the states of the measured system and the measuring device, will put $U$ in a superposition as well. (More precisely, it will put the total quantum system, which consists of the measured system and the measuring device together, into a superposition.)

Some parts of a measurement apparatus must be macroscopic to achieve the 'registration' ( ie irreversibility) . Therefore saying that the apparatus + system are in a superposition brings us back to the OPs question - "can macroscopic objects be in superposition" !

 

[PeterDonis]

Some parts of a measurement apparatus must be macroscopic to achieve the 'registration' ( ie irreversibility) . There fore saying that the apparatus + system are in a superposition brings us back to the OPs question - "can macroscopic objects be in superposition" !

As a matter of modeling in the math, yes, they can.

Whether the "real" state of the "real" system is a superposition like $|U_A\rangle |A\rangle + |U_B\rangle |B\rangle$ after measurement depends on what interpretation of QM you adopt.

 

[Mentz114]

As a matter of modeling in the math, yes, they can.

Whether the "real" state of the "real" system is a superposition like $|U_A\rangle |A\rangle + |U_B\rangle |B\rangle$ after measurement depends on what interpretation of QM you adopt.

That is fair. My interpretation is that a macroscopic system modeled with thermodynamics can be in a metastable state 'between' two different stable states - which has some of the properties of a physical superposition.

 

[stevendaryl]

So, if the state of the measuring apparatus is $|U_A\rangle$, we will have measured outcome $|A\rangle$. But why does unitary evolution require that we get a superposition of states of $|U_x\rangle|x\rangle$?

1. The assumption is that we have a measuring device in state $|U\rangle$ and a system in the state $|A\rangle + |B\rangle$.
2. Put them together, and you have the tensor product $|U\rangle \otimes (|A\rangle + |B\rangle)$.
3. It's a fact about tensor products that they distribute over $+$: $|U\rangle \otimes (|A\rangle + |B\rangle) = |U\rangle \otimes |A\rangle + |U\rangle \otimes |B\rangle$
4. Another fact about quantum mechanics is that time evolution is linear. That is, if the state $|\psi_1\rangle$ evolves into the state $|\psi_1'\rangle$ and $|\psi_2\rangle$ evolves into $|\psi_2'\rangle$, then $\alpha |\psi_1\rangle + \beta |\psi_2\rangle$ evolves into $\alpha |\psi_1'\rangle + \beta |\psi_2'\rangle$
5. So, if $|U\rangle \otimes |A\rangle$ evolves into $|U_A\rangle \otimes |A\rangle$ and $|U\rangle \otimes |B\rangle$ evolves into $|U_B\rangle \otimes |B\rangle$, then it follows that $|U\rangle \otimes |A\rangle + |U\rangle \otimes |B\rangle$ evolves into $|U_A\rangle \otimes |A\rangle + |U_B\rangle \otimes |B\rangle$

 

[entropy1]

Whether the "real" state of the "real" system is a superposition like $|U_A\rangle |A\rangle + |U_B\rangle |B\rangle$ after measurement depends on what interpretation of QM you adopt.

It occurs to me that if you bring irreversability of the measurement evolution process in the picture, a single outcome gets selected in the macroscopic world (the pointer value gets determined). Would it be right to view it that way?

 

[entropy1]

(I) It converts a pure state of S, as given by a linear combination $\sum_r c_r u_r$ of its orthonormal energy eigenstates $u_r$, into a statistical mixture of these states for which $|c_r|^2$
is the probability of finding this system in the state $u_r$.

So, if I understand correctly: decoherence gives us the probabilities of getting certain outcomes rather than leading us to a specific outcome? (probabilities are instrumental here?)

 

[stevendaryl]

It occurs to me that if you bring irreversability of the measurement evolution process in the picture, a single outcome gets selected in the macroscopic world (the pointer value gets determined). Would it be right to view it that way?

Irreversibility doesn't by itself select one outcome out of a number of possibilities. Both the transitions:

1. $|U\rangle |A\rangle \Rightarrow |U_A\rangle |A\rangle$
2. $|U\rangle |B\rangle \Rightarrow |U_B\rangle |A\rangle$

are irreversible, in the thermodynamics sense. That doesn't imply that a single outcome is selected.

 

[stevendaryl]

That is fair. My interpretation is that a macroscopic system modeled with thermodynamics can be in a metastable state 'between' two different stable states - which has some of the properties of a physical superposition.

Yes. For example, a coin balanced on its edge. A tiny push on the "heads" side will cause the coin to land "heads", and a tiny push on the "tails" side will cause it to land "tails".

 

[PeterDonis]

It occurs to me that if you bring irreversability of the measurement evolution process in the picture, a single outcome gets selected in the macroscopic world (the pointer value gets determined). Would it be right to view it that way?

No, because, as @stevendaryl has pointed out, the state transition is irreversible even without selecting a single outcome.

 

[entropy1]

I was scanning this thread again, and I feel I want to thank everyone who took the effort to answer my questions. I apparently had a hard time grasping what I understand better now. I think I got confused because the branches of MWI were not mentioned here. Anyway: many thanks to all! (If I can focus I will read some introductory book on QM again)

 

[jefferywinkler2]

Thought experiments are idealized hypothetical scenarios. If you did the Schrodinger's Cat experiment in real life, the cat would be dead or alive before you opened the box because, from a practical point of view, you can not keep that many particles entangled and isolated from the environment. So the answer to your question is that, in real life, you can't keep macroscopic objects entangled.

 

[EPR]

What if we could in practice keep the atom isolated so it could be in a superposition of decayed/not decayed state? Is the cat alive or dead?

 

[PeterDonis]

What if we could in practice keep the atom isolated so it could be in a superposition of decayed/not decayed state?

You can't do that and also have the atom affect the cat. For the atom to affect the cat, it has to interact with other things, and "isolated" means "not interacting with other things".

 

[EPR]

Indeed. Isolation is not part of the original setup. Can the quantum system be said to be in a definite state, even when not isolated? No. It's in a probabilistic state.
These measurement paradoxes are never trivial.