Do macro objects get entangled?

[entropy1]

Decoherence kills the superposition in any physically realistic macroscopic measuring devices.

I think you are asking: Do observers/measuring apparati Alice and Bob become entangled after measuring each of a pair of entangled particles? I would say the answer is no if that is your question.

If Wigner's friend can be considered in superposition because the object is in superposition, then the object (+friend) can be considered in superposition, right?

 

[stevendaryl]

If Wigner's friend can be considered in superposition because the object is in superposition, then the object (+friend) can be considered in superposition, right?

The lesson of decoherence is that superpositions of subystems tend to "infect" the rest of the universe. So it makes sense to talk about a microscopic system (a single atom, maybe) as being in a superposition of different possibilities. And it makes sense to talk about the entire universe being in a superposition of different possibilities. But it doesn't make sense to talk about a macroscopic system that is smaller than the whole universe being in a superposition.

 

[entropy1]

Surely you've seen Lindley's book "Where does the weirdness go?" recommended here before? It's a good layman's starting point on this question.

Is in the pipeline

Very informally: Decoherence says that even if you were able to prepare a cat in a state that is a quantum superposition of dead and alive, that state would very quickly evolve into the classical state "the cat is definitely either dead or alive; we won't know which unless we look, but it as surely one way or the other as a tossed coin on the floor is either heads-up or heads-down whether it's observed or not".

So if I understand correctly, we, observers, as inhabitants of the universe, become part of the WF of the cat, being in superposition of $|A \rangle |measure A \rangle$ and $|B \rangle |measure B \rangle$, and therefore part of only one of the two possible outcomes?

(If this is correct, it is really fun! )

Could we speak of: "As the measured value tends to A, the measurement outcome tends to A", and: "As the measurement outcome tends to B, say, the measured value tends to B"?

 

[entropy1]

$|\Psi_f\rangle = \alpha |D\rangle |\psi_{environment, D}\rangle + \beta |A\rangle |\psi_{environment, A}\rangle$

Is this a pure state?

 

[stevendaryl]

Is this a pure state?

Yes. However, it becomes a mixed state when the unobservable environmental degrees of freedom are traced out.

 

[stevendaryl]

Could we speak of: "As the measured value tends to A, the measurement outcome tends to A", and: "As the measurement outcome tends to B, say, the measured value tends to B"?

I'm not sure what you mean by that.

 

[Lord Jestocost]

Yes. However, it becomes a mixed state when the unobservable environmental degrees of freedom are traced out.

No! Physics should at least be precise when discussing fundamental and essential questions.

“Decoherence is, formally, never complete. There always remain exponentially small non-diagonal terms in the reduced density matrix, reminding us that an initial pure state remains pure according to basic quantum mechanics.” (Roland Omnes, “Results and Problems in Decoherence Theory“)

 

[stevendaryl]

No! Physics should at least be precise when discussing fundamental and essential questions.

“Decoherence is, formally, never complete. There always remain exponentially small non-diagonal terms in the reduced density matrix, reminding us that an initial pure state remains pure according to basic quantum mechanics.” (Roland Omnes, “Results and Problems in Decoherence Theory“)

I don't think I deserved an exclamation point, because what I said was right. When you start with a pure state, form the density matrix, and trace out some of the degrees of freedom, the result is a mixed state.

 

[entropy1]

I'm not sure what you mean by that.

I mean that the WF evolves to
$|A \rangle|measured A \rangle + |B \rangle|measured B \rangle$
rather than
$|A \rangle|measured B \rangle + |B \rangle|measured A \rangle$
?

 

[stevendaryl]

I mean that the WF evolves to
$|A \rangle|measured A \rangle + |B \rangle|measured B \rangle$
rather than
$|A \rangle|measured B \rangle + |B \rangle|measured A \rangle$
?

Oh. Yeah, I think that there probably would be a tiny amplitude for having the "wrong" measurement result. The way I think of a measurement is in terms of an unstable equilibrium, where a tiny (microscopic) shove in one direction or another makes a big (macroscopic) difference in the outcome. For example, if you carefully balance a telephone pole on one end, a small push can make it fall in one direction or the other. Getting the "wrong" measurement result would be comparable to pushing on a telephone pole and having it fall toward you, instead of away from you. That's certainly possible (if it's not perfectly balanced, or if there are other forces pushing it toward you).

 

[entropy1]

Oh. Yeah, I think that there probably would be a tiny amplitude for having the "wrong" measurement result. The way I think of a measurement is in terms of an unstable equilibrium, where a tiny (microscopic) shove in one direction or another makes a big (macroscopic) difference in the outcome. For example, if you carefully balance a telephone pole on one end, a small push can make it fall in one direction or the other. Getting the "wrong" measurement result would be comparable to pushing on a telephone pole and having it fall toward you, instead of away from you. That's certainly possible (if it's not perfectly balanced, or if there are other forces pushing it toward you).

So the measured value (the cat) is fixed according to you? We have the resulting WF $|A \rangle|measured A \rangle + |B \rangle|measured B \rangle$. So wouldn't that suggest the measured value is not fixed?

 

[stevendaryl]

So the measured value (the cat) is fixed according to you? We have the resulting WF $|A \rangle|measured A \rangle + |B \rangle|measured B \rangle$. So wouldn't that suggest the measured value is not fixed?

I'm not sure what you mean by "fixed". With the analogy of the telephone pole balanced on one end, we know:

• If there is some tiny force that shoves the pole to the left, then the pole will almost certainly fall to the left.
• If there is some tiny force that shoves the pole to the right, then the pole will almost certainly fall to the right.
• Therefore, to the extent that everything is describable using quantum mechanics, which uses linear evolution equations, if you start in a state that is a superposition of a force shoving to the left and a force shoving to the right, then you will end up in a state that is a superposition of a state in which the pole fell to the left and a state in which the pole fell to the right.

 

[entropy1]

I'm not sure what you mean by "fixed". With the analogy of the telephone pole balanced on one end, we know:

• If there is some tiny force that shoves the pole to the left, then the pole will almost certainly fall to the left.
• If there is some tiny force that shoves the pole to the right, then the pole will almost certainly fall to the right.
• Therefore, to the extent that everything is describable using quantum mechanics, which uses linear evolution equations, if you start in a state that is a superposition of a force shoving to the left and a force shoving to the right, then you will end up in a state that is a superposition of a state in which the pole fell to the left and a state in which the pole fell to the right.

So do you mean the amplitude of one of the terms in the WF goes to zero during decoherence?

 

[stevendaryl]

So do you mean the amplitude of one of the terms in the WF goes to zero during decoherence?

No. I am definitely not saying that. It seems to me that the information has been presented again and again in this thread, but the same questions about it keep getting asked.

The terms in a superposition evolve independently. If an amplitude starts off nonzero, then it will always be nonzero.

 

[stevendaryl]

There seems to be misunderstandings that are just not getting cleared up.

Decoherence does not make one branch of the wave function go to zero. What it does is destroy interference between branches. The interference terms are what go to zero. If there is no interference between branches, then for all intents and purposes, you can reason as if we're in one branch or the other. The existence of the other branch has no effect on anything in this branch.

 

[stevendaryl]

I don't know how to make this clearer: If there are two systems, the observer and the cat, there is a distinction between:

$|Me\rangle (|Dead\rangle + |Live\rangle)$

and

$|Me_{dead}\rangle |Dead\rangle + |Me_{live}\rangle |Live\rangle$

In the former case, it would be correct for me to say that "The cat is in a superposition". In the latter case, there are two different "me"s, and for each of them, the cat is in a definite state.

 

[entropy1]

@stevendaryl So in which branch we end up is still a mystery? (measurement problem)

Or is the state of the cat fixed but only unknown? Or is the cat also in two distinct branches?

 

[stevendaryl]

@stevendaryl So in which branch we end up is still a mystery? (measurement problem)

We're venturing into the philosophy of many-worlds here, but in MW, both branches happen. There is a branch in which I see a dead cat, and there is a branch in which I see a live cat. The two branches have no effect on each other, so in each branch, I'm free to act as if it's the only branch.

 

[entropy1]

@stevendaryl Ok, so if you are in the branch in which the cat is dead, you are not going to change that anymore when the interference is gone?

Does that have to do with entropy or with decoherence?

(I hope it is not a too stupid question )

 

[stevendaryl]

@stevendaryl Ok, so if you are in a universe in which the cat is dead, you are not going to change that anymore?

Does that have to do with entropy or with decoherence?

(I hope it is not a too stupid question )

I'm not sure what you mean. The world keeps evolving, and both you and the cat can change with time.

 

[entropy1]

$|Me\rangle (|Dead\rangle + |Live\rangle)$

and

$|Me_{dead}\rangle |Dead\rangle + |Me_{live}\rangle |Live\rangle$

So is this possibly an epistemological matter? The environment doesn't know whether the cat is dead or alive, and so the formalism keeps both possibilities open, until the data comes in and then "both possibilities open" becomes "one of the two true", but the formalism doesn't tell us which one?

If the $|Me\rangle (|Dead\rangle + |Live\rangle)$ part is a matter of lack of knowledge, then why wouldn't the $|Me_{dead}\rangle |Dead\rangle + |Me_{live}\rangle |Live\rangle$ part also be a matter of lack of knowledge, since they are both pure states that represent a superposition?

Or conversely: why wouldn't the $|Dead\rangle + |Live\rangle$ part be an uncertain state, wherein it is possible that the cat is in fact neither dead nor alive?

This paradox could in my eyes possibly be resolved if retrocausality is brought into the picture, where that what is measured (the cat) and the outcome of the measurement are the result of each other, like the measurement outcomes of entangled particles are the result of each other, at least as far as what we know allows!

Compare it to how the measurement itself creates the measured value, for instance in an SG app. Only now it creates it retrocausaly.

 

[stevendaryl]

So is this possibly an epistemological matter? The environment doesn't know whether the cat is dead or alive, and so the formalism keeps both possibilities open, until the data comes in and then "both possibilities open" becomes "one of the two true", but the formalism doesn't tell us which one?

If the $|Me\rangle (|Dead\rangle + |Live\rangle)$ part is a matter of lack of knowledge,

No, it isn't. If an electron is in a superposition of spin-up in the z-direction and spin-down in the z-direction, it isn't just that I don't know whether it's spin-up or spin-down. A superposition is neither spin-up nor spin-down.

...then why wouldn't the $|Me_{dead}\rangle |Dead\rangle + |Me_{live}\rangle |Live\rangle$ part also be a matter of lack of knowledge, since they are both pure states that represent a superposition?

That's the whole point of distinguishing $|Me_{dead}\rangle$ from $|Me_{alive}\rangle$. They are different versions of me. So the cat's state is determined by my state---if you were smart enough, you'd be able in theory to figure out whether the cat was alive or dead by examining me.

Or conversely: why wouldn't the $|Dead\rangle + |Live\rangle$ part be an uncertain state, wherein it is possible that the cat is in fact neither dead nor alive?

That is the case. If the cat is in a superposition, then it is neither alive nor dead. But the point of this whole thread is that a cat CAN'T be in a superposition of alive and dead for more than a tiny fraction of second before the rest of the universe is "infected" by the cat's state.

 

[entropy1]

No, it isn't. If an electron is in a superposition of spin-up in the z-direction and spin-down in the z-direction, it isn't just that I don't know whether it's spin-up or spin-down. A superposition is neither spin-up nor spin-down.

That is the case. If the cat is in a superposition, then it is neither alive nor dead. But the point of this whole thread is that a cat CAN'T be in a superposition of alive and dead for more than a tiny fraction of second before the rest of the universe is "infected" by the cat's state.

What I thought is that the cat is either dead nor alive, or it is one of the two, but we don't know which (as long as it is in the box). Isn't that mutually exclusive?

So, is the solution that the cat decoheres, so that it no longer is in a superposition?

But doesn't that mean it was in a superposition?

That's the whole point of distinguishing $|Me_{dead}\rangle$ from $|Me_{alive}\rangle$. They are different versions of me. So the cat's state is determined by my state---if you were smart enough, you'd be able in theory to figure out whether the cat was alive or dead by examining me.

That part is crisp clear to me.

 

[stevendaryl]

What I thought is that the cat is either dead nor alive, or it is one of the two, but we don't know which (as long as it is in the box). Isn't that mutually exclusive?

I'm sorry that every other thing you say my response is that I'm not sure what you're asking. But I'm not sure what you're asking. What are the two things that are mutually exclusive?

So, is the solution that the cat decoheres, so that it no longer is in a superposition?

But doesn't that mean it was in a superposition?

Well, no. I would say that there is never a point where the cat is in a macroscopic superposition.

A cat is made of many, many particles. For the entire cat to be in a superposition, there would have to be basically no interactions between those particles during the time it takes for a cat to die from whatever it is that Schrödinger is using to kill it.

 

[entropy1]

A cat is made of many, many particles. For the entire cat to be in a superposition, there would have to be basically no interactions between those particles during the time it takes for a cat to die from whatever it is that Schrödinger is using to kill it.

So then, $|Dead\rangle + |Live\rangle$ (from your $|Me\rangle (|Dead\rangle + |Live\rangle)$), is representing a lack of knowledge of the people outside the box?

I'm sorry that every other thing you say my response is that I'm not sure what you're asking. But I'm not sure what you're asking. What are the two things that are mutually exclusive?

Doesn't matter And also, my English is a bit poor. These two things appear mutually exclusive to me:

• "the cat is neither dead nor alive"
  
• "or it is one of the two, but we don't know which"

(as long as it is in the box)

I should have said "either: neither dead nor alive, or: one of the two but we don't know which."

 

[stevendaryl]

So then, $|Dead\rangle + |Live\rangle$ (from your $|Me\rangle (|Dead\rangle + |Live\rangle)$), is representing a lack of knowledge of the people outside the box?

No. It seems that you keep getting implications from what I say which I certainly don't intend. I thought I had said exactly the opposite of that: If you write something as a ket, that definitely does NOT represent a person's lack of knowledge. As I'm pretty sure I already said, being in a superposition of states $A$ and $B$ is definitely not a case of "The system is in state $A$, or it's in state $B$, I just don't know which".

By bringing up the fact that a cat is composed of many, many particles, I'm just saying that it's a huge oversimplification to say that there is a $|Dead\rangle$ state for the cat.

Doesn't matter And also, my English is a bit poor. These two things appear mutually exclusive to me:

• "the cat is either dead nor alive"
  
• "or it is one of the two, but we don't know which"

(as long as it is in the box)

Do you mean "neither" rather than "either" in the first bullet?

I'm assuming that what you mean was

1. The cat is not alive AND the cat is not dead.
2. The cat is either alive or it is dead, but we don't know which.

Those two distinctions make sense. If a cat is in a superposition of a "live" state and a "dead" state, then 1. would be true. Otherwise, 2. would be true.

The claim at the basis of this thread is that 1. is never the case (or at least never for more than a tiny fraction of a second). The cat is almost never in a superposition of alive or dead. 2. is closer to the truth, regardless of whether you open the box, or not.

 

[entropy1]

If you write something as a ket, that definitely does NOT represent a person's lack of knowledge.

Ok, that is clear. A ket is not a mixed state. So my suggestion was wrong. But it is not clear to me what signifies the superposition then, especially in this context.

2. is closer to the truth,

regardless of whether you open the box, or not.

Ok, got it. (I hope )

 

[stevendaryl]

Ok, that is clear. A ket is not a mixed state. So my suggestion was wrong. But it is not clear to me what signifies the superposition then, especially in this context.

I thought I had answered that question many times.

1. If the situation is described by $|observer\rangle (|Dead\rangle + |Alive\rangle)$, then you would say that, from the point of view of the observer, the cat is in a superposition of states.
2. If the situation is described by $(|observer_d\rangle |Dead\rangle) + (|observer_a\rangle |Alive\rangle)$, then you would not say that the cat is in a superposition (from a many-worlds perspective, you could say that the entire world is in a superposition, but not the cat alone)

Situation #1 is never true. The cat is never in a superposition of alive and dead. (Or at least never for more than a fraction of a second.) So situation #2 is a better description of the situation.

 

[Nugatory]

Ok, that is clear. A ket is not a mixed state. So my suggestion was wrong. But it is not clear to me what signifies the superposition then, especially in this context.

If $|A\rangle=|B\rangle+|C\rangle$ then we say that the state A is a superposition of B and C.

 

[entropy1]

I understand that the situation goes quickly to $(|observer_d\rangle |Dead\rangle) + (|observer_a\rangle |Alive\rangle)$ ¹, in which the cat is not in superposition for the observer. Also, nonwithstanding that the cat is isolated in the box, inside the box it is not in superposition but rather already decohered. However, when the observer has not looked in the box, his assessment of the situation is $|observer\rangle (|Dead\rangle + |Alive\rangle)$ ², right?

So I wonder, if the cat is almost never in superposition, why the observer nevertheless is in state ²? Also, when does state ¹ occur? Already in the box, or only when the observer takes a look? It seems to me that the formulations ¹ and ² depend on whether the observer actually observes the cat.

So, the cat is not in superposition in the box because of (quick) decoherence, but it is in superposition at that time, because of state ²? That seems contradictory!