Do macro objects get entangled?

[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 Schrödinger'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.