Entangled system pure state or not?

[stevendaryl]

in terms of Many-Worlds, we only share the same world because we are entangled.
how about in Bohmian, Copenhagen, Objective Collapse, Cramers, Ensemble, how do you describe it in terms of the description "we only share the same world because we are entangled", can we say for example "In terms of Bohmian, we only share the same pilot wave because we are entangled"? how about others? thank you.

So let me illustrate with Schrodinger's cat.

You have a radioactive atom that is in a superposition of a decayed atom and an undecayed atom. The decayed atom triggers the release of poison, killing a cat. The undecayed atom leaves the cat alive.

So in the Many-Worlds interpretation, the state of the universe is a superposition of two possibilities:

1. $\psi_{dead}$: The atom is decayed, the poison is released, the cat is dead, Schrodinger sees a dead cat, etc.
2. $\psi_{alive}$: The atom is not decayed, the poison is not released, the cat is alive, Schrodinger sees a live cat, etc.

So the atom, the poison, the cat and Schrodinger (and basically the rest of the universe) are all entangled. (That is, there is a state for the composite system of atom + poison + cat + Schrodinger, but the composite state doesn't "factor" into a product of the state for the atom $\times$ the state of the poison $\times$ the state of the cat $\times$ the state of Schrodinger.) Note: $\psi_{alive}$ does not describe an entangled composite system, and $\psi_{dead}$ does not describe an entangled composite system, but the superposition $\psi_{alive} + \psi_{dead}$ does describe an entangled composite system.

In the Bohmian interpretation, you have exactly the same state of the universe as in Many-Worlds, so there is still the same entanglement. The difference is that in the Bohmian interpretation, there is more to the state of the universe than the wave function; the world has an ACTUAL state, where all particles have definite positions at all times. So the cat is either actually dead, or actually alive, although the wave function alone doesn't tell you which is the case.

In collapse interpretations, once Schrodinger checks on his cat, the wave function collapses to one of the possibilities $\psi_{alive}$ or $\psi_{dead}$. Since neither of those states is entangled, then after the collapse, nothing is entangled.

So it's really only in collapse interpretations does entanglement ever disappear.

 

[bluecap]

So let me illustrate with Schrodinger's cat.

You have a radioactive atom that is in a superposition of a decayed atom and an undecayed atom. The decayed atom triggers the release of poison, killing a cat. The undecayed atom leaves the cat alive.

So in the Many-Worlds interpretation, the state of the universe is a superposition of two possibilities:

1. $\psi_{dead}$: The atom is decayed, the poison is released, the cat is dead, Schrodinger sees a dead cat, etc.
2. $\psi_{alive}$: The atom is not decayed, the poison is not released, the cat is alive, Schrodinger sees a live cat, etc.

So the atom, the poison, the cat and Schrodinger (and basically the rest of the universe) are all entangled. (That is, there is a state for the composite system of atom + poison + cat + Schrodinger, but the composite state doesn't "factor" into a product of the state for the atom $\times$ the state of the poison $\times$ the state of the cat $\times$ the state of Schrodinger.) Note: $\psi_{alive}$ does not describe an entangled composite system, and $\psi_{dead}$ does not describe an entangled composite system, but the superposition $\psi_{alive} + \psi_{dead}$ does describe an entangled composite system.

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom $\times$ the state of the poison $\times$ the state of the cat $\times$ the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom $\times$ the state of the poison.. etc...: what the $\times$ and "doesn't factor" mean? Thank you.

 

[stevendaryl]

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom $\times$ the state of the poison $\times$ the state of the cat $\times$ the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom $\times$ the state of the poison.. etc...: what the $\times$ and "doesn't factor" mean? Thank you.

If you have two objects, $A$ and $B$, then you have a joint wave function (or state) $\Psi_{A,B}$. Sometimes the joint wave function can be written as a product of component wave functions: $\Psi_{A,B} = \psi_A \cdot \psi_B$, where $\psi_A$ only involves $A$, and $\psi_B$ only involves $B$. In that case, we say that the composite wave function "factors". But now, suppose the joint wave function is a superposition: $\Psi_{A,B} = \alpha \psi_A \cdot \psi_B + \beta \phi_A \cdot \phi_B$. You can't write that as a product of component wave functions. It doesn't factor.

It's analogous to being able to factor multinomials: The expression $xy + x + y + 1$ can be factored into $(x+1) \cdot (y+1)$, where the first factor refers only to $x$ and the second factor refers only to $y$. In contrast, $x+y$ can't be written as something involving only $x$ times something involving only $y$.

 

[stevendaryl]

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom $\times$ the state of the poison $\times$ the state of the cat $\times$ the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom $\times$ the state of the poison.. etc...: what the $\times$ and "doesn't factor" mean? Thank you.

There is an analogous notion of "entanglement" in classical probability theory: You have a joint probability $P(A \wedge B)$ for two events $A$ and $B$. If $A$ and $B$ are independent, then the joint probability factors as follows:

$P(A \wedge B) = P(A) P(B)$

But if the events are not independent, then the joint probability doesn't factor.

For example, for a random person, the probability of being male is $1/2$, and the probability of being blue-eyed is $1/20$ (suppose). Since these are independent, the probability of being a blue-eyed male is $1/2 \cdot 1/20 = 1/40$. On the other hand, if the probability of having a beard is $1/20$, then the probability of being a bearded male is $1/20$, not $1/40$. They aren't independent, since only males have beards (pretty much).

 

[bluecap]

There is an analogous notion of "entanglement" in classical probability theory: You have a joint probability $P(A \wedge B)$ for two events $A$ and $B$. If $A$ and $B$ are independent, then the joint probability factors as follows:

$P(A \wedge B) = P(A) P(B)$

But if the events are not independent, then the joint probability doesn't factor.

For example, for a random person, the probability of being male is $1/2$, and the probability of being blue-eyed is $1/20$ (suppose). Since these are independent, the probability of being a blue-eyed male is $1/2 \cdot 1/20 = 1/40$. On the other hand, if the probability of having a beard is $1/20$, then the probability of being a bearded male is $1/20$, not $1/40$. They aren't independent, since only males have beards (pretty much).

Thank you for your clear explanations. A question that keeps bugging me for days about about page 6 of https://arxiv.org/pdf/quant-ph/0101077v1.pdf

"The second unanswered question in the Everett picture was more subtle but equally important: what physical mechanism picks out the classical states — face up and face down for the card — as special? The problem was that from a mathematical point of view, quantum states like "face up plus face down" (let’s call this "state alpha") or "face up minus face down" ("state beta", say) are just as valid as the classical states "face up" or "face down".

So just as our fallen card in state alpha can collapse into the face up or face down states, a card that is definitely face up — which equals (alpha + beta)/2 — should be able to collapse back into the alpha or beta states, or any of an infinity of other states into which "face up" can be decomposed. Why don’t we see this happen?

Decoherence answered this question as well. The calculations showed that classical states could be defined and identified as simply those states that were most robust against decoherence. In other words, decoherence does more than just make off-diagonal matrix elements go away. If fact, if the alpha and beta states of our card were taken as the fundamental basis, the density matrix for our fallen card would be diagonal to start with, of the simple form

density matrix = [1 0]
--------------------[0 0]

since the card is definitely in state alpha. However, decoherence would almost instantaneously change the state to

density matrix = [1/2 0]
--------------------[0 1/2]

so if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the card in the state "face up", it would stay "face up" in spite of decoherence. Decoherence therefore provides what Zurek has termed a "predictability sieve", selecting out those states that display some permanence and in terms of which physics has predictive power."

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Stevendaryl. What I'd like clarification is the following. It says above that if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. Whereas if it is face-up.. it will stay face up in spite of decoherence.. Does it mean that the alpha and beta state are really still there just we don't see it because it says if you measure it you could still get random outcome? Thanks you.

 

[Mentz114]

Particles become entangled through interactions, and interactions are local. However, once entangled, particles stay entangled even after they get far apart. And if A is entangled with B, and B becomes entangled with C, then A will be entangled with C, as well. So, everything becomes entangled with everything else.

Is it possible for A to become entangled with an earlier version of itself ?

 

[stevendaryl]

Is it possible for A to become entangled with an earlier version of itself ?

Interesting question. I'm not sure exactly what that would mean.

 

[Mentz114]

Interesting question. I'm not sure exactly what that would mean.

If A interacts with B, B with C, C with D in that order and then A interacts with D.

 

[Zafa Pi]

According to atty.. entangled system are pure (can be in superposition)
while according to Bill.. entangled system are not pure (not in superposition)

Here's the prove of what they stated:
atty said:
https://www.physicsforums.com/threa...-states-in-laymens-terms.734987/#post-4642601 message number 11:

"In decoherence, the system consisting of environment + experiment is in a pure state and does not collapse. Here the experiment is a subsystem. Because we can only examine the experiment and not the whole system, the experiment through getting entangled with the environment will evolve from a pure state into an improper mixed state. Since the improper mixed state looks like a proper mixed state that results from collapse as long as we don't look at the whole system, decoherence is said to be apparent collapse."

In a thread Bill said:
"Just to reiterate - because its entangled it is not in a pure state hence not in a superposition which only applies to pure states."

Can someone settle this clearly?
So in an entangled system, is it in pure state (in superposition) or not?
How can one of them (both expert) be wrong in something this basic?

My source is Nielsen & Chuang's "Q-Computation & Q-Information" (highly recommended and easy to read as opposed to Bill's recommendations). They say:
"A quantum system whose state |ψ⟩ is known exactly is said to be in a pure state. In this case the density operator is simply ρ = |ψ⟩⟨ψ|."
A mixed state is when the state is not completely known, is an ensemble of states with associated probabilities and
ρ = Σp_k•|ψ_k⟩⟨ψ_k|. A state is pure if tr(ρ²) = 1, and is mixed if tr(ρ²) < 1.

Using Q- info notation |0⟩ = [1,0] the 2D vector, |1⟩ = [0,1], |0⟩|0⟩ = |00⟩ is the tensor product |0⟩⊗|0⟩ which can be realized as the vector [1,0,0,0] in 4D. The superposition √½(|00⟩ + √½|11⟩) = √½[1,0,0,1] is in the 4D tensor product space, but is not a tensor product and is thus an entangled state, and further it is completely known. It is a pure state. In the "Bell basis" consisting of, √½(|00⟩ + |11⟩), ½(|00⟩ - |11⟩), √½(|10⟩ + |01⟩), √½(|01⟩ - |10⟩) the density operator for √½(|00⟩ + |11⟩) is the 4 by 4 matrix with 1 in the upper left corner and 0s elsewhere.

 

[bluecap]

I'm not going to get into specifics about emotions and civil war, etc. But in terms of Many-Worlds, we only share the same world because we are entangled. Entanglement just means that my state is correlated with your state. If we influence one another, or are influenced by something in common, then we will become correlated/entangled.

For context. Let's get back to this punchline. "But in terms of Many-Worlds, we only share the same world because we are entangled".

Let's say I met Sabine (let's be more practical instead of the more inaccessible Trump) and Stevendaryl also met Sabine. Then we 3 share the same world (o branch) because we are entangled. Our World is one mixed state component. It means if I feel sad or happy. It won't affect Sabine or Stevendaryl because we are already in single outcome (mixed state). Does it mean we are forever in that one mixed state or world. For example. Me and Sabine and Stevendaryl worlds are all happy. So I cna no longer access the other branch of Sabine where she is sad? As well as access the other branch where Stevendaryl is said. I can imagine me having different worlds or branches.. and I know I'm stuck in one world or branch (one mixed state chosen). But does it mean once I entangled with Sabine and we correlate in one world.. it is one world forever correlated and I can't access Sabine or Stevendaryl other branches forever no matter what kind of interaction? I wonder if this is called multi-body entanglement (maybe there is some papers about this).

 

[stevendaryl]

Let's say I met Sabine (let's be more practical instead of the more inaccessible Trump) and Stevendaryl also met Sabine. Then we 3 share the same world (o branch) because we are entangled. Our World is one mixed state component. It means if I feel sad or happy. It won't affect Sabine or Stevendaryl because we are already in single outcome (mixed state). Does it mean we are forever in that one mixed state or world. For example. Me and Sabine and Stevendaryl worlds are all happy. So I cna no longer access the other branch of Sabine where she is sad?

Yes, that's right. The splitting of worlds in Many-Worlds is irreversible, in the same way that if smash a glass bottle, the parts will never form together into a bottle.

As well as access the other branch where Stevendaryl is said. I can imagine me having different worlds or branches.. and I know I'm stuck in one world or branch (one mixed state chosen). But does it mean once I entangled with Sabine and we correlate in one world.. it is one world forever correlated and I can't access Sabine or Stevendaryl other branches forever no matter what kind of interaction? I wonder if this is called multi-body entanglement (maybe there is some papers about this).

For practical purposes, other branches of a macroscopic superposition are forever inaccessible.