# Equivalence of entanglement and decoherence?

• B
Gold Member
Just taking an advance in what I want to learn someday:

I understand that decoherence and entanglement are more or less equivalent. So, I take it decoherence is in principle a process of entanglement.

Consider two particles A and B who are entangled. if A decoheres by interacting with particle C, are then B and C entangled with each other? And if C dehoheres futher by interacting with particle D, are then B and D entangled, and so forth?

If so, suppose D decoheres all the way to the top: particle X, and X is, in some magical way, the readout of the measuring device on the one side. Similarly B decoheres to particle (and readout) Y. Does this mean that if I manipulate some property of particle X, it will have consequences for the equivalent property of particle Y? That is to say, do we have instant correlation between X and Y?

I guess we have to restrict ourselves to a linear chain of particles on both sides, so not diverging into a macroscopic reaout. What I mean is: if we have a chain of entanglement, does manipulating one end put restrictions on the outcomes of the other end in order to keep the value of the correlation constant?

I hope the question is clear.

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jfizzix
Gold Member
Decoherence is simply the loss of coherence.
In quantum mechanics, there is no meaningful distinction between the coherence of a quantum state, and the purity of a quantum state, so that a state that is nearly pure (i.e., can be represented by just one vector) is also highly coherent.

When one system interacts with another, while both remain isolated from the environment, the coherence of their joint state remains constant, but the states of either party may decohere due to the interaction.
It is true that entanglement (and interaction) with an external party of the most commonly understood cause of decoherence.

If A and B are highly entangled, then A in isolation is already decohered.
If A interacts with C, it is possible that after the interaction that C would be entangled with B instead, but this is just one of multiple possibilities.Any possible final state between B and C is possible so long as their joint coherence remains constant.
it is possible to move this entanglement from one party to the next and so on, so long as you know how to do it.
It is also possible for macroscopic devices to share quantum entanglement.
This doesn't allow for instantaneous communication, though, since you can't ever know you have entanglement unless you have data from both parties involved.

entropy1, atyy and bhobba
You should wait for a mentor or someone better qualified than I to be sure, but I think you're mixing up a couple different things in the description of your thought experiment.

Try looking up "entanglement-swapping experiment" or similar because I think that's what you're trying to describe. So what you describe CAN be done, I believe. I'm not sure what that setup gets you (as in re: your reason for asking it in the first place) over regular bell-pair Aspect experiments though, because in that case you still have a chain of particle interactions that eventually results in
extra-strong correlations between whatever macroscopic object is used to record the results of each of the trials.

Also I think you confuse a lot of terms with what you wrote, like decoherence_vs_measurement, entanglement_vs_correlation, entanglement_vs_bell-type_entanglement (maybe, sorry I can't see you're original post atm), and instant_vs_atemporal(?). And trust me when I say I know how hard it is to speak about this stuff in 100% correct and precise language, but more careful use of a few of those key terms in your OP would go a long way toward making your life easier, as far as understanding any answer for your original query.

Gold Member
[..]In quantum mechanics, there is no meaningful distinction between the coherence of a quantum state, and the purity of a quantum state, so that a state that is nearly pure (i.e., can be represented by just one vector) is also highly coherent.

When one system interacts with another, while both remain isolated from the environment, the coherence of their joint state remains constant, but the states of either party may decohere due to the interaction.[..]
Does that mean that the state of the individual particles may become a mixed state?
This doesn't allow for instantaneous communication, though, since you can't ever know you have entanglement unless you have data from both parties involved.
I know

Gold Member
Try looking up "entanglement-swapping experiment" or similar because I think that's what you're trying to describe. So what you describe CAN be done, I believe.
Actually I was thinking of entanglement swapping when writing my OP, so I'm confusing that with decoherence I guess...

A. Neumaier
2019 Award
decoherence and entanglement are more or less equivalent.
This is completely wrong.

Entanglement is simply the fact that most states of a composite system, described by a tensor product, are not themselves representable as tensor products of states - hence their constituents are entangled. it has nothing at all to do with decoherence, it happens in every system whether isolated or not.

Decoherence is that fact that the interaction with a typical environment ensures that an unprotected quantum system will - to high accuracy - always be in (or very quickly turn into) a quasiclassical mixed state, with a density matrix that is diagonal or coherent [not in the sense of pure but of ''coherent states''] in a basis selected by the specific interaction. This happens only in open quantum systems. Decoherence destroys certain kinds of entanglement.

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bhobba
jfizzix
Gold Member
Does that mean that the state of the individual particles may become a mixed state?

Yep.

bhobba
Mentor
You may be referring to the section in Susskinds excellent theoretical minimum QM book on entanglement and density matrices. There he explains that when observing one part of an entangled system it behaves like a mixed state. This is the start of using decoherence to explain the measurement problem (in the sense you need to understand it before looking into decoherence proper). But only a start, and as people here have correctly explained decoherence is not the same as entanglement, although sometimes in chatting about this stuff the distinction is not clearly made. I know I certainly do that a fair bit. Even Schlosshauer falls into that trap in his excellent textbook on decoherence - its very easy to do.

Thanks
Bill

stevendaryl
Staff Emeritus
Does that mean that the state of the individual particles may become a mixed state?

I would like to give an expanded (though still simplistic) answer to that question, beyond "yes" or "no".

Suppose that you have a particle initially in a pure superposition of states $|A\rangle$ and $|B\rangle$: $|\psi\rangle = \alpha |A\rangle + \beta |B\rangle$. Now, include the environment. Realistically, the environment should never be modeled as a pure state, but should always be modeled as a mixed state, but let me naively model it as initially in state $|\psi_0\rangle$, which you can think of as the vacuum, maybe. So considering particle + environment, you have a composite state of the form $(\alpha |A\rangle + \beta |B\rangle) \otimes |\psi_0\rangle$. Now, let the particle interact with the environment, and assume that $|A\rangle$ has a different effect than state $|B\rangle$. Then the composite system will evolve into a system that looks like:

$\alpha |A'\rangle \otimes |\psi_A\rangle + \beta |B'\rangle \otimes |\psi_B\rangle$

At this point, as far as any measurements that could be performed on the particle, the results will be the same as if the particle were in the mixed state:

$\rho = |\alpha|^2 |A'\rangle \langle A'| + |\beta|^2 |B'\rangle \langle B'|$

In particular, the fact that the particle's state has become entangled with the state of the environment means that you can't observe interference between states $|A'\rangle$ and $|B'\rangle$, and this loss of interference is the hallmark of a mixed, rather than pure, state.

You could still view the composite system of particle + environment, as being in a pure state, but that pure state is unobservable and inaccessible for measurements (because measurements typically act on subsystems, such as a single particle, instead of the whole universe). So for practical purposes, people view the particle after interacting with the environment to be in a mixed state.

nrqed
stevendaryl
Staff Emeritus
as people here have correctly explained decoherence is not the same as entanglement, although sometimes in chatting about this stuff the distinction is not clearly made.

Would you say it is accurate to describe decoherence as just entanglement with a system (the environment) with many, many degrees of freedom? Entanglement of one particle with another particle can sometimes be undone, if you are clever. But entanglement with the environment can never be undone ("never" in the sense of "a collection of broken shards of glass will never reassemble itself into an unbroken bottle").

bhobba
Mentor
Would you say it is accurate to describe decoherence as just entanglement with a system (the environment) with many, many degrees of freedom? Entanglement of one particle with another particle can sometimes be undone, if you are clever. But entanglement with the environment can never be undone ("never" in the sense of "a collection of broken shards of glass will never reassemble itself into an unbroken bottle").

That's reasonable.. And indeed it can be undone eg delayed choice experiment.

Thanks
Bill

Gold Member
$|\psi\rangle = \alpha |A\rangle + \beta |B\rangle$ becomes
$\rho = |\alpha|^2 |A'\rangle \langle A'| + |\beta|^2 |B'\rangle \langle B'|$
(paraphrased by entropy1)
I think I mean the initial particle not in a superposition of states, but in a mixed state already, due to entanglement. So the sytem of both particles is pure, not the particles themselves.

Gold Member
Would you say it is accurate to describe decoherence as just entanglement with a system (the environment) with many, many degrees of freedom? Entanglement of one particle with another particle can sometimes be undone, if you are clever. But entanglement with the environment can never be undone ("never" in the sense of "a collection of broken shards of glass will never reassemble itself into an unbroken bottle").
That's why I modelled the system as a series of interactions of a series of single particles.

Gold Member
So I guess I mean to ask if. after all the entanglement/decoherence, is the state of the combined system of particles X and Y still a pure one.

So let me rephrase a bit what I mean:

Suppose we start with particle X and entanglement-swap it with particle A, A with B, B with C,.... etc. and particle Y is the end of the line. Then particles X and Y are entangled. So if we swap the entanglement-swapping bij decoherence, do we have the same result, in other words: is the concept of entanglement the same as that of decoherence in a linear chain of particles? It is the concept of decoherence I wish to understand.

Thanks.

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Gold Member
If A and B are highly entangled, then A in isolation is already decohered.
If A interacts with C, it is possible that after the interaction that C would be entangled with B instead, but this is just one of multiple possibilities.Any possible final state between B and C is possible so long as their joint coherence remains constant
If A and B are fully entangled, wouldn't that property be preserved between A and C after subsequent decoherence of B into C?

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jfizzix
Gold Member
If A and B are fully entangled, wouldn't that property be preserved between A and C after subsequent decoherence of B into C?

Whatever the final state of B and C is depends entirely on the nature of the measurement interaction. Entanglement is only conserved if the parties evolve independently and without interaction.

If A and B are highly entangled, and object C, "measures" B, it could be that the joint state describing A,B, and C becomes jointly entangled (as in three-party entanglement) and the state might be described as a GHZ state (a three-party version of a Bell state). In that case, the entanglement shared is not reducible to entanglement shared by each pair of parties, since the state describing each pair would actually be separable. Again, this is just one of multiple possibilities.

Demystifier
Gold Member
Decoherence destroys certain kinds of entanglement.
Decoherence destroys certain kinds of entanglement by creating new entanglement. The new entanglement is entanglement with the environment. Since environment has a large number of degrees of freedom, in practice one cannot observe the precise state of all these degrees, so the new entanglement cannot be observed in practice. Therefore decoherence replaces measurable entanglement with a new unmeasurable entanglement.

So to speak, decoherence is a destruction of entanglement by more entanglement. Pictorially, this is like destroying marriage by a harem.

Jeronimus, Truecrimson, bhobba and 1 other person
A. Neumaier
2019 Award
Decoherence destroys certain kinds of entanglement by creating new entanglement.
It is not decoherence that creates the entanglement.
With any realistic interaction, the Schroedinger equation always creates entanglement, with or without decoherence.

Gold Member
I just read that the interaction of a particle with the measuring device is unitary time evolution. Collapse, if it occurs at all, can't be identified as taking place somewhere specific. So it seems that the measuring device gets entangled with the particle. How does this fit in with decoherence?

Gold Member
In particular, the fact that the particle's state has become entangled with the state of the environment means that you can't observe interference between states $|A'\rangle$ and $|B'\rangle$, and this loss of interference is the hallmark of a mixed, rather than pure, state.

You could still view the composite system of particle + environment, as being in a pure state, but that pure state is unobservable and inaccessible for measurements (because measurements typically act on subsystems, such as a single particle, instead of the whole universe). So for practical purposes, people view the particle after interacting with the environment to be in a mixed state.
Can the state of the system within the lightcone after interaction still be regarded as pure?

Decoherence destroys certain kinds of entanglement by creating new entanglement. The new entanglement is entanglement with the environment. Since environment has a large number of degrees of freedom, in practice one cannot observe the precise state of all these degrees, so the new entanglement cannot be observed in practice. Therefore decoherence replaces measurable entanglement with a new unmeasurable entanglement.

So to speak, decoherence is a destruction of entanglement by more entanglement. Pictorially, this is like destroying marriage by a harem.
Interesting. Let me take a simple example to see if I understand.
If particles A and B are entangled in state √½(|00⟩ + |11⟩) [I'm using quantum information notation, e.g. Nielsen and Chuang] and we now introduce particle C, destroy the original entanglement and create the entangled state √½(|000⟩ + |111⟩) of all three particles [obtaining what jfizzix in post #16 calls a GHZ state], then have we experienced some decoherence? [a mini-harem]

Demystifier
Gold Member
Interesting. Let me take a simple example to see if I understand.
If particles A and B are entangled in state √½(|00⟩ + |11⟩) [I'm using quantum information notation, e.g. Nielsen and Chuang] and we now introduce particle C, destroy the original entanglement and create the entangled state √½(|000⟩ + |111⟩) of all three particles [obtaining what jfizzix in post #16 calls a GHZ state], then have we experienced some decoherence? [a mini-harem]
Essentially, yes. But 3 degrees of freedom is still a sufficiently small number so that, in practice, one can still measure who is entangled with him. When this is the case, we usually don't call it decoherence. If you want to escape from a marriage, a small harem is not enough.

That reminds me of an old Croatian nerd joke. An economist, an engineer and a physicist talk about women. The issue is what is better - to have a wife or to have a mistress?
- economist: It's better to have a mistress. It's cheaper and it leaves you more freedom.
- engineer: No, it's better to have a wife. It makes your life much more stable.
- physicist: No, you are both wrong. The best is to have both. You tell the mistress that you are with your wife, and tell the wife that you are with your mistress, so you have the whole day to be alone and do physics.

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Demystifier
Gold Member
It is not decoherence that creates the entanglement.
With any realistic interaction, the Schroedinger equation always creates entanglement, with or without decoherence.
Technically, you are right.
But would you be fine if I said that decoherence involves entanglement?

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A. Neumaier