0
$\begingroup$

Consider two polarity-wise entangled photons A and B in an EPR experiment.

The process of measuring the polarization of photon A by Alice is described by the decoherence of the 2-photons system within Alice's apparatus leading to the einselection of a polarization value; let's call this "process A".

Similarly, the measurement by Bob amounts to the decoherence of the same system within Bob's apparatus; this is "process B".

Let's assume that the measurements happen in spacelike separated frames.

If einselection is local, how does decoherence theory explain that processes A and B are correlated ?

$\endgroup$
0
$\begingroup$

When you perform a measurement of some observable, the system couples to the measurement apparatus. This spreads information that formerly was confined in the system to the system plus the measurement apparatus, from whence it spreads into the environment. Some of that information is necessary to bring about interference between different values of unsharp observables, and so this spread of information suppresses interference. The observable measured in this way gives the possible outcomes of that measurement - its eigenvalues are selected: einselection. For both subsystems of an entangled system this process of decoherence takes place locally.

Note that decoherence prevents interference between the different possible measurement outcomes. It does not select one of those outcomes and eliminate the others. Both outcomes take place, but they can't interact with one another although they can still sometimes play a role in explaining experimental results,as they do in experiments involving entanglement. The full description of the system in question, and the measurement apparatus, is still given in terms of a set of Heisenberg picture observables, not a single number representing the measurement outcome. Bell's theorem says that if you can represent the state of a system in terms of stochastic classical variables, then physics must be non-local to match the probabilities predicted by quantum mechanics. But in quantum mechanics, a system is represented by its Heisenberg picture observables, which are not classical stochastic variables.

If you have two spacelike separated systems that are entangled with one another and you measure them, then each system decoheres locally. The correlation is established only after the results are compared. They are established by decoherent systems carrying locally inaccessible quantum information: information that is present in a system but does not affects expectation values of measurements on that system alone. See

http://arxiv.org/abs/quant-ph/9906007

http://arxiv.org/abs/1109.6223.

$\endgroup$
  • $\begingroup$ In the Deutsch & al. papers you linked decoherence only appears with respect to the "decoherent channel" of their very particular take on EPR. This does not really address my question which is about the basics of the decoherence mechanism. More about your answer in another comment. $\endgroup$ – Stéphane Rollandin Mar 8 '16 at 17:42
  • $\begingroup$ When you say "Each system decoheres locally. The correlation is established only after the results are compared", it seems to imply that even after decoherence happened for both systems they are still not classical (or almost classical if you wish wrt the density matrix). Isn't that contradictory with the very idea of decoherence ? $\endgroup$ – Stéphane Rollandin Mar 8 '16 at 17:49
  • 1
    $\begingroup$ Decoherence doesn't make the world classical. It's a consequence of quantum mechanics, it doesn't contradict quantum mechanics, as classical physics does. A system decoheres as a consequence of information about that system spreading to the outside world, which prevents interference. In EPR type experiments some information about the joint system is hidden from the outside world in locally inaccessible information and does not decohere until that locally inaccessible information is accessed. $\endgroup$ – alanf Mar 9 '16 at 1:07
  • $\begingroup$ You say that the paper doesn't address your question about decoherence. It explains why you see a correlation when you compare the results of measurements on A and B. Namely, the processes that underlie comparing the correlations give rise to the correlations between the results. You could model decoherence explicitly in some specific example of such a process and the paper explains what would happen in such a case in section 4 starting at "Let us de­fine locally inaccessible information". See also most of section 6. $\endgroup$ – alanf Mar 9 '16 at 1:23
  • $\begingroup$ "Decoherence doesn't make the world classical". If you think so then I'm afraid we are not talking about the same thing. Just google "decoherence classical" and see how many papers have titles such as "Decoherence and the Quantum to Classical Transition" (this is by Zeh) or "Entanglement, Decoherence and the Quantum/Classical boundary" (this one by Haroche). This is really what I am interested about: how do two spacelike separated local processes lead to "classical/quasi-classical/almost-diagonalized/your-prefered-vocable-here" systems that are correlated, according to decoherence theory ? $\endgroup$ – Stéphane Rollandin Mar 9 '16 at 8:40
0
$\begingroup$

I think your concern might really all be about the meanings of the words. So I'll make a very simplified version of the picture so you can see what local means and doesn't mean.

In a Stern-Gerlach device when you measure $\hat\sigma_z$ the incoming particle's beam has some incoming width transverse to the direction it is going. Then the beam widens and splits and on each branch of the split, the beam the spin has changed to be polarized into an eigenstate of $\hat \sigma_z.$

So now you can look at a system of two particles use the x direction for the width of the beam of one particle and the y direction for the width of the beam of particle two. So it starts out like a square and at each point in the square the wave has a value for this joint spin state, say $$\left[\begin{matrix}1\\0\end{matrix}\right]\otimes\left[\begin{matrix}0\\1\end{matrix}\right]+\left[\begin{matrix}0\\1\end{matrix}\right]\otimes\left[\begin{matrix}1\\0\end{matrix}\right].$$

If we send particle one through a Stern-Gerlach then the square widens splits along a vertical line and the spin state becomes $\left[\begin{matrix}1\\0\end{matrix}\right]\otimes\left[\begin{matrix}0\\1\end{matrix}\right]$ on the left square and becomes $\left[\begin{matrix}0\\1\end{matrix}\right]\otimes\left[\begin{matrix}1\\0\end{matrix}\right]$ on the right square.

If instead we send particle two through a Stern-Gerlach then the square gets taller, splits along a horizontal line and the spin state becomes $\left[\begin{matrix}1\\0\end{matrix}\right]\otimes\left[\begin{matrix}0\\1\end{matrix}\right]$ on the bottom square and becomes $\left[\begin{matrix}0\\1\end{matrix}\right]\otimes\left[\begin{matrix}1\\0\end{matrix}\right]$ on the top square.

Each of those is local in the sense that marginals of the other particle didn't change. If you then do the other measurement afterwards, the the whole square deflects left/right or deflects up/down just like a Stern-Gerlach deflects an eigenstate in a particular direction. And the spin state doesn't change.

When you do both at the same time, the it just heads over the the top-right or the bottom left, again the marginals move the same way.

The whole point is that as the spin state changes, the positions change too. That's where that information is getting spread.

$\endgroup$
  • $\begingroup$ I have had a hard time mapping your answer to my question :) The fact that you use electrons+spins instead of photons+polarization is ok; more troubling is the fact that nowhere is decoherence mentioned, although I take it that the details you give about the measurement via Stern-Gerlach may have something to do with what the decoherence process looks like in that setup (correct me if that's wrong). The conclusion is just surreal: "That's where that information is getting spread"... I have no idea what you are talking about at this point. You lost me :) To be followed in next comment... $\endgroup$ – Stéphane Rollandin Mar 10 '16 at 12:26
  • $\begingroup$ Getting a little more abstract now (and back to the terms of my question): is the gist of your answer that I have wrongly considered that the decoherence process A is local (and respectively B), while it is actually automatically correlated with process B via the entanglement of particles A and B ? $\endgroup$ – Stéphane Rollandin Mar 10 '16 at 12:30
  • $\begingroup$ @StéphaneRollandin I'm saying the Hamiltonian causes different spin eigenstates with position wave packets have the position wave packets deflect differently. So one position wave packet with the spin values not an eigenstate gets split, continuously and locally in the sense that the split happens in part of configuration space representing the position of that particle. Decoherence comes in two places, one the cascade of information about which branch the separated waves are in from the position of the one particle to other parts of the system in a thermodynamically irreversible fashion. And $\endgroup$ – Timaeus Mar 16 '16 at 23:52
  • $\begingroup$ @StéphaneRollandin And whole classical type external field of the device itself leading to the interaction. It's a bit like a classical action reaction pair. A classical type field is causing a split and each branch of the split has classical type information that can freely flow around in the split. $\endgroup$ – Timaeus Mar 16 '16 at 23:54
  • $\begingroup$ So is what you say in agreement with the answer I wrote ? (I'm guessing you favor the MWI in more or less the same way as @alanf by the way you see information flow as the big deal here) $\endgroup$ – Stéphane Rollandin Mar 17 '16 at 0:07
0
$\begingroup$

The question is ill-posed in that it assumes that decoherence explains away the measurement problem, in a local manner. However decoherence only says that probability amplitude phases spread from a quantum system to its environment, effectively entangling the whole thing. So instead of providing a way to get an eventual quasi-classical system, one could say it does the opposite and leads to an eventual quantum system which now includes the environment, with the twist that it may be considered kind of classical in the sense that interference effects are not seen anymore (at least in the limits of experimental error boxes) because phase information is now all other the place, so that probabilities can be handled without having to bother about quantum correlations. But this does not change in any way what EPR has to say about entanglement and nonlocality.

Now as alanf pointed out in its answer, some authors see decoherence as a process in a larger picture where they describe EPR as being local all the way up from the isolated measures to the manifested correlation. See arxiv.org/abs/quant-ph/9906007. But not everyone agrees; see arxiv.org/abs/quant-ph/0312155.

$\endgroup$

protected by Qmechanic Mar 14 '16 at 10:33

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.