atyy said:
It may be correct but it is certainly not standard, not even by the research literature.
It's standard. See the paper I have linked to many many times:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf
See section 3.1:
(4) Ignorance interpretation: The mixed states we find by taking the partial trace over the environment can be interpreted as a proper mixture. Note that this is essentially a collapse postulate.
Whenever decoherence is discussed on this forum, and it has been discussed a lot, it always seems to get around to this factorisation issue.
Yes its a legit issue - but a fringe one.. It really only seems gain traction around here - there doesn't seem to be that much interest in it. The above review article doesn't mention it. Schlosshauer does but it doesn't have a lot on it - in fact I can't even really recall where it is like most of the things discussed here where I can easily find reference to it. The only reason I know its there is someone pointed out to me it was.
The factorisation issue does not disprove decoherence based interpretations. All its saying is the answer you get may depend on factorisation - not that a factorisation can't be found that gives standard QM predictions. I haven't even seen proof that a different factorisation gives a different answer in the cases that generally occur in practice such as detailed in Chapter 3 of Schlosshauer where he examines scattering models like photons decohering dust particles. But such is not the issue in disproving it - that would require showing something totally different - you can't break the system into what's observed and what's doing the observing.
I gave the argument before but for completeness here it is again. We will consider two systems A and B. A can be in state |a1> and |a2>. B in state |b1> abd |b2>. System B is the system being observed and system A is doing the observing. We have arranged things that if system A is in |a1> then B is in |b1> and similarly for |b2>.
Suppose we have the following superposition |p> = 1/√2|b1>|a1> + 1/√2|b2>|a2>. This is obviously an entangled system where system A is entangled with system B ie what is doing the observing is entangled with what is being observed. Obviously I chose that particular superposition purely for ease of exposition - it can be in any kind of superposition. It's a pure state. It remains in a pure state until observed ie until its interacted with.
But now we will do an observation on just system A with the observable A.
E(A) = <p|A|p> = 1/2 <a1|<b1|A|b1>|a1> + 1/2 <a1|<b1|A|b2>|a2> + 1/2 <a2|<b2|A|b1>|a1> + 1/2 <a2|<b2|A|b2>|a2>
Now here is the kicker - since you are only observing system A the observable A has no effect on the B system or its states. So we have:
<p|A|p> = 1/2 <Aa1|<b1|b1>|a1> + 1/2 <Aa1|<b1|b2>|a2> + 1/2 <Aa2|<b2|b1>|a1> + 1/2 <Aa2|<b2|b2>|a2> = 1/2 <a1|A|a1> + 1/2 <a2|A|a2>
= Trace((1/2|a1><a1| + 1/2|a2><a2|) A) = Trace (p' A)
Here p' is the mixed state 1/2|a1><a1| + 1/2|a2><a2|. Thus observing system A is equivalent to observing a system in the mixed state p' - which by definition is the state from |p> by doing a partial trace over B. The observation will of course give |a1> or |a2> and the entanglement will be broken so that if you get |a1> system B will be in |b1> and conversely. We still have collapse if you like that language - but now it has a different interpretation - you are not observing a pure state - but a mixed one. Its not a proper mixed state because its not prepared the way a proper mixed state is prepared - but the state is exactly the same. Any observable A will not be able to tell the difference. This means we, in a sense, can kid ourselves and say, somehow, its a proper mixed state. If it was a proper mixed state then prior to observation it is in state |a1> or state |a2> with probability of half. Prior to observation its in superposition - after it isnt. Until observed it continues in superposition - its simply because of the entanglement it can now be interpreted differently. By observing 'inside' the system - ie only observing system A - it is in a mixed state - not a proper one - but still a mixed state. Because of that it allows a different and clearer interpretation that avoids a lot of problems.
The factorisation issue does not disprove the above. Indeed the math is tight - its difficult to see how it could be disproved. What the factorisation issue is, is the claim that the answer we get depends on factorising the observed system and what is observed ie factoring it into system A and B. But the very existence of observations here in the macro world depends on a macro system interacting with a quantum system. If such was not possible QM in any interpretation would be in deep do do.
Thanks
Bill