Preferred basis and prereduction of density matrices

[comote]

I have seen it written that after a preferred basis (for example |1>, |2>) is chosen a pure state say
[c^2]|1><1|+[(1-c)^2]|2><2|+c(1-c)|1><2|+c(1-c)|2><1| will reduce to the mixed state
c^2|1><1|+(1-c)^2|2><2|.

I wonder about the necessity of this "pre-reduction" postulate. It seems to me that rather than adding this postulate, we can say that once we have chosen what observable it is we are going to measure(and hence the preferred basis) that the information conveyed by the |1><2| and |2><1| terms are inaccessible because the only things that are accessible to us are the eigenstates(|1><1|, |2><2|).

I guess my question is could someone please explain to me what is the need to add this non-unitary pre-reduction of a pure state into a mixed state.

 

[genneth]

Consider a system interacting with a reservoir --- the density matrix for the system will evolve non-unitarily even if the entire system+reservoir states does.

 

[comote]

Thank you for replying. I accept that there is nonunitary evolution in an open system. But this particular form of it just seems unnecessary to me.

Consider the following hypothetical situation.
We have a state |phi>. We have two orthonormal bases
|p1>, |p2> and |o1>,|o2> associated with measurement of A, B respectively.

Let 0<c<1; 0<d<1 and assume that |phi> = c|p1>+(1-c)|p2> = d|o1>+(1-d)|o2>.

The density operator associated with |phi> is
rho = c^2|p1><p1|+(1-c)^2|p2><p2|+c(1-c)|p1><p2|+c(1-c)|p2><p1|
rho = d^2|o1><o1|+(1-d)^2|o2><o2|+d(1-d)|o1><o2|+d(1-d)|o2><o1|

depending on the representation.

Now assuming we care about A first our density matrix will have a prereduction to
the mixed state c^2|p1><p1|+(1-c)^2|p2><p2|, our experimental results
give the eigenvalue corresponding to |p1> with probability c and so on and so forth . . .

Assuming we have reduced to |p1><p1| upon measurement of A we now have the pure state
f^2|o1><o1|+f(1-f)|o1><o2|+f(1-f)|o2><o1|+(1-f)^2|o2><o2|
where f=<o1|p1> and (1-f) = <o2|p1>.

Now we are interested in B, before measurement we have a pre-reduction to a mixed state f^2|o1><o1|+(1-f)^2|o2><o2|
and now upon measurement of B we get the eigenvalue corresponding to |o1><o1|
with probabiltiy f.

My issue with the above situation is that I do not see that I necessarily needed this prereduction anywhere. The whole argument will give me the same probability if I just hold to the (seemingly) reasonable assumption that the only states that are accessible to our measuring devices are of the form |alpha1><alpha1|(or |alpha2><alpha2|) and forget about this prereduction.

The reason I want to forget about the prereduction is while it seems perfectly clear to me that nonunitary evolution does happen in an open system, I am not convinced that it takes this form. In order for the nondiagonal elements to dissappear don't we have to choose beforehand which operator we are going to measure?

While the pure state and the reduced mixed state are not the same, at the level of observation they really do seem to be the same to me. So even if it does take that form, it just seems clearer to do away with it. My point is I am missing something in my reasoning and I hope someone can help me find it.

Thanks for reading . . .

 

[genneth]

What you have said is in fact correct. The pre-selection, or einselection as it is called, is approximately encoding for the interaction. Often, the interaction with the reservoir will be very complicated and a first principles calculation may well be too hard or even rather un-illuminating. So we quite often postulate a particular basis that the reservoir tends to force the system into, and assume a markov-ian interaction (no history). It is possible to show that this is in fact a very good approximation to a large range of interactions.

 

[comote]

Thanks for taking the time to read and reply. Would you be able to point me to a reference where this is discussed?

 

[genneth]

http://arxiv.org/abs/quant-ph/0105127 is the arxiv'ed version of a published paper.

 

[comote]

Thank you