Decoherence and apparent collapse

[stevendaryl]

I was talking about the reduced density matrices of the individual particles which can be written in this form.

You get a reduced matrix when you trace out unobservable degrees of freedom. Both particles are observable in EPR, so tracing out one of the particles to get a reduced matrix for the other particle is throwing away information.

 

[kith]

Both particles are observable in EPR, so tracing out one of the particles to get a reduced matrix for the other particle is throwing away information.

Agreed. But then what are your objections to my post #46? My point was very similar to yours: that we cannot interpret the reduced density matrix of an entangled state as a proper mixture. So your answer to my question there seems to be yes.

 

[A. Neumaier]

In the case of an entangled pair of particles, you CAN'T write the density matrix in the form \sum_i p_i |\psi_i\rangle \langle \psi_i|. The state of an entangled pair is \frac{1}{\sqrt{2}}(|U, D \rangle - |D,U\rangle). If you form the density matrix, you get:
\frac{1}{2}(|U,D\rangle \langle U, D| - |U,D\rangle \langle D, U| - |D, U\rangle\langle U, D| + |D,U\rangle\langle D,U|

Every density matrix $\rho$, and hence in particular yours, can be written in the stated form, by taking the $\psi_i$ as an orthonormal basis of eigenvectors of $\rho$. One gets mixed terms only if one insists on choosing a particular basis. In the case of a pure state, as in your example, the sum consists of a single term only.

 

[Ralph Dratman]

[This link] explains it clearly:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf
It is however technical. To understand it you must learn the technicalities.

That is an excellent paper which delivers on the promise of a clear explanation.