naima said:
in that thread Atyy wrote:
"In an improper mixed state, each system is in the same pure state."
I think I can explain Atyy's quote using a pair of entangled electrons:
The spin-part of the wave function for a pair of electrons with total spin zero looks like this:
\frac{1}{\sqrt{2}}(|U\rangle |D\rangle - |D\rangle |U\rangle)
It's a superposition of two possibilities: (1) The first electron is spin-up, and the second electron is spin-down, or (2) the other way around.
Now, if one of the electrons is unobservable (it's lost down a black hole, or something), we can describe the other, observable electron as being in a mixed state of being 50% spin-up and 50% spin-down by tracing out the degrees of freedom due to the lost electron. But unlike a proper mixed state, the probabilities are not due to ignorance about the true state of the electron, but are due to the mathematical device of tracing out unobservable degrees of freedom.
Mathematically, a two-component system will have a density matrix of the form \rho_{i \alpha j \beta} where i, j are indices referring to the first subsystem, and \alpha, \beta are indices referring to the second subsystem. Tracing out the second subsystem means coming up with an effective one-system density matrix \rho'_{ij} defined by:
\rho'_{ij} = \sum_\alpha \rho_{i \alpha j \alpha}
For any experiment that only involves the first subsystem, \rho' makes exactly the same predictions as \rho.
What is for him the definition of a pure state?
For me a pure state can have a density matrix with only one non null element and diagonal density matrices cannot.
Every density matrix can be written (possibly in more than one way) in the form:
\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |
where |\psi_j\rangle is a column matrix and \langle \psi_j| is its hermitian conjugate, a row matrix.
A pure state is just one that can be written using just one such term (that is, all the p_j are equal to zero except for one). Equivalently, a pure state is one where \rho^2 = \rho.