In Why decoherence solves the measurement problem by Art Hobson:

$|\psi \rangle _{SA} = c_1|s_1 \rangle |a_1 \rangle + c_2 |s_2\rangle |a_2 \rangle$ which is a wavefunction that describes non-local entangled state of system $S$ with measuring apparatus $A$ and reduced density matrix that describes subsystem is $\rho_S = (|s_1 \rangle |c_1|^2 \langle s_1|\,\,+\,\,|s_2 \rangle |c_2|^2 \langle s_2| )/2$ and $\rho_A = (|a_1 \rangle |c_1|^2 \langle a_1|\,\,+\,\,|a_2 \rangle |c_2|^2 \langle a_2| )/2$.

Now then paper says that when $c_1 = c_2$, reduced density matrices have non-unique basis, but have unique basis when $c_1 \neq c_2$. Can anyone present the reason why this is the case when $c_1 \neq c_2$? (So I do get that $c_1 = c_2$ case.)

Add: $\langle s_1 | s_2 \rangle = 0$ and $\langle a_1 | a_2 \rangle = 0$ is assumed.

  • $\begingroup$ Have you thought about what kind of matrices they are when $c_1 = c_2$? (Hint: They're scalar matrices) Also, "basis set" is a term I'm unfamiliar with when applied to matrices. Do you mean a basis in which they are diagonal? $\endgroup$ – ACuriousMind Jul 20 '14 at 12:51
  • $\begingroup$ Presumably he gave the constraint $\langle s_1 | s_2 \rangle = \langle a_1 | a_2 \rangle = 0$, and you just left it out of the question. It's linear algebra. Here, $|s_1\rangle$ and $|s_2 \rangle$ are the eigenvectors of $\rho_S$, and this fact comes from the fact that Hermitian matrices with distinct eigenvalues have a unique set of eigenvectors. $\endgroup$ – Peter Shor Jul 20 '14 at 12:56
  • $\begingroup$ I get it about $c_1 = c_2$, but I don't know completely why $c_1 \neq c_2$ would result in unique basis. And for basis sets, I should have just said basis. $\endgroup$ – user55624 Jul 20 '14 at 13:36
  • $\begingroup$ Oh sorry Peter Shor. I missed the part about you talking about linear algebra. Thanks for help. I get it. $\endgroup$ – user55624 Jul 20 '14 at 14:24

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