Suppose I have a two level system with the states labeled ##|0\rangle## and ##|1\rangle##. In this basis, these correspond to density matrices:(adsbygoogle = window.adsbygoogle || []).push({});

##

\rho_0 =

\begin{pmatrix}

1 & 0 \\

0 & 0

\end{pmatrix}

\quad

\rho_1 =

\begin{pmatrix}

0 & 0 \\

0 & 1

\end{pmatrix}

##

I can create a coherent superposition of my basis states like ##|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)## or any other pure state on the Bloch sphere. This particular one has density matrix:

##

\rho_{\psi} = |\psi\rangle\langle\psi| =

\frac{1}{2}\begin{pmatrix}

1 & 1 \\

1 & 1

\end{pmatrix}

##

If all I had were ##\rho_0## and ##\rho_1## and I decided I wanted to produce to a coherent superposition of those two states, is there any operation I can do to ##\rho_0## and ##\rho_1## directly to produce ##\rho_{\psi}## (or the density matrix of some other state in the Bloch sphere)? Or is it necessary to always first factor the density matrices into the outer products of vectors, produce the superposition with the vectors, and then convert the new state vector back into a density matrix? Clearly it's not just a matter of taking some kind of linear combination of ##\rho_0## and ##\rho_1## because at best that gives you a mixed state and at a worst doesn't even give you something positive, Hermitian, and of trace 1.

Equivalently (more or less): if I handed you ##\rho_0##, ##\rho_1##, and ##\rho_{\psi}## you can easily check that all three are pure states by confirming that they're idempotent without needing to compute their eigenvectors. Can you similarly confirm that ##\rho_{\psi}## is a superposition of the others from the matrices alone?

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# Superposition in the density matrix formalism

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