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States belonging to some space $\mathcal H$ can be described by density operators $\rho\in L(\mathcal H)$ that are positive and have trace one. Pure states are the ones that can be written as $\rho=|\psi\rangle\langle \psi|$ for some $|\psi\rangle\in\mathcal H$, all others are said to be mixed.

My question is twofold:

  • can all mixed states be written as a convex combination $\rho=\sum_j p_j |\psi_j\rangle\langle \psi_j|$ given an ensemble of pure states $\{|\psi_j\rangle\langle\psi_j|\}$ and a probability distribution $p_j$? To elaborate: given $\rho=\sum_j p_j \rho_j$ where $\rho_j$ is not necessarily pure, can it always be written in the form above?

  • if all $\rho$ admit a spectral decomposition $\rho=\sum_k \lambda_k|k\rangle\langle k|$ in an orthonormal basis $\{|k\rangle\}$ of $\mathcal H$, how do I distinguish between pure and mixed states in this form?

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    $\begingroup$ To distinguish between mixed and pure states take the trace of $\rho^2$. It would be less than 1 for mixed states and equal to 1 for pure states. $\endgroup$
    – Mauricio
    Sep 7 at 18:32
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  1. Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble $\{ | 0 \rangle, |1\rangle \}$

But it is always possible to find some ensemble to do this in. As you have mentioned, the spectral decomposition tells us how to do this in the ensemble of the eigenstates of $\rho$

  1. For any density matrix the eigenvalues are always non-negative and sum to one. Now if a state is pure then its largest eigenvalue will be $1$ and all other eigenvalues will be zero. So you can tell apart a pure state from a mixed state just by looking at the eigenvalues.
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  • $\begingroup$ Thank you. This is what I had in mind with regard to the first issue: say someone hands me the operator $\rho=\sum_j p_j \rho_j$ and the $\rho_j$ are mixed. Then I could find an ensemble $\{q_k^{(j)},\rho_k^{(j)}\}$ for each of these mixed states and write $$\rho=\sum_{jk}q_k^{(j)} p_j \rho_k^{(j)}. $$ Now, if the $\rho_k^{(j)}$ are pure, I'm done; otherwise I iterate until $\rho$ is written in some ensemble $\{p_{j_1,...,j_n},\rho_{j_1,...,j_n}\}$ where the $\rho_{j_1,...j_n}$ are pure states. Does this make sense? $\endgroup$
    – Balter 90s
    Sep 11 at 12:14
  • $\begingroup$ Given any $\rho$, there is no unique ensemble decomposition for it. So unless you have some specific reason to choose mixed state ensembles, you can always do a spectral decomposition and find an ensemble of pure states to decompose $\rho$ in. $\endgroup$
    – biryani
    Sep 13 at 4:16

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