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Let $\rho$ a density operator $\rho: H \to H$ representing a mixed state, with $H$ Hilbert space. The operator $\rho$ (together with its corresponding matrix) represents a mixed state; when it additionally satisfies the property $\rho^2 = \rho$, then it is a pure state.

In any case, I was wondering if $\rho^2$ has any physical meaning at all. Clearly $\rho^2$ will still be self-adjoint because $\rho$ is by definition, but does it have any physical meaning at all? Would it make sense to compute the expectation value of $\rho^2$ as an observable?

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Yes, of course it has a meaning. It is the quantity providing the quantum purity when traced, $$ \gamma \, \equiv \, \mbox{tr}(\rho^2), $$ so a pure state has maximal purity $\gamma =1$; whereas a maximally impure state proportional to the identity in a d-dimensional Hilbert space has minimal purity $\gamma =1/d$.

More elegantly, it is related to the von Neumann entropy of the system described, $$ S = \, -\mbox{tr}(\rho \ln \rho) = -\langle \ln \rho \rangle \\ = \langle 1- \rho \rangle + \langle (1- \rho )^2 \rangle/2 + \langle (1- \rho)^3 \rangle /3 + ... . $$ One minus the purity is just the leading term in this Mercator series expansion for the logarithm in powers of $1-\rho$ around the unity. The entropy itself ranges from 0 for pure to ln d for maximally impure.

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  • $\begingroup$ Yes, I am aware of this, but that’s the physical interpretation of its trace. I was asking if $\rho^2$ has a physical meaning on its own. Admitting it has one and remembering that one can always compute the expectation value for any observable (self-adjoint operator with physical meaning), what would it mean to compute the expectation value for $\rho^2$? $\endgroup$
    – catesia
    Commented Mar 26, 2023 at 22:33
  • $\begingroup$ I'm not sure what you expect from a "physical interpretation". $\langle \rho ^2\rangle=\operatorname{tr} \rho^3$ enters in the subleading term of the entropy Mercator expansion, beyond linear entropy. You want a physical interpretation of the cumulant expansion? $\endgroup$
    – Cosmas Zachos
    Commented Mar 26, 2023 at 23:51

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