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If $\rho$ is the density matrix of a system then $Tr(\rho ^2) \leq 1$. If the equality holds the system is in a pure state and it is in a mixed state otherwise. But, what is the physical meaning of $Tr(\rho^2)$ ?

$Tr(\rho) = 1$ for all valid density matrices. This stems from the normalization constraint that total probability must be one. Is there any such interpretation for $Tr(\rho ^2)$ ?

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    $\begingroup$ This quantity can be measured experimentally, you might be interested by this paper arxiv.org/pdf/1509.01160v1.pdf $\endgroup$
    – Adam
    Commented Mar 30, 2016 at 8:41
  • $\begingroup$ What do you mean by "physical meaning"? -- Note that one could well argue that $\mathrm{tr}(\rho)$ has no physical meaning, since states must always be normalized. $\endgroup$
    – Norbert Schuch
    Commented Mar 30, 2016 at 11:34
  • $\begingroup$ @Adam Note that this quantity can only be measured given at least two copies of $\rho$. $\endgroup$
    – Norbert Schuch
    Commented Mar 30, 2016 at 11:34
  • $\begingroup$ One possible interpretation is the overlap of $\rho\otimes \rho$ with the symmetric subspace. Would this be a satisfactory answer?? (After all, $\rho$ has to be pure in order for $\rho\otimes \rho$ to be fully contained in the symmetric subspace.) $\endgroup$
    – Norbert Schuch
    Commented Mar 30, 2016 at 11:34
  • $\begingroup$ It's not clear what you're asking. The inequality is saturated if and only if $\rho$ is pure, what more of a "physical meaning" do you seek? $\endgroup$
    – ACuriousMind ♦
    Commented Mar 30, 2016 at 11:50

2 Answers 2

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You can interpret $1-Tr[\rho^2]$ as a kind of entropy. If you think of $\rho$ as a classical mixture of quantum states, then $Tr[\rho^2] = \sum_i p_i^2$, where $p_i$ is probability of being found in eigenstate $i$. The more dispersed the states, the smaller the quantity. So for a completely mixed state you have $1/n$, whereas for a pure state you have 1.

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If the system is in a pure state, the density operator is just the projector onto that state, and so $\rho^2 = \rho$. Since, $\mathrm{tr}\,\rho=1$, in a pure state clearly $\mathrm{tr}\,\rho^2=1$.

Since probabilities must be non-negative, $\rho$ has only non-negative eigenvalues ($\rho$ is positive semidefinite). With this and the trace/total probability condition, each eigenvalue $\lambda_i$ for a mixed state satisfies $\lambda_i < 1$. Then $\mathrm{tr}\,\rho^2<1$ for a mixed state.

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    $\begingroup$ It's not clear to me how this constitutes an interpretation of the quantity $\mathrm{tr}(\rho^2)$. $\endgroup$
    – joshphysics
    Commented Mar 30, 2016 at 7:35
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    $\begingroup$ It does: Pure states are 1, therefore it measures some degree of "purity". $\endgroup$
    – Martin
    Commented Mar 30, 2016 at 8:52
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    $\begingroup$ @Martin But I understand the OP understands this. He wants some interpretation "beyond that" (whatever that might be -- I'm not saying that the question makes sense, I just don't feel that this provides an answer). $\endgroup$
    – Norbert Schuch
    Commented Mar 30, 2016 at 11:33
  • $\begingroup$ @NorbertSchuch: You are right. I overread that in the original question. $\endgroup$
    – Martin
    Commented Mar 30, 2016 at 11:56

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