It's often written in the QI literature that, for a density operator $\rho$, if $\text{Tr}\left[\rho^{2}\right] < 1$, then $\rho$ describes a mixed state. However, I haven't seen any proofs of this except in the case where the states are in $\rho$ are orthonormal, i.e., if
$$\rho = \sum_{i}p_{i}|\psi_{i}\rangle\langle \psi_{i}|$$
then all the proofs I have seen only apply when $\langle \psi_{i}|\psi_{j}\rangle = \delta_{ij}$. I have written what I think is a proof for the case when that does not hold, but it feels very simple and I wanted feedback on whether or not it is rigorous. Here is the proof (note that $|\phi_{k}\rangle$ just denotes an element of some orthonormal basis on the space of quantum states):
$$\text{Tr}\left[\rho^{2}\right] = \text{Tr}\left[\sum_{i,j}p_{i}p_{j}|\psi_{i}\rangle\langle\psi_{i}|\psi_{j}\rangle\langle\psi_{j}|\right] = \sum_{i,j,k}p_{i}p_{j}\langle\psi_{i}|\psi_{j}\rangle\langle\phi_{k}|\psi_{i}\rangle\langle\psi_{j}|\phi_{k}\rangle$$
$$=\sum_{i,j}p_{i}p_{j}|\langle\psi_{i}|\psi_{j}\rangle|^{2} < \sum_{i,j}p_{i}p_{j} = 1$$
The inequality comes from my assumption that this is a mixed state, not a pure state. I think this must be rigorous, but if the proof is really this simple why can't I find it published anywhere…? Maybe I'm overthinking this, but confirmation that this reasoning is correct (or corrections to it) would be much appreciated.
Best Answer
The (statistical) density matrix describing a system is defined as $$\rho = \sum_ip_i\left|\psi_i\right\rangle\left\langle\psi_i\right|$$ where $p_i$ is the probability that the system is in state $\left|\psi_i\right\rangle$. Notice that it is not necessary for $\left|\psi_i\right\rangle$ to be an eigenstate of the Hamiltonian; they can be a superposition of eigenstates. Hence, in general, one assumes that $\left|\psi_i\right\rangle$ are normalized but not orthogonal.
If the system is in a pure state, all $p_i=0$ except one value. Otherwise, it is in a mixed state, i.e. there exists at least two nonzero values of $p_i$.
The inequality in your solution can be proved using Schwarz inequality $$\sum_{i,j}p_ip_j|\left\langle\psi_i|\psi_j\right\rangle|^2\le\sum_{i,j}p_ip_j\left\langle\psi_i|\psi_i\right\rangle\left\langle\psi_j|\psi_j\right\rangle = \sum_ip_i\sum_jp_j=1.$$ The equality holds only when $\left|\psi_j\right\rangle=\left|\psi_j\right\rangle$ for all $i$ and $j$, which means that there is no summation, or the system is in a pure state. Therefore, for mixed states, $\text{Tr}(\rho^2)<1$. Notice that in this derivation, one only employed the normalization of $\left|\psi_j\right\rangle$, not the orthogonality property.