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.

  • $\begingroup$ A pure state has a density matrix $\rho = \vert \rangle \langle \vert$, which has $\rho ^2=\rho$ and trace $=1$. Am I missing something? $\endgroup$
    – pppqqq
    Nov 10, 2016 at 22:07
  • $\begingroup$ The point is to check that trace($\rho^{2}$) < 1 for mixed states that sum over a set of non-orthonormal quantum states. I understand the pure state case. $\endgroup$
    – miggle
    Nov 10, 2016 at 22:56
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    $\begingroup$ A Hermitian matrix is guaranteed to have a decomposition into orthogonal eigenvectors. Just use that decomposition instead of the non-orthogonal one. $\endgroup$ Nov 11, 2016 at 2:29
  • $\begingroup$ Umm OK, I'll need to think about that a little. I can hit $\rho^{2}$ with unitaries to make it diagonal but I need to think about what the coefficients in the expansion of $\rho^{2}$ will look like. $\endgroup$
    – miggle
    Nov 11, 2016 at 4:34
  • $\begingroup$ @CraigGidney, could you write something a little more detailed? I understand your comment, but I can't see how I can relate the expansion coefficients in the basis where $\rho^{2}$ is diagonal to the original $p_{i}$ which obey the nice normalization condition. $\endgroup$
    – miggle
    Nov 11, 2016 at 5:58

2 Answers 2


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.


Pure States : Consider an ensemble of given objects in the states $\{ |\psi_i\rangle \}$. If all the objects are in the same state, the ensemble is represented by a pure state. To make probabilistic statements the whole ensemble of identically prepared systems must be considered. Let the system be in state $|\psi\rangle$, then the density matrix is given by $$\rho = |\psi\rangle\langle\psi|$$ $$\rho^2 = |\psi\rangle\langle\psi|\psi\rangle\langle\psi| = |\psi\rangle\langle\psi| = \rho$$ Taking trace both sides: $$Tr(\rho^2)=Tr(\rho)=1$$ Mixed States : Let us next study the situation where not all of the N systems (objects) of the ensemble are in the same state, i.e. Ni systems are in the state $|\psi_i\rangle$ respectively, such that 􏰘 $N_i = N$. The probability pi to find an individual system of the ensemble described by the state $|\psi_i\rangle$ is then given by: $$p_i = \frac{N_i}{N}\quad \mbox{where} \quad \sum_ip_i = 1$$ Therefore we can write mixed state in terms of weighted sum of pure states: $$\rho_{mix} = \sum_ip_i\rho_i^{pure} = \sum_ip_i|\psi_i\rangle\langle\psi_i|$$ $$\rho^2 = \sum_ip_i|\psi_i\rangle\langle\psi_i| \sum_jp_j|\psi_j\rangle\langle\psi_j|$$ $$\rho^2=\sum_{i,j}p_ip_j|\psi_i\rangle\langle\psi_i|\psi_j\rangle\langle\psi_j|=\sum_{i,j}p_ip_j|\psi_i\rangle\langle\psi_j|\delta_{ij} = \sum_ip_i^2|\psi_i\rangle\langle\psi_i|\neq \rho$$ Since $p_i^2 < p_i$, therefore $Tr(\rho^2) < Tr(\rho)$

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    $\begingroup$ This is not an answer to the question. $\endgroup$ Aug 24, 2018 at 13:09

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