The purity of a state $\rho$ is $\newcommand{\tr}{\operatorname{Tr}}\tr(\rho^2)$, which is known to equal $1$ iff $\rho$ is pure.

Another quantity that can be used to quantify how close a state $\rho$ is to be pure is its rank. This equals the number of nonzero eigenvalues in the eigendecomposition of $\rho$, or equivalently, the minimum number of pure states that need to be mixed to obtain $\rho$. Again, $\rho$ is pure iff its matrix rank is $1$.

Consider now two states $\rho$ and $\sigma$. Their having the same rank clearly does not imply that they have the same purity. A trivial example of this is given by $\rho=I/2$ and $$\sigma=(1/2-\epsilon)|0\rangle\!\langle0| + (1/2+\epsilon)|1\rangle\!\langle 1|,$$ for small enough $\epsilon>0$.

It is however less trivial to see whether $\rho$ and $\sigma$ having the same purity implies their having the same rank. If these were generic matrices, it would not be true, but it might hold for unit-trace Hermitian operators. Is this the case?

I can see that it is for two-dimensional matrices: if $\rho$ and $\sigma$ have same purity, that means that $p^2+(1-p)^2=q^2 + (1-q)^2$, with $p$ and $q$ one of the two eigenvalues of $\rho$ and $\sigma$, respectively. But in this case $\rho$ and $\sigma$ having different rank means that one of the two (say $\rho$) has full-rank while the other one has rank $1$. This would mean that $q=1$ (or, equivalently, $q=0$), while $0<p<1$. But then the condition on the purity would become $p^2+(1-p)^2=1$, which clearly cannot be satisfied for $0<p<1$ (a nice geometrical way to see it is that the solutions of this equation are the intersection of the unit circle in 2D and the line $x+y=1$).

What about the more general case?

  • $\begingroup$ Qubit density matrices are (up to basis choice) a one-parameter family. So the state (up to basis choice) is fully determined by that number. This is clearly specific to qubits. Thus, checking such kind of statements only for qubits is bound to fail. $\endgroup$ Jan 3, 2020 at 15:49

2 Answers 2


$$ \rho = \begin{pmatrix} a & 0 & 0\\ 0 & 1-a & 0\\ 0 & 0 & 0\\ \end{pmatrix}\\ \sigma = \begin{pmatrix} b & 0 & 0\\ 0 & c & 0\\ 0 & 0 & 1-b-c\\ \end{pmatrix}\\ Tr (\rho^2 ) = 2 a^2 - 2a + 1\\ Tr (\sigma^2 ) = b^2 + c^2 + (1-b-c)^2 $$

Let's pick $a=1/2$ so $2a^2-2a+1=\frac{1}{2}$.

$$ b=\frac{2}{3}\\ c=\frac{1}{6}\\ 1-b-c=\frac{1}{6}\\ Tr (\sigma^2) = \frac{4}{9} + \frac{1}{36} + \frac{1}{36}\\ = \frac{16}{36} + \frac{1}{36} + \frac{1}{36}\\ = \frac{1}{2}\\ $$

  • $\begingroup$ +1. So is purity simply unrelated to rank? $\endgroup$ Dec 11, 2019 at 5:29

The other answer already gave a counterexample.

From a geometrical point of view, the question is about the intersection of hyperplanes with hyperspheres. Indeed, the purity of a state $\rho$ with eigenvalues $(p_i)_i$ is $\sum_i p_i^2=\|\boldsymbol p\|^2$, whereas these are probabilities and therefore also constrained by $\sum_i p_i=1$.

The set of probabilities for states of purity $C^2$ is therefore given by the intersection between the hypersphere with radius $C$, and the hyperplane $\sum_i p_i=1$ (with the additional constraint that $0\le p_i\le 1$ for all $i$). Let us denote the hyperplane $\sum_i p_i=1$ in $N$ dimensions as $\mathcal P_N$, and with $\mathcal S_N(r)\subseteq\mathbb R^N$ the $(N-1)$-sphere with radius $r$. Fixing $C$, there are states with purity $C$ and rank $r$ whenever $\mathcal S_r(C)$ intersects $\mathcal P_N$. This doesn't happen for all values of $C$ and $r$. For example, for the smallest possible purity, $C^2=1/N$, there is a single point of intersection between $\mathcal P_N$ and $S_N(C)$, and no smaller-dimensional hypersphere of radius $c$ intersects $\mathcal P_N$.

Here is a visualisation of this for $N=3$, showing $\mathcal P_3$ and $\mathcal S_3(2/3)$:

We can see from this that for the purity $C^2=(2/3)^2$, we cannot have states with rank smaller than $3$. Indeed, we can go further and say that states with rank $<3$ are only possible when the purity $C^2$ is greater than or equal to $1/2$ (consistently with AHusain's answer).

Generalising this, we find that states with rank $r$ and purity $C^2$ exist whenever $C^2\ge1/r$.

To get two states with the same purity and different rank in dimension $3$, it is therefore sufficient to choose a purity $C^2\in[1/2,1)$. To find a rank-$2$ probability vector $(p,1-p)$ with this purity we then need to solve $p^2+(1-p)^2=C^2$, which has solutions $2p=1\pm \sqrt{2C^2-1}$. Note how this has solutions in $[0,1]$ only for $C^2\in [1/2,1]$. We then need to only take another point in the intersection $S_3(C)\cap \mathcal P_3$ that does not sit on one the planes $p_1=0,p_2=0,p_3=0$, and we get an example of a rank-3 state with same purity as a rank-2 one.

Generally speaking, I think the important observation is that, while states with the same purity all sit at the same distance from the origin of probability space, the distance from the origin of the hyperplanes $\mathcal P_N$ ranges from $1/\sqrt N$ to $1$. It is therefore clear that if we are looking for states with purity $C^2$, these must have distance $C$ from the origin, and therefore cannot also sit on a plane $\mathcal P_n$ (i.e. correspond to states with rank $n$) unless $C\ge 1/\sqrt n$.


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