Rank of a density matix

Question

I was just trying to understand the meaning of rank of a density matrix. I came across the following post, which says that the rank of density matrix is the number of non-zero eigenvalues. And for a pure state is always one. However, I fail to understand that for a two-level system, the general state is given by

$$\rho = \begin{pmatrix} 1-p & x\\ x^* & p \end{pmatrix}. $$

Which represents a pure state and has two eigenvalues ( not one ). So does it mean that this matrix has rank two?

Edit: The state $|\psi> = \alpha |0> + \beta |1>$ is a pure state, with the density matrix

$$\rho_{\psi} = \begin{pmatrix} |\alpha|^2 & \alpha \beta^*\\ \alpha^* \beta & |\beta|^2 \end{pmatrix}. $$

with $|\alpha|^2 + |\beta|^2 = 1$. This case is similar to $\rho$?

Answer 1

An $n\times n$ hermitian matrix has always $n$ eigenvalues (counted with multiplicity).

In case

$$ p=1/2 $$

and

$$ |x|^2=1 $$

the eigenvalues become $0,1$, the matrix $\rho$ has rank one and represents a pure state.

In this case you can check that

$$ \rho = |\psi \rangle \langle \psi | $$

with

$$ |\psi \rangle = \frac{1}{\sqrt{2}} ( |0 \rangle + x |1 \rangle ). $$

For all the other values the matrix has rank 2.

Answer 2

Your "general state" is mixed, not pure.

In response to your edit: the difference is that in the general state, the complex number $x$ is arbitrary and independent of the real number $p$. But in the pure state, its norm is constrained to equal $\sqrt{p(1-p)}$. This additional constraint means that the pure state only has two real degrees of freedom: $p$ and the phase of $x$ (corresponding to the polar and azimuthal coordinates of the Bloch sphere, respectively). While the general state has three: $p$ and the phase and magnitude of $x$ (corresponding to the polar, azimuthal, and radial coordinates of the Bloch ball, respectively).