# Rank of a density matix

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 $$\rho$$?

• The general state is not a pure state - it is a mixed state which is a classical mixture of pure states. If $x=x^*=0$ in your example (you can make this happen by diagonalizing $\rho$) and the basis you have chosen is the computational basis, it means you have with probability $(1-p)$ the state $\vert 0\rangle$ and with probability $p$ the state $\vert 1\rangle$. – user1936752 Dec 23 '18 at 14:18
• Thanks, @user1936752. What about $\rho_{\psi}$ which I mentioned in my Edit? – Zilch Dec 23 '18 at 17:01
• Try diagonalizing it with $\alpha = \cos\theta$ and $\beta =\sin\theta$ (whose squares sum to 1) and you will find that one of the eigenvalues is zero. From this, you see that $\rho_\psi$ is not as general as $\rho$. – user1936752 Dec 23 '18 at 20:12

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.

• @Zilch You will notice that $\rho$ is more general than $\rho_\psi$. In the latter case there is a relation between $x, p$, and $1-p$. (I am not sure if this goes in the direction you are asking). – lcv Dec 23 '18 at 17:17
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).
• Thanks, @tparker. I just added $\rho_{\psi}$, which is a valid pure state, resembles $\rho$. Your comments, please. – Zilch Dec 23 '18 at 17:03