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 $\rho$?

  • 2
    $\begingroup$ 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 $\vert 1\rangle$. $\endgroup$ – user1936752 Dec 23 '18 at 14:18
  • $\begingroup$ Thanks, @user1936752. What about $\rho_{\psi}$ which I mentioned in my Edit? $\endgroup$ – Zilch Dec 23 '18 at 17:01
  • $\begingroup$ 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 $\rho$. $\endgroup$ – user1936752 Dec 23 '18 at 20:12

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

In case

$$ p=1/2 $$


$$ |x|^2=1 $$

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

In this case you can check that

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


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

For all the other values the matrix has rank 2.

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  • $\begingroup$ Thanks, @Icv. I made some edit above. Your comments, please. $\endgroup$ – Zilch Dec 23 '18 at 16:40
  • $\begingroup$ @Zilch What is the question now? $\endgroup$ – lcv Dec 23 '18 at 17:13
  • $\begingroup$ @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). $\endgroup$ – lcv Dec 23 '18 at 17:17

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).

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  • $\begingroup$ Thanks, @tparker. I just added $\rho_{\psi}$, which is a valid pure state, resembles $\rho$. Your comments, please. $\endgroup$ – Zilch Dec 23 '18 at 17:03
  • $\begingroup$ @Zilch Edited.. $\endgroup$ – tparker Dec 23 '18 at 17:50

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