Find an ensemble consisting of three distinct pure states that correspond to the same mixed state

Question

I am trying to solve part b) of the following problem. I solved part a here e Compute a quantum state $\rho$ corresponding to an ensemble

Consider the ensemble consisting of the qubit states $|0⟩⟨0|$ and $|1⟩⟨1|$ occuring with probabilities 2/3 and 1/3, respectively. (a) Compute the quantum state $\rho$ corresponding to this ensemble. Is $ρ $ pure or mixed?

(b) Find an ensemble consisting of three distinct pure states (with non-zero probabilities) that corresponds to the same state $\rho$. Why can such an ensemble not correspond to an eigen-decomposition?

My try: I am not sure if the $\rho$ in part b is the same as in part a. I will assume so since the aim of this problem is to see that a linear combination of pure states can be mixed.

I am also not sure if I understand the question correctly. Does " an ensemble consisting of three distinct pure states taht correspond to the same state" mean the following:?

Let $|\psi_1⟩=(a_1,b_1)^T$, $|\psi_2⟩=(a_2,b_2)^T$,$|\psi_3⟩=(a_3,b_3)^T$ all unit vectors

Then the quantum ensamble would be $|\psi_1⟩⟨\psi_1|, |\psi_2⟩⟨\psi_2|$ and $|\psi_3⟩⟨\psi_3|$ with probabilities $p_1,p_2 and p_3$, such that $p_1+p_2+p_3=1$

I think the quantum state would be the linear combination

$\tilde\rho:=p_1|\psi_1⟩⟨\psi_1|+p_2|\psi_2⟩⟨\psi_2|+p_3|\psi_3⟩⟨\psi_3|$

$=p_1\pmatrix{a_1^2 & a_1b_1 \\ a_1b_1 & b_1^2}+p_2\pmatrix{a_2^2 & a_2b_2 \\ a_2b_2 & b_2^2}+p_3\pmatrix{a_3^2 & a_3b_3 \\ a_3b_3 & b_3^2}=\rho=\pmatrix{2/3 & 0 \\ 0 & 1/3} $

Moreover these matrices have to be hermitian Positive semidefinite with trace=1, since that is states are defined are hermitian operators that are Positive semidefinite with trace=1.. All entries in the matrices are complex numbers and the probabilities are real numbers >0

Then I equated each of the 4 entries of the equation and got 4 equations. I guess I need to assign values such that all restrictions hold, but I haven't been able to do so. How can I complete this? And why wouldn't this ensamble correspond to an eigen decomposition?

Answer 1

Assuming $\rho$ is a single-qubit state. Let $\rho=\sum_{i=1}^2 p_i |i\rangle\!\langle i|$ be the eigendecomposition of $\rho$.

Obviously, a decomposition of $\rho$ as a convex combination of three pure states cannot be an eigendecomposition, as an eigendecomposition is a convex combination of two such states.

Then note that $\rho - p_1 |1\rangle\!\langle 1|= p_2 |2\rangle\!\langle 2|$ is a pure state, and therefore cannot be written as a nontrivial convex combination of two other pure states. This shows that you cannot decompose a qubit state as a convex decomposition of 3 (different) pure states one of which is an eigenstate.

You could have such a convex decomposition one element of which is an eigenstate though, assuming the weight is strictly smaller than the corresponding eigenvalue. For example, we could write $$\rho = \begin{pmatrix}2/3&0\\0&1/3\end{pmatrix} = \frac13\mathbb{P}_0 + \frac13\mathbb{P}_++ \frac13\mathbb{P}_-,$$ where I used the shorthand notation $\mathbb{P}_\psi\equiv |\psi\rangle\!\langle \psi|$.

The most general approach to this kind of question is to notice it's essentially the same as asking about purifications of the state. You can observe in general that writing convex decompositions of the form $\rho=\sum_i \lambda_i \mathbb{P}(u_i)$ is the same as a matrix decomposition of the form $\rho=U\Lambda U^\dagger$. From general considerations, you can conclude that all such decomposition must have the form $\rho=(\sqrt\rho V^\dagger)(\sqrt\rho V^\dagger)^\dagger$ for some isometry $V$. This connects with the question at hand because it amounts to saying that all (convex) decompositions of $\rho$ in terms of rank-1 operators can be written as sums of $a_i a_i^\dagger$ with $a_i$ the columns of such $\sqrt\rho V^\dagger$.

For example, observe that $$\sqrt\rho \underbrace{\frac1{\sqrt3}\begin{pmatrix}1&1&1\\1&\omega_3&\omega_3^2\end{pmatrix}}_{\equiv V^\dagger} = \frac13\begin{pmatrix}\sqrt2 & \sqrt2 & \sqrt2 \\ 1 & \omega_3 & \omega_3^2\end{pmatrix},$$ where $\omega_3\equiv \exp(2\pi i/3)$. This tells you that another valid decomposition is $$\begin{pmatrix}2/3&0\\0&1/3\end{pmatrix} = \frac13 \mathbb{P}\left(\frac{\sqrt2|0\rangle+|1\rangle}{\sqrt3}\right) + \frac13 \mathbb{P}\left(\frac{\sqrt2|0\rangle+\omega_3 |1\rangle}{\sqrt3}\right) + \frac13 \mathbb{P}\left(\frac{\sqrt2|0\rangle+\omega_3^2 |1\rangle}{\sqrt3}\right).$$ Of course, you could have got this kind of decomposition from simply reasoning geometrically in the Bloch sphere: it amounts to taking three points on the Bloch sphere, all at the same height (i.e. $\sigma_z$ coordinate), placed on a regular triangle, so that their balanced mixture gives a point on the N-S axis of the sphere.

Note that while this procedure always works, it doesn't always provide nontrivially different elements. For example, this cannot happen if the initial state $\rho$ is pure. To see it explicitly, let's try the same recipe with $\rho=\mathbb{P}_0$. We get $$\begin{pmatrix}1&0\\0&0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix} \frac1{\sqrt3}\begin{pmatrix}1&1&1\\1&\omega_3&\omega_3^2\end{pmatrix}= \frac{1}{\sqrt3}\begin{pmatrix}1&1&1\\0&0&0\end{pmatrix},$$ which corresponds to the "decomposition" $\mathbb{P}_0 = \frac13 \mathbb{P}_0+\frac13 \mathbb{P}_0 + \frac13 \mathbb{P}_0.$

We could have also got the decomposition I showed in the first equation with this procedure using instead $$V^\dagger = \frac12\begin{pmatrix}\sqrt2 & 1 & 1 \\ 0 & \sqrt2 & -\sqrt2\end{pmatrix}.$$

Some related posts on qc.SE:

• How can pure state ensemble decompositions not be unique?
• Are the pure quantum states that make up a mixed state orthogonal?