Density matrix $\rho = I/2$ implies an ensemble of orthonormal states

Question

Suppose that a density matrix $\rho = I/2$ is obtained as a description of an ensemble of two pure states. How can I show that the ensemble must then be of the form: $$ \{(|\psi\rangle, 1/2), (|\psi^\perp\rangle, 1/2)\} $$ where $|\psi\rangle$ is a pure state and $|\psi^\perp\rangle$ is a pure state orthogonal to $|\psi\rangle$?

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I've tried considering any ensemble $$ \mathsf E = \{(|\psi_1\rangle, p_1), \cdots, (|\psi_n\rangle, p_n)\} $$ and then $$ \rho = \sum_i p_i |\psi_i\rangle \langle \psi_i | $$ Hence $$ \begin{aligned} \langle\phi|\rho |\phi\rangle &= \sum_i p_i \langle \phi|\psi_i\rangle \langle\psi_i | \phi\rangle \\ &= \sum_i p_i |\langle \phi|\psi_i\rangle|^2 \end{aligned} $$ but can't work out where to go from here (or if I should be here in the first place). I only have a very basic understanding of density matrices so would appreciate an elementary explanation. Thanks!

Answer 1

As stated at the beginning of your post let us assume there exist state vectors $\psi_1,\psi_2\in\mathbb C^2$ (i.e. $\langle\psi_1|\psi_1\rangle=\langle\psi_2|\psi_2\rangle=1$) and $p_1,p_2\in[0,1]$ such that $$ \tfrac{I}2=p_1|\psi_1\rangle\langle\psi_1|+p_2|\psi_2\rangle\langle\psi_2|\,.\tag1 $$ Our goal is to show that $p_1=p_2=\frac12$ and $\langle\psi_1|\psi_2\rangle=0$, and we shall do so in two steps.

Step 1: $\langle\psi_1|\psi_2\rangle=0$

Taking the trace of (1) yields \begin{align*} 1={\rm tr}(\tfrac I2)&=p_1{\rm tr}(|\psi_1\rangle\langle\psi_1|)+p_2{\rm tr}(|\psi_2\rangle\langle\psi_2|)\\ &=p_1\langle\psi_1|\psi_1\rangle+p_2\langle\psi_2|\psi_2\rangle\\ &=p_1+p_2\,.\tag2 \end{align*} On the other hand \begin{align*} \tfrac12\langle\psi_1|\psi_2\rangle&=\langle\psi_1|\tfrac I2|\psi_2\rangle\\ &\overset{\text{(1)}}=\big\langle\psi_1\big|\; p_1|\psi_1\rangle\langle\psi_1|+p_2|\psi_2\rangle\langle\psi_2| \;\big|\psi_2\big\rangle\\ &=p_1\langle\psi_1|\psi_1\rangle\langle\psi_1|\psi_2\rangle+p_2\langle\psi_1|\psi_2\rangle\langle\psi_2|\psi_2\rangle\\ &=(p_1+p_2)\langle\psi_1|\psi_2\rangle\\ &\overset{\text{(2)}}=\langle\psi_1|\psi_2\rangle \end{align*} But $\tfrac12\langle\psi_1|\psi_2\rangle=\langle\psi_1|\psi_2\rangle$ is only possible if $\langle\psi_1|\psi_2\rangle=0$, as desired.

Step 2: $p_1=p_2=\frac12$

This is now easy because we can just take expectation values of (1): $$ \tfrac12=\tfrac12\langle\psi_1|\psi_1\rangle=\langle\psi_1|\tfrac{I}2|\psi_1\rangle\overset{\text{(1)}}=p_1+p_2|\langle\psi_1|\psi_2\rangle|^2\overset{\text{Step 1}}=p_1 $$ and similarly for $p_2$.