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

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 $$|\psi\rangle$$?

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!

• – glS
Commented Jun 4 at 9:55

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 $$p_2$$.

• Thanks! Why may we assume that there are only two state vectors in the ensemble? Commented Jun 2 at 20:23
• @OllyBritton That's what you ask to assume in your question! Commented Jun 2 at 20:31
• @NorbertSchuch Ah! Yes of course. Commented Jun 2 at 20:41