# How to show that a density matrix $\rho$ is extreme iff $\rho=|\psi\rangle\!\langle\psi|$?

A density matrix $$ρ$$ is called extreme if the only way to write $$ρ$$ as $$ρ = p σ + (1 − p) τ$$ , with $$σ ∈ S_d$$, $$τ ∈ S_d$$, and $$p ∈ (0, 1)$$ is to have $$ρ = σ = τ$$ . I want to show that a density matrix is extreme if and only if it is of the form $$ρ = |ψ\rangle\langleψ|$$.

How would I show this? I haven't made much progress on either of the directions unfortunately

• – glS
Mar 6, 2022 at 10:48

Here's a sketch which you can fill in the details of.

The spectral theorem says that for any density matrix $$\rho$$ there exists an orthonormal basis $$\{|v_i\rangle\}_i$$ and $$\lambda_i \geq 0$$ with $$\sum_i \lambda_i = 1$$ such that
$$\rho = \sum_i \lambda_i |v_i \rangle \langle v_i |.$$ You can interpret this as saying that $$\rho$$ can represent the experiment where we prepare the state $$|v_i \rangle$$ with probability $$\lambda_i$$.

So if rho is extreme, this means that we can only have a single nonzero coefficient in its spectral decomposition. I.e., $$\rho = |v_i \rangle \langle v_i|$$ for some $$|v_i\rangle$$. This gives us one direction immediately.

For the converse if $$\rho= |\psi\rangle \langle \psi|$$ then $$\rho$$ is a rank one positive semidefinite matrix. If we have two positive semidefinite operators $$\sigma$$ and $$\tau$$ then any convex combination of them cannot decrease their ranks i.e., we have $$\mathrm{rank}(p \tau + (1-p) \sigma) \geq \max\{\mathrm{rank}(\tau), \mathrm{rank}(\sigma)\}.$$ Thus we need $$\tau$$ and $$\sigma$$ to both be rank one, so $$\tau=|v\rangle \langle v|$$ and $$\sigma = |w \rangle \langle w|$$. You can then show that if $$|v\rangle \neq |w\rangle$$ then $$\rho$$ must have rank 2. Hence we need $$|v\rangle = |w\rangle = |\psi\rangle$$.

Another easy method for me. If $$\rho=|\psi\rangle\langle\psi|$$, then we have $$Tr\left( \rho ^2 \right) =1=Tr\left( \left( p\sigma +\left( 1-p \right) \tau \right) ^2 \right) \\ =Tr\left( p^2\sigma ^2+\left( 1-p \right) ^2\tau ^2+p\left( 1-p \right) \left( \sigma \tau +\tau \sigma \right) \right) \\ =p^2Tr\left( \sigma ^2 \right) +\left( 1-p \right) ^2Tr\left( \tau ^2 \right) +p\left( 1-p \right) Tr\left( \sigma \tau +\tau \sigma \right) \\ \le p^2+\left( 1-p \right) ^2+2p\left( 1-p \right) =1$$

where $$Tr\left( \sigma \tau +\tau \sigma \right) \le 2$$ can be shown by mind that $$Tr\left( \sigma \tau +\tau \sigma \right) \\ =2\mathrm{Re}Tr\left( \sigma \tau \right) \\ =2\mathrm{Re}Tr\left( \sum_i{p_i|i\rangle \langle i|\sum_j{q_j|j\rangle \langle j|}} \right) \\ =2\sum_{ij}{p_iq_j\left| \langle i|j\rangle \right|^2} \\ \le 2$$ So we must have $$\tau=\sigma=|\psi\rangle\langle\psi|=\rho$$.

Another direction: if $$\rho$$ can not be written as $$\rho =p\sigma +\left( 1-p \right) \tau$$, then it must be pure state, we can transfer this into if $$\rho$$ is a mixed state, then it must can be written into $$\rho =p\sigma +\left( 1-p \right) \tau$$. Which can be shown by deduction while I will only give an example to show this. If $$\rho =p_1|1\rangle \langle 1|+p_2|2\rangle \langle 2|+p_3|3\rangle \langle 3|$$, then we have $$\rho =p_1|1\rangle \langle 1|+\left( 1-p_1 \right) \left( \frac{p_2}{1-p_1}|2\rangle \langle 2|+\frac{p_3}{1-p_1}|3\rangle \langle 3| \right)$$, and take $$\sigma =|1\rangle \langle 1|$$ and $$\tau =\frac{p_2}{1-p_1}|2\rangle \langle 2|+\frac{p_3}{1-p_1}|3\rangle \langle 3|$$, we finish the proof.

It seems straight forward, since when the equality holds, so in $$p$$ probability $$\rho$$ is happening, and in $$(1-p)$$ there is also $$\rho$$ happening, so $$p$$ probability is meaningless (same density matrix in both cases), and your state is not mixed state (not 2 pure that happen in some probability) but a pure state, which from the definition of density matrix, is exactly $$|\psi\rangle\langle\psi|$$

Just reminding definition of state matrix, is a sum over all density matrices of the pure states that are composing the mixed state, each one is multiplyed by its probabilty.

• Thanks! Any idea how to solve the reverse direction? Mar 6, 2022 at 0:20