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
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.
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 $|v\rangle = |w\rangle = |\psi\rangle$.
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.
