Can we always find a unitary operation connecting qubit states with given eigendecompositions?

Question

Consider the density matrices $\rho_0 = |0 \rangle \langle 0|$ and $\rho_1 = |1 \rangle \langle 1|$. Let $\{p_1, p_2\}$, and $\{p_3, p_4\}$ be two probability distributions, that is, $$0 \leq p_1, p_2, p_3, p_4 \leq 1$$ $$p_1 + p_2 = 1$$ $$~\text{and}~ p_3 + p_4 = 1.$$ These probability distributions refers two mixed states $\rho = p_1 \rho_0 + p_2 \rho_1$ and $\rho' = p_3 \rho_0 + p_4 \rho_1$. Now I have the following questions:

1. Is there is a unitary matrix $U$ such that $\rho' = U^\dagger \rho U$ ?

2. How to calculate $U$ when all $p_1, p_2, p_3$ and $p_4$ are known, if $U$ exists?

3. Can we represent $U$ with a quantum circuit with qubits if $U$ exists?

Answer 1

Since conjugation of the state $\rho$ with unitaries, i. e. $\rho \mapsto U \rho U^\dagger$, preserves eigenvalues, this could only be possible in your case if $p_1 =p_3$ or $p_1 = p_4$. In those cases the unitary transformation required is the identity and $\sigma_x$ respectively.