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

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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?

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.