# Can $U(\rho_1\otimes\rho_2)U^\dagger$ be entangled if either of $\rho_1$ or $\rho_2$ is a maximally mixed state?

Given two qubit states $$\rho_1$$ and $$\rho_2$$. By applying some unitary $$U$$ we get $$\rho = U(\rho_1 \otimes \rho_2)U^\dagger$$. Can $$\rho$$ be entangled if either of $$\rho_1$$ or $$\rho_2$$ is a maximally mixed state?

To see the equivalence, observe that if $$\rho_2=I/d$$ ($$d$$ is here the dimension of the space) then $$\rho=\frac1 d\sum_{k=1}^d U(\rho_1\otimes|k\rangle\langle k|)U^\dagger,$$ and if we don't put further restrictions on $$U$$ and $$\rho_1$$, then the only thing we know about $$U(\rho_1\otimes |k\rangle\langle k|)U^\dagger$$ is that they are orthogonal states (and if we are to hope the mixture is entangled, the states cannot be all separable).
This is not the case. As a counterexample in the two-qubit scenario, consider $$\sqrt2|\Phi_+\rangle \equiv |00\rangle + |11\rangle, \qquad \sqrt3|\Psi\rangle \equiv |00\rangle + |10\rangle - |11\rangle.$$ You can then verify that $$(|\Phi_+\rangle\!\langle\Phi_+|+|\Psi\rangle\!\langle\Psi|)/2$$ is still entangled (using e.g. the PPT criterion).
Translated into the notation of the question, this example would correspond to $$\rho_1=|0\rangle\langle 0|$$ (or any other pure state), and $$U$$ such that $$U|0,0\rangle=|\Phi_+\rangle$$ and $$U|0,1\rangle=|\Psi\rangle$$.