Time evolution by block diagonal unitary I think this might be a fairly "trivial" question.
Given a block diagonal unitary $$\hat{U} = \sum_{\alpha=0}^N \hat{U}_\alpha$$ (where $\hat{U}_\alpha$ is a matrix of all zeros apart from a $k_\alpha\times k_\alpha$ unitary subspace),
we can say $\hat{U}_\alpha = \hat{U}_\alpha \Pi_\alpha$, where $\Pi_\alpha$ is a projection operator onto the $\alpha$-th subspace.
My question is this: if we evolve any density operator through $\hat{U}$ is it correct to assume that the final state will be always block diagonal?
My reasoning is quite simple: given a density operator $$\sigma^0 = \sum_{\eta=0}^dp_\eta |{\phi_\eta}\rangle \langle{\phi'_\eta}|$$ evolving it through $\hat{U}$ means
$$\sigma^f = \hat{U}\sigma^0\hat{U}^\dagger = \sum_{\alpha=0}^N \hat{U}_\alpha \Pi_\alpha\sigma^0\Pi_\alpha\hat{U}_\alpha^\dagger = \sum_{\alpha=0}^N \hat{U}_\alpha \sigma_\alpha^0 \hat{U}_\alpha^\dagger  $$
where $\sum_\alpha^N \sigma^0_\alpha$ is a block diagonal operator comprised of non-normalized density operators $\sigma_\alpha$. Henceforth $\sigma^f$ will be block diagonal.
 A: This is not true. The easiest way to see this is to find a counter example, so if
$$
U = \left(\begin{array}{cccc} 1 &0&0&0\\
0 &\cos\phi & \sin \phi & 0 \\
0 &-\sin\phi & \cos \phi & 0 \\
0&0&0&1
\end{array}\right)
$$
and
$$
\rho = \left(\begin{array}{cccc}
a&b&0&0\\
b^* & c &0&0\\
0&0&d&e\\
0&0&e^*&f
\end{array}\right)
$$
then
$$
 U \rho U^\dagger = \left(\begin{array}{cccc}
a & b\cos\phi & -b\sin\phi&0\\
b^*\cos\phi & c\cos^2\phi+d\sin^2\phi& (d-c)\sin\phi\cos\phi&e\sin\phi\\
-b^*\sin\phi& (d-c)\sin\phi\cos\phi& c\sin^2\phi+d\cos^2\phi&e\cos\phi\\
0&e^*\sin\phi&e^*\cos\phi&f
\end{array}\right)
$$
which is clearly not block diagonal.
The mistake in your reasoning is that
$$
U\sigma^0U^\dagger = \sum_\alpha \sum_\beta U_\alpha\Pi_\alpha \sigma^0 \Pi_\beta U_\beta^\dagger \ne \sum_\alpha U_\alpha\Pi_\alpha \sigma^0 \Pi_\alpha U_\alpha^\dagger.
$$
The only case where that step would be valid would be if
$$
\Pi_\alpha \sigma^0 \Pi_\beta\propto \delta_{\alpha\beta}
$$
that is in the case that $\sigma^0$ is already block diagonal, with the same blocks as $U$.
