Time evolution by block diagonal unitary

Question

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.

Answer 1

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 $U$.