# How do Rényi entropies act under unitary time evolution?

I am trying to find information/ help on Rényi entropies given by $$S_n(\rho) = \frac{1}{1-n} \ln [Tr(\rho^n)]$$

and how it acts under unitary time evolution? Is the entropy independent on the state of $$\rho$$ i.e it doesn't matter if $$\rho$$ is pure or mixed? I am also unsure on how to apply Von Neumann's equation $$\rho(t) = U(t, t_0) \rho(t_0) U^{\dagger} (t, t_0)$$ In order to see how it acts.

Let $$\rho(t)=U\rho U^\dagger,$$ just to simplify notation a bit. Now notice that $$\rho(t)^2=U\rho U^\dagger U\rho U^\dagger=U\rho^2 U^\dagger$$ since $$U^\dagger U=I$$. Thus, similarly, $$\rho(t)^n=U\rho^n U^\dagger.$$ So, take the trace of this, remembering that trace is invariant under permutations: $$\text{Tr}(\rho(t)^n)=\text{Tr}(U\rho^n U^\dagger)=\text{Tr}(\rho^n U^\dagger U)=\text{Tr}(\rho^n).$$ Thus, $$S_n(\rho(t))=S_n(\rho).$$
They are invariant under conjugation of unitaries, i.e. under the mapping $$\rho \to U \rho U^*$$. To see this note that $$(U \rho U^*)^{\alpha} = U \rho^\alpha U^*$$. Then we have \begin{aligned} S_\alpha(U \rho U^*) &= \frac{1}{1-\alpha} \log \mathrm{Tr}[(U \rho U^*)^{\alpha}] \\ &= \frac{1}{1-\alpha} \log \mathrm{Tr}[U \rho^\alpha U^*] \\ &= \frac{1}{1-\alpha} \log \mathrm{Tr}[U^* U \rho^{\alpha}] \\ &= \frac{1}{1-\alpha} \log \mathrm{Tr}[\rho^\alpha] \\ &= S_\alpha(\rho). \end{aligned}
On the second line we used the above identity, on the third line we used cyclicity of the trace and on the fourth line we used $$U^* U = I$$ as $$U$$ is unitary.