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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.

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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, $$ S_n(\rho(t))=S_n(\rho). $$

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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.

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