Consider a $C_0$-semigroup $S_t:\mathscr{B(H)} \to \mathscr{B(H)}$ with generator $U$. Now define $P_t:\mathscr{B_1(H)} \to \mathscr{B_1(H)}$ where $P_t(\rho)=S_t\rho S_t^*$. How can I prove $P_t$ to be $C_0$-semigroup and the generator of $P_t$ is given by $A$, where $A(\rho)=U\rho+\rho U^*$?
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3$\begingroup$ This doesn't look like a research question. Bratteli and Robinson is my favorite source for the basic theory of semigroups of operators. $\endgroup$– Nik WeaverJan 29, 2013 at 11:31
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1$\begingroup$ What is ${\mathcal B}_1({\mathcal H})$? $\endgroup$– Delio MugnoloJan 29, 2013 at 16:30
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1$\begingroup$ If U is bounded, this is rather obvious. However, if U is unbounded, there is in general an issue of domains. It is easy to come up with example where $U\rho$ is an unbounded operator, but $A(\rho)$ is bounded. Hence the generator of $P_t$ will in general be an extension of $A$. $\endgroup$– Michael RenardyJan 29, 2013 at 16:45
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The semigroups you construct is in general only weak-* continuous. Are you looking for so-called implemented semigroups?
See for example
Alber, Jochen, On implemented semigroups, Semigroup Forum 63, No. 3, 371-386 (2001). ZBL1041.47028.
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$\begingroup$ Andras, I do not understand. Would this semigroup be in general only weak*-continuous even if $\mathcal H$ is a Hilbert space? I thought this may happen only if the semigroup acts on a non-reflexive Banach space. $\endgroup$ Jan 30, 2013 at 20:53
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1$\begingroup$ $\mathcal B(\mathcal H)$ is in general a non-reflexive Banach space... $\endgroup$ Jan 31, 2013 at 13:04
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1$\begingroup$ oh. such a stupid mistake. of course, you are right. $\endgroup$ Feb 1, 2013 at 0:49