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Let $(\mathcal A, \tau)$ be a $C^*$-dynamical system and let $\omega$ be a state on $\mathcal A$.

Is the following true, and if yes, why?

$$ |\omega(\tau_t(A)| \le \|A\| \quad \forall t \in \mathbb R, A \in \mathcal A $$

I have a hunch why it should be true for a concrete example.

If the dynamics is given by a Hamiltonian $H$, that is

$$ \tau_t(A) := e^{itH} A e^{-itH} \,,$$

then the result follows from the formula

$$ \omega(B^*AB) \le \omega(B^*B)\|A\| $$

replacing $B$ with $e^{-itH}$.

However, even in this concrete example, I am not sure if $e^{-itH}$ belongs to the algebra $\mathcal A$, given that the Hamiltonian $H$ is already in the algebra. I think it has to do with the fact that $\mathcal A$ is a Banach space by definition. Therefore, the convergent exponential series should have a limit point in the algebra. Right?

Nevertheless, this is just a hunch with a concrete example. I would like to know the more general theorem stated above.

Nanashi No Gombe
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1 Answers1

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A state by definition has norm $1$, and $\tau_t$ is an automorphism for all $t$, hence isometric (any injective $*$-homomorphism between $C^*$-algebras is isometric, see this link). Thus we have $$|\omega\tau_tA|\leq\|\omega\|\|\tau_tA\|=\|\tau_tA\|=\|A\|.$$

Aweygan
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  • I think the first inequality should read: $|\omega\tau_tA|^2 \leq |\omega| \omega ( \tau_t (A^A) ).$ How do you see that $\omega ( \tau_t (A^A) ) = |\tau_tA|^2,?$ – Nanashi No Gombe Jun 24 '17 at 22:21
  • $\omega$ is a bounded linear functional, hence $|\omega A|\leq|\omega||A|$ for all $A$. Now replace $A$ with $\tau_t A$. – Aweygan Jun 24 '17 at 22:25
  • How do you see that $|\omega A|\leq|\omega||A|$? From the Cauchy-Schwarz inequality in a $C^∗$-algebra, one can only conclude that $|\omega A|^2 \leq |\omega| \omega ( A^A ).$ Note that only for a specific pure* state, it holds that $\omega ( A^*A ) = |A|^2.$ – Nanashi No Gombe Jun 24 '17 at 22:33
  • This has nothing to do with $\omega$ being positve, just that it is a bounded linear functional. Remember back from functional analysis. – Aweygan Jun 24 '17 at 22:36
  • Okay, if I take your word for it, I am safe to use intuition from functional analysis. So just to be clear, does this mean the equality $\omega ( A^A ) = |A|^2$ is always* true (for consistency, of course), irrespective of whether or not the state is pure? But my book has lemma on exactly when this equality holds (depending on whether or not the state is pure). What is going on? Thank you very much. – Nanashi No Gombe Jun 24 '17 at 22:44
  • No. For a state we would have $\omega(A^A)\leq|\omega||A^A|=|A^*A|=|A|^2$, but equality cannot be proven. – Aweygan Jun 24 '17 at 22:46
  • Clear. Thanks a lot. – Nanashi No Gombe Jun 24 '17 at 22:48
  • You're welcome, glad to help. – Aweygan Jun 24 '17 at 22:48