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Let $M\subset B(H)$ be a von Neumann algebra, with cyclic separating vector $\xi$.

Then the modular conjugation operator $S$ is defined to be the closure of the operator $$S_{0}:M\xi\to M\xi\text{ defined by }S_{0}(x\xi) = x^{*}\xi$$

Then the modular operator is defined by $\Delta := S^{*}S$.

Now my question is this:

For a real number $t$, how do we make sense of the expression $\Delta^{it}$?

If we were dealing with bounded operators on a Hilbert space, we could use the Analytic, Continuous, or Borel Functional Calculus. But since $S$ need not be bounded, it seems we cannot expect $\Delta$ to be.

So what tools are available to assign meaning to $\Delta^{it}$?

roo
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  • I'm not sure if this is correct, but instinctively my gut says to use the equation $\langle\Delta^{it}x,x\rangle = \langle\Delta x,x\rangle^{it}$.

    Since a bounded operator $T$ on a Hilbert space $H$ is completely determined by the behaviour of $\langle Tx,x\rangle$ for all $x\in H$, I see no reason not to expect this to apply to unbounded operators as well?

     – roo Sep 03 '15 at 21:36

1 Answers1

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Positive unbounded operators satisfy the Spectral Theorem, so Borel functional calculus applies to them.

Martin Argerami
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  • Is it the same definition then? $\Delta^{it} = \int_{\sigma(\Delta)}\lambda^{it}E(d\lambda)$, where $E$ is the spectral measure of $\Delta$? – roo Sep 04 '15 at 16:53
  • I can't find an exact reference, but when you say "satisfy the spectral theorem", do you just mean that they have spectral measures? – roo Sep 04 '15 at 16:53
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    Yes, that there exists $E$ such that $\Delta \xi=\int_{\sigma(\Delta)}\lambda,dE(\lambda)\xi$. for every $\xi$ in the domain of $\Delta$. – Martin Argerami Sep 04 '15 at 21:03