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Consider a unital $C^*$-algebra $\mathfrak A$ and a state $\omega$ on it. Let $\mathcal N := \{A\in\mathfrak A : \omega(A^*A)=0 \}$. Construct a Hilbert space $\mathcal H:= \overline{\mathfrak A /\mathcal N}$ with the inner product defined by: $ \langle \psi_A,\psi_B \rangle \equiv \omega(A^*B) $ where $\psi_A, \psi_B \in \mathcal H$. Construct further a GNS representation $\pi:\mathfrak A \to \mathcal L(\mathcal H)$ defined by: $\pi(A)\psi_B\equiv\psi_{AB}.$ The vacuum state is defined as $\Omega := \psi_1$ where $1 \in \mathfrak A$ is the unit.

Given two such GNS representations $(\mathcal H_i,\pi_i,\Omega_i)$ with $i=1,2$ of $\mathfrak A$, I want to prove that the map $U: \mathcal H_1 \to \mathcal H_2$ defined by $U\pi_1(A)\Omega_1:=\pi_2(A)\Omega_2 \ \forall A \in \mathfrak A$ is unitary. Note that $U\pi_1(A)\Omega_1=\pi_2(A)\Omega_2$ is the same as saying $U\psi^{(1)}_A = \psi^{(2)}_A$ where $\psi^{(1)}_A \in \mathcal H_1$ and $\psi^{(2)}_A \in \mathcal H_2 \, .$

If I could only prove that $\psi^{(1)}_A = U^*\psi^{(2)}_A,$ my job is done. How do I do that?

How would you prove the unitary equivalence of these representations?

Kindly advise.


NOTE:

Plugging $A=1$ in $U\psi^{(1)}_A = \psi^{(2)}_A$ implies that $\Omega_2 = U \Omega_1$ which in turn implies from the original condition that $(U\pi_1(A)-\pi_2(A)U)\Omega_1=0 \Rightarrow U\pi_1(A)=\pi_2(A)U$. Then my question becomes very similar to this one. However, there is one crucial difference. I do not assume irreducibility of the representations at any point.

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  • $\begingroup$ I suggest not using $i$ as an index when dealing with C$^*$-algebras (or anything Hilbert spaces), since we already use $i$ for a certain complex number. $\endgroup$ – Martin Argerami May 8 '17 at 2:49
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That $U$ is linear, isometric, and surjective, is trivial. Irreducibility has nothing to do with it, but note that one requires that the two GNS representations satisfy that $\pi_i(\mathfrak A)\Omega_i$ is dense in $\mathcal H_i$.

The only nontrivial issue is whether $U$ is well-defined. That is, you want to show that if $\pi_1(A)\Omega_1=\pi_1(B)\Omega_1$, then $\pi_2(A)\Omega_2=\pi_2(B)\Omega_2$. This requires that the two GNS representations are minimal, that is, that $\pi_i(\mathfrak A)\Omega_i$ is dense in $\mathcal H_i$.

We have, if $\pi_1(A)\Omega_1=\pi_1(B)\Omega_1$, \begin{align} \|\pi_2(A)\Omega_2-\pi_2(B)\Omega_2\|^2&=\langle \pi_2(A)\Omega_2,\pi_2(A)\Omega_2\rangle+\pi_2(B)\Omega_2,\pi_2(B)\Omega_2\rangle-2\text{Re}\,\langle \pi_2(A)\Omega_2,\pi_2(B)\Omega_2\rangle\\ \ \\ &=\langle\pi_2(A^*A)\Omega_2,\Omega_2\rangle+\langle\pi_2(B^*B)\Omega_2,\Omega_2\rangle-2\text{Re}\,\langle\pi_2(B^*A)\Omega_2,\Omega_2\rangle\\ \ \\ &=\omega(A^*A)+\omega(B^*B)-2\text{Re}\,\omega(B^*A)\\ \ \\ &=\|\pi_1(A)\Omega_1-\pi_1(B)\Omega_1\|^2\\ \ \\ &=0. \end{align} So $\pi_2(A)\Omega_2=\pi_2(B)\Omega_2$, and so $U$ is well-defined.

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  • $\begingroup$ I wanted to prove the unitarity of $U$ which I believe follows directly from construction: $\langle \psi^{(2)}_A,\psi^{(2)}_B \rangle = \omega(A^*B) = \langle \psi^{(1)}_A,\psi^{(1)}_B \rangle.$ Anyway, the representations are minimal by construction as well: $\pi_i(\mathfrak A)\Omega_i = \{ \psi^{(i)}_A : A \in \mathfrak A\}$ is a pre-Hilbert space. But where did you use this denseness property in the above analysis? $\endgroup$ – Nanashi No Gombe May 9 '17 at 8:06
  • $\begingroup$ From the construction you get that $U $ is a symmetry; you still need to show that it is surjective. And the representations are not minimal "by construction": if both were constructed as in your first paragraph, there would be nothing to prove; the point is to show that two representations where the inner product is given by $\omega $ are unitarily equivalent, and you need the minimality for that. $\endgroup$ – Martin Argerami May 9 '17 at 12:31
  • $\begingroup$ I am sorry I still don't see where you used the minimality of the representations? Is this implicit in the first equality you wrote? And if we ignore the first paragraph to avoid triviality, we still use the same argument (as my first comment) to prove unitary right? $\endgroup$ – Nanashi No Gombe May 9 '17 at 15:26
  • $\begingroup$ Your "unitary" is, by definition, an isometry. But you need it to be surjective. If the representation is not minimal, there is no reason for your $U$ to be surjective. $\endgroup$ – Martin Argerami May 9 '17 at 15:39
  • $\begingroup$ Okay. Thanks. :) $\endgroup$ – Nanashi No Gombe May 9 '17 at 15:47

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