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In a previous exercise, I have proven that states $\omega$ on the $C^*$-algebra $M_n(\mathbb C)$ correspond to a unique density matrix $\rho$ by the relation $\omega(A) = \mathrm{Tr}(\rho A)$. I was then asked to construct a GNS-triplet for $\omega$, which I did in the following way:

Equipping $M_n(\mathbb C)$ with its Hilbert-Schmidt sesquilinear form and given a state $\omega$ corresponding to a density matrix $\rho$, divide out by the set $$ N :=\{B \in M_n(\mathbb C) \mid \omega(B^*B) = 0\}$$ and we claim that on the space $M_n(\mathbb C)/N$, there exists a representation $\pi_\omega$ with cyclic vector $\xi_\omega$ for this representation such that $\omega(A) = \langle \pi_\omega(A)\xi_\omega, \xi_\omega\rangle$ on this space. If we take a look at the GNS-condition for the representation and cyclic vector and interpret the Hilbert-Schmidt sesquilinear form, we are asked to find $\pi_\omega$ and $\xi_\omega$ such that $$\forall A \in M_n(\mathbb C)/N: \mathrm{Tr}(\rho A) = \mathrm{Tr}(\xi_\omega^* \pi_\omega(A)\xi_\omega)\, .$$ This seems to ask for $\pi_\omega$ to be the left multiplication operator with $A$. Indeed, since $\rho$ was positive, we can set $\xi_\omega := \rho^{1/2}$ and observe that by the `commutative property' of the trace, $$\mathrm{Tr}(\rho A) = \mathrm{Tr}(\xi_\omega A \xi_\omega) = \mathrm{Tr}(\xi_\omega^*\pi_\omega(A)\xi_\omega)\, ,$$ because $\xi_\omega$ is positive and hence self-adjoint.

It remains to show that $\xi_\omega$ is a cyclic vector for the representation $\pi_\omega$. It suffices to prove the following statement: $$\forall B \in M_n(\mathbb C)/N \, \exists A \in M_n(\mathbb C)/N: \pi_\omega(A)\xi_\omega=B\, ,$$ so we need to find $A \in M_n(\mathbb C)$ with $\omega(A^*A) \neq 0$ such that $A\xi_\omega = B$ [argument under construction].

The question I'm struggling with now goes as follows:

Can you generalize the construction you made in the previous exercise to a possible infinite-dimensional setting, i.e. considering for a Hilbert space $\mathcal H$ and a trace-class operator $\rho$ the positive functional $$\omega_\rho: \mathcal B(\mathcal H) \to \mathbb C: A \mapsto \omega_\rho(A) = \mathrm{Tr}(\rho A)\, ?$$

When asked about this problem, my professor gave the following hint:

In the infinite-dimensional setting, only normal states are of the form $\mathrm{Tr}(\rho A)$.

We hadn't seen the notion of a normal state in the course, so I googled its definition (for every monotone increasing net of operators $H_\alpha$ with upper bound $H$, $\omega_\rho(H_\alpha)$ converges to $\omega_\rho(H)$). I don't really see how to tackle the question, or why it is relevant that $\omega_\rho$ should be normal. Can someone get me started on solving the problem and/or proving the hint?

Edit: additionally, I noticed that the argument I gave for cyclicity of $\xi_\omega$ is wrong: it is possible that $(\det \xi_\omega)^2 = \det \rho = 0$, so $\xi_\omega$ is not necessarily invertible. Therefore, I ask another question: how do I prove that $\xi_\omega$ is a cyclic vector for $\pi_\omega$ on $M_n(\mathbb C)/N$? Assuming that $\omega(A^*A) \neq 0$ is equivalent to saying that $\mathrm{Tr}(\xi_\omega A) \neq 0$.

I was hesitant to tag this post as homework since this is not an assignment, but just an exercise in the course notes of our Operator Algebras course. Please let me know if I violated any rules concerning posting on StackExchange, as this is my first post. Thanks in advance for your answers!

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    $\begingroup$ The issue is that trace-class operators only consist of the predual of $B(H)$. As such, they embed into the dual (and are called normal). The dual contains more elements. In the infinite dimensional case, the various operator topologies are distinct. Normal states are continuous with respect to the WOT, which is a stronger requirement than continuity with respect to the norm topology. $\endgroup$ – Michael Jun 8 '13 at 20:01
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    $\begingroup$ Also, you just mimicked the general GNS construction there. Some general properties are immediate, e.g. the cyclic vector must be the class of $1_{M_n}$. One can be much more specific. The GNS representation in this case is a representation of $M_n$ on a finite dimensional Hilbert space $H$. Using properties of $\rho$, you can say what is the dimension of $H$ and what is the multiplicity of the (faithful) representation. $\endgroup$ – Michael Jun 9 '13 at 6:10
  • $\begingroup$ Thank you for your comments thus far. I am not familiar with the terminology of `multiplicity of a representation,' and in the theory for this course, the only inspiration for constructing GNS-triplets is indeed the constructive proof of the GNS-theorem. My tactic was to prove that the square root $\xi_\omega$ of $\rho$ is a representant of the unit (dividing by $N$), which would immediately prove that $\xi_\omega$ is cyclic for $\pi_\omega$ as in that proof, amounting to proving that $\mathrm{Tr}(\xi\rho) = \mathrm{Tr}(\rho)$, which does not seem true at least. Is there a different tactic? $\endgroup$ – premaut Jun 9 '13 at 7:35
  • $\begingroup$ As for the infinite-dimensional case, I'm also still unsure what your comment implies. Will the construction of the GNS-triplet still go through for normal states and is the only difference that this is no longer possible for all states? $\endgroup$ – premaut Jun 9 '13 at 7:39
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    $\begingroup$ GNS construction holds for any $C^*$-algebra $A$. The question is, when $A = B(H)$, whether the given state can be explicitly identified with a trace-class operator. The answer is yes iff the state is normal. $\endgroup$ – Michael Jun 10 '13 at 2:57
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I contacted my professor with this question and am adding his response here should anyone need it in the future:

  • For the finite-dimensional case, my representation and cyclic vector were correct. In case $\det \xi_\omega = 0$, $\xi_\omega$ is not cyclic on $M_n(\mathbb C)$, but it is cyclic on $$\{ A\xi_\omega \mid A \in M_n(\mathbb C)\}$$ per construction. (My professor said "if $\xi_\omega$ is not invertible, then indeed you do not retrieve all of $M_n(\mathbb C)$" and I'm a bit puzzled: is there some explicit way to describe this set if $\xi_\omega$ is not invertible that I'm not seeing or am I being paranoid?)

  • In the infinite-dimensional setting, as Michael noted there is the issue is that not all operators are associated to a trace-class operator. The exercise my professor meant to give was to construct GNS-triplets for those states that are, so I misunderstood his question as I was confused by the hint. As Hilbert space, one should take the Hilbert-Schmidt operators with not-necessarily-cyclic vector $\rho^{1/2}$: if $\rho$ admits zero as eigenvalue, one should again 'restrict' to a space on which $\rho^{1/2}$ is cyclic per construction.

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