Equivalence between gibbs states representations with different temperatures I'm asked to answer this question: why two Gibbs states with different temperatures give the same (GNS) representation?
Actually, I can't even imagine if this is true and if not how to find a counter example.
So for a Gibbs state $\rho_{\beta}=Z_{\beta}^{-1}exp(-\beta H)$ where $Z_{\beta}= Tr(exp(- \beta H))$ the GNS representation is construted in this way: $r_0=\rho_{\beta}^{\frac{1}{2}}$ is a Hilbert-Schmidt operator. Then take the vector space of Hilbert-Schmidt operators $D_0$ and associate the following scalar product: $(X,Y)  = Tr(X^{+}Y)$. $D_0$ is actually a Hilbert space and we denote $r_0$ with the common $|r_0>$. This will be our cyclic vector in GNS representation. Then define a representation in this way: $\pi(A)|r_0> = |A r_0>$. So the triple ($r_0, D_0, \pi$) is the GNS triple for a Gibbs state and the expectation values are written as $\omega_{\beta}(A)=Tr(\rho_{\beta}A)=<r_0|\pi(A)|r_o>$.
Now I must find for two different Gibbs states, respectively with representations $(r_1, D_1, \pi_1)$ and $(r_2,D_2,\pi_2)$ an unitary operator $U:D_1 \to D_2$ such that $\pi_1(A)= U^{+}\pi_2(A)U$.
But at this point I don't know how to proceed... 
 A: Consider the fact that $$\rho_\beta :=Z^{-1}_\beta e^{-\beta H}$$ is trace class by definition  in a given (separable) Hilbert space $\cal H$, and thus it can be expanded as:
$$\rho_\beta = \sum_{j\in \mathbb N} p_j(\beta) |\psi_j\rangle \langle \psi_j|\quad \mbox{where } 0 \leq p_j(\beta)\leq 1 \mbox{ and} \sum_j p_j(\beta)=1\quad (1)$$
above I use the convention $p_j \leq p_{j+1}$ (where the number of $j$ with a common fixed value $p_j$ is finite because $\rho_\beta$ is compact) and the convergence is in the strong operator topology. 
In the considered case, more strongly (it turns out to be important shortly):
$$p_j(\beta)= e^{-\beta E_j}>0 \quad \forall j\:, \forall \beta >0\:.$$
Fix another Hilbert basis $\{\phi_j\}_{j\in \mathbb N} \subset {\cal H}$ and, in ${\cal K}:= {\cal H}\otimes {\cal H}$, define the normalized vector:
$$\Psi_\beta := \sum_{j \in \mathbb N} \sqrt{p_j(\beta)} \psi_j \otimes \phi_j\:.  \quad (2)$$
The triple $\left({\cal K}, \pi, \Psi_\beta \right)$ is a GNS triple for $\rho_\beta$ if:
$$\pi : B({\cal H}) \ni A \mapsto A \otimes I\quad (3)$$
The only thing to be proved, because not self-evident, is that $\{ A\otimes I \Psi_\beta\:|\: A \in B({\cal H})\}$ is dense in $\cal K$.  For every $j$ and $k$, we may pick out $A_{jk} \in B(\cal H)$ such that $A_{jk}\psi_j = \sqrt{\frac{1}{p_j(\beta)}}\psi_k$ (notice that all $p_j$ are strictly positive) and 
$A_{jk}\psi_{j'}=0$ if $j\neq j'$.  Therefore 
$\pi(B(\cal H)) \Psi_\beta$ includes every product $\psi_k \otimes \phi_j$ and thus every linear combinations of these products. Since $\{\psi_k\otimes \phi_j\}_{k,j \in \mathbb N}$ is a Hilbert basis of ${\cal H}\otimes {\cal H}={\cal K}$, we can infer that $\{ A\otimes I \Psi_\beta\:|\: A \in B({\cal H})\}$ is dense in $\cal K$.
To conclude, notice that I never used the value of $\beta$ barring in the definition of $\Psi_\beta$. Consequently, changing $\beta$ we can use the same GNS Hilbert space and GNS representation. Just the identity map is the wanted unitary operator $U : {\cal K} \to {\cal K}$ verifying $U \pi_\beta(A)= \pi_{\beta'}(A)U$.  
