I have a question about a specific step in the proof of the monotonicity theorem for the relative entropy given in [1, Thm. 5.3, p. 81 f.]; the proof relies on the spatial derivative operator between two states on a von Neumann algebra, introduced by A. Connes in [2]. (The context of relative entropy is actually not relevant for this question, since it is only about a computation involving the spatial derivative.)
I will first define the relevant objects, outline the setup of the proof, and then formulate my actual question.
Setup
Modular Theory
Let $\mathfrak{M}$ be a von Neumann algebra acting on a Hilbert space $\mathcal{H}$, let $\psi$ be a positive normal linear functional on $\mathfrak{M}$, let $D(\mathcal{H}, \psi) := \{x \in \mathcal{H} : \exists C > 0 \ \forall a \in \mathfrak{M} : \Vert ax \Vert^2 \le C \psi(a^\ast a)\}$ be the lineal (or set of $\psi$-bounded vectors) of $\psi$, and let $(\mathcal{H}_{\psi}, \pi, \varPsi)$ be the GNS-triple of $\psi$.
Define first for every $x \in D(\mathcal{H}, \psi)$ a linear operator $R^{\psi}(x) : \mathcal{H}_{\psi} \to \mathcal{H}$ such that $R^{\psi}(x)\bigl(\pi(a) \varPsi\bigr) = a x$ for all $a \in \mathfrak{M}$. Then, set $\varTheta^{\psi}(x) := R^{\psi}(x) \bigl(R^{\psi}(x)\bigr)^\ast$, which is an operator in the commutant $\mathfrak{M}^\prime$. Finally, for a fixed normal state $\varphi^\prime$ on $\mathfrak{M}^\prime$, define $$q(x) := \varphi^\prime\bigl(\varTheta^{\psi}(x)\bigr)$$ which can be shown to be a densely defined lower semi-continuous positive quadratic form on $\mathcal{H}$ (see, for example, [3, 7.3]). By the representation theorem for semi-bounded forms, there exists a positive self-adjoint linear operator $\varDelta(\varphi^\prime / \psi)$ on $\mathcal{H}$ such that $$\Vert \varDelta(\varphi^\prime / \psi)^{1/2} z \Vert^2 = q(z)$$ for all $z \in \mathrm{dom}(q)$, and the domain of $q$ is a core for $\varDelta(\varphi^\prime / \psi)^{1/2}$. This operator is called the spatial derivative of $\psi$ with respect to $\varphi^\prime$. It can be used to define the relative entropy between two states on the von Neumann algebra $\mathfrak{M}$. (I will not recall this definition since it is not needed to formulate my question below.)
Proof of Theorem 5.3 in [1]
Assume now that two von Neumann algebras $\mathfrak{M}_1, \mathfrak{M}_2$ acting on Hilbert spaces $\mathcal{H}_1, \mathcal{H}_2$ are given, and with them positive normal linear functionals $\varphi_1, \omega_1$ on $\mathfrak{M}_1$ and $\varphi_2, \omega_2$ on $\mathfrak{M}_2$, where for $i = 1,2$ the functionals $\omega_i$ are assumed to be vector states induced by $\xi_i \in \mathcal{H}_i$. Suppose further that there exists a Schwarz mapping $\alpha : \mathfrak{M}_1 \to \mathfrak{M}_2$, that is, a linear unital mapping satisfying for all $a \in \mathfrak{M}_1$ $$\alpha(a)^\ast \alpha(a) \le \alpha(a^\ast a) \ ,$$ such that we have the properties $$\varphi_2 \circ \alpha \le \varphi_1 \quad \text{and} \quad \omega_2 \circ \alpha \le \omega_1$$ on $\mathfrak{M}_1$. Consider the mapping $T : a \xi_1 \mapsto \alpha(a) \xi_2$, $a \in \mathfrak{M}_1$, which is a linear contraction by the above properties, and extend it to a map $\mathcal{H}_1 \to \mathcal{H}_2$. Let $\omega_i^\prime$ be the vector state on the commutant $\mathfrak{M}_i^\prime$ induced by $\xi_i$, and let $\varDelta_i$ denote the spatial derivative $\varDelta(\varphi_i / \omega_i^\prime)$. Finally, for $n \in \mathbb{N}$ consider $x_n := a_n \xi_1 + y_n$, where $a_n \in \mathfrak{M}_1$, $y_n \in \overline{\mathfrak{M}_1 \xi_1}^\perp$. (It holds that the space $\mathfrak{M}_1 \xi_1 + (\mathfrak{M}_1 \xi_1)^\perp$ is a core for $\varDelta_i^{1/2}$.)
Question
I do not see how the following equality (first equation on p. 82 in [1]) can be justified: $$\Vert \Delta_2^{1/2} T x_n - \Delta_2^{1/2} T x_m\Vert^2 = \varphi_2\bigl(\alpha(a_n - a_m)^\ast \alpha(a_n - a_m)\bigr) \ .$$ When I was trying to derive it, I obtained the following result: if $q_2$ denotes the quadratic form from which $\varDelta_2^{1/2}$ is obtained, then \begin{align*}\Vert \Delta_2^{1/2} T x_n - \Delta_2^{1/2} T x_m\Vert^2 &= q_2\bigl(T(x_n - x_m)\bigr) \\ &= q_2\bigl(\alpha(a_n - a_m) \xi_2\bigr) \\ &= \varphi_2\Bigl(\varTheta^{\omega_2^\prime}\bigl(\alpha(a_n - a_m) \xi_2\bigr)\Bigr) \ .\end{align*} Now, I would use the fact that $\omega_2^\prime$ is a vector state on $\mathfrak{M}^\prime$ to write the operator $\varTheta^{\omega_2^\prime}(\dotsb)$ as (cf. [1, p. 70] or [2, p. 157]) $$\varTheta^{\omega_2^\prime}\bigl(\alpha(a_n - a_m) \xi_2\bigr) = \alpha(a_n - a_m) \circ [\mathfrak{M}_2^\prime \xi_2] \circ \alpha(a_n - a_m)^\ast \ ,$$ where $[\mathfrak{M}_2^\prime \xi_2]$ denotes the orthogonal projection onto the subspace $\overline{\mathfrak{M}_2^\prime \xi_2}$. At this point, I am stuck; even if the projection were given by the identity (for example, if the vector $\xi_2$ were separating for $\mathfrak{M}_2$ and hence cyclic for its commutant), the operators $\alpha(a_n - a_m)$ and $\alpha(a_n - a_m)^\ast$ are in the wrong order (which is crucial for the following steps of the proof, because one wants to apply the Schwarz inequality and the assumptions on the states).
If more details are needed, I am happy to provide them and apologize in advance for not doing so from the beginning. (I did not want the post to be even longer than it already is.) I thank you for taking the time of reading my question, and I would greatly appreciate any help you could provide!
[1] Ohya, M. and Petz, D. Quantum Entropy and Its Use. Corrected Second Printing. Springer, Berlin, Heidelberg, 2004.
[2] Connes, A. “On the Spatial Theory of von Neumann Algebras”. J. Functional Analysis 35 (1980), pp. 153–164.
[3] Strătilă, Ş. V. Modular Theory in Operator Algebras. Cambridge University Press, Cambridge, 2020.
It appears that in the case in which the functionals $\omega_1$ and $\omega_2$ are additionally assumed to be faithful, a similar identity as given in the book [1] mentioned in the OP can be justified from which the further steps of the proof of monotonicity of relative entropy follow.
As noted in the OP, if the states are faithful then $[\mathfrak{M}_i^\prime \xi_i] = \mathrm{Id}_{\mathcal{H}_i}$, $i = 1, 2$, since the vector representative of a faithful functional is separating for the algebra, hence cyclic for its commutant. Moreover, the Schwarz mapping $\alpha : \mathfrak{M}_1 \to \mathfrak{M}_2$ satisfies the inequality $\alpha(a) \alpha(a)^\ast \le \alpha(a a^\ast)$ for all $a \in \mathfrak{M}_1$; this follows from the fact that Schwarz mappings are positive, and that positive mappings between $C^\ast$-algebras are self-adjoint, i.e., $\alpha(a^\ast) = \alpha(a)^\ast$. Therefore, one has \begin{align*} \Vert \Delta_2^{1/2} T x_n - \Delta_2^{1/2} T x_m\Vert^2 &= \varphi_2\Bigl(\varTheta^{\omega_2^\prime}\bigl(\alpha(a_n - a_m) \xi_2\bigr)\Bigr) \\ &= \varphi_2\bigl(\alpha(a_n - a_m) \alpha(a_n - a_m)^\ast\bigr) \\ &\le \varphi_2\Bigl(\alpha\bigl((a_n - a_m)(a_n - a_m)^\ast\bigr)\Bigr) \\ &\le \varphi_1\bigl((a_n - a_m)(a_n - a_m)^\ast\bigr) \\ &= \varphi_1\Bigl(\varTheta^{\omega_1^\prime}\bigl((a_n - a_m) \xi_1\bigr)\Bigr) \\ &= \Vert \Delta_1^{1/2} x_n - \Delta_1^{1/2} x_m\Vert^2 \ . \end{align*} The second line is the substitute for the identity in question in the OP; the fourth line follows from the assumptions on the states $\varphi_1, \varphi_2$. This result is precisely the inequality needed in the sequel of the proof of Theorem 5.3 in [1].
For physical applications, the states are usually (assumed to be) faithful; however, in the book [1] there is no explicit assumption of faithfulness mentioned. Hence, this can only be considered a partial answer. (I posted it nevertheless, since in the OP, it was not clear how to proceed even under the additional assumption of faithfulness.)