The notion of observable and the moment problem for $*$-algebras and their GNS representations

Abstract

We address some usually overlooked issues concerning the use of $*$-algebras in quantum theory and their physical interpretation. If ${\mathfrak {A}}$ is a $*$-algebra describing a quantum system and $\omega : {\mathfrak {A}}\rightarrow {\mathbb {C}}$ a state, we focus, in particular, on the interpretation of $\omega (a)$ as expectation value for an algebraic observable $a=a^*\in {\mathfrak {A}}$, studying the problem of finding a probability measure reproducing the moments $\{\omega (a^n)\}_{n\in {\mathbb {N}}}$. This problem enjoys a close relation with the selfadjointness of the (in general only symmetric) operator $\pi _\omega (a)$ in the GNS representation of $\omega$ and thus it has important consequences for the interpretation of a as an observable. We provide physical examples (also from QFT) where the moment problem for $\{\omega (a^n)\}_{n\in {\mathbb {N}}}$ does not admit a unique solution. To reduce this ambiguity, we consider the moment problem for the sequences $\{\omega _b(a^n)\}_{n\in {\mathbb {N}}}$, being $b\in {\mathfrak {A}}$ and $\omega _b(\cdot ):=\omega (b^*\cdot b)$. Letting $\mu _{\omega _b}^{(a)}$ be a solution of the moment problem for the sequence $\{\omega _b(a^n)\}_{n\in {\mathbb {N}}}$, we introduce a consistency relation on the family $\{\mu _{\omega _{b}}^{(a)}\}_{b\in {\mathfrak {A}}}$. We prove a 1-1 correspondence between consistent families $\{\mu _{\omega _{b}}^{(a)}\}_{b\in {\mathfrak {A}}}$ and positive operator-valued measures (POVM) associated with the symmetric operator $\pi _\omega (a)$. In particular, there exists a unique consistent family of $\{\mu _{\omega _{b}}^{(a)}\}_{b\in {\mathfrak {A}}}$ if and only if $\pi _\omega (a)$ is maximally symmetric. This result suggests that a better physical understanding of the notion of observable for general $*$-algebras should be based on POVMs rather than projection-valued measure.

Appendix

1.1 Reducing subspaces

Definition 37

If $D(T) \subset {\mathsf {H}}$ is a subspace of the Hilbert space ${{\mathsf {H}}}$, let $T : D(T) \rightarrow {\mathsf {H}}$ be an operator and ${\mathsf {H}}_0 \subset {\mathsf {H}}$ a closed subspace with $\{0\} \ne {\mathsf {H}}_0 \ne {\mathsf {H}}$, $P_0$ denoting the orthogonal projector onto ${\mathsf {H}}_0$. In this case, ${\mathsf {H}}_0$ is said to reduceT if both conditions are true

• (i) $T(D(T) \cap {\mathsf {H}}_0) \subset {\mathsf {H}}_0$ and $T(D(T) \cap {\mathsf {H}}^\perp _0) \subset {\mathsf {H}}^\perp _0$,

• (ii) $P_0(D(T)) \subset D(T)$,

so that the direct orthogonal decomposition holds

(66)

The operator is called the part ofTon${\mathsf {H}}_0$. $\square$

Remark 38

(1) It is worth stressing that (i) does not imply (ii). (A counterexample is given by ${\mathsf {H}}:=L^2([0,+\infty ),\mathrm{d}x)$, $T:=-i\frac{\mathrm{d}}{\mathrm{d}x}$ with $D(T):= C_c^\infty ((0,+\infty ))$ and ${\mathsf {H}}_0:={\text {span}}\{\varphi \}$, where $\varphi (x)=e^{-x}$. Here $D(T)\cap {\mathsf {H}}_0 = \{0\}$ so that the former inclusion in (i) is trivially valid, whereas the latter is valid by integrating by parts. However, it is easy to pick out $\psi \in D(T)$ such that $\langle \psi |\varphi \rangle \ne 0$, so that $(P_0\psi )(x)$ has the support of $\varphi$ itself given by the whole $[0,+\infty )$ and $P_0\psi \not \in D(T)$.) Moreover, without (ii) the direct orthogonal decomposition (66) cannot take place.

(2) Evidently ${\mathsf {H}}_0$ reduces T iff ${\mathsf {H}}_0^\perp$ reduces T, since $I-P_0$ projects onto ${\mathsf {H}}_0^\perp$.

(3) A condition equivalent to (i)–(ii) is $P_0T\subset TP_0$ as the reader immediately proves. $\square$

Proposition 39

If the closed subspace ${\mathsf {H}}_0$ reduces the symmetric (selfadjoint) operator T on ${\mathsf {H}}$, then also and are symmetric (resp. selfadjoint) in ${\mathsf {H}}_0$ and ${\mathsf {H}}_0^\perp$, respectively.

Proof

Direct inspection. $\square$

A useful technical fact is presented in the following proposition [24].

Proposition 40

Let T be a closed symmetric operator on a Hilbert space ${{\mathsf {H}}}$. Let $D_0$ be a dense subspace of a closed subspace ${{\mathsf {H}}}_0$ of ${{\mathsf {H}}}$ such that $D_0 \subset D(T )$ and $T(D_0)\subset {{\mathsf {H}}}_0$. Suppose that is essentially selfadjoint on ${{\mathsf {H}}}_0$. Then ${{\mathsf {H}}}_0$ reduces T and $\overline{T_0}$ is the part of T on ${{\mathsf {H}}}_0$.

Proof

See [24, Prop.1.17]. $\square$

1.2 More on generalized symmetric and selfadjoint extensions

According to definition 21 we have

$$\begin{aligned} D(A)\subset D(B)\cap {\mathsf {H}}\subset D(B), \end{aligned}$$

(67)

where D(B) is dense in ${\mathsf {K}}$ and D(A) is dense in ${\mathsf {H}}$. Generalized extensions B with $B\supsetneq A$ are classified accordingly to the previous inclusions following [1], in particular:

• (i) B is said to be of kind I if $D(A)\ne D(B)\cap {\mathsf {H}}= D(B)$—that is, if B is a standard extension of A;

• (ii) B is said to be of kind II if $D(A)= D(B)\cap {\mathsf {H}}\ne D(B)$;

• (iii) B is said to be of kind III if $D(A)\ne D(B)\cap {\mathsf {H}}\ne D(B)$;

Proposition 41

If $A:D(A) \rightarrow {\mathsf {H}}$ with $D(A) \subset {\mathsf {H}}$ is maximally symmetric and $B \supsetneq A$ is a generalized symmetric extension, then B is of kind II.

Proof

The kind I is not possible a priori since A does not admit proper symmetric extensions in ${\mathsf {H}}$. Let us assume that B is either of kind II or III and consider the operator $P_{{\mathsf {H}}}BP_{{\mathsf {H}}}$, with its natural domain $D(P_{{\mathsf {H}}}BP_{{\mathsf {H}}})= D(BP_{{\mathsf {H}}})$, where $P_{{\mathsf {H}}}\in {\mathscr {L}}({\mathsf {K}})$ is the orthogonal projector onto ${\mathsf {H}}$. Since this is a symmetric extension of A in ${\mathsf {H}}$ which is maximally symmetric, we have $A=P_{{\mathsf {H}}}BP_{{\mathsf {H}}}$, in particular $D(BP_{{\mathsf {H}}})=D(A)$. As a consequence, if $x \in D(B)\cap {\mathsf {H}}$, then $x \in D(BP_{{\mathsf {H}}}) =D(A)$ so that $D(A) \supset D(B)\cap {\mathsf {H}}$ and thus $D(A) = D(B)\cap {\mathsf {H}}$ because the other inclusion is true from (67). We have proved that B is of kind II. $\square$

We have an important technical result

Theorem 42

A non-selfadjoint symmetric operator A always admits a generalized selfadjoint extension B. Such an extension can be chosen of kind II when A is closed.

Proof

See [1, Thm. 1. p.127 Vol II] for the former statement. The latter statement relies on the comment under the proof of [1, Thm. 1. p.127 Vol II] and it is completely proved in [17, Thm.13].¹¹ The fact that the selfadjoint extensions can be chosen in order to satisfy (iii) of Definition 21 is proved in [18, Thm.7]. $\square$

1.3 Proof of some propositions

Lemma 43

Referring to (1) in Example 11, $\psi _\omega \in \mathcal{D}_{\omega _B}$.

Proof

To prove that $\psi _\omega \in \mathcal{D}_{\omega _B}$ observe that, since A and $A^*$ are generators of ${\mathfrak {A}}_{\text {CCR},1}$ satisfying $[A,A^*]=I$, we can always rearrange every element $B \in {\mathfrak {A}}_{\text {CCR},1}$ into the form

$$\begin{aligned} B= \sum _{n,m } c^{(B)}_{n,m}A^{*n}A^m, \end{aligned}$$

where $A^0:=A^{*0}:=I$ and where only a finite number of coefficients $c^{(B)}_{n,m} \in {{\mathbb {C}}}$ depending on B are different from 0. Here, (13) implies

$$\begin{aligned} \psi _{\omega _B} = \sum _{n,m } c^{(B)}_{n,m}A^{*n}A^m\psi _\omega = \sum _{k \in {{\mathbb {N}}}} d^{(B)}_k \psi _k, \end{aligned}$$

(68)

where again only a finite number of coefficients $d^{(B)}_k \in {{\mathbb {C}}}$ depending on B are different from 0 and to pass from the first sum to the second one we took advantage of the standard harmonic oscillator algebra of the Hermite basis where $A\psi _k=\sqrt{k}\psi _{k-1}$, $A^*\psi _k=\sqrt{k+1}\psi _{k+1}$ with $\psi _0:= \psi _\omega$. To conclude, notice that, if $C \in {{\mathfrak {A}}}_{\text {CCR},1}$, (13) also implies that $\pi _{\omega _B}(C)\psi _{\omega _B}= \pi _{\omega }(C)\psi _{\omega _B}$. Therefore, if $k_B\in {{\mathbb {N}}}$ is the largest natural such that $d^{(B)}_k \ne 0$ in (68), choosing $C:= \frac{1}{d^{(B)}_{k_B}\sqrt{k_B!}}A^{k_B}\in {{\mathfrak {A}}}_{\text {CCR},1}$, we have

$$\begin{aligned} \pi _{\omega _B}(C)\psi _{\omega _B} = \sum _{k \in {{\mathbb {N}}}} d^{(B)}_k \frac{1}{d^{(B)}_{k_B}\sqrt{k_B!}} A^{k_B}\psi _k = 0+ \frac{d^{(B)}_{k_B} \sqrt{k_B!}}{d^{(B)}_{k_B} \sqrt{k_B!}}\psi _0= \psi _0 = \psi _\omega . \end{aligned}$$

In other words, $\psi _\omega \in \mathcal{D}_{\omega _B}$ as argued. $\square$

Lemma 44

Referring to (2) in Example 11, $\psi _\omega \in {\mathcal{D}}_{\omega _b}$ if (29) is valid.

Proof

First define the elements of the algebra ${{\mathfrak {A}}}[M,g]$

$$\begin{aligned} A_{\varphi } :=\frac{1}{2} \left( \varPhi [\varphi ] + i \varPhi [\varphi '] \right) \quad \text{ and }\quad A^*_{\varphi } :=\frac{1}{2} \left( \varPhi [\varphi ] - i \varPhi [\varphi '] \right) . \end{aligned}$$

We stress that the elements $A^*_{\varphi }$ and $A_{\varphi }$ form a set of generators of ${{\mathfrak {A}}}[M,g]$, because the $\varPhi [\varphi ]$ are generators and the previous relations can be inverted to

$$\begin{aligned} \varPhi [\varphi ] = A_{\varphi } + A^*_{\varphi } \end{aligned}$$

By using (26), linearity of $x \mapsto a_x^+$, antilinearity of $x \mapsto a_x$ and (29), we immediately find

$$\begin{aligned} \pi _\omega \left( A_{\varphi }\right) = a_{K\varphi }|_{{\mathcal{D}}_\omega } \quad \text{ and } \quad \pi _\omega \left( A^*_{\varphi }\right) = a^+_{K\varphi }|_{{\mathcal{D}}_\omega } \end{aligned}$$

(69)

The algebraic relations reflecting (25) are also valid from (b) and (c) in item (2) of Example  11,

$$\begin{aligned}{}[A_{\varphi } , A^*_{{\tilde{\varphi }}} ] = \langle K\varphi , K{\tilde{\varphi }}\rangle I, \quad [A_{\varphi } , A_{{\tilde{\varphi }}} ] = [A^*_{\varphi } , A^*_{{\tilde{\varphi }}} ] =0. \end{aligned}$$

Exploiting the found results, we can prove that $\psi _\omega \in {\mathcal{D}}_{\omega _b}$ as wanted. The generic element $b \in {{\mathfrak {A}}}[M,g]$, taking advantage of the generators $A_{\varphi }$ and $A^*_{\varphi }$ and their commutation relations, is always of the form (where we adopt the convention that $\prod _{j=1}^{0} A^{(*)}_{\varphi _j}:= {{\mathbb {I}}}$)

$$\begin{aligned} b = \sum _{k,h=0}^{N} c_{kh}^{j_1\ldots j_k i_1\ldots i_h} A^*_{\varphi ^{(k)}_{j_1}}\cdots A^*_{\varphi ^{(k)}_{j_k}} A_{{\tilde{\varphi }}^{(h)}_{i_1}}\cdots A_{{\tilde{\varphi }}^{(h)}_{i_h}} \end{aligned}$$

(70)

for sets of solutions $\{\varphi ^{(k)}_{j}\}_{j\in \{1,\ldots , D_k\}}$ and $\{{\tilde{\varphi }}^{(h)}_{i}\}_{i \in \{1,\ldots , E_h\}}$ in Sol[M, g] and where we adopted Einstein’s summation convention. Obviously, $N,D_h, E_h$, and the complex coefficients $c_{k,h}^{j_1\ldots j_k i_1\ldots i_h}$ depend on b and these coefficients are completely symmetric separately in the indices $j_r$ and in the indices $i_r$ due to the fact that the elements $A^*_{\varphi _{j_1}},\cdots , A^*_{\varphi _{j_k}}$ and $A_{\varphi _{i_1}},\cdots ,A_{\varphi _{i_h}}$ separately pairwise commute. Hence, (69) entails

$$\begin{aligned} \psi _{\omega _b} = \pi _{\omega }(b) \psi _\omega = \sum _{k=0}^{N} c_{k0}^{j_1\ldots j_k} a^+_{\psi ^{(k)}_{j_1}}\cdots a^+_{\psi ^{(k)}_{j_k}}\psi _\omega \end{aligned}$$

(71)

where $\psi ^{(k)}_j := K\varphi ^{(k)}_j$ and we henceforth assume that not all coefficients $c_{N0}^{j_1\ldots j_N}$ vanish and all vectors $\psi ^{(N)}_j = K\varphi ^{(N)}_j$ for $j=1,\ldots , D_N$ do not vanish. (Otherwise, the Nth addend in (71) would give no contribution.) Defining

$$\begin{aligned} d := \overline{c_{N0}^{i_1\cdots i_N}} A_{K\varphi ^{(N)}_{i_1}} \cdots A_{K\varphi ^{(N)}_{i_N}} \in {{\mathfrak {A}}}[M,g] \end{aligned}$$

(25) yields, if $c^{j_1\cdots j_N} := c_{N0}^{j_1\cdots j_N}$ and $\psi _j := \psi ^{(N)}_j$

$$\begin{aligned} {\mathcal{D}}_{\omega _b} \ni \pi _{\omega }(d) \pi _\omega (b) \psi _\omega =\overline{c^{i_1\cdots i_N}}c^{j_1\cdots j_N} a_{\psi _{i_1}} \cdots a_{\psi _{i_N}} a^+_{\psi _{j_1}}\cdots a^+_{\psi _{j_N}} \psi _\omega . \end{aligned}$$

(72)

At this juncture we can fix an orthonormal basis $\{e_l\}_{i=1,\ldots , D}$ in the span of the vectors $\{\psi _{j}\}_{j=1,\ldots , D_N}$ and we can rearrange the identity above as

$$\begin{aligned} \pi _{\omega }(d) \pi _\omega (b) \psi _\omega =\overline{c'^{i_1\cdots i_N}}c'^{j_1\cdots j_N} a_{e_{i_1}} \cdots a_{e_{i_N}} a^+_{e_{j_1}}\cdots a^+_{e_{j_N}} \psi _\omega \end{aligned}$$

where not all the symmetric coefficients $c'^{j_1\cdots j_N}$ vanish (since they are the components in a new basis of the non-vanishing tensor defined by the components $c^{j_1\cdots j_N}$). Expanding the contractions we find, where $P_N$ denotes the group of permutations of N objects

$$\begin{aligned} \pi _{\omega }(d) \pi _\omega (b) \psi _\omega= & {} \sum _{\sigma \in P_N} \overline{c'^{i_1\cdots i_N}}c'^{j_1\cdots j_N} \delta _{i_1j_{\sigma (1)}}\cdots \delta _{i_N j_{\sigma (N)}}\psi _\omega \\= & {} \sum _{\sigma \in P_N} \overline{c'^{i_1\cdots i_N}}c'^{j_{\sigma ^{-1}(1)}\cdots j_{\sigma ^{-1}(N)}} \delta _{i_1j_1}\cdots \delta _{i_N j_N}\psi _\omega \\= & {} \sum _{\sigma \in P_N} \overline{c'^{i_1\cdots i_N}}c'^{j_1\cdots j_N} \delta _{i_1j_1}\cdots \delta _{i_N j_N}\psi _\omega . \end{aligned}$$

(In the second line, $c'^{j_{\sigma ^{-1}(1)}\cdots j_{\sigma ^{-1}(N)}}= c'^{j_1\cdots j_N}$ arises from the symmetry of the coefficients.) That is

$$\begin{aligned} \pi _{\omega }(d) \pi _\omega (b) \psi _\omega = \left( N! \sum _{j_1, \ldots , j_N =1}^{D} |c'^{j_1\cdots c_N}|^2\right) \psi _\omega \in {\mathcal{D}}_{\omega _b}. \end{aligned}$$

(73)

Since the coefficient in front of $\psi _\omega$ in (73) does not vanish, then $\psi _\omega \in {\mathcal{D}}_{\omega _b}$ as wanted. $\square$

Proof of Proposition 22

Let B be a generalized symmetric extension in the Hilbert space ${{\mathsf {K}}}$ of the selfadjoint operator A in the Hilbert space ${{\mathsf {H}}}$ with ${{\mathsf {H}}}\subset {{\mathsf {K}}}$. The closed symmetric operator $B'= {\overline{B}}$ in ${\mathsf {K}}$ extends A. In view of Proposition 39–40, since is selfadjoint on ${\mathsf {H}}$, then ${\mathsf {H}}$ reduces $B'$ and , where the two addends are symmetric operators, respectively, on ${\mathsf {H}}$ and ${\mathsf {H}}^\perp$. Requirement (iii) in Definition 21 imposes that ${\mathsf {H}}^\perp = \{0\}$, so that ${\mathsf {K}}= {\mathsf {H}}$ and B is a standard symmetric extension of the selfadjoint operator A which entails $B=A$. $\square$

Proof of Theorem 23

(a) and (b). If A is symmetric, for each generalized selfadjoint extension B as in Definition 21 (they exist in view of Theorem 42 and, if A is selfadjoint, $B:=A$), one can define a corresponding normalized POVM $Q^{(A)}:{\mathscr {B}}({\mathbb {R}})\rightarrow {\mathfrak {B}}({\mathsf {H}})$ as follows. Let $P:{\mathscr {B}}({\mathbb {R}})\rightarrow {\mathscr {L}}({\mathsf {K}})$ be the unique PVM associated with B

$$\begin{aligned} B := \int _{{\mathbb {R}}}\lambda \mathrm{d}P(\lambda ), \end{aligned}$$

(74)

(see, e.g., [15, Thm. 9.13]) and let $P_{{\mathsf {H}}}\in {\mathscr {L}}({\mathsf {K}})$ denote the orthogonal projection onto ${\mathsf {H}}$ viewed as a closed subspace of ${\mathsf {K}}$. A normalized POVM $Q^{(A)}:{\mathscr {B}}({\mathbb {R}})\rightarrow {\mathfrak {B}}({\mathsf {H}})$ is then defined by setting

(75)

The POVM $Q^{(A)}$ is linked to A by the following identities as the reader easily proves from standard spectral theory:

$$\begin{aligned} \langle \psi |A\varphi \rangle =\int _{{\mathbb {R}}}\lambda \mathrm{d}Q^{(A)}_{\psi ,\varphi }(\lambda ), \quad \Vert A\varphi \Vert ^2=\int _{{\mathbb {R}}}\lambda ^2\mathrm{d}Q^{(A)}_{\varphi ,\varphi }(\lambda ),\quad \forall \psi \in {\mathsf {H}},\varphi \in D(A), \end{aligned}$$

(76)

The above relation implies the following facts, where $\lambda ^k$ henceforth denotes the map ${{{\mathbb {R}}}} \ni \lambda \rightarrow \lambda ^k \in {{{\mathbb {R}}}}$ for $k\in {\mathbb {N}} := \{0,1,2, \ldots \}$,

(77)

the second equality arising from the identity

$$\begin{aligned} \int _{{\mathbb {R}}}\lambda ^2\mathrm{d}P_{\psi ,\psi }(\lambda )= \int _{{\mathbb {R}}}\lambda ^2\mathrm{d}Q^{(A)}_{\psi ,\psi }(\lambda )\,\quad \forall \psi \in {\mathsf {H}}, \end{aligned}$$

which descends from Eq. (75). [1, Thm. 2, p. 129 Vol II] proves that every POVM (it is equivalent to a spectral function used therein, see Remark 14) satisfying (76) arises from a PVM of a generalized selfadjoint extension B of A as above, so that (77) are satisfied. The proof of (a) and (b) is over.

(c) If the selfadjoint operator B extending A as in (b) is a standard extension of A, then ${{\mathsf {H}}}={{\mathsf {K}}}$ so that $P_{{{\mathsf {K}}}}=I$ and $Q^{(A)}=P$ so that $Q^{(A)}$ is a PVM. If $Q^{(A)}$ is a PVM, $B := \int _{{\mathbb {R}}}\lambda \mathrm{d}Q^{(A)}(\lambda )$ with domain $D(B)= \{\psi \in {{\mathsf {H}}}\,|\, \int _{{\mathbb {R}}}\lambda ^2 \mathrm{d}Q^{(A)}_{\psi ,\psi }(\lambda ) <+\infty \}$ is a standard selfadjoint extension of A and the associated trivial dilation triple ${{\mathsf {K}}}:={{\mathsf {H}}}$, $P_{{\mathsf {H}}}:=I$, $P:=Q^{(A)}$ trivially generates $Q^{(A)}$ as in (b).

(d) Referring to the proof of (a) and (b) above, for generalized selfadjoint extensions B of kind II, and this choice for B is always feasible when A is closed in view of Theorem 42, we have $D(A)=D(B)\cap {\mathsf {H}}$ and therefore

(78)

Since (77) is valid for every POVM satisfying (76), we have that the identity $D(A)=\{\psi \in {\mathsf {H}}|\;\lambda \in L^2({\mathbb {R}},P_{\psi ,\psi })\}$ holds if and only if , namely B is an extension of type II of A.

(e) and (f). They are established in [1, Thm. 2, p. 135 Vol II], taking Theorem 42 into account and (d) above. $\square$

Proof of Theorem 26

Consider Naimark’s dilation triple of Q, $({\mathsf {K}}, P_{{\mathsf {H}}}, P)$ and define the selfadjoint operator $B := \int _{{\mathbb {R}}}\lambda \mathrm{d}P(\lambda )$. By hypothesis $D(A^{(Q)})= D(B)\cap {\mathsf {H}}$. Since D(B) is a subspace of ${\mathsf {K}}$, it also holds that $D(A^{(Q)})$ is a subspace of ${\mathsf {H}}$ proving (a). A well-known counterexample due to Naimark [1] proves that, in some cases, $D(A^{(Q)}) =\{0\}$ though Q is not trivial. It is clear that, if an operator $A^{(Q)}: D(A^{(Q)}) \rightarrow {\mathsf {H}}$ satisfies (39) then it is unique due to the arbitrariness of $\varphi \in {\mathsf {H}}$. So we prove (b) and (d) simultaneously just checking that satisfies (39). The proof is trivial since, from standard properties of the integral of a PVM, if $\psi \in D(A^{(Q)})\subset D(B)$ and $\varphi \in {\mathsf {H}}$, then

$$\begin{aligned} \left\langle \varphi \left| P_{{\mathsf {H}}} \int _{{\mathbb {R}}} \lambda \mathrm{d}P(\lambda )\psi \right. \right\rangle = \left\langle P_{{\mathsf {H}}}\varphi \left| \int _{{\mathbb {R}}} \lambda \mathrm{d}P(\lambda )\psi \right. \right\rangle = \int _{{\mathbb {R}}}\lambda \mathrm{d}P_{P_{{\mathsf {H}}}\varphi , \psi } (\lambda )= \int _{{\mathbb {R}}}\lambda \mathrm{d}Q_{\varphi , \psi }(\lambda ). \end{aligned}$$

(c) immediately arises with the same argument taking advantage of the fact that $B=B^*$ and $P_{{\mathsf {H}}} \psi = \psi$ if $\psi \in {\mathsf {H}}$. Regarding (e), it is sufficient observing that (see, e.g., [15]), if $\psi \in D(B)$, then $||B\psi ||^2 = \int _{{\mathbb {R}}} \lambda ^2 \mathrm{d}P_{\psi ,\psi }(\lambda )$ and next taking advantage of $D(A^{(Q)})= D(B)\cap {\mathsf {H}}$ and (d) observing, in particular, that $P_{{\mathsf {H}}}^* P_{{\mathsf {H}}}\varphi = P_{{\mathsf {H}}} P_{{\mathsf {H}}}\varphi = P_{{\mathsf {H}}}\varphi =\varphi$ if $\varphi \in {\mathsf {H}}$. The fact that $A^{(Q)}$ is closed immediately follows from the fact that B is closed (because selfadjoint) and $A= B|_{D(B) \cap {\mathsf {H}}}$ where ${\mathsf {H}}$ is closed. $\square$

Proof of Proposition 32

What we have to prove is just that the right-hand side of (54) is a positive semi definite Hermitian scalar product over X. Indeed, with that definition of the scalar product $(\, |\, )_p$, the identity $p(x) = \sqrt{(x|x)_p}$ is valid (see below) and this fact automatically implies that p is a seminorm. Uniqueness of the scalar product generating a seminorm p is a trivial consequence of the polarization identity. Let us prove that $(\, |\, )_p$ defined in (54) is a positive semidefinite Hermitian scalar product over X. From the definition of $(\,|\,)_p$ and property (i) of $p:X\rightarrow [0,+\infty )$, it is trivial business to prove the following facts per direct inspection:

• 1. $(x|x)_p = p(x)^2 \ge 0$,

• 2. $(x|0)_p=0$,

• 3. $(x|y)_p= \overline{(y|x)_p}$,

• 4. $(x|iy)_p = i(x|y)_p$.

With these identities, we can also prove

• 5. $(x|y+z)_p = (x|y)_p + (x|z)_p$,

• 6. $(x|-y)_p = -(x|y)_p$,

by exploiting property (ii) of p. Actually, (6) immediately arises form (2) and (5). We will prove (5) as the last step of this proof. Iterating property (5), we easily obtain $(x|ny)_p = n(x|y)_p$ for every $n\in {\mathbb {N}}$, so that $(1/n) (x|z)_p = (x|(1/n) z)_p$ when replacing ny for z. As a consequence, $\lambda (x|y)_p = (x|\lambda y)_p$ if $\lambda \in \mathbb {Q}$. This results extends to $\lambda \in {\mathbb {R}}$ if the map ${\mathbb {R}} \ni \lambda \mapsto (x|\lambda y)_p$ is right-continuous because, for every $x\in [0,+\infty )$ there is a decreasing sequence of rationals tending to it. Definition (54) of $(x|y)_p$ proves that map is in fact right-continuous since property (iii) of p implies that, if $\lambda _0 \in [0, +\infty )$,

$$\begin{aligned} p(x + \lambda i^k y) = p( (x+ \lambda _0 i^ky) + (\lambda -\lambda _0) i^k y) \rightarrow p(x+ \lambda _0 i^ky) \quad \text{ for } {\mathbb {R}}\ni \lambda \rightarrow \lambda _0^+. \end{aligned}$$

We can therefore add the further property

• 7. $\lambda (x|y)_p = (x|\lambda y)_p$ if $\lambda \in [0,+\infty )$.

Collecting properties (1), (3), (5), (7), (6), (4) together, we obtain that $X\times X \ni (x,y) \mapsto (x|y)_p$ defined as in (54) is a positive semi definite Hermitian scalar product over X whose associated seminorm is p as wanted.

To conclude the proof, we establish property (5) from requirement (ii) on p.

$$\begin{aligned} 4(x|y+z)_p = \sum _{k=0}^3(-i)^k p(x+i^k(y + z))^2= \sum _{k=0}^3(-i)^k p((x/2+i^ky) + (x/2 + i^k z))^2. \end{aligned}$$

Since, from (i) of the requirements on p,

$$\begin{aligned} \sum _{k=0}^3(-i)^k p((x/2+i^ky) - (x/2 + i^k z))^2 = \sum _{k=0}^3(-i)^k p( i^k (y-z))^2 = \sum _{k=0}^3(-i)^k p( (y-z))^2 =0, \end{aligned}$$

we can rearrange the found decomposition of $4(x|y+z)_p$ as

$$\begin{aligned} 4(x|y+z)_p = \sum _{k=0}^3(-i)^k \left[ p((x/2+i^ky) + (x/2 + i^k z))^2 - p((x/2+i^ky) - (x/2 + i^k z))^2\right] . \end{aligned}$$

From (ii) of the requirements on p,

$$\begin{aligned} 4(x|y+z)_p = \sum _{k=0}^3(-i)^k \left[ 2p(x/2+i^ky)^2 + 2p(x/2+i^kz)^2\right] = 8(x/2|y)_p + 8(x/2|z)_p. \end{aligned}$$

The special case $z=0$ and (2) yield $2(x/2|y)_p= (x|y)_p$ which, exploited again above, yields the wanted result (5) $4(x|y+z)_p = 4(x|y)_p + 4(x|z)_p$. $\square$