Causal localizations of the massive scalar boson

Abstract

The positive operator-valued localizations (POL) of a massive scalar boson are constructed, and a characterization and structural analyses of their kernels are obtained. In the focus of this article are the causal features of the POL. There is the well-known causal time evolution (CT). Recently a POL by Terno and Moretti, which is a kinematical deformation of the Newton–Wigner localization (NWL) and belongs to the here fully analyzed class of finite POL, is shown by V. Moretti to comply with CT. A further POL with CT treated here, which is in the same class, is the only one being the trace of a projection-valued localization (like NWL) with CT. Causality imposes a condition CC, which implies CT but is more restrictive than CT. Extending Moretti’s method it is shown rigorously that the POL of the class introduced by Petzold et al. satisfy CC. Their kernels are called causal kernels, of which a rather detailed description is achieved. One the way there the case of one spatial dimension is solved completely. This case is instructive. In particular it directed Petzold et al. and subsequently Henning, Wolf to find their basic one-parameter family $K_r$ of causal kernels. The causal kernels are, up to a fixed energy factor, normalized positive definite Lorentz invariant kernels. A full characterization of the latter is attained due to their close relation to the zonal spherical functions on the Lorentz group. Finally these considerations discharge into the main result that $K_{3/2}$ is the absolute maximum, viz. $|K|\le K_{3/2}$ for all causal kernels K.

1 Introduction

As well-known relativistic massive particle position would be described in a fully satisfactory manner by the Newton–Wigner localization (NWL), since above all NWL provides an abundance of boundedly localized states, if there were not the requirement of causality. The so-called Einstein causality in the most direct interpretation requires from localization that the probability of localization in a region of influence cannot be less than that in the region of actual localization. NWL violates blatantly this principle as it localizes frame-dependently. There is no state of the particle for which the particle is boundedly localized by NWL in different spacelike hyperplanes.

So one is induced to determine the probability of localization in a region no longer by the expectation value of an orthogonal projection attributed to the region as occurs for NWL, but rather of a positive operator with spectrum in the unit interval thus giving rise to a positive operator-valued localization (POL). By now PO-localization is a settled concept. For a brief discussion and several references see, for instance, the passage in [4, sect. III.F].

In Sect. 4 we construct the POL for a massive scalar boson. The result is the explicit formula in the theorem (2), which furnishes myriads of different POL. In order to recognize their features it is very effective to describe a POL in a concise manner by its kernel. A useful characterization and structural analyses of the POL kernels are given by the theorem (11).

To every POL a Euclidean covariant position operator is attributed, which is the first moment of the POL. The physical relevance of the POL is underlined by the fact that in case of a real smooth POL kernel the position operator coincides with the Newton–Wigner position operator. The proof is due to V. Moretti. Recently in [17] a POL $T^{TM}$ by Terno and Moretti is studied in detail showing furthermore that $T^{TM}$ is a kinematical deformation of NWL with almost localized states for every region with non-empty interior and which obeys causal time evolution (see CT in Sect. 10). From general criteria for POL kernels we obtain rather short proofs of these properties, see Sect. 10.1, (48). What is more, (47) provides a simple explicit formula for a sequence of localized states at any point of the space. Actually (47) is a byproduct of the proof of the easy criterion (46) in Sect. 17 on POL kernels for the existence of point localized sequences of states.

In Sect. 10 we treat also the POL with trace CT $T^{tct}$, which has the noted properties of $T^{TM}$ and which is the unique POL being the trace of a projection-valued localization with CT. $T^{tct}$ is also a kinematical deformation of NWL (15), all of which are given by the formula (9.3).

However one cannot content oneself with POL obeying CT as the latter satisfies only partially the requirements of causality. If T is a POL, $\Delta$ a spacelike region, and $\sigma$ any spacelike hyperplane, then as expounded in Sect. 11 causality imposes the condition

$$\begin{aligned} T(\Delta )\le T(\Delta _\sigma ) \end{aligned}$$

(CC)

were $\Delta _\sigma$ denotes the region of influence of $\Delta$ in $\sigma$, i.e., the set of all points in $\sigma$, which can be reached from some point in $\Delta$ by a signal not moving faster than light. CT concerns just the case that $\Delta$ and $\sigma$ are parallel. POL which satisfy CC are called causal.

The causality principle formulated above let one think of the probability of localization as a conserved quantity reigned by an associated density current. Petzold and his group [8] were apparently the first to study the conserved covariant four-vector currents with a positive definite zeroth component. The outcome on POL kernels is reported in Sect. 11.1. See (55) for a clear-cut result in this regard.

Petzold et al.  [9] argue that their POL obey CT. Probably they were the first to introduce and to treat the concept of CT. However they do not take under consideration the full causality requirement CC. Actually, extending Moretti’s method [17] we succeed in (56) to prove rigorously that their POL are causal. Moreover, we mention particularly a communication by R.F. Werner, which suggests that most probably these are the only causal POL. Therefore we refer quite simply to their kernels (11.1) as the causal kernels.

A small digression (Sect. 12) concerns the case of one spatial dimension which we completely solve by the formula in (22). This case is rather instructive since it directed Petzold et al.  [8] and subsequently Henning, Wolf [11] to find the particular causal kernels $K_r$ (11.3), which turn out to be fundamental.

Multiplying a causal kernel by an element of ${\mathcal {G}}$ yields again a causal kernel. Here ${\mathcal {G}}$ denotes the set of all continuous positive definite Lorentz invariant normalized kernels (28). What is more, any causal kernel is the product of some element of ${\mathcal {G}}$ and a fixed energy factor. Therefore it is important to know ${\mathcal {G}}$. In (32) a complete description of ${\mathcal {G}}$ is attained employing means of group representation theory.

Each $G\in {\mathcal {G}}$ determines by the Reproducing Kernel Hilbert Space (RKHS) construction a continuous unitary representation of $SL(2,\mathbb {C})$ in separable Hilbert space (Sect. 14.1). By this construction G is closely related to the matrix element of a SU(2)-invariant vector. Since $SL(2,\mathbb {C})$ is locally compact tame with countable basis, $SL(2,\mathbb {C})$ admits an integral representation of the latter by the zonal spherical functions on $SL(2,\mathbb {C})$ (14.7).

The question is to find out which $G\in {\mathcal {G}}$ determine a causal kernel. By (42) these are only G’s, which are derived from the principal series and are not extreme. We mention the inversion formula in (35) concerning these elements.

At this point we are ready for the main result (43). Put $m>0$ mass, $k,p\in \mathbb {R}^3$ momentum, $\epsilon (k), \epsilon (p)$ energy. The causal kernel

$$\begin{aligned} K_{3/2}(k,p)=\frac{\epsilon (k)+\epsilon (p)}{2\sqrt{\epsilon (k)\epsilon (p)}}\,\, \Big (\frac{2m^2}{m^2+\epsilon (k)\epsilon (p)-kp}\Big )^{3/2} \end{aligned}$$

is maximal. Actually $|K(k,p)| < K_{3/2}(k,p)$ holds for all $k\ne p$ and all causal kernels $K\ne K_{3/2}$ (44). The relation $|K|\le K_{3/2}\le {{\textbf {1}}}$, where ${{\textbf {1}}}$ is the kernel of the NWL distinguished by its abundance of boundedly localized states, suggests that the POL with kernel $K_{3/2}$ is the causal POL with the best localization features. Further studies should work on this distinguished POL, which satisfies several prerequisites. Above all it should be clarified whether there are point localized sequences of states for this POL.

Presumable, like (24) in the case of one spatial dimension, not all $G\in {\mathcal {G}}$ with $G\le G_{3/2}$ (the second factor of $K_{3/2}$) determine a causal kernel. We rather guess that the causal kernels are exactly

$$\begin{aligned} K_r\,G \; for\; G\in {\mathcal {G}}, r\ge 3/2 \end{aligned}$$

where it suffices to consider $r<2$, since $K_r/K_{3/2}\in {\mathcal {G}}$ for $r\ge 2$ by (26).

Finally here is a brief guide to the article.

Sections 2–7 are concerned with the POL and their kernels. Their content is summarized in (2), (6.1), and (11).

Section 9 deals with the class of finite POL. The concluding result is (9.3). In Sect. 10 causal time evolution CT is introduced and two finite POL with CT are studied.

Section 11 deduces the causality condition CC for POL and introduces the class of causal kernels (11.1). There in 11.2 is also an overview of the known results on causal kernels.

Section 12 treats the case of one spatial dimension, which is completely solved by the formula in (22).

Sections  13–16 are concerned with a detailed description of the causal kernels. The main results in this regard are (42), (34), (35), and (43), (44).

Section 17 introduces the concept of a point localized sequence of states. The criterion (46) and its corollary (47) regard all POL kernels.

The discussion in the final Sect. 18 incites to find answers to two outstanding questions of great relevance.

“Appendices C and D” provide space to some cumbersome technical details, whereas “Appendices A and B” house all which regards the CT criterion (52) and the result on CC (56), respectively.

2 Some facts on representations of the Euclidean and Poincaré group

We recall some facts on the reps¹ of the universal covering groups $\tilde{{\mathcal {E}}}=ISU(2)$ and $\tilde{{\mathcal {P}}}=ISL(2,\mathbb {C})$ of the Euclidean group and the Poincaré group, respectively. $\tilde{{\mathcal {P}}}$ acts on $\mathbb {R}^4$ as

$$\begin{aligned} g\cdot {\mathfrak {x}}:={\mathfrak {a}}+\Lambda (A) {\mathfrak {x}}\quad \text { for } g=({\mathfrak {a}},A)\in \tilde{{\mathcal {P}}}, \, {\mathfrak {x}}\in \mathbb {R}^4 \end{aligned}$$

(2.1)

Here $\Lambda :SL(2,\mathbb {C})\rightarrow O(1,3)_0$ is the universal covering homomorphism onto the proper orthochronous Lorentz group. Identifying $\Lambda (SU(2))\equiv SO(3)$, SU(2) acts on $\mathbb {R}^3$. Representing Minkowski space by $\mathbb {R}^4$ the Minkowski product is given by

$$\begin{aligned} {\mathfrak {a}}\cdot {\mathfrak {a}}':=a_0a'_0-a_1a'_1-a_2a'_2-a_3a'_3 \end{aligned}$$

(2.2)

2.1 Irreps of $\tilde{{\mathcal {E}}}$

Up to unitary equivalence the irreducible mutually inequivalent reps of $\tilde{{\mathcal {E}}}$ are $U^{0,j}$ on $\mathbb {C}^{2j+1}$ for $j\in \mathbb {N}_0/2$ and $U^{\rho ,s}$ on $L^2(S_\rho ^2)$ for $\rho >0$, $s\in {\mathbb {Z}}/2$ with the sphere $S_\rho ^2:=\{p\in \mathbb {R}^3:|p|=\rho \}$ endowed with the rotational invariant measure normalized to 1. Explicitly one has

• $U^{0,j}(b,B):=D^{(j)}(B)$ and

• $(U^{\rho ,s}(b,B)f)(p):={\text {e}}^{-{\text {i}}b p} \kappa (p,B)^{2s}f(B^{-1}\cdot p)$

where ${\text {diag}}\big (\kappa (p,B),\overline{\kappa (p,B)} \,\big ):=B(p)^{-1} B\,B(B^{-1}\cdot p)$. Here $B(p)\in SU(2)$ denotes a cross section satisfying $B(p)\cdot e_3=\frac{p}{|p|}$ for all $p\in \mathbb {R}^3\setminus \{0\}$, whence the Wigner rotation on the right hand side leaves $e_3$ invariant and hence is diagonal.²

2.2 Particular reps of $\tilde{{\mathcal {E}}}$

We will be concerned with the reps $U^{(s)}$, $s\in {\mathbb {Z}}/2$, of $\tilde{{\mathcal {E}}}$ on $L^2(\mathbb {R}^3)$ given by

• $(U^{(s)}(b,B)\varphi )(p):={\text {e}}^{-{\text {i}}b p} \kappa (p,B)^{2s}\varphi (B^{-1}\cdot p)$

for $p\ne 0$. Note that $\kappa (0,B)$ is not defined. The obvious unitary equivalence

$$\begin{aligned} U^{(s)}\simeq \int _0^\infty U^{\rho ,s} 4\pi \rho ^2{\text {d}}\rho \end{aligned}$$

(2.3)

yields the decomposition of $U^{(s)}$ into irreps. It shows in particular that $U^{(s)}$ and $U^{(s')}$ for $s\ne s'$ are disjoint, i.e., that any two subreps of the latter are inequivalent.

2.3 Induced reps of $\tilde{{\mathcal {E}}}$ from SU(2)

A projection-valued measure (PM) on $\mathbb {R}^3$, which is Euclidean covariant, is called a Wightman localization (WL). By Mackey’s imprimitivity theorem every WL is Hilbert space isomorphic to a rep of $\tilde{{\mathcal {E}}}$ induced from the subgroup SU(2) together with the related system of imprimitivity.

On momentum space $L^2(\mathbb {R}^3,\mathbb {C}^{2j+1})$ the rep $D^{(j)}_{\tilde{{\mathcal {E}}}}$ of $\tilde{{\mathcal {E}}}$ induced from the irrep $D^{(j)}$, $j\in \mathbb {N}_0/2$, of SU(2) and the system of imprimitivity E read

• $(D^{(j)}_{\tilde{{\mathcal {E}}}}(b,B)\varphi )(p) ={\text {e}}^{-{\text {i}}bp} D^{(j)}(B)\,\varphi (B^{-1}\cdot p), \quad E={\mathcal {F}}E^{can}{\mathcal {F}}^{-1}$

The PM operator $E^{can}(\Delta )$ multiplies by the indicator function $1_\Delta$. ${\mathcal {F}}$ denotes the unitary Fourier integral transformation with kernel $(2\pi )^{-1/2}{\text {e}}^{-{\text {i}}xy}$ in $L^2$. The WL $(D^{(j)}_{\tilde{{\mathcal {E}}}},{\mathcal {F}}E^{can}{\mathcal {F}}^{-1})$, $j\in \mathbb {N}_0/2$, are irreducible, mutually inequivalent, and complete up to unitary equivalence. The decomposition into subreps is obtained by the unitary transformation $X^{(j)}$ on $L^2(\mathbb {R}^3,\mathbb {C}^{2j+1})$

• $(X^{(j)}\varphi )(p):=D^{(j)}(B(p)^{-1})\varphi (p)$

for $p\ne 0$. Due to $D^{(j)}_{ss'}\big (B(p)^{-1} B\,B(B^{-1}\cdot p)\big ) =\kappa (p,B)^{2\,s} \delta _{ss'}$ one yields

$$\begin{aligned} X^{(j)}D^{(j)}_{\tilde{{\mathcal {E}}}}X^{(j)-1}= \oplus _{s}U^{(s)}, \quad s=-j,-j+1,\dots ,j-1,j \end{aligned}$$

(2.4)

which is completed to a WL by the PM $X^{(j)}{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}X^{(j)-1}$.

One concludes that every sum $\oplus _{\iota }U^{(s_\iota )}$ is the subrep of a rep with an Euclidean covariant PM forming an WL. Hence generally this holds true for every rep, which is unitarily equivalent to such a sum.

2.4 Irreps of $\tilde{{\mathcal {P}}}$ for positive masses

The irreps are labeled by mass $m>0$, spin $j\in \mathbb {N}_0/2$, sign $\eta =\pm$. The sign $\eta =-$ denotes the antiparticle case. Hence up to unitary equivalence we consider the mutually inequivalent irreps of $\tilde{{\mathcal {P}}}$ on momentum space $L^2(\mathbb {R}^3,\mathbb {C}^{2j+1})$ given by³

• $\big (W^{m,j,\eta }({\mathfrak {a}},A)\varphi \big )(p)=\sqrt{\epsilon (q^\eta )/\epsilon (p)}\, {\text {e}}^{{\text {i}}{\mathfrak {a}}\cdot \,{\mathfrak {p}}^{\eta }} \,D^{(j)}\big (R({\mathfrak {p}}^\eta ,A)\big )\,\varphi (q^\eta )$

for $p\ne 0$, where $\epsilon (p)=\sqrt{m^2+p^2}$, ${\mathfrak {p}}^{\eta }=(\eta \epsilon (p),p)$, ${\mathfrak {q}}^\eta = (q_0^\eta ,q^\eta ):=A^{-1}\cdot {\mathfrak {p}}^\eta$, and R the Wigner rotation with respect to the canonical cross section satisfying $R( {\mathfrak {p}}^\eta ,B)=B$ for $B\in SU(2)$, see, e.g., [5, sect. 3.3 (1)].

2.5 Decomposition of $W^{m,j,\eta }|_{\tilde{{\mathcal {E}}}}$

By the foregoing considerations it is easy to verify

$$\begin{aligned} W^{m,j,\eta }|_{\tilde{{\mathcal {E}}}}\simeq \oplus _{s}U^{(s)}, \quad s=-j,-j+1,\dots ,j-1,j \end{aligned}$$

(2.5)

3 How to get PO-Localizations

A positive operator-valued measure T on the Borel sets ${\mathcal {B}}(\mathbb {R}^3)$ (POM), which is Euclidean covariant, is called a PO-localization (POL). Recall that a POM E constituted by orthogonal projections is denoted by PM, and a Euclidean covariant PM is called a Wightman localization (WL).

Given a rep W of $\tilde{{\mathcal {P}}}$ and a POM T such that $(W|_{\tilde{{\mathcal {E}}}},T)$ is a POL, then by [5, (9) Theorem] there is a unique Poincaré covariant extension (W, T). This means that there is just one map on ${\mathfrak {S}}$, the set of all Lebesgue measurable subsets of spacelike hyperplanes of Minkowski space, still denoted by T, which satisfies

$$\begin{aligned} W(g)T(\Delta )W(g)^{-1}=T(g\cdot \Delta ) \quad \forall \; g\in \tilde{{\mathcal {P}}}, \;\Delta \in {\mathfrak {S}} \end{aligned}$$

(3.1)

with $g\cdot \Delta =\{g\cdot {\mathfrak {x}}:{\mathfrak {x}}\in \Delta \}$, see (2.1). The elements of ${\mathfrak {S}}$ are called regions.

Henceforth let W be a rep of $\tilde{{\mathcal {P}}}$ with positive mass spectrum and finite spinor dimension. This means equivalently that W is a finite orthogonal sum of reps from Sect. 2.4.

Regarding parts of (1) see, e.g., [25, sect. II] for the general theory. Let (U, E) be a WL acting on the Hilbert space ${\mathcal {K}}$. Let ${\mathcal {H}}$ be a U-invariant subspace of ${\mathcal {K}}$ and $j:{\mathcal {H}}\rightarrow {\mathcal {K}}$ the identical injection. Then the POL $j^*(U,E)j$ is called the trace (or compression) of (U, E) on ${\mathcal {H}}$.

Theorem 1

Let W be a rep of $\tilde{{\mathcal {P}}}$ on ${\mathcal {H}}$.

• (a) There is a Poincaré covariant POL (W, T).

• (b) For every Poincaré covariant POL (W, T), the POL $(W|_{\tilde{{\mathcal {E}}}},T_{{\mathcal {B}}(\mathbb {R}^3)})$ is the trace on ${\mathcal {H}}$ of a WL on a separable Hilbert space.

• (c) Every trace on ${\mathcal {H}}$ of a WL $(U',E)$, where $U'$ extends $W|_{\tilde{{\mathcal {E}}}}$, determines uniquely a Poincaré covariant POL (W, T).

Proof

• (a) Let $U:=W|_{\tilde{{\mathcal {E}}}}$. By Sect. 2.5, $U\simeq \oplus \, U^{(s)}$, which is a finite sum considering multiplicities. Therefore by Sect. 2.3 there is a WL $(U',E)$ such that U is a subrep of $U'$. So the trace of $(U',E)$ on ${\mathcal {H}}$ is a POL (U, T), which by [5, (9) Theorem] extends uniquely to a Poincaré covariant POL (W, T).

• (b) This is an application of the Principal Theorem on dilations in [20, Appendix sec. 6] (see, e.g.,  [22, 1]).

• (c) See again [5, (9) Theorem].

$\square$

4 POL for a Massive Scalar Boson

We construct all Poincaré covariant POL $(W^{m,0,\eta },T)$, $m>0$ following (1). For the result see (2).

By (2.5), $W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}=U^{(0)}$. According to Sect. 2.3 the extensions $U'$ of $U^{(0)}$ to deal with are Hilbert space isomorphic with $D_{\tilde{{\mathcal {E}}}}:=\oplus _j\, \nu _j D^{(j)}_{\tilde{{\mathcal {E}}}}$, which are countable sums considering multiplicities $\nu _j\in \mathbb {N}_0\cup \{\infty \}$.⁴ Due to (1)(c) it is no restriction to assume at once $\nu _j=\infty$ for all j. Obviously half-integer spins do not contribute, whence $j\in \mathbb {N}_0$ suffices. So $D_{\tilde{{\mathcal {E}}}}$ acts on $L^2(\mathbb {R}^3,{\mathcal {S}})$, where the outer orthogonal sum

$$\begin{aligned} {\mathcal {S}}:=\oplus _{j\in \mathbb {N}_0}\infty \mathbb {C}^{2j+1} \end{aligned}$$

is a countably dimensional Hilbert space called spinor space, by

$$\begin{aligned} (D_{\tilde{{\mathcal {E}}}}(b,B)\varphi )(p)={\text {e}}^{-{\text {i}}bp}D(B)\varphi (B^{-1}\cdot p) \text { with } D:=\oplus _j\, \infty D^{(j)} \end{aligned}$$

Applying (2.4), $D_{\tilde{{\mathcal {E}}}}$ is diagonalized by

$$\begin{aligned} X:=\oplus _j\infty X^{(j)} \end{aligned}$$

yielding $XD_{\tilde{{\mathcal {E}}}}X^{-1}=\oplus _j\infty \oplus _{s=-j}^{j} U^{(s)}$. Hence by Sect. 2.2 its component for $U^{(0)}$ is $\oplus _j\infty U^{(0)}$ with carrier space $L^2(\mathbb {R}^3,{\mathcal {S}}_0)$, where ${\mathcal {S}}_0:= \oplus _j\infty \oplus _{s=0}\mathbb {C}\subset {\mathcal {S}}$. The subspaces of $L^2(\mathbb {R}^3,{\mathcal {S}})$, which are invariant under $XD_{\tilde{{\mathcal {E}}}}X^{-1}$ and reduce $XD_{\tilde{{\mathcal {E}}}}X^{-1}$ to a rep equivalent to $U^{(0)}$, are

$$\begin{aligned} V_{{\text {e}}}:=\{\varphi \in L^2(\mathbb {R}^3,{\mathcal {S}}): \varphi (p)=\phi (p){\text {e}}(|p|),\,\phi \in L^2(\mathbb {R}^3)\} \end{aligned}$$

for every measurable

$$\begin{aligned} {\text {e}}:]0,\infty [\rightarrow {\mathcal {S}}_0 \text { with } \parallel \!\!{\text {e}}(\rho )\!\!\parallel =1 \text { for all } \rho \end{aligned}$$

This is easy to verify because of Sect. 2.2. Let the map ${\text {e}}$ be called a spinor choice.

Obviously the invariant subspaces $V_{{\text {e}}}$ and $V_{{\text {e}}'}$ for the spinor choices ${\text {e}}$ and ${\text {e}}'$ are equal if and only if ${\text {e}}$ and ${\text {e}}'$ are equivalent, i.e., ${\text {e}}'(\rho )=s(\rho ){\text {e}}(\rho ) \text { a.e with } |s(\rho )|=1$.

Now recall $W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}=U^{(0)}$. We embed the carrier space of $U^{(0)}$ in $L^2(\mathbb {R}^3,{\mathcal {S}})$ mapping it onto $V_{{\text {e}}}$ by the injection

$$\begin{aligned} j_{{\text {e}}}:L^2(\mathbb {R}^3)\rightarrow L^2(\mathbb {R}^3,{\mathcal {S}}),\quad j_{{\text {e}}}\phi :=\phi {\text {e}}(|\cdot |) \end{aligned}$$

Recall further that $U^{(0)}$ is a subrep of $U'$. Denote by $P_0$ the projection onto the carrier space of $U^{(0)}$. Then $U^{(0)}=P_0U'P_0^*$. Now let $\iota$ be a Hilbert space isomorphism satisfying $U'=\iota D_{\tilde{{\mathcal {E}}}}\iota ^{-1}$ and put $E':=\iota {\mathcal {F}}E^{can}{\mathcal {F}}^{-1}\iota ^{-1}$. Then $(U',E')$ is a WL, and its trace $(U^{(0)}, T)$ for $T:=P_0 E' P_0^*$ is the POL we are looking for.

Since $U^{(0)}=P_0U'P_0^*=(P_0\iota X^{-1}) \,(XD_{\tilde{{\mathcal {E}}}}X^{-1})\,(X\iota ^{-1}P_0^*)$ it follows $X\iota ^{-1}P_0^*=j_{{\text {e}}}$ for some spinor choice ${\text {e}}$. Indeed, first it follows that $j:=X\iota ^{-1}P_0^*$ is an isometry from $L^2(\mathbb {R}^3)$ into $L^2(\mathbb {R}^3,{\mathcal {S}})$ with final space $V_{{\text {e}}'}$ for some spinor choice ${\text {e}}'$. Hence $j\phi =(S\phi ) {\text {e}}'(|\cdot |)$ for some unitary S on $L^2(\mathbb {R}^3)$. Therefore $(XD_{\tilde{{\mathcal {E}}}}(b,B)X^{-1})\,j\phi =U^{(0)}(b,B)(S\phi ) {\text {e}}'(|\cdot |)$ and further $j^*\,(XD_{\tilde{{\mathcal {E}}}}(b,B)X^{-1})\,j=S^*U^{(0)}(b,B)S$. So S commutes with $U^{(0)}$. This implies $(S\phi )(p)=s(|p|)\phi (p)$ for some $s:]0,\infty [\rightarrow \mathbb {C}$ measurable, $|s(\rho )|=1$ for all $\rho$. One concludes $j=j_{{\text {e}}}$ for ${\text {e}}:=s(|\cdot |){\text {e}}'(|\cdot |)$. Hence T coincides with $T_{{\text {e}}}$ in (4.1). We summarize.

Theorem 2

The POL covariant with respect to $W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}$ are

$$\begin{aligned} T_{{\text {e}}}:=j_{{\text {e}}}^*X{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}X^{-1}j_{{\text {e}}} \end{aligned}$$

(4.1)

where ${\text {e}}$ runs through all spinor choices. Extending $T_{{\text {e}}}$ to ${\mathfrak {S}}$ by [5, (9) Theorem], the Poincaré covariant POL are $(W^{m,0,\eta },T_{{\text {e}}})$.

Definition 3

The finite dimensional spinor spaces, considered as finite dimensional D-invariant subspaces of ${\mathcal {S}}$, are

$$\begin{aligned} {\mathcal {S}}^{fin}:=\oplus _{j=0}^J\oplus _j \nu _j\mathbb {C}^{2j+1} \end{aligned}$$

for $J\in \mathbb {N}_0, \nu _j\in \mathbb {N}_0$ with $\dim ({\mathcal {S}}^{fin}) =\sum _{j=0}^J\nu _j (2j+1)$.

A POL T is called finite if it is the trace of an WL with finite dimensional spinor space. The spinor choice ${\text {e}}$ is called finite if the range of ${\text {e}}$ lies in a finite dimensional spinor space ${\mathcal {S}}^{fin}$.

Obviously a POL T is finite if and only if $T=T_{{\text {e}}}$ for some finite spinor choice ${\text {e}}$. We will be concerned with finite POL in Sect. 9 providing an equivalent characterization of the latter as kinematic deformations of the NWL. This expression is due to Moretti [17] thus calling a POL proposed by Terno [23]. We give a rigorous definition of this term (14). Besides the POL of Terno–Moretti in Sect. 10 a further physically relevant finite POL is treated.

We conclude this section by some remarks on the assignment ${\text {e}}\rightarrow T_{{\text {e}}}$. If ${\text {e}}$ and ${\text {e}}'$ are equivalent, then $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}})$ and $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}'})$ are unitarily equivalent. Indeed, by assumption ${\text {e}}'(\rho )=s(\rho ){\text {e}}(\rho ) \text { a.e. with } |s(\rho )|=1$. Since $s(\rho )=\langle {\text {e}}(p),{\text {e}}'(\rho )\rangle$ a.e., s is measurable. Therefore $(S\phi )(p):=s(|p|)\phi (p)$ defines a unitary operator on $L^2(\mathbb {R}^3)$, which commutes with $U^{(0)}=W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}$. Moreover, $j_{{\text {e}}'}=j_{{\text {e}}}S$. Hence $S^{-1}(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}})S=(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}'})$.

Conversely, let $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}})$ and $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}'})$ be unitarily equivalent. Let the equivalence be effected by the unitary operator S. Then S and $U^{(0)}=W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}$ commute, whence $(S\phi )(p):=s(|p|)\phi (p)$ for some measurable s with $|s(\rho )|=1$ for $\rho >0$. It follows $S^{-1}(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}})S=(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\tilde{{\text {e}}}}})$, where ${\tilde{{\text {e}}}}$ given by ${\tilde{{\text {e}}}}(\rho ):=s(\rho ){\text {e}}(\rho )$ is equivalent with ${\text {e}}$. However this does not imply ${\tilde{{\text {e}}}}={\text {e}}'$ a.e. In other words, ${\text {e}}\mapsto T_{{\text {e}}}$ is not injective. See the following general example and the concrete example in Sect. 10.2.1. A definite answer to this question is given in (8) and (11).

Let C be a unitary operator on $L^2(\mathbb {R}^3,{\mathcal {S}})$, $(C\varphi )(p):=C_0\varphi (p)$, where $C_0$ is unitary on ${\mathcal {S}}$ such that $C_0=\oplus _jC_{0,j}$ with $C_{0,j}$ acting on the multiplicity space of $D^{(j)}_{\tilde{{\mathcal {E}}}}$. Then C and $X{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}X^{-1}$ commute and $Cj_{{\text {e}}}=j_{{\text {e}}'}$ for ${\text {e}}':=C_0{\text {e}}$. Thus $T_{{\text {e}}}=T_{{\text {e}}'}$.

5 WL and NWL

NWL is for Newton–Wigner localization. Let $(W^{m,0,\eta },E)$ be a Poincaré covariant WL. Recall $W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}=D^{(0)}_{\tilde{{\mathcal {E}}}}$. According to the imprimitivity theorem reported in Sect. 2.3, $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}},E)=S( D^{(0)}_{\tilde{{\mathcal {E}}}},{\mathcal {F}}E^{can}{\mathcal {F}}^{-1})S^{-1}$ for some unitary S. This implies that S and $D^{(0)}_{\tilde{{\mathcal {E}}}}=U^{(0)}$ commute, whence $(S\phi )(p)=s(|p|)\phi (p)$ with measurable $s:]0,\infty [\rightarrow \mathbb {C}$, $|s(\rho )|=1$. So $E=S{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}S^{-1}$. Hence referring to (4.1), ${\mathcal {S}}_0\equiv \mathbb {C}$ and

$$\begin{aligned} E=j_{{\text {e}}}^*{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}j_{{\text {e}}} \text { for } {\text {e}}=s \end{aligned}$$

The NWL $E^{NW}$ is given by ${\text {e}}=1$. About the uniqueness of $E^{NW}$ see [5, sect. 2].

6 Kernels of the POL

Let T be any POL of a massive scalar boson. Then there is a spinor choice ${\text {e}}$ with $T=T_{{\text {e}}}$. From (4.1) it follows for $\phi \in L^2(\mathbb {R}^3)\cap L^1(\mathbb {R}^3)$

$$\begin{aligned} \langle \phi ,T(\Delta )\phi \rangle =(2\pi )^{-3}\int _\Delta \int \int \, \textsc {t}(k,p) {\text {e}}^{{\text {i}}(p-k)x}\overline{\phi (k)}\phi (p) \,{\text {d}}^3k\, {\text {d}}^3p\, {\text {d}}^3x\nonumber \\ \end{aligned}$$

(6.1)

for every region $\Delta \subset \mathbb {R}^3$, where the kernel $\textsc {t}=\textsc {t}_{{\text {e}}}$ of $T_{{\text {e}}}$ reads

$$\begin{aligned} \textsc {t}_{{\text {e}}}(k,p):=\langle D(B(k)){\text {e}}(|k|),D(B(p)){\text {e}}(|p|)\rangle =\langle {\text {e}}(|k|),D\big (B(k)^{-1}B(p)\big ){\text {e}}(|p|)\rangle \nonumber \\ \end{aligned}$$

(6.2)

Obviously $\textsc {t}_{{\text {e}}}$ is a measurable positive definite kernel⁵ on $\mathbb {R}^3\setminus \{0\}$ with $\textsc {t}_{{\text {e}}}(p,p)=1$ for all p. It is also rotation invariant, i.e., $\textsc {t}_{{\text {e}}}(Rk,Rp)= \textsc {t}_{{\text {e}}}(k,p)$ for all rotations R, as we will see explicitly below. Clearly T and its kernel $\textsc {t}$ determine each other.

For later use we insert (4) and (5). Recall the definition (3).

Corollary 4

Let $(b^{(l)})_l$ be an orthonormal basis of ${\mathcal {S}}$. Then

$$\begin{aligned} \textsc {t}_{{\text {e}}}(k,p)=\sum _l \overline{f_l(k)}\,f_{l}(p) \end{aligned}$$

(1)

holds for all $k,p\ne 0$ and $f_l(p):=\langle \, b^{(l)},D\big (B(p)\big ){\text {e}}(|p|)\,\rangle$. If ${\text {e}}$ is a finite spinor choice then (1) holds for an orthonormal basis $(b^{(l)})_l$ of ${\mathcal {S}}^{fin}$, and the sum is finite.

Definition 5

Let T be a POL and $\textsc {t}$ its kernel. $\textsc {t}$ is called finite if

$$\begin{aligned} \textsc {t}(k,p)=\sum _l \overline{f_l(k)}\,f_{l}(p) \end{aligned}$$

holds for all $k,p\ne 0$ with finitely many measurable $f_l:\mathbb {R}^3\rightarrow \mathbb {C}$.

For an explicit expression of $\textsc {t}_{{\text {e}}}$ (6.2) choose an orthonormal basis $(b^{(j,n)})_{j,n}$ of ${\mathcal {S}}_0$ according to the orthogonal sum ${\mathcal {S}}_0= \oplus _j\infty \mathbb {C}$ so that ${\text {e}}(\rho )=\sum _{j=0}^\infty \sum _{n=1}^{\infty }e_{j,n}(\rho )\, b^{(j,n)}$ with $e_{j,n}(\rho ):= \langle b^{(j,n)},{\text {e}}(\rho )\rangle$ holds for every spinor choice ${\text {e}}$. If ${\text {e}}$ is a finite spinor choice then, adapting the enumeration, $e_{j,n}=0$ if $j>J$ or $n>\nu _j$ for some j.

Corollary 6

Every set of measurable functions $0<\rho \mapsto e_{j,n}(\rho )\in \mathbb {C}$ with $\sum _j\sum _n|e_{j,n}(\rho )|^2=1$ determines a kernel of a POL for the massive scalar boson represented by $W^{m,0,\eta }$ and vice versa by

$$\begin{aligned} \textsc {t}_{{\text {e}}}(k,p)=\sum _{j=0}^\infty \Big (\sum _{n=1}^{\infty } \overline{e_{j,n}(|k|)}\,e_{j,n}(|p|)\Big )D^{(j)}(B(k)^{-1}B(p))_{00} \end{aligned}$$

(6.3)

for $p\ne 0,k\ne 0$, where

$$\begin{aligned}{} & {} D^{(j)}(B(k)^{-1}B(p))_{00}\nonumber \\{} & {} =2^{-j}\sum _{i=0}^{j} {j\atopwithdelims ()i}^2 \Big (1+\frac{kp}{|k||p|}\Big )^{j-i}\Big (-1+\frac{kp}{|k||p|}\Big )^i= P_j\Big (\frac{kp}{|k||p|}\Big ) \end{aligned}$$

(6.4)

Here Szegö’s notation for the Legendre polynomials $P_n=P^{(0,0)}_n$ [6, 10.10] is used.

Proof

Recall the defining relation ${\text {e}}(\rho )=\sum _{j=0}^\infty \sum _{n=1}^{\infty }e_{j,n}(\rho )\, b^{(j,n)}$ for the first part. It remains to show (6.4). Note $D^{(j)}(M)_{00}=\sum _{i=0}^{j} {j\atopwithdelims ()i}^2(M_{11}M_{22})^{j-i}(M_{12}M_{21})^i$ for $j \in \mathbb {N}_0$ and $M\in M_2(\mathbb {C})$, whence $D^{(j)}(M)_{00}=\sum _{i=0}^{j} {j\atopwithdelims ()i}^2|M_{11}|^{2(j-i)}\big (|M_{11}|^{2}-1\big )^i$ for $M\in SU(2)$. So using the explicit expression for B(p) (see Sect. 2.3), one finds $|M_{11}|^2=\frac{1}{2}\big (1+\frac{kp}{|k||p|}\big )$ for $M:=B(k)^{-1}B(p)$. $\square$

Corollary 7

In case of a WL E (see Sect. 5) one has $\textsc {e}_{{\text {e}}}(k,p)=\overline{s({|k|})}s(|p|)$ with $|s|=1$, and the kernel of the NWL $E^{NW}$ is constant 1.

Recall $|P_n(x)|\le 1$ for $|x|\le 1$. The series $\big (\sqrt{n+\frac{1}{2}}P_n\big )_n$ is an ONB of $L^2(-1,1)$. The first three Legendre polynomials read

$$\begin{aligned} P_0(x)=1,\quad P_1(x)=x,\quad P_2(x)=\frac{3}{2}x^2-\frac{1}{2} \end{aligned}$$

For $j\in \mathbb {N}_0$, $\sigma ,\rho >0$ put

$$\begin{aligned} k_j(\sigma ,\rho ):=\sum _{n=1}^{\infty }\overline{e_{j,n}(\sigma )}\,e_{j,n}(\rho ) \end{aligned}$$

Clearly $k_j$ is a measurable positive definite kernel on $]0,\infty [$. By Cauchy-Schwarz inequality $|k_j(\sigma ,\rho )|\le \sqrt{k_j(\sigma ,\sigma )}\sqrt{k_j(\rho ,\rho )}$, whence again by Cauchy-Schwarz inequality $\sum _j |k_j(\sigma ,\rho )| \le \sum _j k_j(\sigma ,\sigma ) \sum _j k_j(\rho ,\rho )=1$ as $\sum _{j,n}|e_{j,n}(\cdot )|^2=1$. So for all $\sigma ,\rho >0$

$$\begin{aligned} \sum _j k_j(\rho ,\rho )=1,\quad \sum _j |k_j(\sigma ,\rho )|\le 1 \end{aligned}$$

(6.5)

In view of the comment on (2) we note

Corollary 8

Let ${\text {e}}$ and ${\text {e}}'$ be two spinor choices. Suppose $\textsc {t}_{{\text {e}}}=\textsc {t}_{{\text {e}}'}$. Then for every $j\in \mathbb {N}_0$ and all $(\sigma ,\rho )$

$$\begin{aligned} \sum _{n=1}^{\infty }\overline{e_{j,n}(\sigma )}\,e_{j,n}(\rho )= \sum _{n=1}^{\infty }\overline{e'_{j,n}(\sigma )}\,e'_{j,n}(\rho ) \end{aligned}$$

Proof

For fixed $\sigma >0$, $\rho >0$, consider $f:[-1,1]\rightarrow \mathbb {C}$, $f:=\sum _j \lambda _j (j+\frac{1}{2})^{1/2}P_j$ with $\lambda _j:=(j+\frac{1}{2})^{-1/2}k_j(\sigma ,\rho )$ and similarly $f'\,=\sum _j \lambda '_j P_j$. Check $f(x)=\textsc {t}_{{\text {e}}}(k,p)$ and $f'(x)=\textsc {t}_{{\text {e}}'}(k,p)$ for $k:=\sigma \,(1,0,0)$ and $p:=\rho \,(x,\sqrt{1-x^2},0)$ by (6.3), (6.4). Hence by assumption $f=f'$. This implies $\lambda _j=\lambda '_j$ being the coefficients of the expansion in orthonormal Legendre polynomials in $L^2(-1,1)$. $\square$

7 Characterization of the POL kernels

For a map

$$\begin{aligned} K:\mathbb {R}^3\setminus \{0\} \times \mathbb {R}^3\setminus \{0\} \rightarrow \mathbb {C}\end{aligned}$$

(7.1)

consider the properties (i) to be measurable, (ii) to be normalized, i.e., $K(p,p)=1$ for all p, (iii) to be a positive definite kernel, (iv) to be a positive definite separable kernel, i.e., the RKHS associated to K (see, e.g.,  [7]) is separable, and (v) to be rotation invariant, i.e., $K(Rk,R p)=Kk,p)$ for all k, p and rotations R.⁶ Recall that (iii) implies

$$\begin{aligned} |K(k,p)|^2\le K(k,k)K(p,p) \end{aligned}$$

(7.2)

whence if (ii) holds, $|K|\le 1$.

Proposition 9

Let K satisfy (i)-(iv). Then there is a measurable feature map J on $\mathbb {R}^3{\setminus }\{0\}$ in a separable Hilbert space ${\mathcal {H}}_J$ for K, i.e., $K(k,p)=\langle J_k,J_p\rangle$, such that

$$\begin{aligned} T=j^*{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}j \end{aligned}$$

(7.3)

with $j: L^2(\mathbb {R}^3)\rightarrow L^2(\mathbb {R}^3,{\mathcal {H}}_J)$, $(j\phi )(p):=\phi (p)J_p$, is a POM T on $L^2(\mathbb {R}^3)$ with kernel K. It is translation covariant with respect to the rep $(U^{(0)}(b)\phi )(p)={\text {e}}^{-{\text {i}}bp}\phi (p)$ of translations b, i.e., $U^{(0)}(b)T(\Delta )U^{(0)}(b)^{-1}=T(b+\Delta )$.

Proof

Let ${\mathcal {H}}_J$ denote the RKHS associated to the kernel K and put $J_p:=K(\cdot , p)$. Since $\parallel \!\!J_p\!\!\parallel ^2=K(p,p)=1$, j is an injection. Note that j is well defined just because ${\mathcal {H}}_J$ is separable. Hence T is a POM. Its kernel is K, since $\langle \phi ,T(\Delta )\phi \rangle = \langle j \phi ,{\mathcal {F}}E^{can}(\Delta ){\mathcal {F}}^{-1} j\phi \rangle = (2\pi )^{-3}\int _\Delta \int \int \langle J_k,J_p\rangle \,{\text {e}}^{{\text {i}}(p-k)x}\overline{\phi (k)}\phi (p){\text {d}}^3k\,{\text {d}}^3p\, {\text {d}}^3x$ for $\phi \in L^{2}\cap L^1$, $\Delta \subset \mathbb {R}^3$ measurable. Hence translation covariance of T easily follows. $\square$

Obviously every POL kernel $\textsc {t}_{{\text {e}}}$ satisfies (i)-(v). In particular see (6.2). The converse holds true, too.

Corollary 10

Let K satisfy (i)-(v). Then (7.3) is a Euclidean covariant POM with respect to $U^{(0)}$. Hence $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}},T)$ is a POL for the massive scalar boson and K is its kernel.

Proof

Recall $\big (U^{(0)}(B)\phi \big )(p)=\phi (B^{-1}\cdot p)$. It suffices to check rotation covariance $U^{(0)}(B)T(\Delta )U^{(0)}(B)^{-1}=T(B\cdot \Delta )$. This is easy using the formula on $\langle \phi ,T(\Delta )\phi \rangle$ in the proof of (9). $\square$

Let K be rotation invariant. Introduce the map

$$\begin{aligned}{} & {} \textsc {k}:\;]0,\infty [\times ]0,\infty [ \times [-1,1]\rightarrow \mathbb {C},\quad \textsc {k}(\sigma ,\rho ,x)\nonumber \\{} & {} \quad :=K\big (\sigma (0,0,1),\rho (0,\sqrt{1-x^2},x)\big ) \end{aligned}$$

(7.4)

Due to rotation invariance one has $K(k,p)=\textsc {k}\big (|k|,|p|,\frac{kp}{|k||p|}\big )$ for all $k,p\ne 0$.

Theorem 11

The kernels of the POL of the massive scalar boson are exactly the measurable normalized rotational invariant positive definite separable kernels K on $\mathbb {R}^3\setminus \{0\}$. They are given by

$$\begin{aligned} K(k,p)=\sum _{j=0}^\infty k_j(|k|,|p|)P_j\left( \frac{kp}{|k||p|} \right) \end{aligned}$$

(7.5)

for all $k,p\ne 0$ with $(k_j)_{j=0,1,2,\dots }$ any sequence of measurable positive definite kernels $k_j$ on $]0,\infty [$ (not excluding $k_j=0$ for some j) satisfying $\sum _jk_j(\rho ,\rho )=1$ for all $\rho$. K and $(k_j)_j$ determine each other uniquely. The sum converges everywhere. Recall (7.4). One has

$$\begin{aligned} k_j(\sigma ,\rho )=\int _{-1}^1\textsc {k}(\sigma ,\rho ,x) \,(j+\frac{1}{2})P_j(x){\text {d}}x,\; j\in \mathbb {N}_0 \end{aligned}$$

Proof

In view of (10) and (6) it remains to add the following consideration regarding the POL kernels. Choose a measurable feature map $\Phi _j$ for $k_j$ (see, e.g.,  [7]) and let $e_{j,n}(\rho )$ be the coefficients of $\Phi _j(\rho )$ with respect to some ONB. Then $k_j(\sigma ,\rho )=\langle \Phi _j(\sigma ), \Phi _j(\rho )\rangle =\sum _{n=1}^{\infty }\overline{e_{j,n}(\sigma )}\, e_{j,n}(\rho )$ for all $\sigma ,\rho , j$. Further by the assumption on $(k_j)$ one has $\sum _{j,n}|e_{j,n}(\rho )|^2=1$ for all $\rho$. Let $\textsc {t}_{{\text {e}}}$ denote the kernel corresponding to $(e_{j,n})$ according to (6). Note $\textsc {k}(\sigma ,\rho ,\cdot )=\sum _jk_j(\sigma ,\rho )P_j$ in $L^2(-1,1)$, where the sum converges everywhere due to (6.3). So K is the kernel of $T_{{\text {e}}}$ and hence equals $\textsc {t}_{{\text {e}}}$. $\square$

Corollary 12

Fix $\sigma ,\rho >0$. Then $\big (k_j(\sigma ,\rho )\big )_j\in \ell ^1$, $\textsc {k}(\sigma ,\rho ,x)=\sum _jk_j(\sigma ,\rho )P_j(x)$ for $|x|\le 1$, and $\textsc {k}(\sigma ,\rho ,\cdot )$ is continuous on $[-1,1]$.

Proof

Recall (6.5) and $|P_j(x)|\le 1$. Then continuity of $\textsc {k}(\sigma ,\rho ,\cdot )$ holds by dominated convergence. $\square$

Clearly the sequence of the coefficients of the expansion of $\textsc {k}(\sigma ,\rho ,\cdot )$ in orthonormalized Legendre polynomials is $\big ((j+\frac{1}{2})^{-1/2}k_j(\sigma ,\rho )\big )_j$ with $\big (k_j(\sigma ,\rho )\big )_j\in \ell ^1$.

8 Change to shell rep

In place of the momentum reps $W^{m,j,\eta }$ of $\tilde{{\mathcal {P}}}$ from Sect. 2.4 frequently it is convenient, because of the simpler transformation formulae, to use the unitarily equivalent shell reps on $L^2({\mathcal {O}}^{m,\eta },\mathbb {C}^{2j+1})$ of functions on the mass shell ${\mathcal {O}}^{m,\eta }:=\{{\mathfrak {p}}\in \mathbb {R}^4:p_0=\eta \epsilon (p)\}$ equipped with the Lorentz invariant measure

• $\big (W^{shell, m,j,\eta }({\mathfrak {a}},A)\Phi \big )({\mathfrak {p}})={\text {e}}^{{\text {i}}{\mathfrak {a}} \cdot {\mathfrak {p}}}D^{(j)}\big (R({\mathfrak {p}},A)\big )\,\Phi (A^{-1}\cdot {\mathfrak {p}})$

The Hilbert space isomorphism from $L^2({\mathcal {O}}^{m,\eta },\mathbb {C}^{2j+1})$ onto $L^2(\mathbb {R}^3,\mathbb {C}^{2j+1})$

• $(X^{m,\eta }\Phi )(p)=\epsilon (p)^{-1/2}\Phi ({\mathfrak {p}}^\eta )$

satisfies $W^{m,j,\eta }=X^{m,\eta }W^{shell, m,j,\eta }(X^{m,\eta })^{-1}$. Note that for this $X^{m,\eta }$ is unique up to a constant factor of modulus 1.

Hence in the case of the massive scalar Boson the Poincaré covariant POM $(W^{shell, m,0,\eta },T^{shell}_{{\text {e}}})$ are given by (cf. (4.1))

$$\begin{aligned} T^{shell}_{{\text {e}}}:= & {} (X^{m,\eta })^{-1}T_{{\text {e}}} X^{m,\eta }\nonumber \\= & {} (X^{m,\eta })^{-1} j_{{\text {e}}}^*X{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}X^{-1}j_{{\text {e}}}X^{m,\eta }, \quad {\text {e}}\text { spinor choice} \end{aligned}$$

(8.1)

Then for $\Phi , \Phi '$ square integrable and $\sqrt{\epsilon }\, \Phi$, $\sqrt{\epsilon }\, \Phi '$ integrable

$$\begin{aligned}{} & {} \langle \Phi ,T^{shell}_{{\text {e}}}(\Delta )\,\Phi \rangle \nonumber \\{} & {} \quad =(2\pi )^{-3}\int _\Delta \int \int \textsc {t}^{shell}_{{\text {e}}}({\mathfrak {k}}^{\,\eta },{\mathfrak {p}}^\eta )\, {\text {e}}^{{\text {i}}(p-k)x}\overline{\Phi '({\mathfrak {k}}^{\,\eta })}\Phi ({\mathfrak {p}}^\eta ) \frac{{\text {d}}^3k}{\epsilon (k)}\, \frac{{\text {d}}^3p}{\epsilon (p)}\, {\text {d}}^3x\nonumber \\ \end{aligned}$$

(8.2)

determines the kernel $\textsc {t}^{shell}_{{\text {e}}}$. Due to $\langle \Phi ',T^{shell}_{{\text {e}}}[g]\,\Phi \rangle =\langle X^{m,\eta }\Phi ',T_{{\text {e}}}X^{m,\eta }\Phi \rangle$ one gets immediately

$$\begin{aligned} \textsc {t}^{shell}_{{\text {e}}}({\mathfrak {k}},{\mathfrak {p}})= \sqrt{\epsilon (k)\epsilon (p)}\,\,\textsc {t}_{{\text {e}}}(k,p) \end{aligned}$$

(8.3)

In particular with respect to the shell representation the kernel of the NWL $E^{NW}$ reads $\sqrt{\epsilon (k)\epsilon (p)}$.

9 How to get a POL from Newton–Wigner Localization

Moretti [17] proves a formula for the POL proposed by Terno [23] (Sect. 10.1) calling it a kinematic deformation of the NWL. Generalizing we like to use Moretti’s expression for a certain kind of POL (14). It turns out that just the finite POL (3) are of this kind (15).

For the momentum operator $P=(P_1,P_2,P_3)$ and a bounded measurable function $f:\mathbb {R}^3\rightarrow \mathbb {C}$ let f(P) be the related operator defined by the functional calculus. On momentum state space $L^2(\mathbb {R}^3,\mathbb {C})$, f(P) is the multiplication operator by f given by $(f(P)\phi )(p):=f(p)\phi (p)$. Obviously $f(P)^*E^{NW}f(P)$, i.e., $\Delta \rightarrow f(P)^*E^{NW}(\Delta )f(P)$, is a POM with value $||f||_\infty ^2 I$ at $\mathbb {R}^3$.

Lemma 13

The kernel of $f(P)^*E^{NW}f(P)$ reads $(k,p)\mapsto \overline{f(k)}f(p)$.

Proof

As $\langle \phi , f(P)^*E^{NW}(\Delta )f(P)\phi \rangle =\langle f(P)\phi , E^{NW}(\Delta )f(P)\phi \rangle =\langle f\phi , E^{NW}(\Delta )(f\phi )\rangle$ the result follows from (6.1) and (7). $\square$

Definition 14

A POL T is called a kinematic deformation of the NWL if

$$\begin{aligned} T=\sum _l \,f_l(P)^*\,E^{NW}\,f_l(P) \end{aligned}$$

(9.1)

holds for some finitely many measurable bounded $f_l:\mathbb {R}^3\rightarrow \mathbb {C}$.

Theorem 15

Let T be a POL and $\textsc {t}$ its kernel. Then the statements (a), (b), and (c) are equivalent.

• (a) T is finite (3).

• (b) $\textsc {t}$ is finite (5).

• (c) T is a kinematic deformation of the NWL (14).

Proof

Assume (a). Then by (4) $\textsc {t}(k,p)=\sum _l \,\overline{f_l(k)}\,f_l(p)$ for all $p,k\ne 0$ with some finitely many measurable bounded $f_l:\mathbb {R}^3\rightarrow \mathbb {C}$ thus showing (b). (b) implies (c) since (9.1) follows by (13).

Now assume (c). Then (9.1) and (13) imply $\textsc {t}(k,p)=\sum _{l=1}^L \,\overline{f_l(k)}\,f_l(p)$ for all $k,p\ne 0$, where $\textsc {t}$ is the kernel of T. We enter into the proof of (9) for $K=\textsc {t}$. The feature map J reads $J_p=\textsc {t}(\cdot ,p)=(f_1(p),\dots ,f_L(p))\in \mathbb {C}^L$, whence ${\mathcal {H}}_J$ is a subspace of $\mathbb {C}^L$. For $B\in SU(2)$, due to the rotational invariance of $\textsc {t}$, $||\sum _i\lambda _iJ_{B\cdot p_i}||^2= \sum _{i,j}\overline{\lambda _i}\lambda _j\textsc {t}(B\cdot p_i,B\cdot p_j)=\sum _{i,j}\overline{\lambda _i}\lambda _j\textsc {t}( p_i,p_j)=||\sum _i\lambda _iJ_{p_i}||^2$ holds for every linear combination of the $J_p$. Therefore $D'(B)J_p:=J_{B\cdot p}$ determines a unitary operator $D'(B)$ and $B\mapsto D'(B)$ a rep of SU(2) on ${\mathcal {H}}_J$. The latter induces the rep $(U'(b,B)\varphi )(p):={\text {e}}^{-{\text {i}}bp} D'(B)\varphi (B^{-1}\cdot p)$ of $\tilde{{\mathcal {E}}}$ on $L^2(\mathbb {R}^3, {\mathcal {H}}_J)$. Hence T is the POL (7.3) with kernel $\textsc {t}$. Since ${\mathcal {H}}_J$ is finite dimensional, (a) follows. $\square$

In conclusion we specify how to get the kinematical deformations T of the NWL. (16) is an application of (15) and (11).

Corollary 16

Let $(g_{j,n})_{j,n}$, $j=0,1,2,\dots ,J$, $n=1,2,3,\dots ,\nu$ be finitely many measurable complex-valued functions on $]0,\infty [$ satisfying $\sum _{j,n}|g_{j,n}(\rho )|^2=1$. Then

$$\begin{aligned} \textsc {t}(k,p):=\sum _j\sum _n \overline{g_{j,n}(|k|)}g_{j,n}(|p|)\,P_j\left( \frac{kp}{|k||p|}\right) \end{aligned}$$

(9.2)

is a finite POL kernel. Every finite POL kernel is of this kind. Write $P_j\left( \frac{kp}{|k||p|}\right) =\sum _{0\le s_1+s_2+s_3\le j} c^j_{s_1s_2s_3} \prod _{i=1}^3\big (\frac{k_i}{|k|}\big )^{s_i}\prod _{i=1}^3\big (\frac{p_i}{|p|}\big )^{s_i}$. Then the kinematical deformations of the NWL are

$$\begin{aligned} T=\sum _j\sum _n \sum _{0\le s_1+s_2+s_3\le j} c^j_{s_1s_2s_3}\,\,g_{j,n}(|P|)^*\prod _{i=1}^3\Big (\frac{P_i}{|P|}\Big )^{s_i} \,E^{NW}\, \prod _{i=1}^3\Big (\frac{P_i}{|P|}\Big )^{s_i}g_{j,n}(|P|)\nonumber \\ \end{aligned}$$

(9.3)

The first coefficients $c^j_{rst}$ read

• $c^0_{000}=1$

• $c^1_{000}=0$, $c^1_{100}=c^1_{010}=c^1_{001}=1$

• $c^2_{000}=-\frac{1}{2}$, $c^2_{110}=c^2_{101}=c^2_{011}=3$, $c^2_{200}=c^2_{020}=c^2_{002}=\frac{3}{2}$

Section 10 is concerned with two important examples of NWL deformations.

10 POL with causal time evolution

Let (U, T) be a POL and (V, U) a rep of the little kinematical group $\mathbb {R}\otimes \tilde{{\mathcal {E}}}$. Then (V, U, T) is said to be a POL with causal time evolution if

$$\begin{aligned} V(t)T(\Delta )V(t)^{-1}\le T(\Delta _t) \end{aligned}$$

(CT)

holds for all regions $\Delta \subset R^3$ and all times $t\in \mathbb {R}$ Here $\Delta _t:=\{y\in \mathbb {R}^3:\,\exists \,x\in \Delta \text { with } |y-x|\le |t|\}$ is the region of influence of $\Delta$. CT simply means that after time t, respectively, before time t, the probability of localization in $\Delta _t$ is not less than originally in $\Delta$. In case of a POL for a massive scalar boson V(t) is given by $W^{m,0,\eta }(t)$.

Probably the first to introduce and to treat this concept of CT is the Petzold group [9, sect. 3]. There CT follows from the existence of a non-spacelike conserved current associated to the probability density of localization. This relation is adapted by Moretti [17] in an advanced manner thus showing rigorously CT for the POL introduced by Terno [23]. In case of WL, CT is examined in [21] and thoroughly studied in [2] and [4, 5].

The POL with CT for a massive scalar boson known so far in the literature are the (not finite) POL with conserved covariant kernels by Petzold et al. [8] and Henning, Wolf [11], which we study in the following sections, and the POL by Terno–Moretti. We add here the finite POL with trace CT.

Henceforth we will consider the particle case $\eta =+$ only, as the antiparticle case $\eta =-$ is quite analogous. Let H denote the energy operator, which in momentum rep equals the multiplication operator by $\epsilon (p)=\big (m^2+|p|^2\big )^{1/2}$.

10.1 POL by Terno–Moretti

In Moretti [17] it is shown that $T^{TM}$ (10.2) is a POL with CT. Below we give a simplified proof of this based on the general criteria (11), (16), and (52).

Proposition 17

$$\begin{aligned} \textsc {t}^{TM}(k,p):=\frac{1}{2}\left( 1+\frac{m^2+kp}{\epsilon (k)\epsilon (p)}\,\right) \nonumber \\ \end{aligned}$$

(10.1)

is the kernel of the finite POL with CT

$$\begin{aligned} T^{TM}= \frac{1}{2}E^{NW}+\frac{m}{\sqrt{2}H}E^{NW}\frac{m}{\sqrt{2}H}+\sum _{i=1}^3\frac{P_i}{\sqrt{2}H} E^{NW} \frac{P_i}{\sqrt{2}H}\nonumber \\ \end{aligned}$$

(10.2)

Proof

Obviously $\textsc {t}^{TM}$ is a finite continuous rotational invariant positive definite normalized kernel on $\mathbb {R}^3\setminus \{0\}$ and hence a finite POL kernel (11). Recall (16). It remains to show CT. For this we apply (52). Indeed, induced by [17] we show that $\textsc {t}^{TM}$ is conserved timelike definite due to $\textsc {j}=( \textsc {j}_1, \textsc {j}_2, \textsc {j}_3)$ given by

$$\begin{aligned} \textsc {j}(k,p)):=\frac{\epsilon (p)k+\epsilon (k)p}{2\epsilon (k)\epsilon (p)} \end{aligned}$$

(10.3)

Show condition (51)(i): $2\,\epsilon (k)\epsilon (p)\, (k-p)\textsc {j}(k,p)) =(\epsilon (k)-\epsilon (p))(m^2+kp+\epsilon (k)\epsilon (p))$.

Show condition (51)(ii): Since $\textsc {j}$ is real symmetric, it suffices to consider real coefficients $c_a$. Put $C:=\sum _ac_a$, $D_i:=\sum _ac_ap_{a.i}/\epsilon (p_a)$, and $M:=\frac{m^2}{2}\sum _ac_a/\epsilon (p_a)$. Then $\sum _i\big (\sum _{a,b}c_ac_b\textsc {j}_i(p_a,p_b)\big )^2=C^2\sum _iD_i^2$ and $\big (\sum _{a,b}c_ac_b\textsc {t}(p_a,p_b)\big )^2=\big (\frac{1}{2}C^2+\sum _iD_i^2+M^2\big )^2$, whence the claim. $\square$

10.1.1 Spinor choice for $T^{TM}$

From (6) one obtains $T^{TM}=T_{{\text {e}}}$ for

$$\begin{aligned} {\text {e}}(\rho ):=\frac{1}{\sqrt{2}}\,b^{(0,1)} +\frac{m}{\sqrt{2}\epsilon (\rho )}b^{(0,2)}+\frac{\rho }{\sqrt{2}\epsilon (\rho )}b^{(1,1)} \end{aligned}$$

(10.4)

Hence a dilation of $(W^{m,0,+}|_{\tilde{{\mathcal {E}}}}, T^{TM})$ corresponding to (10.4) is the WL $(W'|_{\tilde{{\mathcal {E}}}},E)$ for

$$\begin{aligned} W':=W^{m,0,+}\oplus W^{m,0,+}\oplus W^{m,1,+} \text { and } E:={\mathcal {F}}E^{can}{\mathcal {F}}^{-1} \end{aligned}$$

10.2 POL with trace CT

Proposition 18

$$\begin{aligned}{} & {} \textsc {t}^{tct}(k,p)\nonumber \\{} & {} :=\frac{1}{2}\Big (\epsilon (k)(m+\epsilon (k))\epsilon (p)(m+\epsilon (p))\Big )^{-1/2} \Big ((m+\epsilon (k))(m+\epsilon (p)) +kp \Big )\qquad \nonumber \\ \end{aligned}$$

(10.5)

is the kernel of the finite POL with CT

$$\begin{aligned}{} & {} T^{tct}= \left( \frac{m+H}{2H}\right) ^{1/2}E^{NW}\left( \frac{m+H}{2H}\right) ^{1/2}\nonumber \\{} & {} \quad +\sum _{i=1}^3\frac{P_i}{\big (2H(m+H)\big )^{1/2}} E^{NW}\frac{P_i}{\big (2H(m+H)\big )^{1/2}} \end{aligned}$$

(10.6)

Actually, $(W^{m,0,+}|_{\mathbb {R}\otimes \tilde{{\mathcal {E}}}},T^{tct})$ is the only one which is the trace of a relativistic quantum system of finite spinor dimension with CT by a WL.

The proof is postponed to “Appendix D”.

10.2.1 Two spinor choices for $T^{tct}$

Proposition 19

The spinor choice ${\text {e}}$ for which $T^{tct}=T_{{\text {e}}}$ according to the construction in Sect. 4 is

$$\begin{aligned}{} & {} {\text {e}}(\rho ):=\big (4\epsilon (\rho )(m+\epsilon (\rho )) \big )^{-1/2}\nonumber \\{} & {} \quad \Big [\big (m+\epsilon (\rho )\big )b^{(0,1)}+\big (m+\epsilon (\rho )\big )b^{(0,2)}+\rho \, b^{(1,1)}-\rho \, b^{(1,2)}\Big ] \end{aligned}$$

(10.7)

where $\epsilon (\rho )=\sqrt{m^2+\rho ^2}$.

The proof is postponed to “Appendix D”.

On the other hand, following (6) one obtains immediately $T^{tct}=T_{{\text {e}}'}$ for

$$\begin{aligned} {\text {e}}'(\rho ):=\Big (\frac{m+\epsilon (\rho )}{2\epsilon (\rho )}\Big )^{1/2}\,b^{(0,1)} +\rho \,\big (2\epsilon (\rho )(m+\epsilon (\rho ))\big )^{-1/2}b^{(1,1)} \end{aligned}$$

(10.8)

Hence a dilation of $(W^{m,0,+}|_{\tilde{{\mathcal {E}}}}, T^{tct})$ corresponding to (10.8) is the WL $(W'|_{\tilde{{\mathcal {E}}}},E')$ for

$$\begin{aligned} W':=W^{m,0,+}\oplus W^{m,1,+} \text { and } E':={\mathcal {F}}E^{can}{\mathcal {F}}^{-1} \end{aligned}$$

This is rather interesting, since here $(W',E')$ is the NWL and therefore the time evolution of $(W',E')$ is not causal [5, (2) Theorem].

11 Causal POL

CT for a POL is a requirement of the fact that there is no propagation faster than light. Clearly limited propagation affects all kinematical transformations. For instance, according to NWL, if a massive particle localized in a bounded region is subjected to a boost of, however, small rapidity, then there is a non-vanishing probability to observe the particle arbitrarily far away. This frame-dependence of NWL (already described by Weidlich, Mitra 1964 [24] and still currently discussed, see, e.g., [5, 3]) is an acausal behavior. In fact, causality requires from localization that the probability of localization in a region of influence cannot be less than that in the region of actual localization.

More precisely, let $\Delta$ be a region, i.e., a measurable subset of a spacelike hyperplane of Minkowski space, and let $\sigma$ be any spacelike hyperplane, then

$$\begin{aligned} \Delta _\sigma :=\{{\mathfrak {x}}\in \sigma : \exists \,{\mathfrak {z}}\in \Delta \text { with } ({\mathfrak {x}}-{\mathfrak {z}}\}\cdot ({\mathfrak {x}}-{\mathfrak {z}}\}\ge 0\} \end{aligned}$$

is the region of influence of $\Delta$ in $\sigma$. It is the set of all points ${\mathfrak {x}}$ in $\sigma$, which can be reached from some point ${\mathfrak {z}}$ in $\Delta$ by a signal not moving faster than light. Therefore (W, T) (Sect. 3) is called a causal POL if

$$\begin{aligned} T(\Delta )\le T(\Delta _\sigma ) \end{aligned}$$

(CC)

holds for all $\Delta \in {\mathfrak {S}}$, $\sigma$ spacelike hyperplane. The causality condition CC is thoroughly studied in [5]. There causal POL for the Dirac electron and the Weyl fermions are revealed. Apparently the causality condition has not been analyzed elsewhere. Recently, along a tentative “new and different operational interpretation of the notion of spatial position”  in [17] a causality condition corresponding to CC is shown to hold.

Let us verify explicitly that CT is a special case CC. Indeed, for $\Delta \subset \{0\}\times \mathbb {R}^3$, $t\in \mathbb {R}$, and $\sigma :=\{-t\}\times \mathbb {R}^3$ check $\Delta _\sigma =(-t,0,I_2)\cdot \Delta _t$, whence CT by covariance of T.

11.1 Causal kernels

As argued above CC, which implies CT, is an indispensable property for a localization. For this one is highly interested in the kernels of causal POL. We study the promising class ${\mathcal {K}}$ of conserved covariant POL kernels (B.1) introduced by Petzold et al.  [8]. Indeed, (56) shows that a POL with kernel $K\in {\mathcal {K}}$ is causal. Moreover, most probably ${\mathcal {K}}$ comprises even all kernels of causal POL. Therefore we refer quite simply to their kernels (11.1) as the causal kernels.

One deals with the kernels on $\mathbb {R}^3$ of the kind

$$\begin{aligned} K(k,p):=\frac{\epsilon (k)+\epsilon (p)}{2\sqrt{\epsilon (k)\epsilon (p)}}\,\,g\big (\epsilon (k)\epsilon (p)-kp\big ) \end{aligned}$$

(11.1)

with $g: [m^2,\infty [ \rightarrow \mathbb {R}$ continuous and normalized by $g(m^2)=1$, which are positive definite. Occasionally g is said to determine a causal kernel.

Recall $\epsilon (p)=\big (m^2+|p|^2\big )^{1/2}$. Put $\breve{g}(k,p):=g\big (\epsilon (k)\epsilon (p)-kp\big )$ for $k,p\in \mathbb {R}^3$. Note $\breve{g}(p,p)=1$ for all $p\in \mathbb {R}^3$.

11.2 Known results

The problem is to find the functions g so that K is positive definite. Let us report the results, which in the literature are achieved so far including the personal communications of R.F. Werner with proofs. Put

$$\begin{aligned} g_r(t):=(2m^2)^r(m^2+t)^{-r}, \;t\ge m^2, \;r>0 \end{aligned}$$

(11.2)

• (a) The known solutions for K in (11.1) are deduced in [8] and [11]. They read

  $$\begin{aligned} K_r=K \text { where } g=g_r \,\text { for } r\ge 3/2 \end{aligned}$$

  (11.3)

  Further solutions are the pointwise limits of their convex combinations as, e.g., K determined by $g(t):=\big (\frac{(1+c)m^2}{cm^2+t}\big )^n$ for $-1<c\le 1$, $n=2,3,\dots$. Note that $\frac{(1+c)m^2}{cm^2+t}\le g_1(t)$.

• (b) As observed by R.F. Werner, obviously

  $$\begin{aligned} K\breve{h}\in {\mathcal {K}} \end{aligned}$$

  (11.4)

  for every $K\in {\mathcal {K}}$ and continuous $h:[m^2,\infty [\rightarrow \mathbb {R}$ with $h(m^2)=1$, $\breve{h}$ positive definite. $\breve{h}$ is called a noise factor.

• (c) As pointed out by R.F. Werner, one has the important implication

  $$\begin{aligned} K\in {\mathcal {K}} \Rightarrow \;\breve{g} \text { positive definite} \end{aligned}$$

  (11.5)

  Indeed, $\breve{g}(k,p) =\frac{2\,\sqrt{\epsilon (k)\epsilon (p)}}{\epsilon (k)+\epsilon (p)}K(k,p)$ and by the formula by R.F. Werner

  $$\begin{aligned} \frac{2\sqrt{xy}}{x+y}=\int _0^\infty \big (2\sqrt{x}{\text {e}}^{-\lambda x}\big )\big (2\sqrt{y}{\text {e}}^{-\lambda y}\big ){\text {d}}\lambda \end{aligned}$$

  (11.6)

  the first factor is positive definite.

• (d) For K in (11.1), one has $|K|\le 1$ if and only if $|g|\,\le \,g_{1/2}$. This is a communication by R.F. Werner and follows from (21)(b). Hence

  $$\begin{aligned} K\in {\mathcal {K}} \Rightarrow |g|\,\le \,g_{1/2} \end{aligned}$$

  (11.7)

• (e) In [9] it is argued that a POL with kernel $K\in {\mathcal {K}}$ obeys CT.

Henceforth let $m=1$ without restriction.

12 Causal kernels for one spatial dimension

As recommended by [8, footnote to (3.2)] and since there is still some interest in this topic [13], we look first at the instructive case of one spatial dimension. Here, indeed, $g_{1/2}$ arises naturally and turns out to furnish the maximal solution $K^{1}_{1/2}$ (21)(c). A complete description of the causal kernels is given in (22). Accordingly, the product of $K^{1}_{1/2}$ with a normalized positive definite kernel is a causal kernel and every causal kernel is of this kind. Call

$$\begin{aligned} K^{1}:\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}, \quad K^{1}(k,p)= \frac{\epsilon (k)+\epsilon (p)}{2\sqrt{\epsilon (k)\epsilon (p)}}\,\,g\big (\epsilon (k)\epsilon (p)-kp\big ) \end{aligned}$$

(12.1)

with $\epsilon (p):=\sqrt{1+p^2}$ and $g:[1,\infty [\rightarrow \mathbb {R}$ measurable normalized by $g(1)=1$, a causal kernel if it is positive definite. Put ${\tilde{g}}(k,p):=g\big (\epsilon (k)\epsilon (p)-kp\big )$.

The question is about the functions g of causal kernels on $\mathbb {R}$.

Lemma 20

One has

• (i) $\frac{\epsilon (k)+\epsilon (p)}{2\sqrt{\epsilon (k)\epsilon (p)}}= \left( \frac{1+\epsilon (k)\epsilon (p)+kp}{2\epsilon (k)\epsilon (p)}\right) ^{1/2}\frac{1}{{\tilde{g}}_{1/2}(k,p)}$

With respect to the variables $x:=\sinh ^{-1}(k)$, $y:=\sinh ^{-1}(p)$ (i) reads

• (ii) $\frac{\epsilon (\sinh x)+\epsilon (\sinh y)}{2\sqrt{\epsilon (\sinh x)\epsilon (\sinh y)}}=\,h(x,y)\,\frac{1}{g_{1/2}\big (\cosh (x-y)\big )}$

where the kernel h

• (iii) $h(x,y):=\frac{\cosh x/2}{\sqrt{\cosh x}} \frac{\cosh y/2}{\sqrt{\cosh y}} \,\big (1+\tanh x/2\;\tanh y/2\big )$

is positive definite.

• (iv) $K^1(\sinh x,\sinh y)\;= h(x,y) \,\frac{g\big (\cosh (x-y)\big )}{g_{1/2}\big (\cosh (x-y)\big )}$

Proof

Verify the claim by elementary computations. Positive definiteness of h is obvious. $\square$

Lemma 21

Consider $K^{1}$ in (12.1).

• (a) $(k,p)\rightarrow \frac{2\sqrt{\epsilon (k)\epsilon (p)}}{\epsilon (k)+\epsilon (p)}$ is a positive definite kernel on $\mathbb {R}$. If $K^{1}$ is positive definite, then so is ${\tilde{g}}$.

• (b) If $|K^{1}|\le 1$, then $|g|\le g_{1/2}$.

• (c) $K^{1}_{1/2}:=K^{1}$ for $g=g_{1/2}$ is continuous positive definite. ${\tilde{g}}_{1/2}$ is positive definite. Explicitly one has $K^{1}_{1/2}(k,p)=\left( \frac{1+\epsilon (k)\epsilon (p)+kp}{2\epsilon (k)\epsilon (p)}\right) ^{1/2}$.

Proof

(a) follows by the formula (11.6). (b) Assume $|K^{1}|\le 1$. By (20)(iii) the claim is $\sup _{x,y} h(x,y)=1$ under the constraint that $\cosh (x-y)$ is constant. Since h is symmetric it is no restriction assuming $x\ge y$. Now, the constraint means $y=x+c$ for some constant c, and obviously $\lim _{x\rightarrow \infty }h(x,x+c)=1$. (c) Apply (20)(iv),(iii), part (a), and (20)(i). $\square$

Theorem 22

$K^{1}$ in (12.1) is causal if and only if

$$\begin{aligned} K^{1}=K^{1}_{1/2}\; {\tilde{g}} \end{aligned}$$

where $g:[1,\infty [\rightarrow \mathbb {R}$ is measurable with $g(1)=1$ and ${\tilde{g}}$ positive definite. In particular $K^1\le K^1_{1/2}$ for all causal $K^1$.

Proof

Clearly the condition is sufficient. Now assume that $K^{1}$ is positive definite. Put $k:=\frac{g\,\circ \, \cosh }{g_{1/2\,}\circ \, \cosh }$. By (20)(iv),(iii) it suffices to show that $(x,y)\mapsto k(x-y)$ is positive definite on $\mathbb {R}$.

(a) First we note that a real kernel G(x, y) on $\mathbb {R}$ is positive definite if and only if the kernel $G_1(x,y):= h_1(x)h_1(y)G(x,y)$ for $h_1(x):=\frac{\cosh x/2}{\sqrt{\cosh x}}$ is positive definite. Indeed, as to the less trivial implication let $G_1$ be positive definite. Let $c_1,\dots , c_n \in \mathbb {R}$ and $x_1,\dots , x_n\in \mathbb {R}$. Put $c_i':=c_i/h_1(x_i)\in \mathbb {R}$, $i=1,\dots , n$. Then by assumption $0\le \sum _{i,j}c_i'c_j'G_1(x_i,x_j)=\sum _{i,j}c_ic_jG(x_i,x_j)$. Hence G is positive definite.

(b) Recall (20)(iii),(iv) and apply the result (a) to $G(x,y):=\big (1+h_2(x)h_2(y)\big )k(x-y)$ for $h_2(x):=\tanh x/2$. Hence G is positive definite. Now assume that $(x,y)\mapsto k(x-y)$ is not positive definite. Then there are $c_1,\dots , c_n \in \mathbb {R}$ and $x_1,\dots , x_n\in \mathbb {R}$ such that $\sum _{i,j}c_ic_jk(x_i-x_j)=:\gamma <0$. Let $b\in \mathbb {R}$. It follows $0\le \sum _{i,j}c_ic_jG(x_i+b,x_j+b)= \sum _{i,j}c_ic_j\big (1+h_2(x_i+b)h_2(x_j+b)\big )k(x_i-x_j)\rightarrow \sum _{i,j}c_ic_j(1+1)\,k(x_i-x_j)=2\gamma <0$ for $b\rightarrow \infty$, which is a contradiction. $\square$

By (20)(iv) a measurable normalized function $g:[1,\infty [\rightarrow \mathbb {R}$, $g(1)=1$ yields in (12.1) a causal kernel $K^{1}$ if and only if $h\,\circ \, \cosh$ for $h:=g/g_{1/2}$ is normalized of positive type. The functions $g=g_r$, $r\ge 1/2$ (see (11.2)) are of this kind as follows from (23). Other explicit examples can be obtained applying the corollary (57)(a),(b) to Bochner’s theorem.

Lemma 23

The Fourier transform of $g_{1/2}\circ \cosh = [\cosh \frac{1}{2}(\cdot )]^{-1}$ is infinitely divisible. ${\tilde{g}}_r$ is positive definite for $r>0$.

Proof

$\ln \, [\cosh \,x/2]^{-1}=-\frac{1}{2}\int _0^x\tanh \frac{t}{2}{\text {d}}t =-\int _0^x\int _0^\infty \frac{\sin ty}{\sinh \pi y}{\text {d}}y{\text {d}}t$ by [10, 17.33(25)]. By Fubini this becomes $=-\frac{1}{2}\int _{-\infty }^\infty \int _0^x\frac{\sin ty}{\sinh \pi y}{\text {d}}t{\text {d}}y= \int _{-\infty }^\infty \frac{\cos xy \,-1}{y}\frac{1}{2\sinh \pi y}{\text {d}}y=\int \frac{\cos xy\, -1}{y^2} (1+y^2) {\text {d}}\mu (y)$ for the finite measure ${\text {d}}\mu (y):= \frac{y}{2(1+y^2)\sinh \pi y}{\text {d}}y$, whence the first part of the claim by the Lévy–Khintchine formula implying the second part. $\square$

Example 24

$K^1$ in (12.1) need not be positive definite despite $0<g\le g_{1/2}$ and ${\tilde{g}}$ positive definite. The example is furnished by the Gaussian function with standard deviation $0<\varsigma \le 2$. We apply (57).

Indeed, let $g:=k\circ \cosh ^{-1}$, where $k(x):=\exp (-\frac{x^2}{2\varsigma ^2})$, $\varsigma >0$ is of positive type (since ${\hat{k}}_\varsigma =\varsigma \, k_{1/\varsigma }>0$) and $k(0)=1$.

Put $f(x):=\frac{k(x)}{[\cosh \frac{x}{2}]^{-1}}=\frac{1}{2}\big (\exp (\frac{x}{2}-\frac{x^2}{2\varsigma ^2}) +\exp (-\frac{x}{2}-\frac{x^2}{2\varsigma ^2})\big )$. Then ${\hat{f}}(y)=\varsigma {\text {e}}^{\varsigma ^2/8}\cos \big (\frac{\varsigma ^2}{2}y\big )\exp (-\frac{\varsigma ^2}{2}y^2)$. Therefore, by (57)(c), f is not of positive type.

It remains to check $f\le 1$. Note $f(0)=1$. Verify $f'(x)=0$ $\Leftrightarrow$ $\tanh \frac{x}{2}=\frac{2}{\varsigma ^2}x$. This equation has the only solution $x=0$ if $0<\varsigma \le 2$, whence $f(0)=1$ is the maximum. (If $\varsigma >2$, there are exactly three solutions, whence $f(0)=1$ is a local minimum.)

There is a consequence of (22) for K in (11.1). Choose any $p_0\in \mathbb {R}^3$ with $|p_0|=1$. One observes that $K(ap_0,bp_0)$ equals $K^{1}(a,b)$ for $a,b\in \mathbb {R}$. Therefore $K^{1}$ is positive definite if K is positive definite. By (22) this implies the following necessary condition for K being causal.

Corollary 25

Let K be causal. Then $g=g_{1/2}\,h$, where $h:[1,\infty [\rightarrow \mathbb {R}$ is continuous with $h(1)=1$ and ${\tilde{h}}$ positive definite, i.e., $h\,\circ \,\cosh$ of positive type.

13 Definiteness of $\breve{g}_r$ and $K_r$

We return to causal kernels on $\mathbb {R}^3$. Recall $g_r(t)=2^r(1+t)^{-r}$, $r>0$ and let $K_r$ denote the map K in (11.1) for $g=g_r$. Recall that $K_r$ is positive definite for $r\ge 3/2$ (11.3).

The literature is not clear what about the values $0<r<3/2$. By [8, Eq. (3.10)] $K_1$ is not positive definite. In (27) we show that $K_r$ is not positive definite for $0<r<3/2$.

Similarly there is the question about the definiteness of $\breve{g}_r$. Clearly it is positive definite for $r\ge 3/2$. Actually it holds true exactly for $r\ge 1/2$ as we are going to show simply adapting the proof of [11, Theorem].

Put

$$\begin{aligned} h:=\frac{\sigma \rho }{(1+\epsilon (\sigma ))(1+\epsilon (\rho ))}\text { for } \sigma ,\rho > 0 \end{aligned}$$

with $\epsilon (\rho )=\sqrt{1+\rho ^2}$. Verify $1+\epsilon (\sigma )\epsilon (\rho ) -\sigma \rho \,x=\frac{\sigma \rho }{2\,h}+\frac{h\sigma \rho }{2} -\sigma \rho \,x$, whence for $r\in \mathbb {R}$, $-1\le x\le 1$ [11, (7), (10)]

$$\begin{aligned}{} & {} 2^r\big (1+\epsilon (\sigma )\epsilon (\rho ) -\sigma \rho \,x\big )^{-r}\nonumber \\{} & {} \quad =\big (4h/\sigma \rho \big )^r\big (1-2hx+h^2\big )^{-r}=\big (4h/\sigma \rho \big )^r \,\sum _{n=0}^\infty C_n^r(x)h^n \end{aligned}$$

(13.1)

by the definition of the Gegenbauer polynomials $C_n^r$ [6, 3.15.1].

Lemma 26

$\breve{g}_r$ is positive definite for $r\ge 1/2$, it is not positive definite for $0<r<1/2$. The expansion (7.5) of $\breve{g}_{1/2}$ is

$$\begin{aligned} \breve{g}_{1/2}(k,p)=\sum _{j=0}^\infty \frac{\sqrt{2}\,|k|^j}{ \big (1+\epsilon (k)\big )^{j+1/2} } \frac{\sqrt{2}\,|p|^j}{ \big (1+\epsilon (p)\big )^{j+1/2} } P_j\left( \frac{kp}{|k||p|} \right) \end{aligned}$$

Proof

(a) Let $r=1/2$. Note $C_j^{1/2}=P_j$ [6, 10.10 (3)]. Hence (13.1) yields the claimed expansion (7.5) of $\breve{g}_{1/2}$ by (11), whence $\breve{g}_{1/2}$ is positive definite. This implies that $\breve{g}_1$ is positive definite, too.

(b) So let $r>1/2$ and $r \ne 1$. Polar coordinates $k=\sigma (\sin \theta \cos \phi ,\sin \theta \sin \phi , \cos \theta )$, $p=\rho (\sin \theta '\cos \phi ', \sin \theta ' \sin \phi ', \cos \theta ')$ yield $x=\frac{kp}{|k||p|}=\cos \theta \cos \theta '+\sin \theta \sin \theta '\cos (\phi -\phi ')$. This expression substitutes x in (13.1). One obtains the separation of the primed and unprimed variables applying the addition theorem [6, 10.9(34)] twice:

• $C_n^r(\cos \theta \cos \theta '+\sin \theta \sin \theta '\cos \varphi ) =\sum _{m=0}^n a(r;n,m) (\sin \theta )^mC_{n-m}^{r+m}(\theta )(\sin \theta ')^mC_{n-m}^{r+m}(\theta ')C_m^{r-1/2}(\cos \varphi )$

• $C_m^{r-1/2}(\cos \varphi )=\sum _{l=0}^m a(r-1/2;m,l) (\sin \phi )^l C_{m-l}^{r-1/2 +l}(\phi ) (\sin \phi ')^l C_{m-l}^{r-1/2 +l}(\phi ') C_l^{r-1}(1)$ for $\cos \varphi =\cos (\phi -\phi ')=\cos \phi \cos \phi '+\sin \phi \sin \phi '$.

Finish the proof checking that the coefficients $a(r;n,m) a(r-1/2;m,l) C_l^{r-1}(1)$ are positive: $a(r;n,m)=2^m(n-m)!((r)_m)^2\frac{2r+2m-1}{(2r-1)_{n+m+1}} >0$, and $a(r-1/2;m,l) C_l^{r-1}=2^l(m-l)!((r)_l)^2\frac{1}{l!}\,\frac{(2r+2\,l-2)(2r-2)_l}{(2r-2)_{m+l+1}}$ is positive, too, since for $l=0$ the fraction becomes $\frac{(2r-2)\cdot 1}{(2r-2)\dots (2r-2+m)}$.

(c) Let $0<r<1/2$. Put $a:=1+\epsilon (\sigma )\epsilon (\rho )$, $b:=\sigma \rho$. Referring to the expansion (7.5) of $\breve{g}_r$ it suffices to show the non-positive definiteness of the 0th coefficient $\gamma ^{(r)}_0(\sigma ,\rho ):=\frac{1}{2}\int _{-1}^12^r(a-bx)^{-r}{\text {d}}x=\frac{2^{r-1}}{1-r}b^{-1}\big ((a+b)^{1-r}-(a-b)^{1-r}\big )$. The expansion (13.1) in Gegenbauer polynomials for $1-r$ in place of r and for $x=1,-1$ yields

$$\begin{aligned} \gamma ^{(r)}_0(\sigma ,\rho )=\left( \frac{4h}{\sigma \rho }\right) ^r\left( 1-\sum _{l=1}^\infty a(r;l)h^{2l} \right) , \quad a(r;l):=(1-2r)\frac{(2r)_{2l-1}}{(2l+1)!}\nonumber \\ \end{aligned}$$

(13.2)

Note $a(r;l)>0$ and $h=h_1(\sigma )h_1(\rho )$ for $h_1(\rho ):=\rho /(1+\epsilon (\rho ))$, which is injective. Hence for $\rho _1\ne \rho _2$ and $(c_1,c_2):=(1,-1)$ one easily checks $\sum _{i,j=1}^2c_ic_j\gamma ^{(r)}_0(\rho _i,\rho _j)<0$, thus ending the proof. $\square$

Exploiting a further formula [11, (8)]

$$\begin{aligned} \epsilon (\sigma )+\epsilon (\rho )=\frac{\sigma \rho }{2h}\,\big (1-h^2\big ) \end{aligned}$$

(13.3)

(one verifies it using $\big (\epsilon (\rho )+1\big )^{-1}=\rho ^{-2}(\epsilon (\rho )-1)$ ) we complete the result by

Lemma 27

$K_r$ is not positive definite for $0<r<3/2$.

Proof

Since $\breve{g}_r$ is not positive definite for $0<r<1/2$, so is $K_r$. Hence we may assume $1/2\le r<3/2$.

We use again the expansion (7.5) and show that the 0th coefficient of $K_r$ is not positive definite. By (13.3) the latter equals $k_0^{(r)}(\sigma ,\rho )=\frac{1}{2}\big (\epsilon (\sigma )\epsilon (\rho )\big )^{-1/2} \frac{\sigma \rho }{2\,h}(1-h^2)\gamma ^{(r)}_0(\sigma ,\rho )$. Then (13.2) yields

$$\begin{aligned} k_0^{(r)}(\sigma ,\rho )=\big (\epsilon (\sigma )\epsilon (\rho )\big )^{-1/2}\left( \frac{4h}{\sigma \rho }\right) ^{r-1}\left( 1-\sum _{l=1}^\infty b(r;l)h^{2l}\right) , \end{aligned}$$

(13.4)

where $b(r;1):=1-\frac{(2r-1)r}{3}, \,b(r;l)=(2r-1)\Big (\frac{(2r)_{2\,l-3}}{(2\,l-1)!}-\frac{(2r)_{2\,l-1}}{(2\,l+1)!}\Big )$ for $l\ge 2$. Now check $b(1/2;1)=1$, $b(1/2;l)=0$ for $l\ge 2$, and $b(r;l)>0$ for $1/2<r<3/2$ and $l\ge 1$. (Moreover, $b(3/2;l)=0$ for $l\ge 1$, and $b(r;l)<0$ for $r>3/2$ and $l\ge 1$.) Recall $h=h_1(\sigma )h_1(\rho )$ for $h_1(\rho )=\rho /(1+\epsilon (\rho ))$, which is injective. Hence, again, for $\rho _1\ne \rho _2$ and $(c_1,c_2)=(1,-1)$ one easily checks $\sum _{i,j=1}^2c_ic_jk^{(r)}_0(\rho _i,\rho _j)<0$, thus ending the proof. $\square$

We like to point out that an easy proof of the fact that $\breve{g}_r$, $0<r<\frac{1}{2}$ and $K_r$, $0<r<\frac{3}{2}$ are not positive definite is given in (39).

14 Lorentz invariant kernels

The kernels $\breve{g}$ on $\mathbb {R}^3$ given by $\breve{g}(k,p):=g\big (\epsilon (k)\epsilon (p)-kp\big )$ for some $g:[1,\infty [\rightarrow \mathbb {C}$ are characterized as the functions $G:\mathbb {R}^3\times \mathbb {R}^3\rightarrow \mathbb {C}$, which are Lorentz invariant, i.e., which satisfy $G(L\cdot k,L\cdot p)=G(k,p)$ for $L\in O(1,3)_0$, the set of proper orthochronous Lorentz transformations. Here $L\cdot p$ denotes the spatial vector part of $L{\mathfrak {p}}$ for ${\mathfrak {p}}:=\big (\epsilon (p),p\big )$.

Indeed, for the less trivial implication assume G to be Lorentz invariant. For $k,p\in \mathbb {R}^3$ there is $L\in O(1,3)$ such that $L{\mathfrak {k}}=(1,0,0,0)$ and $L{\mathfrak {p}}=(a,0,0,b)$, $a\ge 0$. By the invariance of the Minkowski one finds ${\mathfrak {k}}\cdot {\mathfrak {p}}=L{\mathfrak {k}}\cdot L{\mathfrak {p}}=a$ and $1={\mathfrak {p}}\cdot {\mathfrak {p}}=L{\mathfrak {p}}\cdot L{\mathfrak {p}}=a^2-b^2$: This shows $G(k,p)=G\big (0, (0,0,\sqrt{ ({\mathfrak {k}}\cdot {\mathfrak {p}})^2-1})\big )$ for all k, p, whence G is symmetric and the claim holds by

$$\begin{aligned} \breve{g}=G \text { for } g(t):=G\big (0, (0,0,\sqrt{ t^2-1})\big ) \end{aligned}$$

(14.1)

Definition 28

Let ${\mathcal {G}}$ denote the set of measurable Lorentz invariant positive definite kernels G on $\mathbb {R}^3$ with $G(0,0)=1$.

• (a) $G\in {\mathcal {G}}$ is automatically continuous. Indeed, continuity of G follows from (14.1) and (12).

• (b) $G\in {\mathcal {G}}$ is automatically real-valued since it is symmetric and positive definite.

• (c) $G\in {\mathcal {G}}$ is normalized, i.e., $G(p,p)=G(0,0)=1$ for all $p\in \mathbb {R}^3$, whence G is bounded by 1 (7.2).

• (d) If G, $G' \in {\mathcal {G}}$, then also $GG'\in {\mathcal {G}}$. Moreover, if G is the pointwise limit of a sequence in ${\mathcal {G}}$, then $G\in {\mathcal {G}}$.

• (e) If g determines $K\in {\mathcal {K}}$, then $\breve{g}\in {\mathcal {G}}$. Indeed, recall (11.5).

• (f) If $K\in {\mathcal {K}}$ and $G\in {\mathcal {G}}$, then $KG\in {\mathcal {K}}$. Indeed, recall (11.4).

  The crucial facts, why we are interested in ${\mathcal {G}}$, are (e) and (f). Up to now the only elements of ${\mathcal {G}}$ we know are the maximal element ${{\textbf {1}}}$ and $\breve{g}_r$, $r\ge 1/2$ and pointwise limits of convex combinations of these.

14.1 Positive definite Lorentz invariant kernels, reps of $SL(2,\mathbb {C})$ with SU(2)-invariant vectors

Let G be a positive definite kernel on $\mathbb {R}^3$. We consider the RKHS ${\mathcal {H}}_G$ associated to the kernel G (see, e.g.,  [7]). ${\mathcal {H}}_G$ consists of functions $f:\mathbb {R}^3\rightarrow \mathbb {C}$. Put $G_p:=G(\cdot , p)$. ${\mathcal {H}}_G$ is the completion of the span of $\{G_p:p\in \mathbb {R}^3\}$, on which the inner product is determined by $\langle G_k,G_p\rangle =G(k,p)$. For $f\in {\mathcal {H}}_G$ and $p\in \mathbb {R}^3$ one has $\langle G_p,f\rangle =f(p)$.

Lorentz invariance of G gives rise to a rep of $SL(2,\mathbb {C})$. For $A\in SL(2,\mathbb {C})$ and $p\in \mathbb {R}^3$ let $A\cdot p$ denote the spatial vector part of $A\cdot {\mathfrak {p}}$ for ${\mathfrak {p}}:=\big (\epsilon (p),p\big )$. Moreover, let $A_\kappa$ denote the boost ${\text {e}}^{\kappa \sigma _3/2}$ in direction (0, 0, 1) with rapidity $\kappa \in \mathbb {R}$. One has $A_\kappa \cdot p=(p_1,p_2,\epsilon (p)\sinh \kappa +p_3\cosh \kappa )$. Put $\cosh ^{-1}(t) =\ln \big (t+\sqrt{t^2-1}\,\big )), \,t\ge 1$.

Lemma 29

Let $G\in {\mathcal {G}}$. Then $\big (U_G(A)f\big )(k):=f(A^{-1}\cdot k)$ for $A\in SL(2,\mathbb {C})$, $f\in {\mathcal {H}}_G$ defines a rep $U_G$ of $SL(2,\mathbb {C})$ on ${\mathcal {H}}_G$, and ${\mathcal {H}}_G$ is separable. Moreover, $G=\breve{g}$ for

$$\begin{aligned} g(t):=\langle G_0, U_G(A_\kappa )G_0\rangle \text { for } \kappa =\cosh ^{-1}t \end{aligned}$$

(14.2)

Proof

Define $U(A)G_p:=G_{A\cdot p}$ for all p. Since $G_{A\cdot p}(k) =G(k,A\cdot p)=G(A^{-1}\cdot k,p)=G_p(A^{-1}\cdot k)$, one has $\big (U(A)G_p\big )(k)=G_p(A^{-1}\cdot k)$. U(A) is norm preserving since $\langle U(A)G_k,U(A)G_p\rangle =\langle G_{A\cdot k}, G_{A\cdot p}\rangle =G( A\cdot k, A\cdot p)=G(k,p)=\langle G_k,G_p\rangle$. One easily checks $U(A)U(A')=U(AA')$. By linear extension U becomes a unitary homomorphism on $V:={\text {span}}\{G_p:p\in \mathbb {R}^3\}$. Since the latter is dense in ${\mathcal {H}}_G$, U extends further to a unitary homomorphism on ${\mathcal {H}}_G$. It follows $U=U_G$. Indeed, let $f_n$ in V converge to $f\in {\mathcal {H}}_G$. Then $U(A)f_n\rightarrow U(A)f$ and also $\big (U(A)f_n\big )(p)=f_n(A^{-1}\cdot p)= \langle G(A^{-1}\cdot p),f_n\rangle \rightarrow \langle G(A^{-1}\cdot p),f\rangle =f(A^{-1}\cdot p)$.

Now, the matrix elements $(k,p,A)\mapsto \langle G_k,U(A)G_p\rangle =G(k,A\cdot p)$ are continuous, whence ${\mathcal {H}}_G$ is separable and U is continuous.

This is quite obvious. Indeed, recall that V is dense in ${\mathcal {H}}_G$. Then even ${\text {span}}\{G_p:p\in {\mathbb {Q}}^3\}$ is dense in ${\mathcal {H}}_G$, since $||G_{p_n}-G_p||^2=G(p_n,p_n)-2G(p_n,p)+G(p,p)\rightarrow 0$ if $p_n\rightarrow p$. Hence ${\mathcal {H}}_G$ is separable. As known (see [15] for more information) it suffices to show the weak continuity of U. First note that obviously $A\mapsto \langle f,U(A)g\rangle$ is continuous for f, g in V. Let $f_0\in {\mathcal {H}}_G$, There is $f_n\in V$, $f_n\rightarrow f_0$. Let $A_k\rightarrow A_0$ in $SL(2,\mathbb {C})$. Then $|\langle f_0, U(A_k)g\rangle -\langle f_0, U(A)g\rangle |\le |\langle f_0-f_n, U(A_k)g-U(A_0)g\rangle |+|\langle f_n,U(A_k)g-U(A_0)g\rangle |\le 2||g||\,||f_0-f_n||+|\langle f_n,U(A_k)g\rangle -\langle f_n,U(A_0)g\rangle |$. Now choose $n_0$ such that $2||g||\,||f_0-f_{n_0}||\le \epsilon /2$. Note $|\langle f_{n_0},U(A_k)g\rangle -\langle f_{n_0},U(A_0)g\rangle |\rightarrow 0$ for $k\rightarrow \infty$. Thus the continuity of $A\mapsto \langle f_0,U(A)g\rangle$ for all $f_0\in {\mathcal {H}}_G$ and $g\in V$ follows. Similarly one replaces g by $g_0\in {\mathcal {H}}_G$ showing the assertion.

In view of (14.1) note that $A_\kappa \cdot 0=(0,0,\sinh \kappa )$, whence (14.2). $\square$

One reminds that the matrix element in (14.2) is invariant under unitary equivalence transformations and that $U_G(B)G_0=G_0$ holds for all $B\in SU(2)$, i.e., $G_0$ is SU(2)-invariant.

Lemma 30

Suppose that U is a rep of $SL(2,\mathbb {C})$ with an SU(2)-invariant normalized vector $\Gamma _0$. Put

$$\begin{aligned} g(t):=\langle \Gamma _0, U(A_\kappa )\Gamma _0\rangle \text { for } \kappa =\cosh ^{-1}t \end{aligned}$$

(14.3)

Then $\breve{g}(0,0)=1$ and $\breve{g}$ is real continuous positive definite.

Proof

Note $g(1)=1$ and that g is continuous. We show at once that g is real. The matrix $\tau :=\left( \begin{array}{cc} 0 &{} 1\\ -1 &{} 0 \end{array}\right) \in SU(2)$ satisfies $U(\tau )\Gamma _0=\Gamma _0$ and $\tau A_\kappa \tau ^{-1}=A_{-\kappa }$. So $\overline{\langle \Gamma _0, U(A_\kappa )\Gamma _0\rangle }=\langle U(A_\kappa )\Gamma _0, \Gamma _0\rangle =\langle \Gamma _0, U(A_{-\kappa })\Gamma _0\rangle =\langle \Gamma _0, U(\tau ^{-1})U(A_{-\kappa })U(\tau )\Gamma _0\rangle =\langle \Gamma _0, U(A_\kappa )\Gamma _0\rangle$. It remains to show the positive definiteness of $\breve{g}$. Let $k\in \mathbb {R}^3$. Then there is $C_k\in SL(2,\mathbb {C})$ with $C_k\cdot 0=k$. Set $\Gamma _k:=U(C_k)\Gamma _0$. Here $\Gamma _k$ is well defined, since $A\cdot 0=k$ implies $A^{-1}C_k\cdot 0=0$, whence $A^{-1}C_k\in SU(2)$ and hence $U(A)\Gamma _0=U(A)U(A^{-1}C_k)\Gamma _0=U(C_k)\Gamma _0$. Put $G(k,p):=\langle \Gamma _k,\Gamma _p\rangle$. Clearly G is positive definite. G is also Lorentz invariant, since $G(A\cdot k,A\cdot p)=G(AC_k\cdot 0,AC_p\cdot 0)= \langle U(AC_k)\Gamma _0,U(AC_p)\Gamma _0\rangle = \langle U(C_k)\Gamma _0,U(C_p)\Gamma _0\rangle =G(k,p)$. Now note $G\big (0,(0,0,\sqrt{t^2-1})\big )= G(0,A_\kappa \cdot 0)=\langle \Gamma _0, \Gamma _{A_\kappa \cdot 0}\rangle =\langle \Gamma _0,U(A_\kappa )\Gamma _0\rangle =g(t)$ for $\kappa =\cosh ^{-1}t$, whence the claim by (14.1). $\square$

In (30) the carrier space ${\mathcal {H}}_U$ of U is not assumed to be separable. One concludes

Corollary 31

${\mathcal {G}}=\{\breve{g}: \,g \text { in } {(14.3)}, \,{\mathcal {H}}_U \text { separable}\}$

The matrix elements g in (14.3) are closely related to the zonal (or elementary) spherical functions on $SL(2,\mathbb {C})$, i.e., the SU(2)-bi-invariant functions $\phi$ on $SL(2,\mathbb {C})$ associated to irreps U on $SL(2,\mathbb {C})$ given by $\phi (A):=\langle \Gamma _0,U(A)\Gamma _0\rangle$, where up to normalization $\Gamma _0$ is the unique SU(2)-invariant vector (see below). Recall that for every $A\in SL(2,\mathbb {C})$ there are $B,B'\in SU(2)$ with $A=B'A_\kappa B$. The rapidity $\kappa \ge 0$ is uniquely determined by A, since $A^*A=B^*A^2_\kappa B$, whence ${\text {e}}^\kappa$, ${\text {e}}^{-\kappa }$ are the eigenvalues of $A^*A$, and since $\tau A_\kappa \tau ^{-1}=A_{-\kappa }$ (see the proof of (30)).

14.2 ${\mathcal {G}}$ is the set of convex combinations of its irreducible elements

The theory of reps of $SL(2,\mathbb {C})$ is treated exhaustively in [18], to which we will refer in the sequel. The irreps of $SL(2,\mathbb {C})$ of the principal series are characterized by a pair $(m,\lambda )\in {\mathbb {Z}}\times \mathbb {R}$ of invariant numbers. For $m\ne 0$ they do not contain SU(2)-invariant vectors $\ne 0$, and for $(0,\lambda )$ there is up to a phase just one SU(2)-invariant normalized vector. $(0,\lambda )$ and $(0,-\lambda )$ determine equivalent reps. The matrix element (14.3) reads

$$\begin{aligned} g^{\textsc {p}}_{\lambda }(t):=\frac{\sin \lambda \kappa }{\lambda \sinh \kappa } \text { for } \kappa =\cosh ^{-1}t \text { and } \lambda \in [0,\infty [ \end{aligned}$$

(14.4)

See [18, III §11.5  formula after Eq. (10)] for $\langle \Gamma _0, U(A_{2\kappa })\Gamma _0\rangle$ with $\rho =2\lambda$. For $\lambda \ne 0$ it is not positive and hence $\breve{g}_\lambda ^{\textsc {p}}$ is not positive.

The irreps of $SL(2,\mathbb {C})$ of the supplementary series are characterized by one invariant number $\lambda \in ]0,1[$. They all contain just one one-dimensional SU(2)-invariant subspace. The matrix element (14.3) reads

$$\begin{aligned} g^\textsc {s}_{\lambda }(t):=\frac{\sinh \lambda \kappa }{\lambda \sinh \kappa } \text { for } \kappa =\cosh ^{-1}t \text { and } \lambda \in ]0,1[ \end{aligned}$$

(14.5)

See [18, III §12,7 (3)] for $\rho =2\lambda$. It is positive. Hence, by (31), $\breve{g}_\lambda ^{\textsc {s}}$ is positive for $0\le \lambda \le 1$. The cases $\lambda = 0,1$ hold by continuity. Actually

$$\begin{aligned} g^\textsc {s}_0(t)=g^\textsc {p}_0(t)=\frac{\kappa }{\sinh \kappa }, \quad \kappa =\cosh ^{-1}t \end{aligned}$$

for the case $\lambda =0$, and the trivial rep yields the case $\lambda = 1$. Note the ordering

$$\begin{aligned} |g^\textsc {p}_\lambda |\le g^\textsc {p}_0= g^\textsc {s}_0\le g^\textsc {s}_{\lambda _1}\le g^\textsc {s}_{\lambda _2}\quad \forall \;\lambda \ge 0, \;\;0\le \lambda _1\le \lambda _2\le 1 \end{aligned}$$

(14.6)

We call $g^\textsc {p}_\lambda$ and $g^\textsc {s}_\lambda$ irreducible and $\breve{g}^{{\textsc {p}}}_{\lambda }$, $\breve{g}^{{\textsc {s}}}_{\lambda }$ the irreducible elements of ${\mathcal {G}}$.

Theorem 32

Every $G\in {\mathcal {G}}$ is given by $G=\breve{g}$ for

$$\begin{aligned} g(t)=\int _{[0,\infty [}\frac{\sin \lambda \kappa }{\lambda \sinh \kappa }{\text {d}}\mu ^\textsc {p}(\lambda )+\int _{]0,1]} \frac{\sinh \lambda \kappa }{\lambda \sinh \kappa }{\text {d}}\mu ^\textsc {s}(\lambda ) \end{aligned}$$

(14.7)

with $\kappa =\cosh ^{-1}t$, where $\mu ^\textsc {p}$ and $\mu ^\textsc {s}$ are Borel measures such that $\mu ^\textsc {p}([0,\infty [)+\mu ^\textsc {s}(]0,1])=1$.

Proof

We apply (31). So let U be a rep of $SL(2,\mathbb {C})$ in a separable Hilbert space having normalized SU(2)-invariant vectors. Being semisimple connected, $SL(2,\mathbb {C})$ is a locally compact tame group with countable basis. Therefore by [14, Second Part § 8.4 Theorem 3] it allows the direct integral (continuous sum) decomposition into its primary components such that

$$\begin{aligned} U=\int _{\Lambda }W_\lambda {\text {d}}\mu (\lambda ) \end{aligned}$$

up to unitary equivalence. Here $\mu$ is a probability measure on the set $\Lambda$ of equivalence classes of irreps of $SL(2,\mathbb {C})$, $W_\lambda$ is a primary component written in the form $U_\lambda \otimes I_\lambda$. $U_\lambda$ belongs to the class $\lambda \in \Lambda$, and $I_\lambda$ is the trivial rep of dimension $n(\lambda )\in \mathbb {N}\cup \{0,\infty \}$.

First we describe the SU(2)-invariant normalized vectors $\Gamma _0$ of U. $\Gamma _0$ decomposes into SU(2)-invariant normalized vectors of the primary components $W_\lambda$. Let $\Lambda _0$ denote the set of equivalence classes of reps of the supplementary series and the principal series with $m= 0$. For $\lambda \in \Lambda \setminus \Lambda _0$ there is no such vector for $W_\lambda$. For $\lambda \in \Lambda _0$ they read $\gamma _\lambda \otimes v_\lambda$ with $\gamma _\lambda$ the unique (up to a phase) SU(2)-invariant normalized vector of $U_\lambda$, and $v_\lambda$ any normalized vector in the carrier space of $I_\lambda$. Hence $\Gamma _0(\lambda )=0$ if $\lambda \not \in \Lambda _0$, and $\Gamma _0(\lambda )=C\,\gamma _\lambda \otimes v_\lambda$ with the normalization constant $C:=\mu (\Lambda _0)^{-1/2}$ otherwise.

Determine (14.3): $\langle \Gamma _0, U(A_\kappa )\Gamma _0\rangle =\int _{\Lambda _0}C^2 \langle \gamma _\lambda \otimes v_\lambda ,W_\lambda (A_\kappa )\gamma _\lambda \otimes v_\lambda \rangle {\text {d}}\mu (\lambda )=\int _{\Lambda _0}C^2 \langle \gamma _\lambda \otimes v_\lambda ,\big (U_\lambda (A_\kappa )\gamma _\lambda \big )\otimes v_\lambda \rangle {\text {d}}\mu (\lambda )=\int _{\Lambda _0}C^2 \langle \gamma _\lambda ,U_\lambda (A_\kappa )\gamma _\lambda \rangle {\text {d}}\mu (\lambda )$. Now, $C^2\mu$ is a probability measure on $\Lambda _0$, which we write as the sum of two orthogonal measures $\mu ^\textsc {p}$ and $\mu ^\textsc {s}$ on the disjoint sets $\Lambda ^\textsc {p}:=\{[(0,\lambda )]:\lambda \ge 0\}$ and $\Lambda ^\textsc {s}:=\{[\lambda ]: 0<\lambda \le 1\}$. We identify $\lambda$ with the invariant number determining the irreps of the principal and the supplementary series. The result (14.7) follows by the formulae (14.4), (14.5). $\square$

Several more or less immediate results concerning (32) and (14.7) are gathered in (33). For the following the conclusions (a) - (d) are indispensable. The remaining items illustrate the ordering on ${\mathcal {G}}$.

Lemma 33

• (a) Let $G\in {\mathcal {G}}$ be extreme. Then g is irreducible, i.e., $g=g^\textsc {p}_\lambda$ or $g=g^\textsc {s}_\lambda$ for some $0\le \lambda <\infty$ and $0<\lambda \le 1$, respectively.

Proof

For G extreme, obviously either $\mu ^\textsc {p}=0$ or $\mu ^\textsc {s}=0$. Consider first $\mu ^\textsc {s}=0$. Let $B\subset [0,\infty [$ be a Borel set and $B'$ its complement. Using an obvious notation, $g=g_B+g_{B'}$. So one summand must vanish, say $g_{B'}=0$. This implies $0=g_{B'}(1)=\mu ^\textsc {p}(B')$ and hence $\mu ^\textsc {p}(B)=1$. Thus every Borel set has either measure 1 or 0. Since $\lim _{N\rightarrow \infty }\mu ^\textsc {p}([0,N[)=\mu ^\textsc {p}([0,\infty [)=1$, there is some N with $\mu ^\textsc {p}([0,N[)=1$. Then [0, N[ contains an interval $I_1$ of length N/2 with $\mu ^\textsc {p}(I_1)=1$, and $I_1$ contains an interval $I_2$ of length N/4 with $\mu ^\textsc {p}(I_2)=1$, and so on. The intersection of these intervals consists of one point $\lambda _0$. Hence $\mu ^\textsc {p}=\delta _{\lambda _0}$. The case $\mu ^\textsc {p}=0$ follows analogously. $\square$

• (b) Let $\mu ^\text {p}(]0,\infty [)=0$. Then obviously G is positive. Moreover, $g\ge g^\textsc {s}_0= \frac{\kappa }{\sinh \kappa }$, whence $\lim _{t \rightarrow \infty } t\, g(t)\rightarrow \infty$.

• (c) If $\mu ^\textsc {s}= 0$, then $|g(t)|\le \frac{\kappa }{\sinh \kappa }$. If $\mu ^\textsc {s}\ne 0$, then $\lim _{t \rightarrow \infty } t^r\, g(t)\rightarrow \infty$ for some $r<1$. Finally, $\mu ^\textsc {s}= 0$ if and only if $|g(t)|\le \frac{\kappa }{\sinh \kappa }$.

Proof

Write $g=g^\textsc {p}+g^\textsc {s}$ in the obvious way. Keep the ordering (14.6) in mind. Then $|g^\textsc {p}(t)|\le \frac{\kappa }{\sinh \kappa }$, which implies the first part of the claim. Now let $\mu ^\textsc {s}\ne 0$. Then there is $0<\delta <1$ with $\mu ^\textsc {s}([\delta ,1])>0$. Hence $g^\textsc {s}(t)\ge \frac{\sinh \delta \kappa }{\delta \sinh \kappa }\mu ^\textsc {s}([\delta ,1])$. So $t^r\,g(t)\ge \frac{\cosh ^r\kappa }{\sinh \kappa }(\alpha \sinh \delta \kappa -\kappa )$ for some $\alpha >0$, whence the result for $r=1-\delta /2$. The last part of the claim is an easy consequence of the prior results. $\square$

• (d) Note $g_{1^/2}=g^\textsc {s}_{1/2}$. In view of (11.7) and (25) we show that $\breve{g}\in {\mathcal {G}}$ satisfies

  $$\begin{aligned} g\le g_{1/2}\;\Rightarrow \, \mu ^\textsc {s}(]1/2,1])=0\;\Rightarrow \, |g|\le g_{1/2} \end{aligned}$$

Proof

Write $g=g^\textsc {p}+g^\textsc {s}$ in the obvious way. Keep the ordering (14.6) in mind.

($\alpha$) Let $g\le g_{1/2}$. Assume $\mu ^\textsc {s}(]1/2,1])>0$. Then $|g^\textsc {p}(t)|\le \frac{\kappa }{\sinh \kappa }\mu ^\textsc {p}([0,\infty [)$, and there is $\lambda _0> 1/2$ with $\mu ^\textsc {s}([\lambda _0,1])>0$, whence $g^\textsc {s}(t)\ge \frac{\sinh \lambda _0\kappa }{\lambda _0\sinh \kappa }\mu ^\textsc {s}([\lambda _0,1])>0$. Hence $\alpha \frac{\sinh \lambda _0\kappa }{\sinh \kappa }\le \beta \frac{\kappa }{\sinh \kappa } +\frac{1}{\cosh \frac{\kappa }{2}}$ for some $\alpha >0$ and $\beta \ge 0$. However this inequality obviously does not hold for large $\kappa$. One concludes $\mu ^\textsc {s}(]1/2,1])=0$.

($\beta$) Let $\mu ^\textsc {s}(]1/2,1])=0$. Then $|g^\textsc {p}(t)|\le \frac{\kappa }{\sinh \kappa }\mu ^\textsc {p}([0,\infty [)=g^\textsc {s}_0(t)\,\mu ^\textsc {p}([0,\infty [)$, whence $|g^\textsc {p}|\le g^\textsc {s}_{1/2}\,\mu ^\textsc {p}([0,\infty [)$. Further $g^\textsc {s}\le g^\textsc {s}_{1/2}\,\mu ^\textsc {s}([0,1/2])$. It follows $|g|\le g^\textsc {s}_{1/2}$. Recall $g^\textsc {s}_{1/2}= g_{1/2}$. $\square$

• (e) For every $G\in {\mathcal {G}}\setminus \{1\}$ there is $G_1\ne 1$ with $\mu ^\textsc {p}(]0,\infty [)=0$ such that $G\le G_1$.

Proof

The claim holds by the ordering (14.6). $\square$

• (f) For every $G\in {\mathcal {G}}\setminus \{1\}$ there is $\lambda _0\in ]0,1[$ such that $\breve{g}^{\textsc {s}}_{\lambda _0}\not \le G$.

Proof

By (e) let without restriction $\mu ^\textsc {p}(]0,\infty [)=0$. Fix $\kappa >0$. Recall ${\text {sinch}}(y)=\frac{\sinh y}{y}$ for $y\ge 0$ is strictly increasing. It suffices to show $\int _{]0,1]}\frac{{\text {sinch}}x \kappa }{{\text {sinch}}\lambda _0 \kappa }{\text {d}}\mu ^\textsc {s}(x)<1$ for some $0<\lambda _0<1$. Let $0< \lambda _n<1$, $\lambda _n\rightarrow 1$. Since $\frac{{\text {sinch}}x \kappa }{{\text {sinch}}\lambda _n \kappa }\le {\text {sinch}} \kappa$ $\forall$ n, x, dominated convergence yields $\lim _{n\rightarrow \infty }\int _{]0,1]}\frac{{\text {sinch}}x \kappa }{{\text {sinch}}\lambda _n \kappa }{\text {d}}\mu ^\textsc {s}(x)=\int _{]0,1]}\frac{{\text {sinch}}x \kappa }{{\text {sinch}} \kappa }{\text {d}}\mu ^\textsc {s}(x)=:A$. The integrand is $< 1$ for $x<1$ and $=1$ for $x=1$. Hence $A< 1$ since otherwise $\mu ^\textsc {s}(\{1\})=1$, whence the contradiction $G=1$. The result follows. $\square$

• (g) On the other hand there are $G\in {\mathcal {G}}_+$ satisfying  $G\,\not \le \,\breve{g}^{\textsc {s}}_\lambda \quad \forall \;\;0< \lambda <1$.

For an example take ${\text {d}}\mu ^\textsc {s}(\lambda ):=2\lambda \, 1_{]0,1]}(\lambda ){\text {d}}\lambda$, $\mu ^\textsc {p}=0$ yielding $g(t)=\frac{\tanh \kappa /2}{\kappa /2}$. Then for $\lambda <1$ and $\kappa \rightarrow \infty$ one has $\frac{\kappa }{2}\frac{\sinh \lambda \kappa }{\lambda \sinh \kappa }=\frac{\kappa }{2\lambda }{\text {e}}^{-(1-\lambda )\kappa }\frac{1-\exp (-2\lambda \kappa )}{1-\exp (-2\kappa )}\rightarrow 0$, whereas $\tanh \frac{\kappa }{2}\rightarrow 1$. $\square$

15 The Lorentz invariant kernels due to the principal series

It is easy to see that there are positive G with $\mu ^\text {s}=0$. As an example take ${\text {d}}\mu ^\textsc {p}(\lambda ):=1_{[0,1]}(\lambda ){\text {d}}\lambda$. Then $g(t)=\frac{{\text {Si}}(\kappa )}{\sinh \kappa }>0$ with ${\text {Si}}$ the sine integral. Actually one has the following characterization of

$$\begin{aligned} {\mathcal {G}}^\textsc {p}:=\{\breve{g}\in {\mathcal {G}}: \mu ^\textsc {s}=0 \text { in } (14.7) \} \end{aligned}$$

Proposition 34

Let g denote a function on $[1,\infty [$ and put $\kappa =\cosh ^{-1} t$. The following statements $(\alpha ),(\beta ), (\gamma )$ are equivalent.

$(\alpha )$:

$\breve{g}\in {\mathcal {G}}$ with $|g(t)|\le \frac{\kappa }{\sinh \kappa }$ for all $t\ge 1$.

$(\beta )$:

$\breve{g}\in {\mathcal {G}}^\textsc {p}$, i.e., $g(t)=\int _{[0,\infty [}\frac{\sin \lambda \kappa }{\lambda \sinh \kappa }{\text {d}}\mu (\lambda )$ with $\mu$ a probability Borel measure on $[0,\infty [$.

$(\gamma )$:

$g(t)= \frac{\Psi (\kappa )}{\sinh (\kappa )}$ for $\Psi (\kappa ):=\int _0^\kappa \psi (x){\text {d}}x$ with $\psi$ a real continuous function on $\mathbb {R}$ of positive type normalized by $\psi (0)=1$.

By [19, Theorem IX.9 (Bochner’s theorem)] there is a unique even probability measure $\mu _{even}$ on $\mathbb {R}$ such that

$$\begin{aligned} \psi (x)=\int _\mathbb {R}{\text {e}}^{{\text {i}}\lambda \,x}{\text {d}}\mu _{even}(\lambda ) \end{aligned}$$

(15.1)

Then g, $\psi$, $\mu$, and $\mu _{even}$ determine uniquely each other. In particular one has

$(\delta )$:

$\mu (B)=\mu _{even}(B\cap \{0\})+2\,\mu _{even}(B\setminus \{0\})$ for all Borel sets $B\subset [0,\infty [$.

Proof

$(\alpha )$ $\Leftrightarrow$ $(\beta )$ holds by (32) and (33)(c).

As to $(\beta )$ $\Rightarrow$ $(\gamma )$,  $g(t)\sinh \kappa =\int _{[0,\infty [}\frac{\sin \lambda \kappa }{\lambda }{\text {d}}\mu (\lambda )=\int _{[0,\infty [}\frac{1}{2}\int _{-\kappa }^\kappa {\text {e}}^{{\text {i}}\lambda x}{\text {d}}x {\text {d}}\mu (\lambda )=\int _0^\kappa \int _\mathbb {R}{\text {e}}^{{\text {i}}\lambda x}{\text {d}}\mu _{even}(\lambda ){\text {d}}x$, where $\mu _{even}$ is the even probability measure on $\mathbb {R}$ given by $\mu _{even}(B):=\frac{1}{2}\mu (B)$ for a Borel set $B\subset ]0,\infty [$, $\mu _{even}(B):=\frac{1}{2}\mu (-B)$ for a Borel set $B\subset ]-\infty ,0[$, and $\mu _{even}(\{0\}):=\mu (\{0\})$. Now put $\psi (x):=\int _\mathbb {R}{\text {e}}^{{\text {i}}\lambda x}{\text {d}}\mu _{even}(\lambda )$, $x\in \mathbb {R}$. Clearly $\psi$ is real, $\psi (0)=1$, continuous and, by Bochner’s theorem, of positive type.

We turn to $(\beta )$ $\Leftarrow$ $(\gamma )$ and ($\delta$). Recall (15.1). Then going backward the prior proof, $\frac{\Psi (\kappa )}{\sinh (\kappa )}$ equals g(t) in $(\beta )$ for $\mu$ given in $(\delta )$.

Finally, obviously g and $\psi$ determine each other. Hence so do g, $\psi$, $\mu _{even}$ and, by ($\delta )$, also $\mu$. $\square$

Referring to (34), the characterization of g in $(\gamma )$ gives rise to an inversion formula of the formula in $(\beta )$.

Corollary 35

Let $\breve{g}\in {\mathcal {G}}$ with g a function on $[1,\infty [$. Suppose $|g(t)|\le \frac{\kappa }{\sinh \kappa }$ and $\psi \in L^1(\mathbb {R})$ for $\psi (x):=\frac{{\text {d}}}{{\text {d}}x}\big (\sinh x \;g(\cosh x)\big )$, $x\in \mathbb {R}$. Put

$$\begin{aligned} w(\lambda ):= \frac{1}{\pi }\int _{-\infty }^\infty \psi (x){\text {e}}^{-{\text {i}}\lambda x}{\text {d}}x \end{aligned}$$

(15.2)

Then

$$\begin{aligned} w\ge 0, \;1=\int _0^\infty w(\lambda ){\text {d}}\lambda , \text { and } g(t)=\int _0^\infty \frac{\sin \lambda \kappa }{\lambda \sinh \kappa }w(\lambda ){\text {d}}\lambda \end{aligned}$$

(15.3)

Proof

By (34)($\gamma$) $\psi$ is real continuous bounded by $1=\psi (0)$ of positive type. So by Bochner’s theorem $\psi (x)=\int _\mathbb {R}{\text {e}}^{{\text {i}}\lambda x}{\text {d}}\mu _{even}(\lambda )$, $x\in \mathbb {R}$ for some even probability measure $\mu _{even}$. As by assumption $\psi$ is integrable, (57)(d) applies. Accordingly $w_{even}:=(2\pi )^{-1/2}{\mathcal {F}}\psi$ is integrable nonnegative. It satisfies $\psi =(2\pi )^{1/2}{\mathcal {F}}^{-1} w_{even}$. Thus the Fourier transforms of the finite measures $\mu _{even}$ and $w_{even}m$ with m the Lebesgue measure on $\mathbb {R}$ coincide. Thus the measures itself coincide.

Now recall (34)($\delta$). Put $w:=2\, w_{even}|_{[0,\infty [}$. The claim holds by (34)$(\beta )$. $\square$

Example 36

Let $g=g_1$, i.e., $g(t)=\frac{2}{1+t}$.

We apply (35) to compute the weight function w satisfying (15.3). First check that $\psi (x)=2(1+\cosh x)^{-1}$. The integral (15.2) can be routinely evaluated by the theorem of residues. The poles of the integrand $\psi$ are at $z_k:={\text {i}}(2k+1)\pi$, $k\in {\mathbb {Z}}$. They are all of second order and give rise to the residues $2{\text {i}}\lambda {\text {e}}^{-{\text {i}}\lambda z_k}$. One finds

$$\begin{aligned} w(\lambda )=\frac{4\lambda }{\sinh \pi \lambda } \end{aligned}$$

(One easily confirms the result by the formulae from Gradshteyn [10, 3.521(1)] and [10, 17.34(28))] for $a:={\text {i}}\kappa$, $b:=\pi$, $x:=\lambda$, $\xi :=0$.)

Similarly one finds

$$\begin{aligned} w(\lambda )=\frac{8\lambda ^3}{\sinh \pi \lambda } \end{aligned}$$

for $g=g_2$, i.e., $g(t)=4(1+t)^{-2}$. (Confirm the normalization of w by [10, 3.523(6)].)

In view of (11.3), (43) the next example is particularly interesting.

Example 37

Let $g=g_{3/2}$, i.e., $g(t)=\big (\frac{2}{1+t}\big )^{3/2}$.

We proceed as in (36). One finds $\psi (x)=\big (2-\cosh ^2\frac{\kappa }{2}\big )\cosh ^{-3}\frac{\kappa }{2}$. Note that $\psi \not \ge 0$ (!) despite the fact that $g_{3/2}>0$.

The poles of $\psi$ are at $z_k:={\text {i}}(k+\frac{1}{2})\pi$, $k\in {\mathbb {Z}}$. They are all of third order and give rise to the residues $-4{\text {i}}\lambda ^2 (-1)^k{\text {e}}^{-2{\text {i}}\lambda z_k}$. One finds

$$\begin{aligned} w(\lambda )=\frac{8\lambda ^2}{\cosh \pi \lambda } \end{aligned}$$

(Confirm the normalization of w by [10, 3.523(5)].)

A further result using decisively (34) is

Corollary 38

Let $G\in {\mathcal {G}}^\text {p}$. Then

$$\begin{aligned} G \, is\, extreme\, in\, {\mathcal {G}}^\text {p}\, \Leftrightarrow \, G\, is\, extreme\, in\, {\mathcal {G}}\, \Leftrightarrow \, G\, is\, irreducible \end{aligned}$$

Proof

As to the first $\Rightarrow$, let $G=\alpha G_a+\beta G_b$ with $\alpha , \beta > 0$, $\alpha +\beta =1$, and $G_i\in {\mathcal {G}}$, $i=a,b$. Then by (33)(c) it follows immediately $G_i\in {\mathcal {G}}^\text {p}$. So by assumption $G=G_a=G_b$. The first $\Leftarrow$ is trivial.

The second $\Rightarrow$ holds by (33)(a). We turn to the second $\Leftarrow$. Let $G= \breve{g}^{\text {p}}_{\lambda _0}$ for $\lambda _0\in [0,\infty [$ and let $g^\text {p}_{\lambda _0}=\alpha g_a+\beta g_b$, with $\alpha , \beta > 0$, $\alpha +\beta =1$, and $\breve{g}_a, \breve{g}_b \in {\mathcal {G}}$. Then again, by (33)(c) it follows $\breve{g}_i\in {\mathcal {G}}^\text {p}$, $i=a,b$. Hence due to the uniqueness result in (34) one has $\delta _{\lambda _0}=\alpha \mu ^\textsc {p}_a+\beta \mu ^\textsc {p}_b$. This implies $\mu ^\textsc {p}_a= \mu ^\textsc {p}_b=\delta _{\lambda _0}$. $\square$

16 The maximal causal kernel

It turns out that the positive definite Lorentz invariant kernels $\breve{g}\in {\mathcal {G}}$, which determine causal kernels, are due to the principal series only, i.e., belong to ${\mathcal {G}}^\textsc {p}$. The result is obtained by checking the necessary condition NC below. It discards also all irreducible elements of ${\mathcal {G}}$. (39) reveals NC to be rather sharp. It is the main tool in proving the important result that $K_{3/2}$ is the greatest element of ${\mathcal {K}}$ (43).

We deal with the causal kernels on $\mathbb {R}^3$ (11.1). Referring to the expansion of K (7.5), we exploit the fact that by (11)

$$\begin{aligned} k_0(\sigma ,\rho )=\frac{\epsilon (\sigma )+\epsilon (\rho )}{2\sqrt{\epsilon (\sigma ) \epsilon (\rho )}}\int _{-1}^1g\big (\epsilon (\sigma )\epsilon (\rho )-\sigma \rho \, x\big )\frac{{\text {d}}x}{2} \end{aligned}$$

is positive definite on $[0,\infty [$. Hence it satisfies the inequality $k_0(\sigma ,\rho )^2\le k_0(\sigma ,\sigma )k_0(\rho ,\rho )$ (7.2). For $\sigma =0$ this reads

$$\begin{aligned} \frac{\big (1+\epsilon (\rho )\big )^2}{4\,\epsilon (\rho )}\,g\big (\epsilon (\rho )\big )^2\le \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t \end{aligned}$$

(NC)

So, if g determines a causal kernel, then g satisfies NC.

Effectiveness of NC 39

The necessary condition for positive definiteness of rotational invariant normalized kernels introduced above

$$\begin{aligned} k_0(0,\rho )^2\le k_0(0,0)\,k_0(\rho ,\rho ) \end{aligned}$$

is shown to be not satisfied by $\breve{g}_r$ for $0<r<\frac{1}{2}$ and by $K_r$ for $0<r<\frac{3}{2}$ thus furnishing straightforward alternative proofs to (26) and (27), by which these kernels are not positive definite. Moreover, what is important, the necessary condition holds by equality for the kernels $\breve{g}_{1/2}$ and $K_{3/2}$.

• (a) $\breve{g}_r$ for $0<r<\frac{1}{2}$ is not positive definite

Proof

The left hand side LS of NC reads $2^{2r}\big (1+\epsilon (\rho )\big )^{-2r}$ (there is no prefactor here), the right hand side RS of NC is $\frac{1}{1-r}\frac{1}{\rho ^2}\left( \epsilon (\rho )^{2-2r}-1\right)$, $0\le r\ne 1$. Hence NC imposes

$$\begin{aligned} f(r)\le R(\rho ) \end{aligned}$$

with $f(r):=2^{2r}(1-r)$ and $R(\rho ):=\frac{\epsilon (\rho )^2}{\rho ^2}\left( \frac{1+\epsilon (\rho )}{\epsilon (\rho )} \right) ^{2r}-\frac{\big (1+\epsilon (\rho )\big )^{2r}}{\rho ^2}\rightarrow 1$ for $\rho \rightarrow \infty$. Hence $f(r)\le 1$. However this does not hold, since $f(0)=f(\frac{1}{2})=1$, $f(\frac{1}{4})=\frac{3}{4}\sqrt{2}>1$, and $f'(r)=0$ has exactly one solution ($r=1-\frac{1}{\ln 4}$). $\square$

• (b) $g_{1/2}\big (\epsilon (\rho )\big )^2 = \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g_{1/2}(t){\text {d}}t$, which is NC by equality for $\breve{g}_{1/2}$

Proof

The claim is easily checked. $\square$

• (c) $K_r$ for $0<r<\frac{3}{2}$ is not positive definite

Proof

Here LS has the additional prefactor $\frac{\big (1+\epsilon (\rho )\big )^2}{4\epsilon (\rho )}$. Hence for $r\ne 1$ NC imposes

$$\begin{aligned} L(\rho )\le R(\rho ) \end{aligned}$$

with $L(\rho ):=2^{2r-2}\rho ^2\,\epsilon (\rho )^{-1}\big (1+\epsilon (\rho )\big )^{2-2r}\rightarrow \infty$, $R(\rho ):=\frac{1}{1-r}\left( \epsilon (\rho )^{2-2r}-1\right) \rightarrow \frac{1}{r-1}$ for $\rho \rightarrow \infty$ if $r>1$, and $L(\rho ):=2^{2r-2}\big (1+\epsilon (\rho )\big )^{2-2r}\epsilon (\rho )^{2r-2}\rightarrow 2^{2r-2}$, $R(\rho ):=\frac{1}{1-r}\rho ^{-2}\epsilon (\rho )\left( 1-\epsilon (\rho )^{2r-2}\right) \rightarrow 0$ for $\rho \rightarrow \infty$ if $r<1$. Hence NC is not satisfied for $r\ne 1$.

For $r=1$, LS equals $\epsilon (\rho )^{-1}$ and RS reads $2\rho ^{-2}\,\ln \epsilon (\rho )$. Hence NC imposes $\rho \,\epsilon (\rho )^{-1}\le 2\rho ^{-1} \ln \epsilon (\rho )$, which does not hold. $\square$

• (d) $\frac{\big (1+\epsilon (\rho )\big )^2}{4\,\epsilon (\rho )}\,g_{3/2}\big (\epsilon (\rho )\big )^2= \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g_{3/2}(t){\text {d}}t$, which is NC by equality for $K_{3/2}$

Proof

The claim is easily checked. $\square$

Lemma 40

Let g be irreducible, i.e., equal to $g^\textsc {p}_\lambda$ or $g^\textsc {s}_\lambda$ for some $\lambda$. Then

$$\begin{aligned} g\big (\epsilon (\rho )\big )^2\;= \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t \end{aligned}$$

(1)

Explicitly (1) equals

$$\begin{aligned} \left( \frac{\sin \lambda \, l(\rho )}{\lambda \rho }\right) ^2 \text { for } g^\textsc {p}_\lambda \text { and } \left( \frac{\sinh \lambda \, l(\rho )}{\lambda \rho }\right) ^2 \text { for } g^\textsc {s}_\lambda \end{aligned}$$

(2)

with $l(\rho ):=\ln \big (\epsilon (\rho )+\rho \big )$. The limiting case $\lambda =0$ reads $\big (\frac{l(\rho )}{\rho }\big )^2$. Finally, g does not satisfy NC. So an irreducible g does not determine a causal kernel.

Proof

As to $g\big (\epsilon (\rho )\big )^2$, note $\kappa =\cosh ^{-1}\epsilon (\rho )=\ln \big (\epsilon (\rho )+\sqrt{\epsilon (\rho )^2-1}\big )=l(\rho )$ and $2\sinh l(\rho )=\epsilon (\rho )+\rho -(\epsilon (\rho )+\rho )^{-1}=\epsilon (\rho )+\rho -(\epsilon (\rho )-\rho )=2\rho$, whence (2).

We turn to the integral $\frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t$, which for $g=g^\textsc {p}_\lambda$, $\lambda >0$, becomes $\frac{1}{2\rho ^2\lambda }\int _0^{2\,l(\rho )}\sin \lambda x {\text {d}}x$ using the substitution $x=\cosh ^{-1}t$, whence (2). Analogously one obtains (2) for $\lambda =0$ and for $g=g^\textsc {s}_\lambda$.

Finally, g does not satisfy NC because of (1),(2) and the fact $\frac{1}{\rho }\, \frac{\big (1+\epsilon (\rho )\big )^2}{4\,\epsilon (\rho )}\rightarrow \frac{1}{4}>0$ for $\rho \rightarrow \infty$, $\square$

Corollary 41

Put $l(\rho ):=\ln \big (\epsilon (\rho )+\rho \big )$. For g in (14.7) one has

$$\begin{aligned} g(\epsilon (\rho ))=\int _{[0,\infty [}\frac{\sin \lambda \,l(\rho )}{\lambda \,\rho }{\text {d}}\mu ^\textsc {p}(\lambda )+\int _{]0,1]}\frac{\sinh \lambda \,l(\rho )}{\lambda \,\rho }{\text {d}}\mu ^\textsc {s}(\lambda ) \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t =\int _{[0,\infty [}\left( \frac{\sin \lambda \,l(\rho )}{\lambda \,\rho }\right) ^2{\text {d}}\mu ^\textsc {p}(\lambda )+\int _{]0,1]}\left( \frac{\sinh \lambda \,l(\rho )}{\lambda \,\rho }\right) ^2{\text {d}}\mu ^\textsc {s}(\lambda ) \end{aligned}$$

Proposition 42

Let g determine a causal kernel. Then $\breve{g}\in {\mathcal {G}}^\textsc {P}$, but $\breve{g}$ is not extreme. Recall (34).

Proof

There is the representation (14.7). Put $g=g^\textsc {p}+g^\textsc {s}$ in the obvious way. Assume $g^\textsc {s}\ne 0$. Keep in mind (41).

As to the left hand side LS of NC, $g^\textsc {p}(\epsilon (\rho ))=\int _{[0,\infty [}\frac{\sin \lambda \,l(\rho )}{\lambda \,\rho }{\text {d}}\mu ^\textsc {p}(\lambda )$, whence $|g^\textsc {p}(\epsilon (\rho ))|\le \alpha \frac{l(\rho )}{\rho }$ with $\alpha :=\mu ^\textsc {p}([0,\infty ])\ge 0$. Further $g^\textsc {s}(\epsilon (\rho ))=\int _{]0,1]}\frac{\sinh \lambda \,l(\rho )}{\lambda \,\rho }{\text {d}}\mu ^\textsc {s}(\lambda )$. By the mean value theorem of integrals there is $\lambda _{l,\rho }\in ]0,1]$ satisfying $\beta \,\frac{\sinh \lambda _{l,\rho }\,l(\rho )}{\lambda }=\int _{]0,1]}\frac{\sinh \lambda \,l(\rho )}{\lambda }{\text {d}}\mu ^\textsc {s}(\lambda )$ with $\beta :=\mu ^\textsc {s}(]0,1])>0$. Since $\frac{\sinh \lambda \,l(\rho )}{\lambda }$ is strictly increasing regarding $\lambda$ and $\rho$, the position $\lambda _{l,\rho }$ is uniquely determined and is strictly increasing with respect to $\rho$. Thus the estimation $g^\textsc {p}(\epsilon (\rho ))^2\ge \big (-\alpha \frac{l(\rho )}{\rho }+ \beta \,\frac{\sinh \lambda _{l,\rho }\,l(\rho )}{\lambda \,\rho }\big )^2$ holds. Put $\lambda _l:=\lim _{\rho \rightarrow \infty } \lambda _{l,\rho }$.

The right hand side RS $=\int _{[0,\infty [}\left( \frac{\sin \lambda \,l(\rho )}{\lambda \,\rho }\right) ^2{\text {d}}\mu ^\textsc {p}(\lambda )+\int _{]0,1]}\left( \frac{\sinh \lambda \,l(\rho )}{\lambda \,\rho }\right) ^2{\text {d}}\mu ^\textsc {s}(\lambda )$ of NC is treated similarly getting RS $\le \alpha \big (\frac{l(\rho )}{\rho }\big )^2+\beta \left( \frac{\sinh \lambda _{r,\rho }\,l(\rho )}{\lambda \,\rho }\right) ^2$. Put $\lambda _r=\lim _{\rho \rightarrow \infty } \lambda _{r,\rho }$.

Note $\lambda _{l,\rho }\le \lambda _{r,\rho }$ and $0<\lambda _l\le \lambda _r$. Indeed, this holds because $\left( \beta \,\frac{\sinh \lambda _{l,\rho }\,l(\rho )}{\lambda }\right) ^2=\left( \int _{]0,1]}\frac{\sinh \lambda \,l(\rho )}{\lambda }{\text {d}}\mu ^\textsc {s}(\lambda )\right) ^2\le \beta \int _{]0,1]}\left( \frac{\sinh \lambda \,l(\rho )}{\lambda \,\rho }\right) ^2{\text {d}}\mu ^\textsc {s}(\lambda )= \beta ^2\left( \frac{\sinh \lambda _{r,\rho }\,l(\rho )}{\lambda \,\rho }\right) ^2$ by the Cauchy-Schwarz inequality.

Due to NC one infers

$$\begin{aligned} \frac{\big (1+\epsilon (\rho )\big )^2}{4\,\epsilon (\rho )}\,\left( -\alpha \frac{l(\rho )}{\rho }+ \beta \,\frac{\sinh \lambda _{l,\rho }\,l(\rho )}{\lambda \,\rho }\right) ^2\le \alpha \left( \frac{l(\rho )}{\rho }\right) ^2+\beta \left( \frac{\sinh \lambda _{r,\rho }\,l(\rho )}{\lambda \,\rho }\right) ^2 \end{aligned}$$

Hence one easily verifies

$$\begin{aligned} \rho ^{1-2 (\lambda _{r,\rho }- \lambda _{l,\rho })}L(\rho )\le R(\rho ) \end{aligned}$$

(1)

with $L(\rho ):=\rho ^{-1}\; \frac{\big (1+\epsilon (\rho )\big )^2}{4\,\epsilon (\rho )}\;\rho ^{-2 \lambda _{l,\rho }}\,\left( -\alpha \,l(\rho )+ \beta \,\frac{\sinh \lambda _{l,\rho }\,l(\rho )}{\lambda }\right) ^2\rightarrow \frac{1}{4}\left( \beta \frac{2^{\lambda _l}}{2\lambda _l}\right) ^2>0$ for $\rho \rightarrow \infty$, and similarly $R(\rho )\rightarrow \beta \left( \frac{2^{\lambda _r}}{2\lambda _r}\right) ^2$ for $\rho \rightarrow \infty$.

Now one remembers (25), (33)(d), according to which $\mu ^\textsc {s}(]1/2,1])=0$. This implies $\lambda _r\le \frac{1}{2}$. It follows $2 (\lambda _{r,\rho }- \lambda _{l,\rho })\le 1-\delta$ for some $\delta >0$ and all $\rho \ge \rho _0$ for some $\rho _0$. Thus (1) requires $\beta =0$ contradicting the assumption $g^\textsc {s}\ne 0$.

Finally recall (38) and (40). $\square$

Theorem 43

$|K|\le K_{3/2}$ pointwisely for all causal kernels K.

Proof

Let g determine a causal kernel (11.1). For $x\ge 1$ put $\alpha (x):=\frac{(1+x)^2}{4x}$.

(a) g is integrable. Indeed, g satisfies NC and, by (42), $|g|\le g_0^\textsc {p}$. Hence $\alpha \big (\epsilon (\rho )\big )\,g\big (\epsilon (\rho )\big )^2\le \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t\le \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g_0^\textsc {p}(t){\text {d}}t= g_0^\textsc {p}\big (\epsilon (\rho )\big )^2$ by (40)(1), whence $|g\big (\epsilon (\rho )\big )|\le \alpha \big (\epsilon (\rho )\big )^{-1/2}\,g_0^\textsc {p}\big (\epsilon (\rho )\big )$. The latter means $|g(t)|\le \frac{2\sqrt{\cosh \kappa }}{1+\cosh \kappa }\; \frac{\kappa }{\sinh \kappa }$ with $\kappa =\cosh ^{-1}t$. It implies

$$\begin{aligned} A:=\int _1^\infty |g(t)|{\text {d}}t<\infty \end{aligned}$$

(1)

since $\int _1^\infty |g(t)|{\text {d}}t=\int _0^\infty |g(\cosh x)|\sinh x {\text {d}}x\le \int _0^\infty \frac{2x\sqrt{\cosh x}}{1+\cosh x} {\text {d}}x\le \int _0^\infty \frac{x{\text {e}}^{x/2}}{\cosh ^2x/2} {\text {d}}x\le \int _0^\infty 4x{\text {e}}^{-x/2}{\text {d}}x=8$.

(b) g is dominated by a multiple of $g_{3/2}$. Indeed, the right hand side of NC is bounded by 1, since $|g|\le 1$. This implies $|g\big (\epsilon (\rho )\big )|^2\le \alpha \big (\epsilon (\rho )\big )^{-1}=g_{3/2}\big (\epsilon (\rho )\big )^2\;\frac{1}{2} \epsilon (\rho )\big (1+\epsilon (\rho )\big )\le 2\,g_{3/2}\big (\epsilon (\rho )\big )^2$ for $\rho \le 1$. By (1) the right hand side of NC is bounded by $\rho ^{-2}\,A$, whence $|g\big (\epsilon (\rho )\big )|^2\le g_{3/2}\big (\epsilon (\rho )\big )^2\;A\,\frac{1}{2} \frac{\epsilon (\rho )}{\rho }\big (1+\frac{\epsilon (\rho )}{\rho }\big ) \le 2A\, g_{3/2}\big (\epsilon (\rho )\big )^2$ for $\rho \ge 1$. Thus

$$\begin{aligned} |g|\le C\;g_{3/2} \end{aligned}$$

(2)

for some constant $1\le C<\infty$.

(c) By NC and (2) one has due to (39)(d): $\alpha \big (\epsilon (\rho )\big )\,g\big (\epsilon (\rho )\big )^2\le \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g(t){\text {d}}t\le C\; \frac{1}{2\rho ^2}\int _1^{1+2\rho ^2}g_{3/2}(t){\text {d}}t=C\;\alpha \big (\epsilon (\rho )\big )\,g_{3/2}\big (\epsilon (\rho )\big )^2$. Therefore $|g|\le C^{1/2}\;g_{3/2}$. Iterating this step the result follows. $\square$

The result in (43) is improved by

Corollary 44

$|K(k,p)|<K_{3/2}(k,p)$ for all $k\ne p$ and all $K\in {\mathcal {K}}\setminus \{K_{3/2}\}$.

Proof

Consider $h:[1,\infty [\rightarrow \mathbb {R}$ with $\breve{h}=K/K_{3/2}$. h is continuous and by (43) $h(1)=1$, $|h|\le 1$ holds. The claim is $|h(t)|<1$ for $t>1$.

Since $g_{3/2}h$ determines K, it satisfies NC. Therefore

$$\begin{aligned} h\big (\epsilon (\rho )\big )^2\le h(t_\rho ) \end{aligned}$$

(1)

where $t_\rho \in [1,1+2\rho ^2]$ satisfies

$$\begin{aligned} \int _1^{1+2\rho ^2}h(t)\,g_{3/2}(t){\text {d}}t= h(t_\rho )\int _1^{1+2\rho ^2}g_{3/2}(t){\text {d}}t \end{aligned}$$

(2)

Indeed, (1) follows from NC for $g_{3/2}\, h$ with (2) according to the mean value theorem of integrals. Recall (39)(d).

Now assume $h(t^*)=1$ for some $t^*>1$. Then for $\rho ^*>0$ satisfying $\epsilon (\rho ^*)=t^*$ one has $h(t_{\rho ^*})=1$ by (1), whence $h(t)=1$ for $1\le t\le 1+2\rho ^{*2}$ due to (2). Note that $t^*<1+2\rho ^{*2}$. Therefore this result implies that $\{t\ge 1: h(t)=1\}$ is connected and unbounded and hence equal to $[0,\infty [$ contradicting $h\ne {{\textbf {1}}}$. Now assume $h(t^*)=-1$ for some $t^*>1$. Then still $h(t_{\rho ^*})=1$ by (3). Apply the foregoing result. $\square$

17 Localization in bounded regions

As known a POL T of the massive scalar boson with CT, i.e., for which time evolution is causal, does not localize the boson in any bounded region $\Delta \subset \mathbb {R}^3$, which means that $T(\Delta )$ has no eigenstate with eigenvalue 1. One has the general result

Theorem 45

[5, (8) Theorem] Let T be a POL with CT. Let the relativistic relation $H\ge |P|$ hold. Suppose that there is a state localized in the region $\Delta$. Then $\Delta$ is essentially dense, i.e., $\overline{\Delta \setminus N}=\mathbb {R}^3$ for every Lebesgue null set N.

However, the property $\parallel \!\!T(\Delta )\!\!\parallel =1$ may be regarded as physically equivalent to the presence of the eigenvalue 1. Indeed, as

$$\begin{aligned} ||T(\Delta )||= \sup \left\{ \langle \phi , T(\Delta )\varphi \rangle :\, ||\phi ||=1\right\} \end{aligned}$$

norm 1 means that the system can be localized within that region $\Delta$ by a suitable preparation, not strictly but as accurately as desired. So for obvious physical reasons one is interested in POL with $||T(B)||=1$ for every, however, small open ball $B\ne \emptyset$. We call them separated. In this case for every $b\in \mathbb {R}^3$ there is a sequence $(\phi _n)$ of states satisfying

$$\begin{aligned} \big \langle \phi _n,T(B )\, \phi _n\big \rangle \rightarrow 1, \quad n\rightarrow \infty \end{aligned}$$

(17.1)

for every open ball B around b. This means that by a suitable preparation the system can be localized around b as good as desired, thus distinguishing b from any other point. According to [4, sec. G], any $(\phi _n)$ satisfying (17.1) is called a sequence of states localized at b.

The main result of this section is the following criterion on the separateness of a POL.

Theorem 46

Let K be the kernel on $\mathbb {R}^3\setminus \{0\}$ of a POL T for a massive scalar boson. Suppose that

• (i) $\lim _{\lambda \rightarrow \infty } K(k, \lambda p)$  exists for every k and almost all p

• (ii) $\lim _{\lambda ,\lambda '\rightarrow \infty } K(\lambda p,\lambda ' p)=1$ for almost all p

• (iii) there is $k_0$ such that, for any $\lambda _n\rightarrow \infty$, $K(k_0,\lambda _n p)$ does not vanish for almost all p.

Then T is separated.

Proof

We use the representation (7.3) of T. Let ${\mathcal {H}}_J$ be the RKHS associated to the kernel K (see, e.g.,  [7]) and put $J_p:=K(\cdot , p)$. The orthogonal projection P on $L^2(\mathbb {R}^3,{\mathcal {H}}_J)$, $(P\varphi )(p):=\langle J_p,\varphi (p)\rangle J_p$ maps onto the subspace $j\big (L^2(\mathbb {R}^3)\big )$. There is the representation D of the dilation group acting on $L^2(\mathbb {R}^3,{\mathcal {H}}_J)$ by $(D_\lambda \varphi )(p)=\lambda ^{3/2}\varphi (\lambda p)$. It satisfies $D_\lambda E(\Delta )D_\lambda ^{-1}=E(\lambda \Delta )$ for $E:= {\mathcal {F}}E^{can}{\mathcal {F}}^{-1}$. Put $P_\lambda :=D_\lambda P D_\lambda ^{-1}$. The result holds by [4, Theorem 7] if the strong limit $Q:=\lim _{\lambda \rightarrow \infty }P_\lambda$ exists with $Q\ne 0$.

Check $(P_\lambda \varphi )(p)=\langle J_{\lambda p},\varphi (p)\rangle J_{\lambda p}$. For the Cauchy criterion compute $\parallel \!\!P_\lambda \varphi -P_{\lambda '}\varphi \!\!\parallel ^2=\int {\text {d}}^3p\, | \langle J_{\lambda p},\varphi (p)\rangle - \langle J_{\lambda ' p},\varphi (p)\rangle |^2 + 2 {\text {Re}} \big [(1-K(\lambda p,\lambda ' p)) \langle J_{\lambda ' p},\varphi (p)\rangle \overline{ \langle J_{\lambda p},\varphi (p)\rangle }\,\big ]$.

We evaluate this formula for $\varphi (p)=1_\Delta (p) J_k$ with $\Delta$ a Borel set of finite measure. The integrand $1_\Delta (p)\,\Big (|K(\lambda p,k)-K(\lambda ' p,k)|^2+2{\text {Re}}\big [\big (1-K(\lambda p,\lambda ' p)\big ) K(\lambda ' p,k) \overline{K(\lambda p,k)}\,\big ]\Big )$ vanishes for almost every p when $\lambda ,\lambda '\rightarrow \infty$ due to (i), (ii) and since $|K|\le 1$. So the integral vanishes by dominated convergence. We conclude $\parallel \!\!P_\lambda \varphi -P_{\lambda '}\varphi \!\!\parallel ^2\rightarrow 0$ for $\lambda ,\lambda '\rightarrow \infty$.

Actually this result follows for all $\varphi \in L^2(\mathbb {R}^3,{\mathcal {H}}_J)$ since the set of all $\varphi$ of the above kind is total in $L^2(\mathbb {R}^3,{\mathcal {H}}_J)$ as $\{J_k:k\in \mathbb {R}^3\setminus \{0\}\}$ is total in ${\mathcal {H}}_J$. So $P_\lambda$ converges strongly to some projection Q.

Finally show $Q\ne 0$. Let $\varphi _0(p):={\text {e}}^{-|p|^2}J_{k_0}$ for $k_0$ in (iii). Then $\parallel \!\!P_\lambda \varphi _0\!\!\parallel ^2=\int {\text {e}}^{-|p|^2} |K(\lambda p, k_0)|^2{\text {d}}^3p$ converges to $\parallel \!\!Q\varphi _0\!\!\parallel ^2$ for $\lambda \rightarrow \infty$. Hence $Q\varphi _0 \ne 0$ due to (iii). $\square$

Corollary 47

Let K be from (46). Let $k_0$ satisfy (46)(iii) and let $b\in \mathbb {R}^3$. Then

$$\begin{aligned} \phi _n(p):=c_n {\text {e}}^{-{\text {i}}bp}{\text {e}}^{-|p|^2/n^2}K(k_0,p) \end{aligned}$$

with $c_n>0$ the normalizing constant constitutes a sequence of states localized at b with respect to T.

Proof

Consider at once $b=0$ [5, (18)]. Then for $\varphi _0={\text {e}}^{-|\cdot |^2}J_{k_0}$in the proof of (46) one has $PD_\lambda ^{-1}\varphi _0(p)=\lambda ^{-3/2}{\text {e}}^{-|p|^2/\lambda ^2}K(p,k_0)J_p$, whence the claim by [4, Theorem 7]. $\square$

The foregoing criterion on the separateness of POL is applied to the POL $T^{tct}$, $T^{TM}$, and POL with causal kernel.

Corollary 48

The POL with CT $T^{TM}$ (Sect. 10.1) and $T^{tct}$ (Sect. 10.2), are separated. Every $k_0\in \mathbb {R}^3\setminus \{0\}$ yields a point localized sequence of states by (47).

Proof

(a) Check easily (46)(i)-(iii) for $\textsc {t}^{TM}$ (10.1). In particular $\lim _{\lambda \rightarrow \infty }\textsc {t}^{TM}(k,\lambda p)= \frac{1}{2}\big (1+\frac{kp}{\epsilon (k)|p|}\big )\ge \frac{1}{2}\big (1-\frac{|k|}{\epsilon (k)}\big )>0$ for all k, p.

(b) Similarly, $\textsc {t}^{tct}$ (10.5) satisfies (46)(i)-(iii). In particular $\lim _{\lambda \rightarrow \infty }\textsc {t}^{tct}(k,\lambda p)=\frac{1}{2}\big (1+m/\epsilon (k)\big )^{1/2}+\frac{1}{2|p|}\big (\epsilon (k)(m+\epsilon (k))\big )^{-1/2}kp>0$ for all k, p. $\square$

As to causal kernels recall (11.1) and the remarks following it. One has

Lemma 49

• (a) There is no causal kernel K, which satisfies the condition (46)(ii).

• (b) Let K be a map (11.1), where $g:[m^2,\infty [\rightarrow \mathbb {C}$ is continuous. Then K satisfies (46)(ii) if and only if $g=g_{1/2}$. In view of (46)(i),(iii) check $\lim _{\lambda \rightarrow \infty }K_{1/2}(k, \lambda p)=m\big (2 \epsilon (k)(\epsilon (k)-\frac{1}{|p|}kp)\big )^{-1/2}$.

Proof

Let $m=1$. (b) Check (46)(i)-(iii) for $K_{1/2}$. Now, conversely let $p\ne 0$, $\alpha \ge 1$. Check $\lim _{\lambda \rightarrow \infty } (\epsilon (\lambda p) \epsilon (\alpha \lambda p)-\alpha \lambda ^2 |p|^2 )=\frac{1+\alpha ^2}{2\alpha }$ and $\lim _{\lambda \rightarrow \infty } \frac{\epsilon (\lambda p)+\epsilon (\alpha \lambda p)}{2\sqrt{\epsilon (\lambda p)\epsilon (\alpha \lambda p)}}=\frac{1+\alpha }{2\sqrt{\alpha }}$. Fix $t\in [1,\infty ]$. Then $t=\frac{1+\alpha ^2}{2\alpha }$ for $\alpha =t+\sqrt{t^2-1}$, and for this $\alpha$ one has $\frac{1+\alpha }{2\sqrt{\alpha }}= \sqrt{(1+t)/2}$. Hence continuity of g and condition (46)(ii) imply $\sqrt{(1+t)/2} \,g(t)=1$, as asserted.

(a) follows from (b), since $K_{1/2}$ is not a positive definite kernel. $\square$

Hence a POL T of the massive scalar boson with causal kernel (11.1) is a candidate for a non-separated POL, i.e., $\parallel \!\!T(B)\!\!\parallel <1$ might hold for some small ball $B\ne \emptyset$.

18 Discussion

A POL can be regarded as a physically accessible data set, namely the set of the probabilities of localization of the particle in every state in every region. Therefore among the infinitely many POL for a massive scalar boson (2) only a particular one describes truly the position of the boson. The question is how to identify this POL.

Many, as, e.g., NWL, are ruled out by the requirement of causality CC (Sect. 11). The remaining causal POL are still infinitely many. We studied thoroughly the causal POL related to a conserved density current analyzing the set ${\mathcal {K}}$ of causal kernels (11.1).

In search of the best POL we recall the observation that $KG\in {\mathcal {K}}$ if $K\in {\mathcal {K}}$ and $G\in {\mathcal {G}}$ (28). Probably multiplying K by G deteriorates the localization features of the corresponding POL. More generally, if $K,K'\in {\mathcal {K}}$ with $K'\le K$, then K is better than $K'$? Take as evidence the fact $K'\le K\le {{\textbf {1}}}$ with ${{\textbf {1}}}$ the kernel of NWL distinguished by its abundance of boundedly localized states. If it were true, the POL with kernel $K_{3/2}$, being the greatest element (43), would be the only valid POL.

For the localization features of the POL $T_{3/2}$ with kernel $K_{3/2}$ it is decisive to know whether $||T_{3/2}(B)||=1$ for all balls $B\ne \emptyset$. Assume this. Then $T_{3/2}$ is separated (Sect. 17) and for every point $b\in \mathbb {R}^3$ there is a sequence of states localized at b (17.1), which is very satisfactory from a physical point of view.

However, also the case $||T_{3/2}(B_*)||=\delta <1$ for some ball $B_*\ne \emptyset$ were rather interesting. Assume this. By translation covariance $B_*$ may be centered at 0. The probability for the boson of being localized in $B_*$ is at most $\delta$ in every state. Note $||T_{3/2}(mB_*)||\rightarrow 1$ for $m\rightarrow \infty$ since $mB_*\uparrow _m\mathbb {R}^3$.

Now let $T^m$ denote the POL for the boson with mass $m>0$ with kernel (11.1). T means $T^1$. It is easy to check

$$\begin{aligned} T(m\Delta )=D_mT^m(\Delta )D_m^{-1} \end{aligned}$$

for all regions $\Delta$ with $(D_m\phi )(p):=m^{3/2} \phi (mp)$ unitary on $L^2(\mathbb {R}^3)$. Hence $|| T(m\Delta )||=||T^m(\Delta )||$, whence in particular $||T^m_{3/2}(B_*)||\rightarrow 1$ for $m\rightarrow \infty$.

The physically reasonable result would be that a more massive scalar boson is better localizable in bounded regions than a less massive one.

Appendices

Criterion for CT

This section is concerned with a criterion for CT. It is based on the existence of a non-spacelike conserved current for the probability density of localization. Originally this is an approach due to Petzold et al. [8] when treating their covariant kernels of the density and proving CT for the spatial probability [9]. Moretti [17] improves the technique considerably showing rigorously CT for the POL proposed by Terno [23]. From Moretti’s proof we distill the general criterion for CT (52).

The following lemma, which is an extension of a result in [17], essentially says that CT and CC, respectively, already holds, if it holds for regions $\Delta$ being the union of finitely many disjoint closed balls.

Lemma 50

Let $\varepsilon =\{0\}\times \mathbb {R}^3\equiv \mathbb {R}^3$ be the standard spacelike hyperplane and let $\sigma \subset \mathbb {R}^4$ be any other spacelike hyperplane of Minkowski space. Let T and $T'$ be POM on $\varepsilon$ and $\sigma$, respectively, vanishing at Lebesgue null sets. Let V be a fixed open subset of $\mathbb {R}^3$. Suppose that

$$\begin{aligned} T(\Delta )\le T'(\Delta _\sigma ) \end{aligned}$$

(1)

holds for every $\Delta \subset V$ being a union of finitely many disjoint closed balls. Then (1) holds for all measurable $\Delta \subset V$.

Proof

Provide a total set of unit vectors $\{\phi _j:j\in \mathbb {N}\}$. Then $\mu (\Delta ):=\sum _j 2^{-j}\langle \phi _j,T(\Delta )\phi _j\rangle$ defines a probability measure controlling T. The latter means that

$$\begin{aligned} \mu (\Delta ^{(n)})\rightarrow 0 \Leftrightarrow T(\Delta ^{(n)})\rightarrow 0 \text { strongly} \end{aligned}$$

for every sequence $(\Delta ^{(n)})$ in ${\mathcal {L}}^d$. Plainly, $\Leftarrow$ holds by dominated convergence. As to $\Rightarrow$, note $||T(\Delta )\phi ||^2\le ||\sqrt{T(\Delta )}\sqrt{T(\Delta )}\phi ||^2\le ||\sqrt{T(\Delta )}\phi ||^2=\langle \phi ,T(\Delta )\phi \rangle$. One recalls that $\mu$ is a tight Radon measure (see, e.g., [3, Chapter 2 (67)]). $\mu '$ with respect to $T'$ is defined quite analogously.

Henceforth all sets are subsets of V.

• (a) Let (1) hold for $\Delta ^{(n)}$ with $\Delta ^{(n)} \uparrow \Delta$. Then (1) holds for $\Delta$. Indeed, note first $\Delta ^{(n)}_\sigma \uparrow \Delta _\sigma$. Since $\mu (\Delta \setminus \Delta ^{(n)})\rightarrow 0$, one has $T(\Delta ^{(n)})=T(\Delta )-T(\Delta {\setminus }\Delta ^{(n)})\rightarrow T(\Delta )$. Similarly $T'(\Delta _\sigma ^{(n)})\rightarrow T'(\Delta _\sigma )$ follows.

• (b) Let (1) hold for $\Delta ^{(n)}$ with $\Delta ^{(n)} \downarrow \Delta$ and suppose that $\Delta _\sigma ^{(n)} \downarrow \Delta _\sigma$. Then, arguing as in (a) analogously, (1) holds for $\Delta$.

• (c) Now consider an open bounded set U. Then according to Vitali’s covering theorem [16, 2.8 Theorem] there are countably many disjoint closed balls $B^{(i)}\subset U$ with $\mu (U{\setminus } C)=0$ for $C:=\bigcup _iB^{(i)}$. By assumption (1) holds for $C^{(n)}:=\bigcup _{i=1}^nB^{(i)}$, whence (1) holds for C by (a). Moreover, since $\mu (U\setminus C^{(n)})\rightarrow 0$, one has $T(U\setminus C^{(n)})\downarrow 0$. Therefore, $T(U)=T(C)+T(U{\setminus } C) \le T'(C_\sigma )+T(U{\setminus } C)\le T'(U_\sigma )+T(U{\setminus } C^{(n)})\rightarrow T'(U_\sigma )$. Hence (1) holds for U.

• (d) Next let K be compact. Choose bounded open sets $U^{(n)}$ satisfying $\overline{U^{(n+1})}\subset U^{(n)}$, $U^{(n)}\downarrow K$. Then $U^{(n)}_\sigma \downarrow K_\sigma$. Hence (1) holds for K by (b).

• (e) Finally let $\Delta$ be any measurable set. Since $\mu$ is tight, there are compact $K^{(i)}\subset \Delta$ satisfying $\mu (\Delta {\setminus } C)=0$ for $C:=\bigcup _i K^{(i)}$. By (d), (1) holds for $C^{(n)}:=\bigcup _{i=1}^nK^{(i)}$. Now proceed as in (c). Hence (1) holds for $\Delta$.

$\square$

The applications of (50) in (52) and (56) regard the case $T'=T$ as they concern a POL T of the massive scalar boson, which due to the Poincaré covariance (3.1) is defined on all spacelike hyperplanes. Moreover the particular case in (50) of parallel hyperplanes $\varepsilon$ and $\sigma =\{t\}\times \mathbb {R}^3$ for $t\in \mathbb {R}$ and for $V=\mathbb {R}^3$ is used for the CT criterion below.

Definition 51

Let $\textsc {t}$ be the kernel of POL T of a massive scalar boson. Suppose that $\textsc {t}$ is continuous. $\textsc {t}$ is said to be (i) conserved, (ii) timelike definite, if there are continuous Hermitian kernels $\textsc {j}_i:\mathbb {R}^3\setminus \{0\}\times \mathbb {R}^3\setminus \{0\}\rightarrow \mathbb {C}$, $i=1,2,3$ satisfying, respectively

• (i) $\big (\epsilon (k)-\epsilon (p)\big )\textsc {t}(k,p)=\sum _{i=1}^3(k_i-p_i)\textsc {j}_i(k,p)$   for all $k,p\ne 0$

• (ii) $\big (\sum _{a,b=1}^n{\overline{c}}_ac_b\textsc {t}(p_a,p_b)\big )^2\ge \sum _{i=1}^3\big (\sum _{a,b=1}^n{\overline{c}}_ac_b\textsc {j}_i(p_a,p_b)\big )^2$ for all $(c_1,\dots ,c_n)\in \mathbb {C}^n$, $(p_1,\dots ,p_n)\in (\mathbb {R}^3{\setminus }\{0\})^n$, $n\in \mathbb {N}$

From (51)(ii) it follows that $|\textsc {j}_i(p,p)|\le 1$, $\textsc {t}-\textsc {j}_i$ is positive definite, and in particular boundedness $|\textsc {j}_i(k,p)|\le 3$ for $k,p\ne 0$, $i=1,2,3$.

Theorem 52

Let T be a POL of a massive scalar boson. Suppose that its kernel $\textsc {t}$ is continuous and conserved timelike definite. Then time evolution is causal.

Proof

We treat the particle case $\eta =+$ (the antiparticle case is analogous) and use the shell rep in Sect. 8. Let ${\text {d}}{o}({\mathfrak {p}})$ denote the Lorentz invariant measure on the mass shell. Put $\textsc {j}_0:=\textsc {t}$. Define for $i=0,1,2,3$

$$\begin{aligned} J_i(\Phi ,{\mathfrak {x}}):=(2\pi )^{-3}\int \int \, \sqrt{\epsilon (k)\epsilon (p)} \textsc {j}_i(k,p)\, {\text {e}}^{{\text {i}}({\mathfrak {k}}-{\mathfrak {p}})\cdot {\mathfrak {x}}}\overline{\Phi ({\mathfrak {k}})}\Phi ({\mathfrak {p}})\,{\text {d}}o({\mathfrak {k}}) {\text {d}}o(\mathfrak {{\mathfrak {p}}})\nonumber \\ \end{aligned}$$

(A.1)

for ${\mathfrak {x}}\in \mathbb {R}^4$, and $\Phi$ square integrable and $\sqrt{\epsilon }\, \Phi$ integrable. Recall that $\textsc {j}_i$ is bounded. Put $J:=(J_1,J_2,J_3)$ and ${\mathfrak {J}}:=(J_0,J)$.

• (a) Let $\Delta \subset \mathbb {R}^3\equiv \{0\}\times \mathbb {R}^3$ be measurable, $t\in \mathbb {R}$. Then $\langle W(t)^{-1}\Phi ,T(\Delta )W(t)^{-1}\Phi \rangle = \langle \Phi , W(t)T(\Delta )W(t)^{-1}\Phi \rangle =\langle \Phi ,T(\{t\}\times \Delta )\Phi \rangle$. The latter equals the surface integral $\int _{\{t\}\times \Delta }J_0(\Phi ,{\mathfrak {x}}) {\text {d}}S({\mathfrak {x}})$.

Indeed, by (8.2) $\langle W(t)^{-1}\Phi ,T(\Delta )W(t)^{-1}\Phi \rangle =\int _\Delta J_0(W(t)^{-1}\Phi ,(0,x)){\text {d}}^3x$. Since $\big (W(t)^{-1}\Phi \big )({\mathfrak {p}})={\text {e}}^{-{\text {i}}(t,0)\cdot {\mathfrak {p}}}\Phi ({\mathfrak {p}})$, the former equals $\int _\Delta J_0(\Phi ,(t,x)){\text {d}}^3x$, whence the claim.

• (b) ${\text {div}}\, {\mathfrak {J}}(\Phi ,\cdot )=0$

This follows immediately from (51)(i).

• (c) $J_0(\Phi ,{\mathfrak {x}})\ge |J(\Phi ,{\mathfrak {x}})|$

Put $\phi (p):=\Phi \big (\epsilon (p),p\big )$. Clearly it suffices to prove the claim for $\Phi$, for which $\phi$ is continuous with compact support in $\mathbb {R}^3\setminus \{0\}$. This follows from (51)(ii) just approximating the integrals in (A.1) by appropriate Riemann sums.

(d) Now following [17] the main step toward the result is an application of the divergence theorem, i.e., Gauss’s or Ostrogradsky’s theorem. Assume $t>0$. The case $t<0$ is analogous. Let $B\subset \mathbb {R}^3$ be a closed ball of radius R and with center at the origin 0. Let M be the compact subset of $\mathbb {R}^4$, the boundary of which is the union of the base and top $\{0\}\times B_t$, $\{t\}\times B$, and the lateral surface $L:=\{(s,x)\in \mathbb {R}^4:|R+t-s|=|x|,0\le s\le t\}$, which is the piece of the past light cone with apex $(R+t,0)$ between the base and the top. Note that ${\mathfrak {n}}({\mathfrak {x}}):=\frac{1}{\sqrt{2}}\big (1,x/|x|\big )$ is outward-pointing orthonormal at ${\mathfrak {x}}=(s,x)\in L$.

By (b) the volume integral $\int _M {\text {div}}{\mathfrak {J}}(\Phi ,{\mathfrak {x}}) \,{\text {d}}V({\mathfrak {x}})$ vanishes, whence using (a) one has

$$\begin{aligned} \langle \Phi , W(t)T(B)W(t)^{-1}\Phi \rangle - \langle \Phi , T(B_t)\Phi \rangle =- \int _L {\mathfrak {J}}(\Phi ,{\mathfrak {x}})\,{\mathfrak {n}}({\mathfrak {x}}) {\text {d}}S({\mathfrak {x}})\le 0 \end{aligned}$$

(1)

since $\sqrt{2}\, {\mathfrak {J}}(\Phi ,{\mathfrak {x}})\,{\mathfrak {n}}({\mathfrak {x}})=J_0(\Phi , {\mathfrak {x}})+J(\Phi ,{\mathfrak {x}})\,\frac{x}{|x|}\ge 0$ by (c). So the causality condition (CT) holds for B and hence due to translation covariance for any ball.

We conclude the proof still following [17]. In view of (50) it suffices to verify (CT) for $\Delta$ being the union of finitely many disjoint closed balls $B_i$. The boundary of the compact set $M:=\bigcup M_i$ is the union of the base and top $\{0\}\times \Delta _t$, $\{t\}\times \Delta$, and a piecewise smooth lateral surface L. Note $L\subset \bigcup L_i$. So, as L is composed by finitely many pieces of light cones, (1) holds true also for $\Delta$ in place of B. $\square$

POL with conserved covariant kernel satisfies CC

Extending Moretti’s method we show the result on causality (56) based on the existence of a conserved covariant current for the probability density of localization. The corollary (55) on conserved covariant POL kernels for a massive scalar boson essentially is due to Petzold et al. [8].

Put $m=1$ without restriction, and let ${\mathfrak {p}}=\big (\epsilon (p),p\big )$ with $\epsilon (p)=\sqrt{1+|p|^2}$ for $p\in \mathbb {R}^3$.

Definition 53

${\mathfrak {v}}: \mathbb {R}^3{\setminus } \{0\}\times R^3{\setminus } \{0\} \rightarrow \mathbb {C}^4$ is called (i) conserved, (ii) (Lorentz) covariant, if

• (i) $({\mathfrak {k}}-{\mathfrak {p}}) \cdot {\mathfrak {v}}(k,p)=0$

• (ii) $L\,{\mathfrak {v}}(k,p)={\mathfrak {v}}(L\cdot k,L\cdot p)$ for all $p,k\ne 0$, $L\in O(1,3)_0$

Here $L\cdot p$ denotes the spatial vector part of $L{\mathfrak {p}}$.

Lemma 54

Let ${\mathfrak {v}}: \mathbb {R}^3{\setminus }\{0\}\times \mathbb {R}^3{\setminus }\{0\} \rightarrow \mathbb {C}^4$ be covariant. Then

• (a) $v_0$ determines uniquely ${\mathfrak {v}}$.

• (b) there are unique $\alpha ,\beta : [1,\infty ]\rightarrow \mathbb {C}$ such that ${\mathfrak {v}}(k,p)=\alpha ({\mathfrak {k}}\cdot {\mathfrak {p}}) \,{\mathfrak {k}} + \beta ({\mathfrak {k}}\cdot {\mathfrak {p}})\, {\mathfrak {p}}$.

• (c) ${\mathfrak {v}}$ is conserved, iff   ${\mathfrak {v}}(k,p)=\gamma ({\mathfrak {k}}\cdot {\mathfrak {p}}) \,({\mathfrak {k}} + {\mathfrak {p}})$ for some unique $\gamma :[1,\infty [\rightarrow \mathbb {C}$.

• (d) $v_0$ is Hermitian, iff  ${\mathfrak {v}}(k,p)= g({\mathfrak {k}}\cdot {\mathfrak {p}})({\mathfrak {k}} + {\mathfrak {p}}) +{\text {i}}f({\mathfrak {k}}\cdot {\mathfrak {p}})({\mathfrak {k}} - {\mathfrak {p}})$ with unique $g,f:[1,\infty [\rightarrow \mathbb {R}$.

• (e) ${\mathfrak {v}}$ is conserved if $v_0$ is real symmetric.

• (f) if $v_0$ is positive definite, then $v_0$ is even timelike definite (see (51)(ii)). Cf. [9, sect. 3].

Proof

(a) Verify $v_i(k,p)=\sqrt{2}\,v_0(p,k) - v_0\big (L_i^{-1}\cdot k, L_i^{-1} \cdot p \big )$ with $L_i$ the boost along the ${\text {i}}$th direction with velocity $1/\sqrt{2}$ evaluating the equation $v_0(k,p)=L(1,0,0,0)\,\cdot {\mathfrak {v}}(L\cdot k,L\cdot p)$ for $L=L_i$.

(b) It is no restriction to assume that ${\mathfrak {v}}$ is real, since covariance does not mix the real and the imaginary parts. Consider first the case $k=p$. By covariance the fixed point subgroup of ${\mathfrak {p}}$ leaves ${\mathfrak {v}}(p,p)$ invariant. Therefore ${\mathfrak {v}}(p,p)=\gamma (p){\mathfrak {p}}$ for some scalar $\gamma (p)\in \mathbb {C}$. Then covariance implies that $\gamma$ is constant. Now let $k\ne p$. By a Lorentz transformation map ${\mathfrak {k}}$ to (1, 0, 0, 0) and ${\mathfrak {p}}$ to $(a',0,0,b')$ with $a',b'\ge 0$ and subsequently by the boost in the third direction, (1, 0, 0, 0) to $(\sqrt{2},0,0,1)$ and $(a',0,0,b')$ to (a, 0, 0, b). So there is $L\in O(1,3)_0$ with $L{\mathfrak {k}}=(\sqrt{2},0,0,1)$, $L{\mathfrak {p}}=(a,0,0,b)$. One has $b=b({\mathfrak {k}}\cdot {\mathfrak {p}})$ for $b(t):=t+\sqrt{2}\sqrt{t^2-1}$ and similarly for $a=a({\mathfrak {k}}\cdot {\mathfrak {p}})$. Hence $L{\mathfrak {v}}(k,p)= {\mathfrak {s}}({\mathfrak {k}}\cdot {\mathfrak {p}})$ with ${\mathfrak {s}}(t):={\mathfrak {v}} \big ((0,0,1),(0,0,b(t))\big )$. Note $b\ne 1$ since $k\ne p$ and, by covariance, $R\,{\mathfrak {s}}(t)={\mathfrak {s}}(t)$ for all rotations R around the third axis. Therefore ${\mathfrak {s}}(t)=\alpha (t)(\sqrt{2},0,0,1)+\beta (t)(a(t),0,0,b(t))$ with unique $\alpha (t),\beta (t)$, whence the result.

(c), (d), and (e) follow immediately from (b).

(f) Since $v_0$ is positive definite, the zero component of the four vector

$$\begin{aligned} {\mathfrak {z}}:=\sum _{a,b}{\overline{c}}_ac_b\,{\mathfrak {v}} (p_a, p_b) \end{aligned}$$

is nonnegative for every choice of $(c_1,\dots ,c_n)$ and $(p_1,\dots ,p_n)$. Consider any Lorentz transformation $L\in O(1,3)_0$. Put $\tilde{{\mathfrak {z}}}:=L{\mathfrak {z}}$, $\tilde{{\mathfrak {p}}}_a:=L{\mathfrak {p}}_a$. Then $\tilde{{\mathfrak {z}}}=\sum _{a,b}{\overline{c}}_ac_b\, {\mathfrak {v}}({\tilde{p}}_a,{\tilde{p}}_b)$ by Lorentz covariance, whence the zero component of $L{\mathfrak {z}}$ is nonnegative. $\square$

Corollary 55

Let $\textsc {t}$ be the kernel of a POL of a massive scalar boson. Let $\textsc {t}$ be conserved covariant, which means that there is a conserved covariant ${\mathfrak {v}}$ with $v_0(k,p)=\textsc {t}^{shell} ({\mathfrak {k}},{\mathfrak {p}}) =\sqrt{\epsilon (k)\epsilon (p)}\textsc {t}(k,p)$. Then there is a continuous $g:[1,\infty ] \rightarrow \mathbb {R}$ with $g(1)=1$ such that

$$\begin{aligned} \textsc {t}(k,p) =\frac{\epsilon (k)+\epsilon (p)}{2\sqrt{\epsilon (k)\epsilon (p)}}\,\,g({\mathfrak {k}}\cdot {\mathfrak {p}}) \end{aligned}$$

(B.1)

The right hand side is continuous on $\mathbb {R}^3\times \mathbb {R}^3$ and by continuity it stays positive definite on $\mathbb {R}^3$. The extension is still denoted by $\textsc {t}$.

Conversely let $g:[1,\infty ] \rightarrow \mathbb {R}$ be measurable with $g(1)=1$ such that the right hand side of (B.1) is positive definite. Then the latter is a conserved covariant POL kernel and g is continuous.

Proof

g is continuous by (12). For the second part of the assertion recall (11). $\square$

Theorem 56

Let T be a POL of a massive scalar boson. Suppose that its kernel $\textsc {t}$ is conserved covariant (B.1). Then T is causal.

Proof

By (55), (54)(f) $\textsc {t}$ is continuous timelike definite. So (52) applies with $v_i(k,p)= \sqrt{\epsilon (k)\epsilon (p)} \textsc {j}_i(k,p)$. Recall the beginning of the proof of (52), in particular (A.1).

• (a) Let $\Gamma \in {\mathfrak {S}}$. Then

  $$\begin{aligned} \langle \Phi ,T(\Gamma )\Phi \rangle =\int _\Gamma {\mathfrak {J}}(\Phi , {\mathfrak {x}})\,{\mathfrak {n}}_\Gamma \,{\text {d}}S({\mathfrak {x}}) \end{aligned}$$

  (B.2)

  Regarding the surface integral there is the (Euclidean) scalar product of ${\mathfrak {J}}$ with the (Euclidean) future-pointed orthonormal ${\mathfrak {n}}_\Gamma$ on $\Gamma$.

Indeed, there is $g=({\mathfrak {a}},A)\in \tilde{{\mathcal {P}}}$ and $\Delta \subset \mathbb {R}^3\equiv \{0\}\times \mathbb {R}^3$ with $\Gamma =g\cdot \Delta$. Then $\langle \Phi ,T(\Gamma )\Phi \rangle =\langle W(g)^{-1}\Phi ,T(\Delta ) W(g)^{-1}\Phi \rangle =\int _\Delta J_0(W(g)^{-1}\Phi ,(0,x)){\text {d}}^3x$. Inserting $\big (W(g)^{-1}\Phi \big )({\mathfrak {p}})={\text {e}}^{-{\text {i}}L^{-1}{\mathfrak {a}}\,\cdot {\mathfrak {p}}}\,\Phi (L{\mathfrak {p}})$, $L:=\Lambda (A)$, in (A.1) for $i=0$ one obtains $J_0(W(g)^{-1}\Phi ,(0,x))=(2\pi )^{-3}\int \int \textsc {j}_0^{shell}(L^{-1}{\mathfrak {k}},L^{-1}{\mathfrak {p}}) {\text {e}}^{{\text {i}}({\mathfrak {k}}-{\mathfrak {p}})\cdot (g\cdot (0,x)) } \overline{\Phi (\mathfrak {k)}} \Phi (\mathfrak {p)} {\text {d}}o({\mathfrak {k}}) {\text {d}}o(\mathfrak {{\mathfrak {p}}})$ by Lorentz invariant integration. Now, since $\textsc {t}$ is covariant, the latter becomes $\big (L^{-1}{\mathfrak {J}}(\Phi ,g\cdot (0,x))\big )_0$, and further using the normal ${\mathfrak {e}}=(1,0,0,0)$, $=\big (L^{-1}{\mathfrak {J}}(\Phi ,g\cdot (0,x))\big )\cdot {\mathfrak {e}}={\mathfrak {J}}(\Phi ,g\cdot (0,x))\cdot {\mathfrak {e}}_L$, ${\mathfrak {e}}_L:= L{\mathfrak {e}}$. Hence one has $\langle \Phi ,T(\Gamma )\Phi \rangle =\int _\Delta {\mathfrak {J}}(\Phi ,g\cdot (0,x))\cdot {\mathfrak {e}}_L \,{\text {d}}^3x$. The Minkowski product by ${\mathfrak {e}}_L$ becomes the scalar product by ${\mathfrak {e}}^-_L$ with the negative spatial part of ${\mathfrak {e}}_L$. Note that ${\mathfrak {e}}^-_L$ is future-oriented orthogonal to $\Gamma$. Therefore (B.2) holds.

(b) Let $\Delta \in {\mathfrak {S}}$ and let $\sigma$ be a spacelike hyperplane. We are going to show causality

$$\begin{aligned} T(\Delta )\le T(\Delta _\sigma ) \end{aligned}$$

(1)

Due to the Poincaré covariance of T it suffices to treat the case $\Delta \subset \varepsilon = \{0\}\times \mathbb {R}^3 \equiv \mathbb {R}^3$ and $\sigma =\{{\mathfrak {x}}\in \mathbb {R}^4: x_0=\alpha x_3\}$ for $0<\alpha <1$. This excludes only the case of parallel hyperplanes $\varepsilon$ and $\sigma$ already treated in (52).

Put $H:=\{x\in \mathbb {R}^3:x_3=0\}$, $H^\pm :=\{x\in \mathbb {R}^3:x_3\gtrless 0\}$, $\Delta ^{\pm }:=\{x\in \Delta :x_3\gtrless 0\}$. Note that H is a null set and that $\Delta ^\pm _\sigma \subset H^\pm$. So it suffices to have (1) for $\Delta ^+$ and $\Delta ^-$, since, assuming this, $T(\Delta )=T(\Delta ^-\cup \Delta ^+)=T(\Delta ^+)+T(\Delta ^-)\le T(\Delta ^+_\sigma )+T(\Delta ^-_\sigma )=T(\Delta ^+_\sigma \cup \Delta ^-_\sigma )\le T(\Delta _\sigma )$. Therefore by the reduction achieved so far one may assume in (1) that $\Delta \subset H^+$ or $\Delta \subset H^-$. We do the case $\Delta \subset H^+$ applying (50) for $V:=H^+$. The case $\Delta \subset H^-$ is analogous.

First let $\Delta$ be a closed ball B with center $(0,0,\beta )$ and radius $R>0$ so that $0<R<\beta$. Let M be the compact subset of $\mathbb {R}^4$, the boundary of which is the union of the base and top B, $B_\sigma$ and the lateral surface L, which is the piece of the future light cone with apex $(-R,0,0,\beta )$ between the base and the top. Exploit the volume integral $\int _M {\text {div}}{\mathfrak {J}}(\Phi ,{\mathfrak {x}}) \,{\text {d}}V({\mathfrak {x}})$ as in part (d) of the proof of (52) applying the formula (B.2). The relation (1) for $\Delta =B$ follows. Now achieve the result still following the proof of (52). $\square$

On functions of positive type

Here some results on functions of positive type are provided.

Lemma 57

(a) Let g be from (12.1), i.e., $g:[1,\infty [\rightarrow \mathbb {R}$ measurable with $g(1)=1$. The three properties $(\alpha )$ ${\tilde{g}}$ is positive definite,⁷$(\beta )$ $(x,y)\rightarrow g\big (\cosh (x-y)\big )$ is positive definite, i.e., $g\circ \cosh$ is a function of positive type, and $(\gamma )$ $g\circ \cosh$ equals a.e. the Fourier transform of an even finite measure on $\mathbb {R}$, are equivalent.⁸

(b) Let $k\in L^1(\mathbb {R})$ with its Fourier transform ${\hat{k}}\in L^1(\mathbb {R})$, ${\hat{k}}\ge 0$; then k is even and ${\tilde{g}}$ for $g:=k\circ \cosh ^{-1}$ is positive definite.⁹

(c) Consider $k\in L^1(\mathbb {R})\cap L^2(\mathbb {R})$. Then k is equal a.e. to a continuous function of positive type if and only if its Fourier transform is real nonnegative.

(d) Let $k\in L^1(\mathbb {R})$ be a continuous function of positive type. Then its Fourier transform ${\hat{k}}$ is integrable and nonnegative.

Proof

(a) $(\alpha )$ $\Leftrightarrow$ $(\beta )$ is obvious. $(\beta )$ $\Leftrightarrow$ $(\gamma )$ holds by [19, Proposition after Theorem IX.10 and Theorem IX.9 (Bochner’s theorem)].

(b) holds also by Bochner’s theorem.

(c) Suppose $k \in L^1\cap L^2$. For u, v in the Schwartz space ${\mathcal {S}}$, consider $\phi :={\mathcal {F}}u$, $\psi :={\mathcal {F}}v$ in ${\mathcal {S}}$. By [12, 21.32], $k*\psi \in L^2$ and, by [12, 21.54(a)], $k*\psi ={\mathcal {F}}\big (({\mathcal {F}}^{-1}k)({\mathcal {F}}^{-1}\psi )\big )$. Hence $\langle \phi ,k*\psi \rangle =\langle {\mathcal {F}}^{-1}\phi ,{\mathcal {F}}^{-1}(k*\psi )\rangle =\langle u, ({\mathcal {F}}^{-1}k)v\rangle$. For $v=u$ this implies $(2\pi )^{-1/2}\int \int k(x-y)\overline{\phi (y)}\phi (x){\text {d}}x{\text {d}}y=\int |u(x)|^2\,({\mathcal {F}}^{-1}k)(x){\text {d}}x$. The right hand side is nonnegative for all u if and only if ${\mathcal {F}}^{-1}k\ge 0$ a.e. Therefore, looking at the left hand side, k is of weak positive type if and only if ${\mathcal {F}}^{-1}k\ge 0$ a.e. or equivalently ${\mathcal {F}}k\ge 0$ a.e. Recall [19, Proposition after Theorem IX.10].

(d) Since k is bounded by $k(0)\ge 0$, k is also square integrable so that (c) applies. So ${\hat{k}}\ge 0$. Note $k={\mathcal {F}}^{-1}{\hat{k}}$. We introduce the measure $\nu :={\hat{k}}\,m$, where m denotes the Lebesgue measure on $\mathbb {R}$. Since $\nu ([-D,D])\le (2D)^{1/2}\parallel \!\!{\hat{k}}\!\!\parallel _2$, the measure $\nu$ is polynomially bounded [19, V: Example 4] so that $S(\varphi ):=\int \varphi {\text {d}}m$ defines a tempered distribution. Its Fourier transform is given by ${\hat{S}}(\varphi )=S({\hat{\varphi }})=\int {\hat{\varphi }}{\text {d}}\nu =\int {\hat{\varphi }}\,{\hat{k}}{\text {d}}m=\int \varphi \,({\mathcal {F}}^{-1}{\hat{k}})\,{\text {d}}m=\int \varphi \,k\,{\text {d}}m$.

On the other hand by [19, Theorem IX.9 (Bochner’s theorem)], there is a finite measure $\mu$ on $\mathbb {R}$ such that its Fourier transform ${\hat{\mu }}$ equals k. This means that the Fourier transform of the tempered distribution given by $T(\varphi ):=\int \varphi \,{\text {d}}\mu$ reads ${\hat{T}}(\varphi )=\int {\hat{\mu }}\,\varphi \,{\text {d}}m=\int \varphi \,k\,{\text {d}}m$ [19, after Theorem IX.8].

Thus ${\hat{S}}={\hat{T}}$, whence $S=T$ [19, Theorem IX.2]. This obviously means $\nu =\mu$ implying ${\hat{k}}\in L^1(\mathbb {R})$. $\square$

Proofs concerning $T^{tct}$

Proof of (10.5), (10.6)

First recall (16). The second part of the assertion follows from [4, sec. E, Cor. 5,6,7] (for a concise summary see [5, sec. 28.1]) and [4, sec. F, Cor. 9]. Accordingly, $(W^{m,0,+},T^{tct})$ is obtained as the trace of the 2.nd system with CT $(W_2,{\mathcal {F}}E^{can}{\mathcal {F}}^{-1})$ on the carrier space of the subrep $W^{m,0,+}$ of the rep $W_2$ of $\tilde{{\mathcal {P}}}$ on $L^2(\mathbb {R}^3,\mathbb {C}^4\otimes \mathbb {C}^2)$. For $(t,b,A) \in \tilde{{\mathcal {P}}}$ the latter is given by

$$\begin{aligned} \big (W_2(t,b,A)\varphi \big )(p):={\text {e}}^{-{\text {i}}bp} \sum _{\eta =\pm 1}\left( {\text {e}}^{{\text {i}}t\eta \epsilon (p)} \pi ^\eta (p)s(A)^{*-1} \otimes R({\mathfrak {p}}^\eta ,A) \right) \varphi (q^\eta )\nonumber \\ \end{aligned}$$

(D.1)

for $p\ne 0$, where $s(A):={\text {diag}}(A,A^{* -1})$, $\pi ^\eta (p):=\frac{1}{2}\big (I_2+\frac{\eta }{\epsilon (p)}h(p)\big )$, and $h(p):=\sum _{k=1}^3\alpha _k p_k + \beta m$ with $\beta ,\alpha _k$ the Dirac matrices in the Weyl representation (see also [5, sec. 13 and sec. 28.1]).

We are going to compute $T^{tct}$ explicitly. Change to the energy rep $Y_2^{-1}W_2Y_2$ by means of the unitary transformation $Y_2:=Y\otimes I_2$

$$\begin{aligned} (Y\varphi )(p)=Y(p)\varphi (p), \;Y(p):=\left( \frac{m}{2\epsilon (p)}\right) ^{1/2} \left( \begin{array}{cc} Q({\mathfrak {p}}^+)&{} Q({\mathfrak {p}}^+)^{-1}\\ Q({\mathfrak {p}}^+)^{-1} &{} -Q({\mathfrak {p}}^+)\end{array}\right) \end{aligned}$$

One obtains $\oplus _\eta W^{\eta }_2$ with $\big (W^{\eta }_2(t,b,A)\varphi \big )(p):=\sqrt{\epsilon (q^\eta )/\epsilon (p)}\,{\text {e}}^{{\text {i}}(\eta t \epsilon (p)-bp)} R({\mathfrak {p}}^\eta ,A)\otimes R({\mathfrak {p}}^\eta ,A) \;\varphi (q^\eta )$, where $\varphi \in L^2(\mathbb {R}^3,\mathbb {C}^2\otimes \mathbb {C}^2)$, cf. [5, sec. 30.4]. By $D^{(1/2)}\otimes D^{(1/2)}\simeq D^{(0)}\oplus D^{(1)}$ one has $W^+_2\simeq W^{m,0,+}\oplus W^{m,1,+}$. So the carrier space of the subrep of $W_2$ equivalent to $W^{m,0,+}$ follows to be $Y_2 L^2(\mathbb {R}^3,\mathbb {C}{\text {d}})$ with ${\text {d}}:=\frac{1}{\sqrt{2}}\big ((1,0,0,0)^T\otimes (0,1)^T-(0,1,0,0)^T\otimes (1,0)^T\big )$. Let $j_{{\text {d}}}:L^2(\mathbb {R}^3)\rightarrow L^2(\mathbb {R}^3,\mathbb {C}^4\otimes \mathbb {C}^2)$ be the injection $j_{{\text {d}}}\phi :=\phi {\text {d}}$. The result is

$$\begin{aligned} T^{tct}=j_{{\text {d}}}^*\,Y_2^{-1}{\mathcal {F}}E^{can}{\mathcal {F}}^{-1}Y_2\,j_{{\text {d}}} \end{aligned}$$

(D.2)

From this one gets $\textsc {t}^{tct}(k,p):= \langle Y_2(k){\text {d}},Y_2(p){\text {d}}\rangle$ for the kernel of $T^{tct}$. Then a straight forward computation, using the explicit expression for $Q({\mathfrak {k}})$ [5, Eq. (3.5)], yields (10.5). $\square$

Proof of (10.7)

Regarding the spinor space it is appropriate to switch from the tensor basis to the standard basis. So let $l:\mathbb {C}^4\otimes \mathbb {C}^2\rightarrow \mathbb {C}^8$ be an isomorphism such that $D:= D^{(0)}\oplus D^{(1)}\oplus D^{(0)}\oplus D^{(1)}=l\,(D^{(1/2)}\oplus D^{(1/2)})\otimes D^{(1/2)}l^{-1}$. In particular let $l{\text {d}}=e_1$, $l{\text {d}}_+=e_3$ with ${\text {d}}_+:=\frac{1}{\sqrt{2}}\big ((1,0,0,0)^T\otimes (0,1)^T+(0,1,0,0)^T\otimes (1,0)^T\big )$, and similarly $l^{-1}e_5$, $l^{-1}e_7$. Accordingly one has the Hilbert space isomorphism $L:L^2(\mathbb {R}^3,\mathbb {C}^4\otimes \mathbb {C}^2)\rightarrow L^2(\mathbb {R}^3,\mathbb {C}^8)$, $(L\varphi )(p):=l\varphi (p)$. It satisfies $D_{\tilde{{\mathcal {E}}}}=L \,W_2|_{\tilde{{\mathcal {E}}}}L^{-1}$. Further put $P_0:L^2(\mathbb {R}^3,\mathbb {C}^8)\rightarrow L^2(\mathbb {R})$, $P_0\varphi :=\varphi _1$. Note $P_0^*\phi =\phi \,e_1$ and $P_0D_{\tilde{{\mathcal {E}}}}P_0^*=U^{(0)}$. Finally recall that $Y_2$ and $W_2|_{\tilde{{\mathcal {E}}}}$ commute.

Now, the extension of $W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}=U^{(0)}$ considered above for the construction of $T^{tct}$ is $U':=L\,Y_2\,W_2|_{\tilde{{\mathcal {E}}}}\,Y_2^{-1}L^{-1}$. Of course it equals $D_{\tilde{{\mathcal {E}}}}$ and hence satisfies $P_0 U' P_0^*=U^{(0)}$. However $U'=\iota \,D_{\tilde{{\mathcal {E}}}}\iota ^{-1}$ for $\iota :L^2(\mathbb {R}^3,\mathbb {C}^8)\rightarrow L^2(\mathbb {R}^3,\mathbb {C}^8)$, $\iota :=L\,Y_2^{-1}L^{-1}$ so that $U'$ is endowed with the covariant PM $E'=\iota {\mathcal {F}}E^{can}{\mathcal {F}}^{-1}\iota ^{-1}$. It gives rise to the POL $(W^{m,0,\eta }|_{\tilde{{\mathcal {E}}}}, T_{{\text {e}}})$ with $T_{{\text {e}}}=P_0E'P_0^*$ for some spinor choice ${\text {e}}$ as shown in Sect. 4.

One easily verifies that $T_{{\text {e}}}=P_0L\,Y_2^{-1}L^{-1} {\mathcal {F}}E^{can}{\mathcal {F}}^{-1}L\,Y_2L^{-1}P_0^*$ coincides with $T^{tct}$ in (D.2). It remains to compute ${\text {e}}$ determined by $j_{{\text {e}}}=XL\,Y_2L^{-1}P_0^*$. Put $X':=L^{-1}X\,L$, $(X'\varphi )(p)=\big ((B(p)^{-1}\oplus B(p)^{-1}) \otimes B(p)^{-1}\big )\varphi (p)$, and $Y'_2:=Y'\otimes I_2$ with

$(Y'\varphi )(p)=Y'(p)\varphi (p)$ and

$$\begin{aligned} Y'(p):=\big (4\epsilon (p)(m+\epsilon (p)) \big )^{-1/2} \left( \begin{array}{cc} m+\epsilon (p)+|p|\sigma _3&{} m+\epsilon (p)-|p|\sigma _3\\ m+\epsilon (p)-|p|\sigma _3 &{} -m-\epsilon (p)-|p|\sigma _3\end{array}\right) \end{aligned}$$

Then $X'Y_2=Y'_2X'$ due to the relation Eq.  [5, (27.9)] concerning the helicity cross section. Hence $j_{{\text {e}}}\phi =L\,Y'_2L^{-1}XP_0^*\phi =L\,Y'_2L^{-1}X\phi \,e_1=L\,Y'_2L^{-1}\phi \,e_1=LY'_2\phi \,{\text {d}}$, whence ${\text {e}}(|p|)=lY'_2(p){\text {d}}$, and (10.7) follows. $\square$