1 Introduction

The momentum \(P=(P^0,\dots P^{D-1})\) generates the unitary representation \(U_a = \textrm{e}^{\textrm{i}\, P \,a}\) of translations in spacetime. This makes models tempting which contain in addition a spacetime position operator \(X=(X_0,\dots X_{D-1})\), which Lorentz transforms as a D-vector and which is translated,

$$\begin{aligned} \textrm{e}^{\textrm{i}\, a\, P} X_n \,\textrm{e}^{-\textrm{i}\, a\, P}= X_n - a_n\ ,\ a \in \mathbb R^D\ . \end{aligned}$$
(1)

Functions of X such as \(V_b(X ) = \textrm{e}^{\textrm{i}\,b\, X}= :V_b\) are shifted, \(U_a \textrm{e}^{\textrm{i}\,b\, X} U_a{}^{-1}=\textrm{e}^{\textrm{i}\,b\, (X-a)} \). These are the Weyl relations

$$\begin{aligned} U_a V_b =V_b U_a\, \textrm{e}^{-\textrm{i}\, a\, b}\ ,\ U_a U_b =U_{a+b}\ ,\ V_a V_b =V_{a+b}\ . \end{aligned}$$
(2)

On the dense subspace of smooth, rapidly decreasing wave functions, the Schwartz space \(\mathcal S(\mathbb R^D) \subset L^2(\mathbb R^D)\), their generators constitute an algebra and represent the Heisenberg Lie algebra

$$\begin{aligned}{}[P^m, P^n]=0 = [X_m, X_n]\ , \ [ P^n,X_m]= \textrm{i}\,\delta ^n{}_m\ ,\ m,n \in \{0,1,\dots D-1\} \ . \end{aligned}$$
(3)

Differentiating \(V_b U_a V_b{}^{-1}=U_a\,\textrm{e}^{\textrm{i}a b}\) with respect to \(a_m\) at \(a=0\) shows

$$\begin{aligned} \textrm{e}^{\textrm{i}\,b\, X} P^m \textrm{e}^{-\textrm{i}\,b\, X} =P^m + b^m\ ,\ b \in \mathbb R^D\ . \end{aligned}$$
(4)

Thus by (1) and (4) the operators \(X_n\) and \(P^m\) are unitarily equivalent to the shifted operators. So their spectra are invariant under shifts and consist of the continuum \(\mathbb R^D\).

However, the possible momenta of relativistic particles do not fill a D-dimensional continuum but are restricted to mass shells

$$\begin{aligned} \mathcal M_m = \left\{ p\in \mathbb R^D :p^0 = \sqrt{m^2 + \textbf{p}^2}\right\} \ . \end{aligned}$$
(5)

The particle states constitute Poincaré multiplets with discrete masses.Footnote 1 Though this discrepancy in covariant quantum string theory, which contains (3) was observed [2, 3, 8] it was not considered a severe fault. Lecture notes as e.g. [1, 10] and books claimed “The physical space of allowed string states is a subspace of the complete Fock space” [7, p.76] with a unitary representation of the Weyl relations.

We show: This is untenable. The spacetime Heisenberg Lie algebra (3) excludes any subspace with a definite mass m whether m vanishes or not.

To constrain in canonically quantized world line models the continuous momentum spectrum to mass shells yields zero, because the volume measure of a lower dimensional submanifold vanishes. This is very much different from constraints which select from discrete possibilities.

In particular by this reason of vanishing measure the Stone-von Neumann theorem excludes mass shells in the continuous momentum spectrum.

If theories, different from the covariantly quantized worldline models or the covariant quantum string, contain only the spatial part of the Heisenberg Lie algebra then this is consistent with massive particles, \(m > 0\). This is compatible with Lorentz covariance, even though of an unusual kind. Covariance does not require \(\textbf{X}\) be the spatial components of a D-vector.

Because Lorentz generators, which are constructed using (3), do not act on particles with a definite mass, we specify the generators which do.

2 Absence of Mass Multiplets

Lemma

A Poincaré multiplet \(\mathcal V\) of states \(\Psi \) of a definite mass m

$$\begin{aligned} (P^2 - m^2)\Psi = 0\ ,\ P^2 = (P^0)^2 - \sum _{i=1}^{D-1}(P^i)^2\ , \end{aligned}$$
(6)

cannot be a subspace of a space with a nondegenerate scalar product in which Heisenberg pairs \((i,j \in \{1,\dots D-1\})\) satisfy

$$\begin{aligned}{}[P^i, P^j]=0 = [X^i, X^j]\ , \ [ P^i,X^j]= -\textrm{i}\,\delta ^{ij}\ , \end{aligned}$$
(7)

and commute with \(P^0\)

$$\begin{aligned}{}[P^0, X^i] \overset{?}{=}\ 0 . \end{aligned}$$
(8)

Proof

The space is the orthogonal sum of the mass multiplet and its complement. All arbitrarily chosen states \(\Psi \) and \(\Phi \) of the multiplet are orthogonal to the complement and have a vanishing matrix element of the commutator \([(P^2 - m^2),X^1]= 2\textrm{i}P^1 \)

$$\begin{aligned} \begin{aligned} \langle {\Phi \, |\, [(P^2 - m^2),X^1]\, \Psi }\rangle&= \langle {\,(P^2 - m^2)\Phi \,|\, X^1 \Psi }\rangle - \langle {\Phi \,| \,X^1\, (P^2 - m^2) \Psi }\rangle \\&= 0 - 0 = 2 \textrm{i}\langle {\Phi \, |\, P^1 \Psi }\rangle \end{aligned} \end{aligned}$$
(9)

\(P^1 \Psi \) is in the Poincaré multiplet \(\mathcal V\) and therefore orthogonal to the complement of the multiplet. By (9) it is also orthogonal to the multiplet. So all scalar products of \(P^1 \Psi \) vanish. But the scalar product is nondegenerate, hence

$$\begin{aligned} P^1 \Psi = 0\ . \end{aligned}$$
(10)

Exchanging \(\Phi \) and \(\Psi \) in the argument, one also has \(P^1 \Phi = 0\). As \(P^1\) is hermitian their matrix element of the commutator \([P^1,X^1]= - \textrm{i}\) vanishes

$$\begin{aligned} \langle {\Phi \,|\,[P^1,X^1]\, \Psi }\rangle = \langle {P^1 \Phi \, | \,X^1 \Psi }\rangle - \langle {\Phi \,| \,X^1 P^1 \Psi }\rangle = 0 - 0 = -\textrm{i}\langle {\Phi |\Psi }\rangle \ . \end{aligned}$$
(11)

All scalar products of \(\Psi \) vanish, thus

$$\begin{aligned} \Psi =0\ . \end{aligned}$$
(12)

The state \(\Psi \) was arbitrarily chosen from the mass multiplet, so there is none.

The argument needs only the nondegeneracy of the scalar product, not its positivity, and does not even require \(X^1\) be hermitian. It needs no assumptions about wavefunctions, which represent a state, nor their explicit scalar product. For our conclusion it is sufficient that they exist.

Our lemma excludes quantized models [1, 9, 13] of free relativistic particles which classically traverse worldlines \(t\mapsto x(t)\) with action given by their length. Canonical quantization yields the Heisenberg algebra. The mass shell condition \((P^2-m^2)\Psi =0\) arises as constraint because of the reparametrization invariance \(t\mapsto t'(t)\). But, no matter how suggestive, aesthetical and geometrical a classical system may be, this does not guarantee that canonical quantization yields a quantum model in which the constraint has a nonvanishing solution.

These models contain the algebra (3) and declare to contain a multiplet of definite mass. They lay claim to the name ‘relativistic particle’ but contain none. This justifies to rename them ‘worldline models’ to avoid the confusing statement that relativistic particles do not exist. Experimentally, relativistic particles are verified beyond doubt, but the covariantly quantized worldline models fail to describe them whatever their denomination pretends. Our lemma applies also to the covariant quantum string which postulates (3).Footnote 2 Its Hilbert space is claimed to be an orthogonal sum of mass multiplets – the physical states – and a complement. Our lemma excludes any multiplet of definite mass. The covariant quantum string has no physical states.Footnote 3

In particular, one cannot constrain the absolutely continuous spectrum of a quantum model to a lower dimensional submanifold: each integral of a projection valued measure on a set of vanishing measure yields zero. This is what vanishing projection measure means.

In a quantum model there is no solutions to a constraint, which restricts a continuous spectrum to a submannifold, even if the solutions of the corresponding classical constraint define a symplectic submanifold with Dirac brackets [4] and with a consistent corresponding quantum model. Its Hilbert space is not a subspace of the unconstrained model. In the constrained model the operator relations of the unconstrained model need not hold.

If one does not want to give up reason altogether then we see only two options to evade our lemma as it excludes also the idea that some of the operators \(X^n\) are not self-adjoint. In this case the Stone-von Neumann theorem would not apply. But our lemma holds also if some \(X^n\) are not self-adjoint.

As first option one can give up the interpretation that \(X^n\) and \(P^m\) are operators in a space with a nondegenerate scalar product. Thereby one gives up the interpretation that they act on quantum particles. For example, they could act on off-shell Feynman graphs. These are not elements of a space of quantum states though one speaks about them in terms of virtual particles. In diagrams with loops virtual particles turn out to have the momentum spectrum \(\mathbb R^D\).

The second option gives up the algebra (3). This is the only option in theories which allow for multiplets of relativistic particles.

3 The Stone-von Neumann Theorem

By the Stone-von Neumann theorem [12, Theorem XI.84]Footnote 4 each unitary representation of the Weyl relations (2) is unitarily equivalent to the one in a Hilbert space \(L^2(\mathbb R^D)\times \mathcal N\) of states \(\Psi : p \mapsto \Psi (p)\) which map \(p\in \mathbb R^D\) almost everywhere to \(\Psi (p)\) in some Hilbert space \(\mathcal N\). The unitary representation acts multiplicatively and by translation

$$\begin{aligned} \begin{gathered} (U_a \Psi )(p)= \textrm{e}^{\textrm{i}\, a\, p}\,\Psi (p)\,,\ (V_b \Psi )(p)= \Psi (p+b)\ ,\\ \langle {\Phi | \Psi }\rangle = \int \! {{\,\textrm{d}\,}}^{D}\!\!\! p\, \langle {\Phi (p)|\Psi (p)}\rangle _{\mathcal N}\ . \end{gathered} \end{aligned}$$
(13)

By the theorem one is not free to choose a different scalar product which integrates not over \(\mathbb R^D\) but only over a mass shell with measure \({{\,\textrm{d}\,}}^{D-1}p / \sqrt{m^2 + \textbf{p}^2}\).

The scalar product of \(L^2(\mathbb R^D)\) implies that each multiplet of definite mass vanishes: \(L^2(\mathbb R^D)\) is the space of equivalence classes of square integrable wave functions, which are equivalent if the D-dimensional measure of the support of their difference vanishes,

$$\begin{aligned} \Psi = 0 \Leftrightarrow \forall \Phi : \int \! {{\,\textrm{d}\,}}^D\!\!p\, \langle {\Phi (p)|\Psi (p)}\rangle _{\mathcal N} = 0\ . \end{aligned}$$
(14)

All wave functions \(\Psi _{\text {phys}}\in L^2(\mathbb R^D)\) with definite mass m

$$\begin{aligned} \Psi _{\text {phys}}(p)= 0\ \text { if }\ p^0 \ne \sqrt{m^2+\textbf{p}^2} \end{aligned}$$
(15)

only have a \((D-1)\)-dimensional support of vanishing D-dimensional measure. They are equivalent to 0 and vanish.

Physical states are not elements of \(L^2(\mathbb R^{D-1})\) obtained from \(L^2(\mathbb R^{D})\) by restriction to the mass shell: restriction of equivalence classes is a linear map, it vanishes if applied to 0. On \(L^2(\mathbb R^{D})\) restriction to a mass shell vanishes altogether. To realize the algebra in a space with a different measure is excluded by the Stone-von Neumann theorem.

If it needed another argument: the Heisenberg algebra (3) is represented by the hermitian operators

$$\begin{aligned} P^m \Psi (p) = p^m \Psi (p)\ ,\ X_n \Psi (p) = -\textrm{i}\, \partial _{p^n}\Psi (p)\ . \end{aligned}$$
(16)

They generate an algebra which is defined on and maps to itself the Schwartz space \(\mathcal S(\mathbb R^D, \mathcal N)\) of smooth functions \(\Psi :\mathbb R^D \rightarrow \mathcal N\) which together with each of their derivatives decrease rapidly [14]. The only smooth function of \(\mathbb R^D\) which vanishes outside mass shells is \(\Psi = 0\).

4 Consistent Spatial Position Operator

The disastrous, innocent looking relation \( [X^i,P^0] \overset{?}{=}\ 0\) (8) in covariantly quantized worldline models is incompatible with the Schrödinger equation \(\textrm{i}\, \partial _t \Psi (t) = P^0 \Psi (t)\) for the motion of a massive \((m>0)\) relativistic particle.

For its expected position \(x^i(t)=\langle {\Psi (t) | X^i \Psi (t) }\rangle \) to change in the course of the time  t by the expected velocity \(\partial _t x^i = v^i = \langle {\Psi |\bigl (P^i/P^0\bigr ) \Psi }\rangle \) one has to have not (8) but different operators acting on \(L^2(\mathcal M_m)\), \(m > 0\), which satisfy (in slight notational abuse we denote these different operators in this section again by their conventional names)

$$\begin{aligned}{}[X^i,P^0]=\textrm{i}\frac{P^i}{P^0} \Leftrightarrow [X^i, P^2 - m^2] = 0\ . \end{aligned}$$
(17)

It is this value which the commutator of \(X^i\) with \(P^0\) must have in order to commute with the mass shell condition. Moreover, (17) is required to justify the denomination ‘position operator’. It entails the notion that in the course of time the position of a particle changes with its velocity.

Poincaré covariance does not require the position operator \(\textbf{X}\) be the spatial part of a D-vector: \(\textbf{X}=\textbf{X}_{\underline{u}}\) is the position operator used by an observer at rest with four-velocity \(\underline{u} = (1,0,\dots )\). Under spacetime translations a and rotations R it transforms linear inhomogeneouslyFootnote 5

$$\begin{aligned} \textrm{e}^{\textrm{i}a P}\, \textbf{X} \,\textrm{e}^{-\textrm{i}a P} = \textbf{X} + \textbf{a} + a^0 \frac{\textbf{P}}{P^0}\ ,\ U_{R} \textbf{X} U_{R}{}^{-1}= R^{-1} \textbf{X}\ , \end{aligned}$$
(18)

where \(U_{R}\) represents the rotation R in Hilbert space. Observers boosted by

$$\begin{aligned} L_u = \begin{pmatrix} \sqrt{1+\textbf{u}^2}&{} \textbf{u}^{{{\,\textrm{T}\,}}}\\ \textbf{u} &{}\quad {\textbf{1}}+ \frac{ \textbf{u}\, \textbf{u}^{{{\,\textrm{T}\,}}}}{1+\sqrt{1+\textbf{u}^2}} \end{pmatrix}\ , \end{aligned}$$
(19)

to four-velocity u, measure position with

$$\begin{aligned} \textbf{X}_u = U_{L_u}\textbf{X}_{\underline{u}} U_{L_u}{}^{-1}\ . \end{aligned}$$
(20)

Under Lorentz transformations \(\Lambda \) the position operators transform by Wigner rotation

$$\begin{aligned} U_\Lambda \textbf{X}_u U_\Lambda {}^{-1}= W^{-1}(\Lambda ,u) \textbf{X}_{\Lambda u}\ , \ W(\Lambda , u) = L_{\Lambda u}{}^{-1} \Lambda L_u\in \text {SO}(D-1)\ , \end{aligned}$$
(21)

in a Poincaré covariant way, even though \(\textbf{X}\) is not the spatial part of a D-vector.

5 Lorentz Generators of Particles

In terms of the algebra (3) one can specify generators \(M_\omega = \omega ^{mn} M_{mn}/2\), \(M_{mn}= - M_{nm}\), of Lorentz transformations \(U_{\textrm{e}^\omega } = \textrm{e}^{-\textrm{i}M_{\omega }}\) [10],

$$\begin{aligned} -\textrm{i}M^{mn}\Psi \overset{?}{=}\ \textrm{i}(P^m X^n - P^n X^m)\Psi + \Gamma ^{mn} \Psi \ , \end{aligned}$$
(22)

where \(\Gamma ^{mn}\) are skew hermitian matrices which commute with X and P and represent the Lorentz Lie algebra

$$\begin{aligned}{}[\Gamma ^{mn}, \Gamma ^{rs}] = -\eta ^{mr}\Gamma ^{ns}+\eta ^{ms}\Gamma ^{nr}+\eta ^{nr}\Gamma ^{ms}-\eta ^{ns}\Gamma ^{mr} \end{aligned}$$
(23)

as do the operators \(l^{mn}= \textrm{i}(P^m X^n - P^n X^m)\).

Nonvanishing \(\Gamma ^{mn}\) can occur only in case the scalar product in spin space is indefinite, otherwise there are no finite dimensional, skew hermitian matrices which generate the Lorentz group.

However as our lemma shows, the operators P and X do not act on particles with a definite mass but in a space in which P has the continuous spectrum \(\mathbb R^D\) (4) with unbounded and also negative energies. This is not the space of relativistic particles.

To show that \(X^n\) and \(P^m\) are not required for the construction of Lorentz generators we specify the ones which act on massive one-particle states. These physical states, to which the generators can can be applied, correspond to rapidly decreasing momentum wave functions \(\Psi : \mathcal M_m \rightarrow \mathbb C^{d} \) which map the massive shell \(\mathcal M_m\), \(m > 0,\) (5) smoothly to some space \(\mathbb C^{d}\), in which skew hermitian matrices \(\Gamma _{ij}=-\Gamma _{ji}\) generate a d-dimensional unitary representation of SO\((D-1)\), \((i,j,k,l\in \{1,\dots ,D-1\})\),

$$\begin{aligned}{}[\Gamma _{ij}, \Gamma _{kl}] = \delta _{ik}\Gamma _{jl}-\delta _{jk}\Gamma _{il}-\delta _{il}\Gamma _{jk}+\delta _{jl}\Gamma _{ik}\ . \end{aligned}$$
(24)

The generators of the Poincaré group [11] and the position operator map byFootnote 6

$$\begin{aligned} \begin{aligned} (-\textrm{i}\, P^n \Psi )(p)&= -\textrm{i}\, p^n \Psi (p)\ ,\\ \bigl (-\textrm{i}M_{ij}\Psi \bigr )(p)&= -\bigl (p^i \partial _{p^j} - p^j \partial _{p^i}\bigr )\Psi (p) + \Gamma _{ij}\Psi (p)\ ,\\ \bigl (-\textrm{i}M_{0i}\Psi \bigr )(p)&= p^0\,\partial _{p^i}\Psi (p) + \Gamma _{ij}\frac{p^j}{p^0 + m}\Psi (p)\ ,\\ \bigl (-\textrm{i}X^i \Psi \bigr )(p)&= \partial _{p^i}\Psi (p) + \frac{p^i}{2(p^0)^2}\Psi (p)\ , \end{aligned} \end{aligned}$$
(25)

this Schwartz space \(\mathcal S(\mathcal M_m, \mathbb C^d)\subset L^2(\mathcal M_m, \mathbb C^d)\) of smooth states of rapid decrease to itself. The generators are skew hermitian with respect to the Lorentz invariant measure \(({{\,\textrm{d}\,}}^{D-1}\!p)/p^0\).

They are equivariant: observers, Lorentz boosted by \(L_u\) (19) to four-velocity u, use the generators (20, 21) and

$$\begin{aligned} U_{L_u} M_{mn} U_{L_u}{}^{-1}=\bigl (L_u\bigr )^r{}_m \bigl (L_u\bigr )^s{}_n M_{rs}\ . \end{aligned}$$
(26)

The massless case is not obtained by simply specifying \(m=0\) in the energy \(p^0=\sqrt{\textbf{p}^2}\). Its inverse \(1/p^0\) and therefore the invariant measure \(({{\,\textrm{d}\,}}^{D-1}\!p)/p^0\), the generators \(M_{0i}\) and \(X^i\) are singular at \(\textbf{p}=0\). There the energy is only continuous, not smooth. The distinguished momentum \(p=0\) is a fixed point of Lorentz transformations and not invariant under translations. This excludes generators \(X^i\) of such translations [6]: massless states do not allow the spatial Heisenberg algebra (7).

The Lorentz generators, acting on massless, physical states turn out not to act on smooth functions of \(\mathbb R^{D-1}\) but on smooth sections of a vector bundle over \(S^{D-2}\times \mathbb R\) which carries a representation of SO\((D-2)\) with generating matrices \(\Gamma _{ij}\).

In the coordinate patch \(\mathcal U_N = \left\{ p: p^0=\sqrt{\textbf{p}^2}, |\textbf{p}| + p_z > 0\right\} \) the sections are smooth functions \(\Psi _N\) which the generators map to (\(p_z := p^{D-1}\), \(i,j,k\in \{1,\dots D-2\}\))Footnote 7

$$\begin{aligned} \begin{aligned} \bigl (-\textrm{i}M_{ij}\Psi \bigr )_N(p)&= - \bigl (p^i\partial _{p^j} - p^j\partial _{p^i}\bigr )\Psi _N(p) + \Gamma _{ij}\,\Psi _N(p)\ ,\\ \bigl (-\textrm{i}M_{zi}\Psi \bigr )_N(p)&= - \bigl (p_z\partial _{p^i} - p^i\partial _{p_z}\bigr )\Psi _N(p) + \Gamma _{ik}\frac{p^k}{|\textbf{p}|+p_z}\,\Psi _N(p)\ ,\\ \bigl (-\textrm{i}M_{0i}\Psi \bigr )_N(p)&= |\textbf{p}|\partial _{p^i}\Psi _N(p) + \Gamma _{ik}\frac{p^k}{|\textbf{p}|+p_z}\,\Psi _N(p)\ ,\\ \bigl (-\textrm{i}M_{0z}\Psi \bigr )_N(p)&= |\textbf{p}|\partial _{p_z}\Psi _N(p)\ . \end{aligned} \end{aligned}$$
(27)

The detailed discussion of massless states, e.g. the relation \(\Psi _S(p)=h_{SN}(p)\Psi _N(p)\) with \(h_{SN}(p)=\Bigl (\frac{p_x+\textrm{i}p_y}{\sqrt{p_x^2+p_y^2}}\Bigr )^{2h}\) where \(\Psi _S\) is smooth in \(\mathcal U_S = \left\{ p: p^0=\sqrt{\textbf{p}^2}, |\textbf{p}| - p_z > 0\right\} \) will be given elsewhere [5]. Here it is only important that in contrast to the covariantly quantized worldline particles they and their Poincaré transformations exist.

6 Conclusions

The spacetime Heisenberg Lie algebra excludes relativistic particles with a definite mass. This important result does not depend on this or that method of quantization. The lemma follows in each quantum theory by elementary algebra. The same conclusion follows from the Stone-von Neumann theorem.

More generally, each solution of a constraint which restricts an absolutely continuous spectrum to a lower dimensional submanifold vanishes.

In quantum physics not only the algebra of operators is important but also the domain on which they act. As the spacetime Heisenberg Lie algebra does not allow a subspace of relativistic particles we specify the generators of Lorentz transformations which do.

Though our lemma has far reaching implications its proof is astonishingly simple. It excludes D Heisenberg pairs as they originate from canonical quantization of classical point particles traversing worldlines which maximize a diffeomorphism invariant action. To impose the corresponding constraint, the mass shell condition, on the momenta of the states, has no solution.