On the Kinematics of the Last Wigner Particle

Abstract

Wigner’s particle classification provides for ‘continuous spin’ representations of the Poincaré group, corresponding to a class of (as yet unobserved) massless particles. Rather than building their induced realizations by use of “Wigner rotations” in the textbooks’ way, here we exhibit a scalar-like first-quantized form of those (bosonic) Wigner particles directly, by combining wave equations proposed by Wigner long ago with a recent prequantized treatment employing Poisson structures.

Appendices

Appendix 1: Poincaré Group Conventions

Our metric on the Minkowski space $\mathbb {M}$ is mostly-negative. The inner product of two vectors $x \equiv x^\mu$, $p \equiv p^\nu$ of spacetime is denoted with parentheses: $(xp) = x^\mu p_\mu$. When (we hope) it does not cause confusion, we often write $p^2 = (pp)$, say.

The Lie algebra $\mathfrak {p}$ of $\mathcal {P}$ has a basis of ten elements $\{P^0,P^a,L^a,K^a : a = 1,2,3\}$, corresponding respectively to time translations, space translations, rotations and boosts. The commutation relations for the Lorentz subgroup are as follows:

$$\begin{aligned}{}[L^a,L^b] = \varepsilon ^{ab}{}_{\!c}\, L^c, \qquad [L^a,K^b] = \varepsilon ^{ab}{}_{\!c}\, K^c, \qquad [K^a,K^b] = -\varepsilon ^{ab}{}_{\!c}\, L^c. \end{aligned}$$

The commutation relations are realized⁸ by $K^a = \frac{1}{2} \sigma ^a$ and $L^a = -\frac{i}{2} \sigma ^a$.

In the real four-dimensional representation:

$$\begin{aligned} J^{01}&\equiv K^1 = \begin{pmatrix} &{} 1 &{}&{} \\ 1 &{}&{}&{} \\ &{}&{} 0 &{} \\ &{}&{}&{} 0 \end{pmatrix}; \quad J^{02} \equiv K^2 = \begin{pmatrix} &{}&{} 1 &{} \\ &{} 0 &{}&{} \\ 1 &{}&{}&{} \\ &{}&{}&{} 0 \end{pmatrix}; \quad J^{03} \equiv K^3 = \begin{pmatrix} &{}&{}&{} 1 \\ &{}&{} 0 &{} \\ &{} 0 &{}&{} \\ 1 &{}&{}&{} \end{pmatrix}; \\ J^{23}&\equiv L^1 = \begin{pmatrix} 0 &{}&{}&{} \\ &{} 0 &{}&{} \\ &{}&{}&{} -1 \\ &{}&{} 1 &{} \end{pmatrix}; J^{31} \equiv L^2 = \begin{pmatrix} 0 &{}&{}&{} \\ &{}&{}&{} 1 \\ &{}&{} 0 &{} \\ &{} -1 &{}&{} \end{pmatrix}; J^{12} \equiv L^3 = \begin{pmatrix} 0 &{}&{}&{} \\ &{}&{} -1 &{} \\ &{} 1 &{}&{} \\ &{}&{}&{} 0 \end{pmatrix}, \end{aligned}$$

with the same commutation relations. Remark that

$$\begin{aligned} (L^1 + K^2)^2 = \begin{pmatrix} 1 &{}&{}&{} -1 \\ &{} 0 &{}&{} \\ &{}&{} 0 &{} \\ 1 &{}&{}&{} -1 \end{pmatrix} = (L^2 - K^1)^2 \end{aligned}$$

and $(L^1 + K^2)^3 = (L^2 - K^1)^3 = 0$.

It is advisable to pull these generators together in matrix form:

$$ J^{\mu \nu } = \begin{pmatrix} &{} K^1 &{} K^2 &{} K^3 \\ -K^1 &{}&{} L^3 &{} -L^2 \\ -K^2 &{} - L^3 &{}&{} L^1 \\ -K^3 &{} L^2 &{} -L^1 &{} \end{pmatrix} \quad \text {or}\quad J_{\mu \nu } = \begin{pmatrix} &{} -K^1 &{} -K^2 &{} -K^3 \\ K^1 &{}&{} L^3 &{} -L^2 \\ K^2 &{} - L^3 &{}&{} L^1 \\ K^3 &{} L^2 &{} -L^1 &{} \end{pmatrix}. $$

The general expression is $(J_{\rho \sigma })^\alpha _{\;\beta } = \delta ^\alpha _\rho \,g_{\sigma \beta } - \delta ^\alpha _\sigma \,g_{\rho \beta }$, and the commutation relations are summarized as:

$$\begin{aligned}{}[J_{\rho \sigma }, J_{\mu \nu }] = - g_{\rho \mu }J_{\sigma \nu } - g_{\sigma \nu } J_{\rho \mu } + g_{\sigma \mu }J_{\rho \nu } + g_{\rho \nu }J_{\sigma \mu }. \end{aligned}$$

(12.16)

The dual tensor:

$$ J^{*\rho \mu } := -\tfrac{1}{2} \varepsilon ^{\rho \mu \nu \tau } J_{\nu \tau } = \begin{pmatrix} &{} -L^1 &{} -L^2 &{} -L^3 \\ L^1 &{}&{} K^3 &{} -K^2 \\ L^2 &{} - K^3 &{}&{} K^1 \\ L^3 &{} K^2 &{} -K^1 &{} \end{pmatrix} $$

plays a role in the theory of the WP. Notice that $\varvec{K} \cdot \varvec{L} = \frac{1}{2} J_{\rho \mu } J^{*\rho \mu }$ is a relativistic invariant; as is $\varvec{K}^2 - \varvec{L}^2 = \frac{1}{2} J_{\rho \mu } J^{\rho \mu } = -\frac{1}{2} J^*_{\rho \mu } J^{*\rho \mu }$. These are just the Casimirs of the Lorentz group. A generic infinitesimal Lorentz transformation is of the form

$$ \varLambda \simeq 1 + \tfrac{1}{2} \omega ^{\rho \sigma } J_{\rho \sigma } \,, \quad \text {or}\quad \varLambda ^\mu {}_{\!\nu } = \delta ^\mu _\nu + \omega ^\mu {}_{\!\nu } \,, $$

where $\omega ^{\rho \sigma }$ must be skewsymmetric.

The $P^\mu$ mutually commute. The remaining nonvanishing commutation relations for $\mathcal {P}$ are given by:

$$ [L^a, P^b] = \varepsilon ^{ab}{}_{\!c}\, P^c, \quad [K^a, P^b] = -\delta ^{ab} P^0, \quad [K^a, P^0] = -P^a; $$

that is, $[J^{\kappa \rho }, P^\mu ] = g^{\mu \rho } P^\kappa - g^{\mu \kappa } P^\rho$.

Let $U(\varLambda )$ be the unitary operator acting on one-particle states, corresponding to a Lorentz transformation $\varLambda$. As discussed for instance in [36, Sect. 2.4], one finds that

$$ U^\dagger (\varLambda ) \,\mathbb {P}^\mu \, U(\varLambda ) = \varLambda ^\mu {}_{\!\nu } \,\mathbb {P}^\nu ; \quad U^\dagger (\varLambda ) \,\mathbb {J}^{\mu \nu }\, U(\varLambda ) = \varLambda ^\mu {}_{\!\rho } \varLambda ^\nu {}_{\!\sigma } \,\mathbb {J}^{\rho \sigma }, $$

where by $\mathbb {P}$ and $\mathbb {J} = \{\mathbb {K},\mathbb {L}\}$ we denote hermitian generators on Hilbert space, with commutation relations:

$$ [\mathbb {L}^a, \mathbb {L}^b] = i\varepsilon ^{ab}{}_{\!c}\, \mathbb {L}^c; \quad [\mathbb {L}^a, \mathbb {K}^b] = i\varepsilon ^{ab}{}_{\!c}\, \mathbb {K}^c; \quad [\mathbb {K}^a, \mathbb {K}^b] = -i\varepsilon ^{ab}{}_{\!c}\, \mathbb {L}^c; $$

that is, (12.16) leads to

$$ [\mathbb {J}_{\rho \sigma }, \mathbb {J}_{\mu \nu }] = i\bigl ( -g_{\rho \mu } \mathbb {J}_{\sigma \nu } - g_{\sigma \nu } \mathbb {J}_{\rho \mu } + g_{\sigma \mu } \mathbb {J}_{\rho \nu } + g_{\rho \nu } \mathbb {J}_{\sigma \mu } \bigr ). $$

Appendix 2: The Lorentz Decompositions of Null Rotations

The unique decomposition of an arbitrary (proper orthochronous) Lorentz matrix S into the product of a rotation and a boost is well known [19, Chap. 1]. It becomes

$$\begin{aligned} S&= \begin{pmatrix} \alpha &{} \varvec{a}^t \\ \varvec{c} &{} N \end{pmatrix} = \begin{pmatrix} 1 &{} 0 \\ 0 &{} N - \varvec{c}\varvec{a}^t/(1 + \alpha ) \end{pmatrix} L_{\varvec{a}/\alpha } \\&=: \begin{pmatrix} 1 &{} 0 \\ 0 &{} N - \varvec{c}\varvec{a}^t/(1 + \alpha ) \end{pmatrix} \begin{pmatrix} \alpha &{} \varvec{a}^t \\ \varvec{a} &{} 1_3 + \frac{\varvec{a}\varvec{a}^t}{1 + \alpha } \end{pmatrix}, \end{aligned}$$

where $\alpha ^2 = 1 + \varvec{a}^2$. Since S and $S^t$ are Lorentz, which implies $N\varvec{a} = \alpha \varvec{c}$, $N^t\varvec{c} = \alpha \varvec{a}$ and $N^t N = 1_3 + \varvec{a}\varvec{a}^t$, one checks that $R := N - \varvec{c}\varvec{a}^t/(1 + \alpha )$ is a rotation and that $R\varvec{a} = \varvec{c}$, and thus also $R + R\varvec{a}\varvec{a}^t/(1 + \alpha ) = N$.

We want to decompose null rotations in $G_p$. Note that there is an infinity of spacelike surfaces, of timelike, null or spacelike vectors, which are orbits of $G_p$ in $\mathbb {M}$, each isometric to the group of motions of a plane [37]. Consider those null rotations which leave invariant the standard momentum $k = (1,0,0,1)$. Denoting when convenient $b_1^2 + b_2^2$ by $|b|^2$, a general null rotation fixing k is given by:

$$ S(b_1,b_2) := \begin{pmatrix} 1 + \frac{1}{2} |b|^2 &{} -b_2 &{} b_1 &{} -\frac{1}{2} |b|^2 \\ -b_2 &{} 1 &{} 0 &{} b_2 \\ b_1 &{} 0 &{} 1 &{} -b_1 \\ \frac{1}{2} |b|^2 &{} -b_2 &{} b_1 &{} 1 - \frac{1}{2} |b|^2 \end{pmatrix} =: \begin{pmatrix} \alpha &{} \varvec{a}^t \\ \varvec{c} &{} N \end{pmatrix}. $$

Simplifying further, we work out first the case $S(0,-b)$, with $b > 0$.

Here $\alpha ^2 = 1 + b^2 + \frac{1}{4} b^4 = (1 + \frac{1}{2} b^2)^2$ so that $1 + \alpha = \frac{1}{2} (4 + b^2)$, and $S(0,-b)$ factorizes as

$$\begin{aligned}&\begin{pmatrix} 1 + \frac{1}{2} b^2 &{} b &{} 0 &{} -\frac{1}{2} b^2 \\ b &{} 1 &{} 0 &{} -b \\ 0 &{} 0 &{} 1 &{} 0 \\ \frac{1}{2} b^2 &{} b &{} 0 &{} 1 - \frac{1}{2} b^2 \end{pmatrix} = \begin{pmatrix} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} \frac{4 - b^2}{4 + b^2} &{} 0 &{} -\frac{4b}{4 + b^2} \\ 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} \frac{4b}{4 + b^2} &{} 0 &{} \frac{4 - b^2}{4 + b^2} \end{pmatrix} \! \begin{pmatrix} 1 + \frac{1}{2} b^2 &{} b &{} 0 &{} -\frac{1}{2} b^2 \\ b &{} 1 + \frac{2b^2}{4+b^2} &{} 0 &{} -\frac{b^3}{4+b^2} \\ 0 &{} 0 &{} 1 &{} 0 \\ -\frac{1}{2} b^2 &{} -\frac{b^3}{4+b^2} &{} 0 &{} 1 + \frac{b^4}{2(4+b^2)} \end{pmatrix} \\&\quad =: RL = (R L R^{-1}) R =: L'R. \end{aligned}$$

We see clearly that R is a rotation around the y-axis, of positive angle $\theta$ turning anticlockwise from the positive z-axis towards the positive x-axis, with $\theta = 2\arctan (b/2)$. The velocity associated with the boost $L'$ is:

$$ \varvec{v} = \bigl ( 2b/(2+b^2), 0, -b^2/(2+b^2) \bigr ); $$

therefore its rapidity parameter is given by $\zeta = {\text {arcsinh}} \bigl ( \frac{1}{2} b\sqrt{4 + b^2} \bigr )$; the direction of the boost forms an angle $\arctan (b/2)$ with the x-axis, tilted towards the negative z-axis. For small angles, it is intuitive that the boost undoes the turn effected by the rotation. The result reproduces the one indicated without proof in [32].