\input phyzzx
\input pstricks

\def\pd{{\vphantom{\dagger}}}
\def\a{\hat a^\pd}
\def\ad{\hat a^\dagger}
\def\b{\hat b^\pd}
\def\bd{\hat b^\dagger}
\def\bx{{\bf x}}
\def\bp{{\bf p}}
\def\bsigma{\setbox0=\hbox{\fourteeni\char27}\setbox2=\hbox{\twelvei\char27}%
        \vbox to \ht0{\vss\hbox to\wd0{\hss\box2\hss}\vss}\kern -\wd0\box0}
\def\D{\!\not\!}
\def\A{{\bar A}}
\def\B{{\bar B}}
\def\eqdef{\buildrel\rm def\over=}

\hsize=6.5in	\hoffset=0in
\vsize=8.5in	\voffset=0in
\parindent=0pt

\centerline{\seventeenrm Spin--Statistics Theorem}

Relativistic causality requires quantum fields at two spacetime points
$x$ and $y$ separated by a space-like interval $(x-y)^2<0$ to either
commute or anticommute with each other.
The {\bf spin--statistics theorem} says that the fields of integral
spins commute (and therefore must be quantized as bosons) while the
fields of half-integral spin anticommute (and therefore must be
quantized as fermions).
The spin-statistics theorem applies to all quantum field theories which have:
\pointbegin
Special relativity, \ie\ Lorentz invariance and relativistic causality;
\point
Positive energies of all particles;
\point
Hilbert space with positive norms of all states.
\par\noindent
The theorem is valid for both free or interacting quantum field theories,
and in any spacetime dimension $D>2$.
In these notes I shall prove the theorem for the free fields in four dimensions
and outline its generalization to $D\neq4$;
proving the theorem for the interactive fields is too complicated for
this class.

\medskip\centerline{$\red\star\quad\star\quad\star$}\smallskip

Consider a generic Lorentz multiplet $\phi_A(x)$ of complex quantum fields
describing particles of spin~$j$ and mass~$M$.
\foot{The spin-statistics theorem applies to massive and massless particles
    alike, but for simplicity of discussion we assume $M\neq0$.}
In general, the multiplet could be reducible
$A\in(j^+_1,j^-_1)\oplus(j^+_2,j^-_2)\oplus\cdots{}$,
but all the irreducible components must have
$$
\left|j^+-j^-\right|\ \le\ j\ \le\ (j^++j^-)\quad{\rm and}\quad
(-1)^{2j^+}(-1)^{2j^-}\ =\ (-1)^{2j} .
\eqn\Spins
$$
Free fields satisfy some kind of linear equations of motion which have
plane-wave solutions with $p^2=M^2$.
Let $p^0=+E_\bp=+\sqrt{\bp^2+M^2}$ and let
$$
e^{-ipx} f_A(\bp,s)\quad{\rm and}\quad e^{+ipx}h_A(\bp,s)
\eqn\PlaneWaves
$$
be respectively the positive-frequency and negative-frequency plane-wave
solutions.
By the CPT theorem
$$
h^{}_A(\bp,s)\ =\ f^{}_A(+\bp,-s)\times i^{2s} (-1)^{2j^-(A)}
\eqn\CPT
$$
where the $i^{2s}$ factor accompanies the spin reversal
and the $(-1)^{2j^-(A)}$ sign is the $(j^+,j^-)$ representation of
the proper-but-not-orthochronous
Lorentz transform~${\bf PT}:x^\mu\to -x^\mu$.
\foot{%
    For example, for the $(\half,0)\oplus(0,\half)$ multiplet of Dirac
    spinor fields, the constant spinors $u_A(\bp,s)\equiv f_A(\bp,s)$
    and $v_A(\bp,s)\equiv h_A(\bp,s)$ satisfy
    $$
    v_A(\bp,s)\ =\ +\left(\sqrt{E-\bp\bsigma}\,\eta_s\right)_A\
    =\ +\left(\sqrt{E-\bp\bsigma}\,(i^{2s}\xi_{-s})\right)_A
    =\ +i^{2s}\,u_A(\bp,-s)
    $$
    for $A\in(\half,0)$ (the left-handed Weyl spinor components), but
    $$
    v_A(\bp,s)\ =\ -\left(\sqrt{E+\bp\bsigma}\,\eta_s\right)_A\
    =\ -\left(\sqrt{E+\bp\bsigma}\,(i^{2s}\xi_{-s})\right)_A
    =\ -i^{2s}\,u_A(\bp,-s)
    $$
    for $A\in(0,\half)$ (the right-handed Weyl spinor components);
    in both cases, the $i^{2s}$ factor comes from $\eta_s=i^{2s}\xi_{-s}$
    while the chirality-dependent sign between the $u_A$ and the $v_A$
    components is the $(-1)^{2j^-(A)}$ factor.
    }
For the complex conjugate plane waves, we have
$$
h^*_{\bar A}(\bp,s)\
=\ f^*_{\bar A}(+\bp,-s)\times (-i)^{2s} (-1)^{2j^+(\bar A)}
\eqn\CPTC
$$
where the last factor is $(-1)^{2j^+(\bar A)}=(-1)^{2j^-(A)}$
because the conjugation exchanges the $j^+$ and the $j^-$ 
of a Lorentz multiplet.

The relation between particle's spin and statistics
follows from the spin sums
$$
{\cal F}_{A\B}(p)\ \eqdef\,\sum_s f^{}_A(\bp,s)\,f^*_\B(\bp,s)
\quad{\rm and}\quad
{\cal H}_{A\B}(p)\ \eqdef\,\sum_s h^{}_A(\bp,s)\,h^*_\B(\bp,s) .
\eqn\SpinSums
$$
In a Lorentz--invariant QFT, these sums must be Lorentz--covariant
functions of the particle's momentum, thus
$$
\eqalign{
{\cal F}_{A\B}(Lp)\ &
=\,\sum_{C,\bar D}M_A^{\,\,C}(L)\,M_\B^{*\bar D}(L)\,{\cal F}_{C\bar D}(p),\cr
{\cal H}_{A\B}(Lp)\ &
=\,\sum_{C,\bar D}M_A^{\,\,C}(L)\,M_\B^{*\bar D}(L)\,{\cal H}_{C\bar D}(p).\cr
}\eqn\FHLorentz
$$
And since we are only interested in the on-shell momenta with
fixed $p^\mu p_\mu=M^2$, the functional form these covariant functions
of $p^\mu$ is determined by the $\rm Spin(3,1)$ analogue
of the Wigner--Eckard theorem.

In three Euclidean dimensions, the Wigner--Eckard theorem usually
concerns the rotational properties of
matrix elements of vector or tensor operators
between states of given angular momenta,
but it can be recast in terms of rotationally-covariant functions
of a vector~$\bf v$.
Consider a covariant matrix of functions $Q_{a,b}({\bf v})$
where the indices $a$ and $b$ run over components of some
(possibly reducible) spin multiplet, $a,b\in(j_1)\oplus(j_2)\oplus\cdots{}$.
According to the Wigner--Eckard theorem,
$$
Q_{a,b}({\bf v}=v{\bf n})\
=\,\sum_{\ell=|j(a)-j(b)|}^{j(a)+j(b)} q_\ell(v)\sum_{m=-\ell}^{+\ell}
v^\ell Y_{\ell,m}({\bf n})\times{\rm Clebbsch}(a,b|\ell,m),
\eqn\WET
$$
where $q_\ell(v)$ depend only on $\ell$ and the magnitude $v$ of the vector
and the spherical harmonics $v^\ell Y_{\ell,m}({\bf n})$ are homogeneous
polynomials (degree~$\ell$) of the Cartesian components $v_x$, $v_y$
and~$v_z$.
For a vector of fixed magnitude ${\bf v}^2=v^2$ the $q_\ell$ coefficients
are constants, hence each $Q_{a,b}$ is effectively
a polynomial of $(v_x,v_y,v_z)$ comprising terms of net degree~$\ell$
ranging from $|j(a)-j(b)|$ to $j(a)+j(b)$.

In four Minkowski dimensions we have a similar situation, except
for the spin group being $SL(2,{\bf C})$ instead of $SU(2)$,
hence $A,\B\in(j_1^+,j_1^-)\oplus(j_2^+,j_2^-)\oplus\cdots{}$.
Also, the Lorentz vector multiplet has $j^+=j^-=\half$ (unlike the
3D vector multiplet which has $\ell=1$) and consequently the Minkowski
analogues ${\cal Y}_{J,m^+,m^-}(p^\mu/M)$
of the spherical harmonics do not have separate integer-valued
indices $\ell^+$ and $\ell^-$ but rather a common index
$J=j^+=j^-$ which takes both integer and half-integer values.
Hence, the Wigner--Eckard theorem for Lorentz-covariant matrices
${\cal F}_{A\B}(p)$ and ${\cal H}_{A\B}(p)$ says:
$$
\eqalign{
{\cal F}_{A\B}(p)\ &
=\,\sum_{J=J_{\rm min}}^{\rm J_{\rm max}}\! f^{}_J(M)\!
    \sum_{-J\le m^+\le J\atop -J\le m^-\le J}
    M^{2J}{\cal Y}_{J,m^+,m^-}(p^\mu/M)
    \times{\rm Clebbsch}(A,\B|J,m^+,J,m^-),\cr
{\cal H}_{A\B}(p)\ &
=\,\sum_{J=J_{\rm min}}^{\rm J_{\rm max}}\! h^{}_J(M)\!
    \sum_{-J\le m^+\le J\atop -J\le m^-\le J}
    M^{2J}{\cal Y}_{J,m^+,m^-}(p^\mu/M)
    \times{\rm Clebbsch}(A,\B|J,m^+,J,m^-),\cr
}\eqn\MinkWET
$$
where $M$ is the particle's mass ($p^\mu p_\mu=M^2$),
the indices $J$, $m^+$ and $m^-$ are all integral or all half-integral
according to
$$
(-1)^{2J}\ =\ (-1)^{2m^+}\ =\ (-1)^{2m^-}\
=\ (-1)^{2j^+(A)}(-1)^{2j^+(\B)}\ =\ (-1)^{2j^-(A)}(-1)^{2j^-(\B)},
\eqn\Integrality
$$
and in the sum over $J$,
$$
\eqalign{
J_{\rm min}\ &=\ \max\left(|j^+(A)-j^+(\B)|,|j^-(A)-j^-(\B)|\right),\cr
J_{\rm max}\ &=\ \min\left((j^+(A)+j^+(\B)),(j^-(A)+j^-(\B))\right).\cr
}\eqn\Jrules
$$
Similar to their 3D counterparts, the 4D ``spherical harmonics''
$M^{2J}{\cal Y}_{J,m^+,m^-}(p^\mu/M)$ are homogeneous polynomials
of the 4--Momentum components $p^0,p^x,p^y,p^z$, although in 4D
the polynomial degree is $2J$ rather than $\ell$.
Consequently, for a fixed particle mass $M$, all spin sums
${\cal F}_{A\B}(p)$ and ${\cal H}_{A\B}(p)$ can be written
as polynomials of the  $p^0,p^x,p^y,p^z$.

Now, once we have written the spin sums~\SpinSums\ as polynomials
of the $p^\mu$ components, we can analytically continue these
polynomials to negative energies $p^0=-E_\bp$ or even
to complex 4--momenta satisfying $p^\mu p_\mu=M^2$.
This analytic continuation allows us to compare the spin sums at
opposite 4--momenta $+p^\mu=(+E,+\bp)$ and $-p^\mu=(-E,-\bp)$,
and because every term in each particular polynomial~\MinkWET\
has the same degree $2J$ modulo~2,
it follows that the whole polynomial is either odd
or even according to eq.~\Jrules, thus
$$
\eqalign{
{\cal F}_{A\B}(-p^\mu)\ &
=\ (-1)^{2j^-(A)}(-1)^{2j^-(\B)}{\cal F}_{A\B}(+p^\mu),\cr
{\cal H}_{A\B}(-p^\mu)\ &
=\ (-1)^{2j^-(A)}(-1)^{2j^-(\B)}{\cal H}_{A\B}(+p^\mu).\cr
}\eqn\FHsigns
$$

Finally, for physical momenta (real $p^\mu=(+E_\bp,\bp)$), the CPT theorem
(\cf\ eqs.~\CPT\ and \CPTC) relates the  positive and the negative frequency
spin sums to each other according to
$$
{\cal H}_{A\B}(p^\mu)\
=\ {\cal F}_{A\B}(p^\mu)\times(-1)^{2j^-(A)}(-1)^{2j^+(\B)}.
\eqn\FHrel
$$
Analytic continuation of the spin sums as polynomials of $p^\mu$
extends eq.~\FHrel\ to any complex momenta, hence in light of eqs.~\FHsigns,
$$
{\cal H}_{A\B}(-p^\mu)\ =\ {\cal F}_{A\B}(+p^\mu)\times
(-1)^{2j^-(\B)}(-1)^{2j^+(\B)}.
\eqno\eq
$$
According to eq.~\Spins, the sign factor in the above formula does not
depend on a particular field component $\phi^\dagger_\B$ but only
on the particle's spin:
$$
\vcenter{\blue\openup 1\jot\ialign{
$\displaystyle{{\cal H}_{A\B}(-p^\mu)\
	=\ {\red #}{\cal F}_{A\B}(+p^\mu)}$\hfil &
\quad for particles of #\hfil\cr
+ & integral spin,\cr
- & half-integral spin.\cr
}}\eqn\SpinRule
$$
It turns out that this little red spin-dependent sign makes
a big difference for the particles' statistics.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip\centerline{$\red\star\quad\star\quad\star$}\smallskip

A free quantum field is a superposition of plane-wave solutions
with operatorial coefficients, thus
$$
\eqalign{
\hat\phi^{}_A(x)\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\sum_s\,
    \left[ e^{-ipx}f^{}_A(\bp,s)\,\a(\bp,s)\
	+\ e^{+ipx}h^{}_A(\bp,s)\,\bd(\bp,s) \right]_{p^0=+E_\bp}\,,\cr
\hat\phi^\dagger_\B(y)\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\sum_s\,
    \left[ e^{-ipy}h^*_\B(\bp,s)\,\b(\bp,s)\
	+\ e^{+ipy}f^*_\B(\bp,s)\,\ad(\bp,s) \right]_{p^0=+E_\bp}\,.\cr
}\eqn\QF
$$
(Without loss of generality we assume complex fields and charged particles;
for the neutral particles we would have $\b\equiv\a$ and $\bd\equiv\ad$.)
{\it Regardless of statistics}, positive particle energies require $\ad(p,s)$
and $\bd(p,s)$ to be creation operators while $\a(p,s)$ and $\b(p,s)$
are annihilation operators, thus
$$
\ad(\bp,s)\ket0\ =\ \ket{1(\bp,s,+)},\quad\
\bd(\bp,s)\ket0\ =\ \ket{1(\bp,s,-)},\quad\
\a(\bp,s)\ket0\ =\ \b(\bp,s)\ket0\ =\ 0,
\eqno\eq
$$
and hence, in a Fock space of positive-definite norm
$$
\bra0\a(\bp,s)\,\ad(\bp',s')\ket0\ =\ \bra0\b(\bp,s)\,\bd(\bp',s')\ket0\
=\ {\red+}2E_\bp(2\pi)^3\delta^{(3)}(\bp-\bp')\delta_{s,s'}
\eqn\Positive
$$
while all the other ``vacuum sandwiches'' of two creation or annihilation
operators vanish identically.
Consequently, {\it regardless of particles' statistics}, vacuum
expectation values of products of two fields at distinct points $x$ and
$y$ are given by
$$
\eqalignno{
\bra0 \hat\phi_A^{}(x)\,\hat\phi^\dagger_\B(y)\ket 0\ &
=\ +\int\!{d^3\bp\over(2\pi)^3}\,{e^{-ip(x-y)}\over 2E_\bp}\times
    \sum_s f^{}_A(\bp,s)\,f^*_\B(\bp,s) &
\eqname\SemiPropA\cr
\noalign{\leftline{and}}
\bra0 \hat\phi^\dagger_\B(y)\,\hat\phi_A^{}(x)\ket 0\ &
=\ +\int\!{d^3\bp\over(2\pi)^3}\,{e^{+ip(x-y)}\over 2E_\bp}\times
    \sum_s h^{}_A(\bp,s)\,h^*_\B(\bp,s). &
\eqname\SemiPropB\cr
}$$
And at this point, we can use the spin sums \SpinSums\
and their polynomial dependence on the particle's 4--momenta to calculate
$$
\bra0 \hat\phi_A^{}(x)\,\hat\phi^\dagger_\B(y)\ket 0\ 
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\,
    \left.e^{-ip(x-y)}{\cal F}_{A\B}(p)\right|_{p^0=+E_\bp}\
=\ {\cal F}_{A\B}(+i\partial_x)\,D(x-y)\quad
\eqn\SemiA
$$
where
$$
D(x-y)\ =\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\,
    \left.e^{-ip(x-y)}\right|_{p^0=+E_\bp}\,,
$$
and likewise
$$
\bra0 \hat\phi^\dagger_\B(y)\,\hat\phi_A^{}(x)\ket 0\ 
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\,
    \left.e^{+ip(x-y)}{\cal H}_{A\B}(p)\right|_{p^0=+E_\bp}\
=\ {\cal H}_{A\B}(-i\partial_x)\,D(y-x).\quad
\eqn\SemiB
$$
As explained in class, for a space-like distance between points
$x$ and $y$, $D(y-x)=+D(x-y)$.
At the same time, the differential operators ${\cal F}_{A\B}(+i\partial_x)$
and ${\cal H}_{A\B}(-i\partial_x)$ are related to each other
according to eq.~\SpinRule.
Therefore, {\blue regardless of particles' statistics}, for $(x-y)^2<0$
$$
\vcenter{\blue\openup 1\jot\ialign{
$\displaystyle{
    	\bra0 \hat\phi_A^{}(x)\,\hat\phi^\dagger_\B(y)\ket 0\
    	=\ {\red #}\bra0 \hat\phi^\dagger_\B(y)\,\hat\phi_A^{}(x)\ket 0
    	}$\hfil &
\quad for particles of #\hfil\cr
+ & integral spin,\cr
- & half-integral spin.\cr
}}\eqn\SpinRule
$$

On the other hands, \psframebox[linecolor=green]{\it relativistic causality}
requires for $(x-y)^2<0$
$$
{\blue\left.\vcenter{\blue\openup 1\jot\ialign{
    $\displaystyle{\hat\phi_A^{}(x)\,\hat\phi^\dagger_\B(y)
    	=\ {\red #}\hat\phi^\dagger_\B(y)\,\hat\phi_A^{}(x)}$\hfil &
    \quad for #\unskip\space fields,\hfil\cr
    + & bosonic\cr
    - & fermionic\cr
    }}\right\}
\hbox{regardless of particle's spin.}
}\eqn\Statistics
$$
%{\blue\it regardless of particle's spin}.
And the only way eqs.~\SpinRule\ and \Statistics\ can both hold true
at the same time if {\it\red all particles of integral spin are bosons
and all particles of half-integral spin are fermions}.

Indeed, for bosonic particles, the creation and annihilation operators
commute with each other except for
$$
\eqalign{
[\a(\bp,s),\ad(\bp',s')]\ &
=\ {\red+}2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp')\delta_{s,s'}\,,\cr
[\bd(\bp,s),\b(\bp',s')]\ &
=\ {\red-}2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp')\delta_{s,s'}\,,\cr
}\eqno\eq
$$
and therefore the quantum fields commute or do not commute according to
$$
\eqalign{
\left[ \hat\phi^{}_A(x),\hat\phi^\dagger_\B(y)\right]\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\sum_s\left(
    	e^{-ip(x-y)}f^{}_A(p,s)f^*_\B(p,s)\
    	{\red-}\ e^{-ip(x-y)}h^{}_A(p,s)h^*_\B(p,s)
    	\right)\cr
&=\ {\cal F}_{A\B}(i\partial_x)\,D(x-y)\
    {\red-}\ {\cal H}_{A\B}(-i\partial_x)\,D(y-x)\cr
&=\ {\cal F}_{A\B}(i\partial_x)
    \left( D(x-y)\ {\red-}\ {\blue(-1)^{2j}}\,D(y-x)\right)\cr
}\eqn\BoseComm
$$
where $j$ is the particle's spin, \cf\ eq.~\SpinRule.
For particles of integral spin, this commutator duly vanishes
when points $x$ and $y$ are separated by a space-like distance.
But for particles of half-integral spin, the two terms on the
last line of eq.~\BoseComm\ add up instead of canceling each other,
and the fields $\hat\phi_A^{}(x)$ and $\hat\phi_\B^\dagger(y)$
fail to commute --- which violates relativistic causality.
To avoid this violation, {\it bosonic particles must have
integral spins only}.

Likewise, for fermionic particles, the creation and annihilation operators
anticommute with each other except for
$$
\eqalign{
\{\a(\bp,s),\ad(\bp',s')\}\ &
=\ {\red+}2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp')\delta_{s,s'}\,,\cr
\{\bd(\bp,s),\b(\bp',s')\}\ &
=\ {\red+}2E_\bp\,(2\pi)^3\delta^{(3)}(\bp-\bp')\delta_{s,s'}\,,\cr
}\eqno\eq
$$
and therefore the quantum fields anticommute or do not anticommute
according to
$$
\!\eqalign{
\!\left\{ \hat\phi^{}_A(x),\hat\phi^\dagger_\B(y)\right\}\ &
=\int\!{d^3\bp\over(2\pi)^3}\,{1\over 2E_\bp}\sum_s\left(
    	e^{-ip(x-y)}f^{}_A(p,s)f^*_\B(p,s)\
    	{\red+}\ e^{-ip(x-y)}h^{}_A(p,s)h^*_\B(p,s)
    	\right)\cr
&=\ {\cal F}_{A\B}(i\partial_x)\,D(x-y)\
    {\red+}\ {\cal H}_{A\B}(-i\partial_x)\,D(y-x)\cr
&=\ {\cal F}_{A\B}(i\partial_x)
    \left( D(x-y)\ {\red+}\ {\blue(-1)^{2j}}\,D(y-x)\right).\cr
}\!\eqn\FermiComm
$$
This anticommutator vanishes when $(x-y)^2<0$ for half-integral~$j$
but not for integral~$j$.
Hence, to maintain relativistic causality,
{\it fermionic particles must have half-integral spins only}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip\centerline{$\red\star\quad\star\quad\star$}\smallskip

I would like to conclude these notes with a few words about
spin-statistics relations in spacetime dimensions other than four.
In any dimension~$D$, quantum fields form multiplets of the
${\rm Spin}(D-1,1)$ Lorentz symmetry while massive particles form multiplets
of the spin symmetry ${\rm Spin}(D-1)$.
For $D>4$, the multiplets are more complicated then in $D=4$, but they
fall into the same two broad classes according to their behavior
under rotations $R(2\pi)$ by $2\pi$ under any spatial axis:
The single-valued tensor multiplets for which $R(2\pi)=+1$,
and the double-valued spinor multiplets for which $R(2\pi)=-1$.
The relation between spin sums~\SpinSums\ follows this distinction:
$$
{\cal H}_{A\B}(-p^\mu)\ =\ {\cal F}_{A\B}(+p^\mu)\times{\red R(2\pi)},
\eqn\SumsD
$$
although the proof is more complicated in higher dimensions.
But but in any dimension, the statistics follow the sign in eq.~\SumsD,
thus {\it\red particles invariant under $2\pi$ rotations must be bosons
while particles for which $R(2\pi)=-1$ must be fermions}.

For $D=3$ (two space dimensions) the situation is more complicated.
The Lorentz symmetry ${\rm Spin}(2,1)=SL(2,{\bf R})$ has finite multiplets
of quantized spin $J=0,\half,1,\coeff32,2,\ldots,$ but the space rotation
group $SO(2)$ is abelian (1 generator only), so its multiplets are singlets
of arbitrary, un-quantized $m_j$.
{\bf If} $m_j$ happens to be an integer or half-integer, {\bf then} this
particle species can be quantized as a free quantum field
of definite $J=m_j$ modulo 1, and
the spin--statistics theorem works as usual:
{\it Particles with integral $m_j$ are bosons while particles
with half-integral $m_j$ are fermions}.
The particles with fractional spins $m_j$ are more difficult to quantize;
they are neither bosons nor fermions but {\it\blue anyons} obeying fractional
statistics where 
$\blue\ket{\alpha,\beta}=\ket{\beta,\alpha}\times e^{\pm2\pi i m_j}$,
depending on how the two particles are exchanged.
But even in this case, the statistics follows the spin: When the spin
is fractional, the statistics has the same fractional phase.



\bye
