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# Relativistic spin sum rules and the role of the pivot

Cédric Lorcé  CPHT, CNRS, Ecole Polytechnique, Institut Polytechnique de
Paris, Route de Saclay, 91128 Palaiseau, France

###### Abstract

Spin sum rules depend on the choice of a pivot, i.e. the point about which the
angular momentun is defined, usually identified with the center of the
nucleon. The latter is however not unique in a relativistic theory and has led
to apparently contradictory results in the literature. Using the recently
developed phase-space approach, we compute for the first time the contribution
associated with the motion of the center of the nucleon, and we derive a
general spin sum rule which reduces to established results after appropriate
choices for the pivot and the spin component.

##  I Introduction

After more than three decades of intense study on both theoretical and
experimental sides, the spin structure of the nucleon remains one of the most
vivid topics in hadronic physics. Reviews about key developments and
achievements in this field can be found in Refs. Filippone:2001ux ;
Bass:2004xa ; Aidala:2012mv ; Leader:2013jra ; Wakamatsu:2014zza ; Liu:2015xha
; Deur:2018roz ; Ji:2020ena . A fundamental question is the decomposition of
the nucleon spin into its constituents’ intrinsic and orbital angular momentum
contributions. While longitudinal spin (or helicity) sum rules derived long
ago Jaffe:1989jz ; Ji:1996ek  are well-established, various transverse spin
sum rules have been discussed over the years and led to a rather confused
situation Burkardt:2002hr ; Bakker:2004ib ; Burkardt:2005hp ; Leader:2011cr ;
Ji:2012sj ; Ji:2012vj ; Leader:2012md ; Harindranath:2012wn ; Leader:2012ar ;
Hatta:2012jm ; Harindranath:2013goa ; Ji:2013tva . Deriving a transverse spin
sum rule is a delicate and subtle problem for the simple reason that
transverse rotations (unlike longitudinal rotations) do not commute with
longitudinal boosts. In the review Leader:2013jra , a whole section has been
devoted to discuss in detail these spin sum rules and to clarify their status.

In a recent paper Ji:2020hii  a number of results about the transverse angular
momentum (AM) in a transversely polarized nucleon are derived, and seem to
contradict those obtained earlier in Refs. Bakker:2004ib ; Leader:2011cr . The
authors argue that the old results are “incorrect” because they contain, in a
moving frame, contributions from the center of mass motion. To get rid of
these contributions, they observe that some terms in the expectation value of
the transverse AM do not contribute to the Pauli-Lubański pseudovector, and
accordingly remove them by hand. This prescription appears however somewhat ad
hoc and requires rigorous justification. In particular, the contribution from
the center-of-mass motion is actually never calculated.

In this paper, we show that the old Bakker:2004ib ; Leader:2011cr  and the new
Ji:2020hii  results are in fact both correct. They just refer to different
definitions for the “center” of the nucleon, and hence for what is meant by
“spin” or “internal AM”. The key role played by this center was recognized
long ago Bakker:2004ib ; Burkardt:2002hr ; Burkardt:2005hp ; Lorce:2011kn  and
motivated a recent discussion of the concept of relativistic center of mass in
the context of hadronic physics Lorce:2018zpf , which clarified the leading-
twist picture of AM in the light-front formalism addressed in Refs.
Hoodbhoy:1998yb ; Burkardt:2002hr ; Burkardt:2005hp ; Ji:2012sj ; Ji:2012vj .
Here we adopt the phase-space approach developed in Refs. Lorce:2018zpf ;
Lorce:2017wkb ; Lorce:2018egm  to identify and compute explicitly the
contribution from the center-of-mass motion. In the case of the transverse AM,
we show that adding this contribution is equivalent to the somewhat ad hoc
prescription used in Ref. Ji:2020hii . We observe however that this
prescription applies only to the AM operators and leads to incorrect results
for the transverse boost operators.

The rest of this paper is organized as follows. In Sec. II we review the
notion of spin in Quantum Field Theory and the problem of defining the center
of the nucleon. Then we explain in Sec. III how to compute rigorously the
expectation values for operators of the form $\int\mathrm{d}^{3}r\,r^{i}O(r)$
and apply the formalism to the Poincaré generators in Sec. IV, followed by a
derivation of the general spin sum rule and a comparison with former results
found in the literature. Finally, we summarize our findings in Sec. V.

##  II What do we mean by “spin”?

An AM operator is usually understood as a generator of rotations about some
point, the pivot. In practice, once a Lorentz frame is fixed, the origin of
the coordinate system is often chosen to coincide with the pivot. But the
choice of the pivot is arbitrary. It is in general an abstract concept unless
it is given some physical meaning. For example, in the simplest non-
relativistic treatment of the hydrogen atom the pivot is placed at the
nucleus, which is taken to be an infinitely heavy point charge, with the
spinless electron moving in its Coulomb potential.

Problems with understanding the magnetic moment in the Stern-Gerlach
experiment led to the introduction of “spin” in Quantum Mechanics to represent
some new intrinsic property of the electron, eventually recognized as a new
type of AM with no classical analog. The concept of spin was then extended to
composite systems and identified with the internal AM111Generally the internal
AM will contain a mixture of contributions from both orbital motion and
intrinsic AM of the constituents., defined as

$\boldsymbol{S}=\boldsymbol{J}-\boldsymbol{R}\times\boldsymbol{P},$ (1)

where $\boldsymbol{J}$ is the total AM about the origin, $\boldsymbol{R}$ is
the “position” of the system with respect to that origin, and $\boldsymbol{P}$
is its total momentum.

If one requires $\boldsymbol{S}$ to generate internal rotations, i.e.
rotations about the position of the system, then the external AM operator
$\boldsymbol{L}=\boldsymbol{R}\times\boldsymbol{P}$ must generate external
rotations about the origin, i.e. it must satisfy the following commutation
relations

$[R^{i},L^{j}]=i\epsilon^{ijk}R^{k},\qquad[P^{i},L^{j}]=i\epsilon^{ijk}P^{k},$
(2)

and hence $[L^{i},L^{j}]=i\epsilon^{ijk}L^{k}$. The necessary and sufficient
conditions for these to hold are that the operators $\boldsymbol{R}$ and
$\boldsymbol{P}$ obey the standard canonical commutation relations

$[R^{i},R^{j}]=0,\qquad[P^{i},P^{j}]=0,\qquad[R^{i},P^{j}]=i\delta^{ij}.$ (3)

They imply in particular

$[R^{i},S^{j}]=0,\qquad[P^{i},S^{j}]=0,\qquad[S^{i},S^{j}]=i\epsilon^{ijk}S^{k}.$
(4)

In other words, the internal AM operator $\boldsymbol{S}$ satisfies the usual
$su(2)$ algebra and is independent of the position and the total momentum of
the system, and therefore of the external AM operator $\boldsymbol{L}$.

The question of interest is: what point defines the position of the system? In
non-relativistic Quantum Mechanics, it is defined by the center-of-mass
position operator

$\boldsymbol{R}:=\boldsymbol{R}_{\text{NR}}=\frac{1}{m}\sum_{n=1}^{N}m_{n}\boldsymbol{R}_{n},$
(5)

where $m=\sum_{n}m_{n}$ is the mass of a system made of $N$ particles with
masses $m_{n}$. Since the particles’ position and momentum operators satisfy
by definition
$[R^{i}_{n},P^{j}_{n^{\prime}}]=i\delta^{ij}\delta_{nn^{\prime}}$ with
$[R^{i}_{n},R^{j}_{n^{\prime}}]=0$ and $[P^{i}_{n},P^{j}_{n^{\prime}}]=0$, one
sees that the non-relativistic spin operator
$\boldsymbol{S}_{\text{NR}}=\boldsymbol{J}-\boldsymbol{R}_{\text{NR}}\times\boldsymbol{P}$
is the generator of internal rotations. If the system consists of a single
structureless particle, then the non-relativistic spin operator simply
coincides with the intrinsic AM operator.

In a relativistic theory, the situation is more complicated for two reasons:

  * •

There exist different expressions for the AM operators in terms of the fields
associated with the particles. The two principal versions are the Canonical
and the Belinfante (or symmetrized) forms, which differ by a spatial integral
over fields evaluated at infinity. As explained in Section 2.6.3 of Ref.
Leader:2013jra , it would be impossible to construct a consistent theory if
one could not ignore such terms in the matrix elements of these AM operators
taken between normalizable states. Hence we may choose which version to use,
and in this paper, as is done in the papers we are commenting on, we shall
only use the Belinfante versions.

  * •

In particle physics it is often adequate to describe the motion of a particle
as being in a momentum eigenstate, i.e. a plane wave, but clearly in that case
the density in coordinate space is constant so that the particle is totally
delocalized. So, to discuss internal AM some form of localized wave packet is
essential. In addition, even when dealing with a localized system, there still
exist several possibilities for the definition of the position of the system,
which are identical in the rest frame of the system, but generally differ in
other frames. The main three definitions are given by the so-called
relativistic centers of energy (or inertia), mass, and spin Pryce:1948pf ;
Moller:1949 ; Fleming:1965 ; Choi:2014nea ; Lorce:2018zpf , which we briefly
review in the following.

###  II.1 The relativistic center of energy or inertia

The relativistic center of energy or inertia is the field theoretical version
of Eq. (5), where the role of inertial mass is played by the energy. Its
position and time component are defined, in any reference frame, by
Fokker:1929 ; Born:1935ap

$R^{\mu}_{E}=\frac{1}{P^{0}}\int\mathrm{d}^{3}r\,r^{\mu}\,T^{00}(r),$ (6)

where the operator $T^{\mu\nu}(r)$ is the energy momentum tensor (EMT) and
$P^{\mu}=\int\mathrm{d}^{3}r\,T^{0\mu}(r)$ is the total four-momentum
operator. Acting on a four-momentum eigenstate with eigenvalue $p^{\mu}$, the
operator $1/P^{0}$ simply means multiplication by $1/p^{0}$. Also, ambiguities
in the ordering of Hermitian operators are resolved by considering implicitly
symmetrized products $AB\equiv\frac{1}{2}(AB+BA)$ Born:1935ap ; Pryce:1948pf .

In a field theory, the AM generators $J^{i}=\frac{1}{2}\epsilon^{ijk}M^{jk}$
and the generators of boosts $K^{i}=M^{0i}$ are defined in terms of the
generalized AM momentum tensor, which reads in Belinfante form Belinfante:1940

$M^{\mu\nu}=-M^{\nu\mu}=\int\mathrm{d}^{3}r\left[r^{\mu}T^{0\nu}(r)-r^{\nu}T^{0\mu}(r)\right].$
(7)

It follows that the position operator for the relativistic center of energy is
directly related to the generators of boosts as follows

$\boldsymbol{R}_{E}=t\,\frac{\boldsymbol{P}}{P^{0}}-\frac{\boldsymbol{K}}{P^{0}}.$
(8)

Clearly, $R^{\mu}_{E}$ does not transform as a Lorentz four-vector. Using the
following commutation relations

$[K^{i},K^{j}]=-i\epsilon^{ijk}J^{k},\qquad[P^{0},K^{i}]=iP^{i},\qquad[P^{i},K^{j}]=i\delta^{ij}P^{0},$
(9)

and the fact that $\boldsymbol{J}$ generates the rotations, one can check that
$\boldsymbol{R}_{E}$ satisfies the canonical commutation relations
$[R^{i}_{E},P^{j}]=i\delta^{ij}$ and
$[R^{i}_{E},J^{j}]=i\epsilon^{ijk}R^{k}_{E}$, so that it qualifies to some
extent as a position operator, albeit with mutually non-commuting components.
This last property implies that $\boldsymbol{R}_{E}\times\boldsymbol{P}$ will
not satisfy the $su(2)$ algebra and cannot be considered as a genuine orbital
AM operator. Nonetheless, as is customary, we shall continue to refer to it as
the AM of the center of energy about the origin.

The nearest one can get to a spin operator transforming as a four-vector is
the Pauli-Lubański pseudovector Lubanski:1942 ; Lubanski:1942bis

$W^{\mu}=\frac{1}{2}\,\epsilon^{\mu\alpha\beta\lambda}M_{\alpha\beta}P_{\lambda},$
(10)

which satisfies the commutation relations

$[W^{\mu},P^{\nu}]=0,\qquad[W^{\mu},W^{\nu}]=-\epsilon^{\mu\nu\alpha\beta}W_{\alpha}P_{\beta}$
(11)

in the convention $\epsilon_{0123}=1$. For a momentum eigenstate with
$\boldsymbol{p}=\boldsymbol{0}$, the action of $\boldsymbol{W}$ coincides with
that of $m\boldsymbol{J}$. Hence one can define a relativistic spin operator
by

$\boldsymbol{S}_{E}=\frac{\boldsymbol{W}}{P^{0}}=\boldsymbol{J}+\frac{\boldsymbol{K}\times\boldsymbol{P}}{P^{0}}=\boldsymbol{J}-\boldsymbol{R}_{E}\times\boldsymbol{P},$
(12)

showing that $\boldsymbol{S}_{E}$ corresponds to the internal AM about the
relativistic center of energy. Note however that, like the external AM
$\boldsymbol{R}_{E}\times\boldsymbol{P}$, it is not a genuine AM operator. An
explicit calculation gives indeed Born:1935ap ; Pryce:1948pf

$[S^{i}_{E},S^{j}_{E}]=i\epsilon^{ijk}\left(\delta^{kl}-\frac{P^{k}P^{l}}{(P^{0})^{2}}\right)S_{E}^{l}$
(13)

and shows that $\boldsymbol{S}_{E}$ does not obey the $su(2)$ algebra, except
when acting on a momentum eigenstate with $\boldsymbol{p}=\boldsymbol{0}$.

###  II.2 The relativistic center of mass

As already mentioned, the relativistic center of energy $\boldsymbol{R}_{E}$
does not behave as the spatial part of a Lorentz four-vector. This is not a
fundamental issue since the center of a system is just a representative point
to which one can attach global properties like mass, momentum and spin. It
does not need to behave as an actual physical point. As a result, different
observers will generally locate the center of energy at different places
inside the convex hull of the system Moller:1949 ; Moller:1949bis . One may
then wonder whether it is possible to define a natural relativistic center
whose coordinates would transform as part of a Lorentz four-vector.

Clearly, the center of energy defined in the rest frame of the system plays a
special role. Since in that frame the energy coincides with the mass of the
system, we will refer to it as the relativistic center of mass. At the
classical level, if $r^{\mu}_{E}|_{\text{rest}}$ represents the space-time
coordinates of the relativistic center of mass in the rest frame, its
components in an arbitrary frame will be defined by
$r^{\mu}_{M}=\Lambda^{\mu}_{\phantom{\mu}\nu}r^{\nu}_{E}|_{\text{rest}}$,
where $\Lambda^{\mu}_{\phantom{\mu}\nu}$ represents the Lorentz transformation
from the rest frame to the moving frame, i.e. such that
$p^{\mu}=\Lambda^{\mu}_{\phantom{\mu}\nu}p^{\nu}_{\text{rest}}$ with
$p^{\mu}_{\text{rest}}=(m,\boldsymbol{0})$. Transposing this construction to
an operator level is not so trivial. First, one needs to clarify what is meant
by “a system at rest” in a quantum theory. Since $\boldsymbol{P}$ is an
operator, the rest frame will simply mean the frame in which the expectation
value222The precise meaning of the expectation value will be given in Sec.
III. of the total momentum operator vanishes

$\langle\boldsymbol{P}\rangle|_{\text{rest}}=\boldsymbol{0}.$ (14)

Second, since $R^{\mu}_{E}$ involves symmetrized products of operators, the
above classical requirements are transposed in a quantum theory into
requirements on the expectation values, namely

$\langle R^{\mu}_{M}\rangle=\Lambda^{\mu}_{\phantom{\mu}\nu}\langle
R^{\nu}_{E}\rangle|_{\text{rest}},\qquad\langle
P^{\mu}\rangle=\Lambda^{\mu}_{\phantom{\mu}\nu}\langle
P^{\nu}\rangle|_{\text{rest}}.$ (15)

A manifestly covariant generalization of Eq. (8) is given by Pryce:1948pf ;
Fleming:1965

$R_{\star}^{\mu}=\tau\,\frac{P^{\mu}}{M}-\frac{P_{\nu}M^{\nu\mu}}{M^{2}},$
(16)

where $\tau$ is the proper time, $P_{\nu}M^{\nu\mu}/M$ is the covariant form
of the boost generators defined by a comoving observer, and $M=\sqrt{P^{2}}$
is the mass operator. In the rest frame, proper time coincides with the time
coordinate333Note that proper time is defined by $\tau=\langle
R^{\mu}_{\star}\rangle u_{\mu}$ with $u^{\mu}=p^{\mu}/m$, whereas the time
coordinate is defined by $t=\langle R^{0}_{\star}\rangle$. so that one has as
required $\langle R^{\mu}_{\star}\rangle|_{\text{rest}}=\langle
R^{\mu}_{E}\rangle|_{\text{rest}}$. The only problems are that
$[R^{i}_{\star},P^{j}]=i\left(\delta^{ij}+\frac{P^{i}P^{j}}{M^{2}}\right)$ and
$R^{0}_{\star}\neq t$, and have to do with the fact that the operator
$\boldsymbol{R}_{\star}$ represents a position at a fixed proper time $\tau$
and not at a fixed time coordinate $t$. The solution is to use another
operator Pryce:1948pf ; Fleming:1965

$R_{M}^{\mu}=\left(t+\frac{\boldsymbol{P}\cdot\boldsymbol{K}}{M^{2}}\right)\frac{P^{\mu}}{P^{0}}-\frac{P_{\nu}M^{\nu\mu}}{M^{2}}$
(17)

that satisfies $[R^{i}_{M},P^{j}]=i\delta^{ij}$ and $R^{0}_{M}=t$. Despite its
lack of manifest covariance, it describes in any frame the same world line as
$R^{\mu}_{\star}$, namely

$\\{\langle R^{\mu}_{M}\rangle\,|\,t\in\text{Reals}\\}=\\{\langle
R^{\mu}_{\star}\rangle\,|\,\tau\in\text{Reals}\\}.$ (18)

In other words, the expectation value $\langle R^{\mu}_{M}\rangle$ transforms
as a Lorentz four-vector. Moreover, one has obviously $\langle
R^{\mu}_{M}\rangle|_{\text{rest}}=\langle R^{\mu}_{E}\rangle|_{\text{rest}}$.

Combining Eqs. (17) and (8), one finds that the shift between the relativistic
centers of energy and mass is given by

$\boldsymbol{R}_{E}-\boldsymbol{R}_{M}=\frac{\boldsymbol{P}\times\boldsymbol{W}}{P^{0}M^{2}}.$
(19)

The internal AM operator about the relativistic center of mass
$\boldsymbol{S}_{M}=\boldsymbol{J}-\boldsymbol{R}_{M}\times\boldsymbol{P}$ can
then be expressed in terms of the Pauli-Lubański pseudovector as follows

$\boldsymbol{S}_{M}=\frac{P^{0}\boldsymbol{W}-\boldsymbol{P}W^{0}}{M^{2}}.$
(20)

Under Lorentz transformations, it behaves in the same way as the total AM
$\boldsymbol{J}$, i.e. as the axial-vector part
$S^{i}_{M}=\frac{1}{2}\,\epsilon^{ijk}S^{jk}_{M}$ of a rank-two antisymmetric
Lorentz tensor
$S^{\mu\nu}_{M}=-\epsilon^{\mu\nu\rho\sigma}W_{\rho}P_{\sigma}/M^{2}$. Note
however that, like $\boldsymbol{R}_{E}$, the components of
$\boldsymbol{R}_{M}$ do not commute with each other. It is also easy to check
that $\boldsymbol{S}_{M}$ does not satisfy the $su(2)$ algebra Pryce:1948pf

$[S^{i}_{M},S^{j}_{M}]=i\epsilon^{ijk}\left(\delta^{kl}+\frac{P^{k}P^{l}}{M^{2}}\right)S_{M}^{l}$
(21)

and cannot be considered as a genuine AM operator, except when acting on a
momentum eigenstate with $\boldsymbol{p}=\boldsymbol{0}$.

###  II.3 The relativistic center of spin

As stressed above, neither the components of $\boldsymbol{R}_{E}$ nor those of
$\boldsymbol{R}_{M}$ commute with each other Born:1935ap ; Pryce:1948pf ;
Moller:1949 ; Fleming:1965 . The corresponding spin operators
$\boldsymbol{S}_{E}$ and $\boldsymbol{S}_{M}$ do not satisfy the $su(2)$
algebra of AM operators, and hence cannot be considered as generators of
rotations. Remarkably, it has been observed that a position operator with
commuting components can be obtained by an appropriate weighted average of
$R^{\mu}_{E}$ and $R^{\mu}_{M}$ Pryce:1935ibt ; Pryce:1948pf ; Moller:1949

$R^{\mu}_{c}=\frac{P^{0}R^{\mu}_{E}+MR^{\mu}_{M}}{P^{0}+M},$ (22)

proved to be unique by Newton and Wigner Newton:1949cq . Clearly, like
$\boldsymbol{R}_{E}$ and $\boldsymbol{R}_{M}$, the operator
$\boldsymbol{R}_{c}$ satisfies the commutations relations
$[R^{i}_{c},P^{j}]=i\delta^{ij}$ and
$[R^{i}_{c},J^{j}]=i\epsilon^{ijk}R^{k}_{c}$, and therefore qualifies as a
position operator.

The corresponding internal AM operator
$\boldsymbol{S}_{c}=\boldsymbol{J}-\boldsymbol{R}_{c}\times\boldsymbol{P}$
reads Pryce:1935ibt ; Pryce:1948pf ; Moller:1949 ; Bogolyubov:1975ps ;
Choi:2014nea

$\boldsymbol{S}_{c}=\frac{P^{0}\boldsymbol{S}_{E}+M\boldsymbol{S}_{M}}{P^{0}+M}=\frac{\boldsymbol{W}}{M}-\frac{\boldsymbol{P}W^{0}}{M(P^{0}+M)}.$
(23)

It satisfies the familiar commutation relations

$[S^{i}_{c},S^{j}_{c}]=i\epsilon^{ijk}S^{k}_{c},\qquad[S^{i}_{c},R^{j}_{c}]=0,\qquad[S^{i}_{c},P^{j}]=0,$
(24)

and can therefore be interpreted as the generator of internal rotations. For
these reasons, the point located at $\langle\boldsymbol{R}_{c}\rangle$ has
been called the relativistic center of spin in Refs. Fleming:1965 ;
Schwartz:2020lys  and the canonical reference point in Ref. Lorce:2018zpf .

Since $P\cdot W=0$, the canonical spin operator can be put in the form
Keister:1991sb

$S^{\mu}_{c}=(L^{-1}_{c}(P))^{\mu}_{\phantom{\mu}\nu}\,\frac{W^{\nu}}{M}$ (25)

with $S^{0}_{c}=0$. The operator

$(L^{-1}_{c}(P))^{\mu}_{\phantom{\mu}\nu}=\delta^{\mu}_{\nu}-\frac{(P^{\mu}+M\delta^{\mu}_{0})(P_{\nu}+M\delta^{0}_{\nu})}{M(P^{0}+M)}+2\frac{\delta^{\mu}_{0}P_{\nu}}{M}$
(26)

satisfies the properties

$(L^{-1}_{c}(P))^{\mu}_{\phantom{\mu}\nu}P^{\nu}=M\delta^{\mu}_{0},\qquad
P_{\nu}=M\delta^{0}_{\mu}\,(L^{-1}_{c}(P))^{\mu}_{\phantom{\mu}\nu}=(L_{c}(P))_{\nu}^{\phantom{\nu}\mu}M\delta^{0}_{\mu},$
(27)

and can be understood as the operator analog of
$(L^{-1}_{c}(p))^{\mu}_{\phantom{\mu}\nu}$, the canonical (rotationless) four-
vector boost from the moving frame to the rest frame Krause:1977yr ;
Fahnline:1982 . Sandwiching Eq. (25) between momentum eigenstates, one gets

$\displaystyle\langle p|S^{i}_{c}|p\rangle$
$\displaystyle=\frac{1}{m}\,(L^{-1}_{c}(p))^{i}_{\phantom{i}\nu}\langle
p|W^{\nu}|p\rangle$ (28) $\displaystyle=\frac{1}{m}\,\langle
p|U^{-1}(L^{-1}_{c}(p))W^{i}U(L^{-1}_{c}(p))|p\rangle$ $\displaystyle=\langle
p_{\text{rest}}|J^{i}|p_{\text{rest}}\rangle.$

Since momentum eigenstates are covariantly normalized, one can write $\langle
p|p\rangle=\langle p_{\text{rest}}|p_{\text{rest}}\rangle$ and therefore
conclude that the expectation value of canonical spin in an arbitrary frame
gives the same value as the rest-frame AM

$\frac{\langle p|S^{i}_{c}|p\rangle}{\langle p|p\rangle}=\frac{\langle
p_{\text{rest}}|J^{i}|p_{\text{rest}}\rangle}{\langle
p_{\text{rest}}|p_{\text{rest}}\rangle}.$ (29)

In particular, a spin-$1/2$ momentum eigenstate with canonical polarization
described by the unit vector $\boldsymbol{s}$ is an eigenstate of the
canonical spin component along $\boldsymbol{s}$

$\boldsymbol{s}\cdot\boldsymbol{S}_{c}|p,\boldsymbol{s}\rangle=-\frac{\mathcal{S}_{\mu}W^{\mu}}{m}|p,\boldsymbol{s}\rangle=\tfrac{1}{2}\,|p,\boldsymbol{s}\rangle,$
(30)

where the (canonical) covariant polarization vector is defined as

$\mathcal{S}^{\mu}(p,\boldsymbol{s})=(L_{c}(p))^{\mu}_{\phantom{\mu}\nu}\mathcal{S}^{\nu}_{\text{rest}}=\left(\frac{\boldsymbol{p}\cdot\boldsymbol{s}}{m},\boldsymbol{s}+\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{m(p^{0}+m)}\right)$
(31)

with $\mathcal{S}^{\mu}_{\text{rest}}=(0,\boldsymbol{s})$.

As a final remark, it is interesting to express the generators of boosts in
terms of the canonical position and spin operators

$\boldsymbol{K}=\boldsymbol{P}t-P^{0}\boldsymbol{R}_{c}-\frac{\boldsymbol{P}\times\boldsymbol{S}_{c}}{P^{0}+M}.$
(32)

This relation clearly shows that the operators $\boldsymbol{K}$ generate not
only boosts of the system (first two terms), but also rotations in internal or
spin space (last term) known as the Wigner rotations Wigner:1939cj ;
Polyzou:2012ut , which are responsible for the non-commutativity of the
components of $\boldsymbol{K}$.

##  III Expectation values

Now that the notion of spin and the question of the pivot have been reviewed,
we turn our attention to the computation of the expectation values of the
Lorentz generators. They require particular care because matrix elements of
the type444The operator $O(r)$ is required not to involve any explicit factor
of $r^{i}$. $\langle
p,\boldsymbol{s}|\int\mathrm{d}^{3}r\,r^{i}O(r)|p,\boldsymbol{s}\rangle$ are
usually ill-defined, being either infinite or, by symmetry, zero. We present
in the following the two approaches designed to deal with this problem.

We will always work with covariantly normalized momentum eigenstates $\langle
p_{f},\boldsymbol{s}|p_{i},\boldsymbol{s}\rangle=(2\pi)^{3}\,2p^{0}_{i}\,\delta^{(3)}(\boldsymbol{p}_{f}-\boldsymbol{p}_{i})$,
so that the completeness relation (resolution of the identity) becomes

$I=\sum_{\sigma=\pm}\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}2p_{0}}\,|p,\sigma\boldsymbol{s}\rangle\langle
p,\sigma\boldsymbol{s}|.$ (33)

For later convenience, we introduce the average four-momentum and four-
momentum transfer variables

$\bar{p}=\tfrac{1}{2}(p_{f}+p_{i}),\qquad\Delta=p_{f}-p_{i}$ (34)

which obey the constraints

$\bar{p}\cdot\Delta=0,\qquad\bar{p}^{2}=m^{2}-\frac{\Delta^{2}}{4}$ (35)

owing to the onshell relations $p^{2}_{i}=p^{2}_{f}=m^{2}$.

###  III.1 Standard approach

The standard approach is a prescription which consists in considering first
the matrix element with nonzero momentum transfer $\boldsymbol{\Delta}$, and
then taking the forward limit $\boldsymbol{\Delta}\to\boldsymbol{0}$ at the
end of the calculation Jaffe:1989jz ; Shore:1999be ; Bakker:2004ib ;
Leader:2013jra ; Lowdon:2017idv ; Cotogno:2019xcl ; Lorce:2019sbq

$\langle
p,\boldsymbol{s}|\int\mathrm{d}^{3}r\,r^{i}O(r)|p,\boldsymbol{s}\rangle\equiv\lim_{\boldsymbol{\Delta}\to\boldsymbol{0}}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|\int\mathrm{d}^{3}r\,r^{i}O(r)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle.$
(36)

Note that $\bar{\boldsymbol{p}}$ is kept fixed so that
$\bar{\boldsymbol{p}}=\boldsymbol{p}$, and from Eq. (35)

$\bar{p}^{0}=\frac{1}{2}\left(\sqrt{(\boldsymbol{p}+\tfrac{\boldsymbol{\Delta}}{2})^{2}+m^{2}}+\sqrt{(\boldsymbol{p}-\tfrac{\boldsymbol{\Delta}}{2})^{2}+m^{2}}\right)$
(37)

whereas $p^{0}=\sqrt{\boldsymbol{p}^{2}+m^{2}}$.

For convenience, we will also work at $t=0$ in the following. One can write
using translation symmetry and the Leibniz rule555Instead of the Leibniz rule,
one can use directly the distributional identity Strichartz:1994
$f(x)\nabla\delta(x)=f(0)\nabla\delta(x)-\left[\nabla
f(x)\right]_{x=0}\delta(x)$.

$\displaystyle\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|$
$\displaystyle\int\mathrm{d}^{3}r\,r^{i}O(r)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle$
(38)
$\displaystyle=\int\mathrm{d}^{3}r\,e^{-i\boldsymbol{\Delta}\cdot\boldsymbol{r}}r^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle$
$\displaystyle=\left[(2\pi)^{3}i\nabla^{i}\delta^{(3)}(\boldsymbol{\Delta})\right]\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle$
$\displaystyle=i\nabla^{i}\\!\left[(2\pi)^{3}\delta^{(3)}(\boldsymbol{\Delta})\,\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]-\left[i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right](2\pi)^{3}\delta^{(3)}(\boldsymbol{\Delta})$
$\displaystyle=\langle
p,\boldsymbol{s}|O(0)|p,\boldsymbol{s}\rangle\,(2\pi)^{3}i\nabla^{i}\delta^{(3)}(\boldsymbol{\Delta})+\left[-i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]_{\boldsymbol{\Delta}=\boldsymbol{0}}(2\pi)^{3}\delta^{(3)}(\boldsymbol{\Delta}),$

where $\nabla^{i}=\frac{\partial}{\partial\Delta^{i}}$. One then finds for the
expectation value

$\frac{\langle
p,\boldsymbol{s}|\int\mathrm{d}^{3}r\,r^{i}O(r)|p,\boldsymbol{s}\rangle}{\langle
p,\boldsymbol{s}|p,\boldsymbol{s}\rangle}=\frac{\langle
p,\boldsymbol{s}|O(0)|p,\boldsymbol{s}\rangle}{\langle
p,\boldsymbol{s}|p,\boldsymbol{s}\rangle}\,(2\pi)^{3}i\nabla^{i}\delta^{(3)}(\boldsymbol{0})+\frac{1}{2p^{0}}\left[-i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]_{\boldsymbol{\Delta}=\boldsymbol{0}}.$
(39)

Note that in the second term one should not forget that both $\Delta^{0}$ and
$\bar{p}^{0}$ are understood as functions of $\boldsymbol{\Delta}$, derived
from the onshell constraints (35).

In order to make sense of the highly divergent term in Eq. (39), one has to
work with normalizable states and hence introduce some wave packet.
Calculations with explicit forms for the wave packets are pretty lengthy and
cumbersome, but they show that the highly divergent term corresponds actually
to the contribution from the center of the wave packet Jaffe:1989jz ;
Bakker:2004ib . Since one is interested in the internal structure of the
nucleon and not in the details of the wave packet, this term is usually
discarded.

###  III.2 Phase-space approach

Discarding the highly divergent term in Eq. (39) is tantamount to throwing the
baby out with the bathwater. Indeed, even though we are not interested in the
details of the wave packet, it is important to retain the information about
the position of its center. This can easily be achieved within the phase-space
approach developed in Refs. Lorce:2018zpf ; Lorce:2017wkb ; Lorce:2018egm ,
and used in Ref. Lorce:2020onh  to elucidate the momentum dependence of
relativistic charge distributions.

In a quantum theory, states localized around the position $\boldsymbol{r}$ at
time $t=0$ are defined as the following Fourier transform of momentum
eigenstates

$|\boldsymbol{r}\rangle\equiv\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}\sqrt{2p^{0}}}\,e^{-i\boldsymbol{p}\cdot\boldsymbol{r}}\,|p\rangle$
(40)

and are normalized as
$\langle\boldsymbol{r}_{f}|\boldsymbol{r}_{i}\rangle=\delta^{(3)}(\boldsymbol{r}_{f}-\boldsymbol{r}_{i})$.
Wave packets associated with physical states normalized as
$\langle\psi|\psi\rangle=1$ are then defined by

$\tilde{\psi}(\boldsymbol{p})\equiv\frac{\langle
p|\psi\rangle}{\sqrt{2p^{0}}}\qquad\text{so
that}\qquad\psi(\boldsymbol{r})\equiv\langle\boldsymbol{r}|\psi\rangle=\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}}\,e^{i\boldsymbol{p}\cdot\boldsymbol{r}}\,\tilde{\psi}(\boldsymbol{p}).$
(41)

A physical state cannot be simultaneously in an eigenstate of both position
and momentum. But it was shown long ago how to define a quantity, analogous to
a density matrix, which gives the quantum weight that a system in a definite
state $|\psi\rangle$ will be found to have average position
$\boldsymbol{\mathcal{R}}$ and average momentum $\boldsymbol{p}$. This quantum
phase-space or Wigner distribution is given by Wigner:1932eb ; Hillery:1983ms

$\displaystyle\rho_{\psi}(\boldsymbol{\mathcal{R}},\boldsymbol{p})$
$\displaystyle\equiv\int\mathrm{d}^{3}z\,e^{-i\boldsymbol{p}\cdot\boldsymbol{z}}\,\psi^{*}(\boldsymbol{\mathcal{R}}-\tfrac{\boldsymbol{z}}{2})\psi(\boldsymbol{\mathcal{R}}+\tfrac{\boldsymbol{z}}{2})$
(42)
$\displaystyle=\int\frac{\mathrm{d}^{3}q}{(2\pi)^{3}}\,e^{-i\boldsymbol{q}\cdot\boldsymbol{\mathcal{R}}}\,\tilde{\psi}^{*}(\boldsymbol{p}+\tfrac{\boldsymbol{q}}{2})\tilde{\psi}(\boldsymbol{p}-\tfrac{\boldsymbol{q}}{2})$

and is not positive-definite because of the Heisenberg uncertainty relations.
It is thus a _quasi_ -probabilistic density matrix. It can be seen that
$\boldsymbol{\mathcal{R}}$ is the average of the initial and final position
vectors and $\boldsymbol{p}$ is the average of the initial and final momenta,
respectively, in Eq. (42). One recovers a probabilistic interpretation after
integration over either average position or average momentum

$\int\mathrm{d}^{3}\mathcal{R}\,\rho_{\psi}(\boldsymbol{\mathcal{R}},\boldsymbol{p})=|\tilde{\psi}(\boldsymbol{p})|^{2},\qquad\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}}\,\rho_{\psi}(\boldsymbol{\mathcal{R}},\boldsymbol{p})=|\psi(\boldsymbol{\mathcal{R}})|^{2}.$
(43)

One can now express the expectation of any operator $O$ as a phase-space
integral Wigner:1932eb ; Hillery:1983ms  over $\boldsymbol{\mathcal{R}}$ and
$\boldsymbol{p}$ of the Wigner distribution times the expectation value of $O$
in a state having average position $\boldsymbol{\mathcal{R}}$ and average
momentum $\boldsymbol{p}$, namely

$\langle\psi|O|\psi\rangle=\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}}\,\mathrm{d}^{3}\mathcal{R}\,\rho_{\psi}(\boldsymbol{\mathcal{R}},\boldsymbol{p})\,\langle
O\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}},$ (44)

where666We note in passing that the covariant formulation of the phase-space
approach proposed in Ref. Lorce:2018egm  is not totally rigorous in the
implementation of the onshell constraints since
$\delta(p^{2}_{f}-m^{2})\,\delta(p^{2}_{i}-m^{2})\neq\delta(2\bar{p}\cdot\Delta)\,\delta(\bar{p}^{2}+\frac{\Delta^{2}}{4}-m^{2})$,
leading to a factor $2\bar{p}^{0}$ in the denominator of Eq. (45) instead of
$\sqrt{2p^{0}_{f}}\sqrt{2p^{0}_{i}}=\sqrt{4(\bar{p}^{0})^{2}-(\Delta^{0})^{2}}$.
In the context of Ref. Lorce:2018egm  this did not matter since the results
were derived in Lorentz frames where $\Delta^{0}=0$.

$\langle
O\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}}=\int\frac{\mathrm{d}^{3}\Delta}{(2\pi)^{3}}\,e^{i\boldsymbol{\Delta}\cdot\boldsymbol{\mathcal{R}}}\,\frac{\langle\bar{p}+\tfrac{\Delta}{2}|O|\bar{p}-\tfrac{\Delta}{2}\rangle}{\sqrt{4\left(\bar{p}^{0}\right)^{2}-\left(\Delta^{0}\right)^{2}}}\qquad\textrm{with}\qquad\bar{\boldsymbol{p}}=\boldsymbol{p}.$
(45)

Although originally developed in the non-relativistic context, this formalism
carries over to Quantum Field Theory BialynickiBirula:1991tx . Requiring the
localization777In Quantum Field Theory, it is often stated that a particle
cannot be localized over distances smaller than the Compton wavelength. It
does not mean that one cannot define a position eigenstate
$|\boldsymbol{r}\rangle$ for a system, but it just expresses the limit of the
single onshell particle picture. Note also that if the average momentum is
non-zero, then the bound is determined by $1/\langle p^{0}\rangle$ and not
$1/m$ Burkardt:2000za . of a system simultaneously in the three spatial
dimensions leads automatically to a position operator with mutually commuting
components, i.e. to $\boldsymbol{R}_{c}$ which, as mentioned, is the only
relativistic operator with that property Pryce:1948pf ; Newton:1949cq ;
Foldy:1949wa . It has indeed been shown in Refs. Newton:1949cq ;
Pavsic:2017orp  that the states $|\boldsymbol{r}\rangle$ defined by Eq. (40)
are eigenstates of the canonical position operator

$\boldsymbol{R}_{c}|\boldsymbol{r}\rangle=\boldsymbol{r}|\boldsymbol{r}\rangle.$
(46)

Since wave-packet details are encoded in
$\rho_{\psi}(\boldsymbol{\mathcal{R}},\boldsymbol{p})$ in Eq. (44), one can
interpret $\langle O\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}}$ as the
part associated with the internal structure of the system localized in the
Wigner sense around the (average) canonical position
$\langle\boldsymbol{R}_{c}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}}=\boldsymbol{\mathcal{R}}$
with (average) momentum
$\langle\boldsymbol{P}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}}=\boldsymbol{p}$.
As noted in Ref. Lorce:2018zpf , the standard expectation value (36)
corresponds simply to an average over $\boldsymbol{\mathcal{R}}$ of the phase-
space expectation value (45)

$\int\frac{\mathrm{d}^{3}\mathcal{R}}{(2\pi)^{3}\delta^{(3)}(\boldsymbol{0})}\,\langle
O\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p}}=\int\mathrm{d}^{3}\Delta\,\delta^{(3)}(\boldsymbol{\Delta})\,\frac{\langle\bar{p}+\tfrac{\Delta}{2}|O|\bar{p}-\tfrac{\Delta}{2}\rangle}{(2\pi)^{3}2p^{0}\delta^{(3)}(\boldsymbol{0})}=\frac{\langle
p|O|p\rangle}{\langle p|p\rangle}.$ (47)

Proceeding similarly to Eq. (38), one finds including now polarization
Lorce:2018zpf

$\langle\int\mathrm{d}^{3}r\,r^{i}O(r)\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}}=\mathcal{R}^{i}\,\frac{\langle
p,\boldsymbol{s}|O(0)|p,\boldsymbol{s}\rangle}{2p^{0}}+\frac{1}{2p^{0}}\left[-i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]_{\boldsymbol{\Delta}=\boldsymbol{0}}.$
(48)

Comparing with Eq. (39), one observes that the highly divergent term is
replaced by a well-defined expression in the phase-space approach, which
confirms its interpretation as the contribution associated with the center of
the wave packet. According to Eq. (47) one should recover Eq. (39) from an
average of Eq. (48) over $\boldsymbol{\mathcal{R}}$. Indeed, the second term
in Eq. (48) is independent of $\boldsymbol{\mathcal{R}}$ and is therefore not
affected by the average over $\boldsymbol{\mathcal{R}}$. The first term
however is linear in $\boldsymbol{\mathcal{R}}$ and gives the ambiguous
contribution in Eq. (39), since one can formally write
$\int\mathrm{d}^{3}\mathcal{R}\,\mathcal{R}^{i}=\lim_{\boldsymbol{\Delta}\to\boldsymbol{0}}\int\mathrm{d}^{3}\mathcal{R}\,e^{-i\boldsymbol{\Delta}\cdot\boldsymbol{\mathcal{R}}}\mathcal{R}^{i}=(2\pi)^{3}\,i\nabla^{i}\delta^{(3)}(\boldsymbol{0})$.

As mentioned earlier, in the literature one is usually interested in the
internal part and accordingly considers only the second term in Eq. (48) or
Eq. (39) Jaffe:1989jz ; Shore:1999be ; Bakker:2004ib ; Leader:2013jra . This
can be justified by choosing the (average) relativistic center of spin as the
origin of the coordinate system, i.e. by setting
$\boldsymbol{\mathcal{R}}=\boldsymbol{0}$

$\langle\int\mathrm{d}^{3}r\,r^{i}O(r)\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}=\frac{1}{2p^{0}}\left[-i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]_{\boldsymbol{\Delta}=\boldsymbol{0}}.$
(49)

The drawback is that when we change the Lorentz frame, a shift of the origin
is required for the new origin to coincide with the relativistic center of
spin in the new Lorentz frame, because the coordinates of the latter do not
transform as part of a four-vector. In fact, we can obtain the internal part
independently of the choice for the origin as follows

$\langle\int\mathrm{d}^{3}r\,(r^{i}-\mathcal{R}^{i})\,O(r)\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}}=\frac{1}{2p^{0}}\left[-i\nabla^{i}\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|O(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle\right]_{\boldsymbol{\Delta}=\boldsymbol{0}}$
(50)

using the relation
$\langle\int\mathrm{d}^{3}r\,O(r)\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}}=\langle
p,\boldsymbol{s}|O(0)|p,\boldsymbol{s}\rangle/2p^{0}$.

##  IV AM sum rules

All the spin sum rules are based on a decomposition of the QCD EMT into quark
and gluon parts

$\displaystyle T^{\mu\nu}(r)$
$\displaystyle=T^{\mu\nu}_{q}(r)+T^{\mu\nu}_{g}(r).$ (51)

In the Belinfante form, they are given by

$\displaystyle T^{\mu\nu}_{q}(r)$
$\displaystyle=\overline{\psi}(r)\gamma^{\\{\mu}\tfrac{i}{2}\overset{\leftrightarrow}{D}\\!\\!\\!\\!\\!\phantom{D}^{\nu\\}}\psi(r),$
(52) $\displaystyle T^{\mu\nu}_{g}(r)$
$\displaystyle=-F^{\mu\lambda}(r)F^{\nu}_{\phantom{\nu}\lambda}(r)+\frac{1}{4}\,g^{\mu\nu}F^{\alpha\beta}(r)F_{\alpha\beta}(r)$

with
$\overset{\leftrightarrow}{D}\\!\\!\\!\\!\\!\phantom{D}_{\mu}=\overset{\rightarrow}{\partial}\\!\\!\\!\\!\phantom{\partial}_{\mu}-\overset{\leftarrow}{\partial}\\!\\!\\!\\!\phantom{\partial}_{\mu}-2igA_{\mu}(r)$
the symmetric covariant derivative, and
$a^{\\{\mu}b^{\nu\\}}=\frac{1}{2}(a^{\mu}b^{\nu}+a^{\nu}b^{\mu})$. The
corresponding decompositions of the other tensors follow automatically.

For a spin-$1/2$ state with mass $m$, the matrix elements of the Belinfante
EMT can be parametrized in general as Kobzarev:1962wt ; Pagels:1966zza ;
Ji:1996ek

$\langle
p_{f},\boldsymbol{s}_{f}|T^{\mu\nu}_{a}(0)|p_{i},\boldsymbol{s}_{i}\rangle=\overline{u}(p_{f},\boldsymbol{s}_{f})\Gamma^{\mu\nu}_{a}(\bar{p},\Delta)u(p_{i},\boldsymbol{s}_{i}),\qquad\qquad
a=q,g$ (53a) with
$\Gamma^{\mu\nu}_{a}(\bar{p},\Delta)=\frac{\bar{p}^{\\{\mu}\gamma^{\nu\\}}}{m}\,A_{a}(\Delta^{2})+\frac{\bar{p}^{\\{\mu}i\sigma^{\nu\\}\rho}\Delta_{\rho}}{2m}\,B_{a}(\Delta^{2})+\frac{\Delta^{\mu}\Delta^{\nu}-g^{\mu\nu}\Delta^{2}}{m}\,C_{a}(\Delta^{2})+mg^{\mu\nu}\bar{C}_{a}(\Delta^{2}).$
(53b)

Here $\boldsymbol{s}_{i}$ ($\boldsymbol{s}_{f}$) is the unit rest-frame
polarization vector associated with the initial (final) state. The EMT form
factors $A_{a},B_{a},C_{a},\bar{C}_{a}$ are Lorentz-invariant functions of
$\Delta^{2}$, which is the only independent Lorentz scalar formed with four-
momenta as indicated by the onshell constraints (35).

Using this parametrization, the standard expectation value of the quark and
gluon parts of the four-momentum operator reads Ji:1997pf

$\frac{\langle p,\boldsymbol{s}|P^{\mu}_{a}|p,\boldsymbol{s}\rangle}{\langle
p,\boldsymbol{s}|p,\boldsymbol{s}\rangle}=p^{\mu}A_{a}(0)+\frac{m^{2}}{p^{0}}\,g^{\mu
0}\bar{C}_{a}(0).$ (54)

The same expression is obtained using the phase-space expectation value
$\langle
P^{\mu}_{a}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}}$,
in agreement with Eq. (47). Four-momentum conservation then implies two
constraints

$\sum_{a}A_{a}(0)=1,\qquad\sum_{a}\bar{C}_{a}(0)=0.$ (55)

Note that the constraint for $\bar{C}$ holds also when $\Delta^{2}\neq 0$ as a
consequence of the conservation of $T^{\mu\nu}$.

Even though $|p,\boldsymbol{s}\rangle$ is in general not an eigenstate of the
AM operator888Surprisingly, it is stated in Refs. Ji:2020ena ; Ji:2020hii
that transverse AM does not commute with the QCD Hamiltonian. This is
obviously incorrect for otherwise the Hamiltonian would not be invariant under
rotations, breaking therefore Poincaré symmetry. The actual reason is that
transverse AM does not commute with the three-momentum operator
$[P^{i},J^{j}]=i\epsilon^{ijk}P^{k}$., the expectation value understood in the
sense of Eq. (48) is well-defined. Performing an expansion in $\Delta$ of the
parametrization (53) for the case
$\boldsymbol{s}_{f}=\boldsymbol{s}_{i}=\boldsymbol{s}$, one gets

$\displaystyle\langle\bar{p}+\tfrac{\Delta}{2},\boldsymbol{s}|T^{\mu\nu}_{a}(0)|\bar{p}-\tfrac{\Delta}{2},\boldsymbol{s}\rangle$
$\displaystyle=2\left[p^{\mu}p^{\nu}A_{a}(0)+m^{2}g^{\mu\nu}\bar{C}_{a}(0)\right]$
(56)
$\displaystyle\quad+\frac{i\Delta_{\lambda}}{m}\left[\left(p^{\mu}\epsilon^{\lambda\nu\alpha\beta}+p^{\nu}\epsilon^{\lambda\mu\alpha\beta}\right)\frac{A_{a}(0)+B_{a}(0)}{2}\right.$
$\displaystyle\quad+\left.\epsilon^{0\lambda\alpha\beta}\,\frac{p^{\mu}p^{\nu}A_{a}(0)+m^{2}g^{\mu\nu}\bar{C}_{a}(0)}{p^{0}+m}\right]\mathcal{S}_{\alpha}p_{\beta}+\mathcal{O}(\Delta^{2}),$

where we remind that
$p^{\mu}=(\sqrt{\boldsymbol{p}^{2}+m^{2}},\boldsymbol{p})$ with
$\boldsymbol{p}=\bar{\boldsymbol{p}}$, and
$\mathcal{S}^{\mu}(p,\boldsymbol{s})$ is the covariant polarization vector
defined in Eq. (31). We then obtain from Eq. (48) our master formula

$\displaystyle\langle\int\mathrm{d}^{3}r\,r^{i}T^{\mu\nu}_{a}(r)\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}}$
$\displaystyle=\mathcal{R}^{i}\,\frac{p^{\mu}p^{\nu}A_{a}(0)+m^{2}g^{\mu\nu}\bar{C}_{a}(0)}{p^{0}}$
(57)
$\displaystyle\quad+(\delta^{i}_{\lambda}-\delta^{0}_{\lambda}\,\tfrac{p^{i}}{p^{0}})\,\frac{p^{\\{\mu}\epsilon^{\nu\\}\lambda\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{p^{0}m}\,\frac{A_{a}(0)+B_{a}(0)}{2}$
$\displaystyle\quad-\frac{\epsilon^{0i\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{2m(p^{0}+m)}\,\frac{p^{\mu}p^{\nu}A_{a}(0)+m^{2}g^{\mu\nu}\bar{C}_{a}(0)}{p^{0}}.$

The first line corresponds to the contribution from the motion of the center
of the wave packet, the second line gives the tensorial part of the canonical
internal contribution, and the last line gives the non-tensorial part arising
due to the Wigner rotation Bakker:2004ib ; Lorce:2017isp . As stressed
earlier, in deriving this formula it is important to keep in mind that
$\Delta^{0}=\bar{\boldsymbol{p}}\cdot\boldsymbol{\Delta}/\bar{p}^{0}$ when
differentiating w.r.t. $\Delta^{i}$.

###  IV.1 Relativistic positions and spins

At $t=0$, the Poincaré generators read

$K^{i}=-\int\mathrm{d}^{3}r\,r^{i}T^{00}(r),\qquad
J^{i}=\epsilon^{ijk}\int\mathrm{d}^{3}r\,r^{j}T^{0k}(r).$ (58)

Using our master formula (57) in the rest frame defined by
$\boldsymbol{p}=\boldsymbol{0}$, we get

$\langle\boldsymbol{K}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{0},\boldsymbol{s}}=-m\boldsymbol{\mathcal{R}}\sum_{a}\left[A_{a}(0)+\bar{C}_{a}(0)\right],\qquad\langle\boldsymbol{J}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{0},\boldsymbol{s}}=\frac{\boldsymbol{s}}{2}\,\sum_{a}\left[A_{a}(0)+B_{a}(0)\right].$
(59)

It then follows from the constraints (55) derived from four-momentum
conservation that
$\langle\boldsymbol{K}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{0},\boldsymbol{s}}=-m\boldsymbol{\mathcal{R}}$.
Moreover, by definition of the rest-frame AM we must have
$\langle\boldsymbol{J}\rangle_{\boldsymbol{\mathcal{R}},\boldsymbol{0},\boldsymbol{s}}=\tfrac{1}{2}\boldsymbol{s}$,
so that AM conservation implies the additional constraint Kobzarev:1962wt ;
Ji:1997pf ; Bakker:2004ib

$\sum_{a}\left[A_{a}(0)+B_{a}(0)\right]=1,$ (60)

and hence the absence of anomalous gravitomagnetic moment $\sum_{a}B_{a}(0)=0$
Teryaev:1999su ; Brodsky:2000ii  when combined with Eq. (55). The same
constraints can alternatively be obtained from the requirement that
$\boldsymbol{J}$ and $\boldsymbol{K}$ be the generators of Lorentz
transformations in any frame Bakker:2004ib ; Lowdon:2017idv ; Cotogno:2019xcl
; Lorce:2019sbq .

Expectation values in an arbitrary frame, where
$\boldsymbol{p}\neq\boldsymbol{0}$, will always be understood in the sense of
Eq. (45). For convenience we will drop from now on the explicit phase-space
labels $\boldsymbol{\mathcal{R}},\boldsymbol{p},\boldsymbol{s}$. We find
Lorce:2018zpf

$\langle\boldsymbol{K}\rangle=-p^{0}\boldsymbol{\mathcal{R}}-\frac{\boldsymbol{p}\times\boldsymbol{s}}{2(p^{0}+m)},\qquad\langle\boldsymbol{J}\rangle=\boldsymbol{\mathcal{R}}\times\boldsymbol{p}+\tfrac{1}{2}\boldsymbol{s},$
(61)

Since we work with four-momentum eigenstates and implicitly symmetrized
products, we can write

$\langle
P^{\mu}\boldsymbol{K}\rangle\equiv\langle\tfrac{1}{2}(P^{\mu}\boldsymbol{K}+\boldsymbol{K}P^{\mu})\rangle=\langle\bar{p}^{\mu}\boldsymbol{K}\rangle=p^{\mu}\langle\boldsymbol{K}\rangle,$
(62)

and similarly for $\langle P^{\mu}\boldsymbol{J}\rangle$. We recover therefore
the well-known result that the expectation value of the Pauli-Lubański
pseudovector is proportional to the covariant polarization vector defined in
Eq. (31)

$\langle W^{\mu}\rangle=\frac{m}{2}\,\mathcal{S}^{\mu}(p,\boldsymbol{s}).$
(63)

Finally, we can express the external contribution for any choice of position
vector $\boldsymbol{R}$ as

$\langle\boldsymbol{R}\times\boldsymbol{P}\rangle=\langle\boldsymbol{R}\rangle\times\boldsymbol{p}=\langle\boldsymbol{R}\rangle\times\langle\boldsymbol{P}\rangle=\langle\langle\boldsymbol{R}\rangle\times\boldsymbol{P}\rangle.$
(64)

The expectation values of the three sets of position and spin operators
discussed in Sec. II can then easily be computed. The results are collected in
Table 1 and the relative positions of the three relativistic centers are
represented in Fig. 1.

Table 1: Expectation values of the main three position operators and of the corresponding spin operators. Relativistic center of | $\langle\boldsymbol{R}_{X}\rangle$ | $\langle\boldsymbol{S}_{X}\rangle=\langle\boldsymbol{J}\rangle-\langle\boldsymbol{R}_{X}\times\boldsymbol{P}\rangle$
---|---|---
energy ($X=E$) | $\boldsymbol{\mathcal{R}}+\frac{\boldsymbol{p}\times\boldsymbol{s}}{2p^{0}(p^{0}+m)}$ | $\frac{m}{2p^{0}}\left(\boldsymbol{s}+\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{m(p^{0}+m)}\right)$
mass ($X=M$) | $\boldsymbol{\mathcal{R}}-\frac{\boldsymbol{p}\times\boldsymbol{s}}{2m(p^{0}+m)}$ | $\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)$
spin ($X=c$) | $\boldsymbol{\mathcal{R}}$ | $\tfrac{1}{2}\boldsymbol{s}$
Figure 1: Illustration of the relative positions of the relativistic centers
of energy, mass and spin inside a transversely polarized nucleon (the arrows
indicate the local momentum seen by the observer). For simplicity, Lorentz
contraction effects are not represented. In the rest frame of the nucleon, the
three relativistic centers coincide. In a general moving frame they all
differ. In the infinite-momentum frame, the relativistic centers of energy and
spin merge and are found half a Compton wavelength away from the relativistic
center of mass.

There are many interesting observations to make. First, the explicit
calculation confirms that the center of the wave packet is nothing but the
relativistic center of spin Lorce:2018zpf , which lies on the segment joining
the relativistic centers of mass and energy. In the rest frame defined by
$\boldsymbol{p}=\boldsymbol{0}$, the three sets of center and spin coincide.
If we decompose three-vectors into longitudinal and transverse components
relative to $\boldsymbol{p}$

$\boldsymbol{V}=(\boldsymbol{V}\cdot\hat{\boldsymbol{p}})\,\hat{\boldsymbol{p}}+\boldsymbol{V}_{\\!\perp}$
(65)

with $\hat{\boldsymbol{p}}=\boldsymbol{p}/|\boldsymbol{p}|$, one sees that the
longitudinal components of position and spin (i.e. helicity) also coincide in
any frame

$\displaystyle\langle\boldsymbol{R}_{E}\rangle\cdot\hat{\boldsymbol{p}}$
$\displaystyle=\langle\boldsymbol{R}_{M}\rangle\cdot\hat{\boldsymbol{p}}=\langle\boldsymbol{R}_{c}\rangle\cdot\hat{\boldsymbol{p}}=\boldsymbol{\mathcal{R}}\cdot\hat{\boldsymbol{p}},$
(66) $\displaystyle\langle\boldsymbol{S}_{E}\rangle\cdot\hat{\boldsymbol{p}}$
$\displaystyle=\langle\boldsymbol{S}_{M}\rangle\cdot\hat{\boldsymbol{p}}=\langle\boldsymbol{S}_{c}\rangle\cdot\hat{\boldsymbol{p}}=\tfrac{1}{2}\boldsymbol{s}\cdot\hat{\boldsymbol{p}}.$

Differences appear only in the corresponding transverse components and
illustrate why transverse spin is always more complicated than longitudinal
spin. In particular, there are three different possible definitions of
transverse spin:

  * •

If one defines spin as the internal AM about the relativistic center of
energy, then transverse spin is Lorentz _contracted_

$\langle\boldsymbol{S}_{E\perp}\rangle=\gamma^{-1}\,\tfrac{1}{2}\boldsymbol{s}_{\perp}$
(67)

with $\gamma=p^{0}/m$ the Lorentz boost factor.

  * •

If one defines instead spin as the internal AM about the relativistic center
of mass, then transverse spin is Lorentz _dilated_

$\langle\boldsymbol{S}_{M\perp}\rangle=\gamma\,\tfrac{1}{2}\boldsymbol{s}_{\perp}$
(68)

This is expected since in this case internal AM transforms in the same way as
total AM, i.e. as the axial-vector part of an antisymmetric rank-two tensor,
whose transverse components are known to increase linearly with energy
Landau:1951 .

  * •

Finally, if one defines spin as the internal AM about the relativistic center
of spin, then transverse spin is _invariant_ under Lorentz boosts

$\langle\boldsymbol{S}_{c\perp}\rangle=\tfrac{1}{2}\boldsymbol{s}_{\perp}$
(69)

just like longitudinal spin, which is in agreement with the result found in
Ref. Bakker:2004ib .

It is seen that the different results for the expectation values of the three
kinds of spin operators, in the case of transverse spin, is simply due to the
fact that they correspond to different versions of what is meant by the
internal AM or “spin”.

Note that in the infinite-momentum frame $|\boldsymbol{p}|\to\infty$, the
relativistic center of spin merges with the relativistic center of energy and
is found half a Compton wavelength away from the relativistic center of mass
when the polarization is purely transverse Burkardt:2002hr ; Burkardt:2005hp ;
Lorce:2018zpf

$\lim_{|\boldsymbol{p}|\to\infty}\left(\boldsymbol{\mathcal{R}}-\langle\boldsymbol{R}_{M}\rangle\right)=\frac{\hat{\boldsymbol{p}}\times\boldsymbol{s}}{2m}=\frac{\hat{\boldsymbol{p}}\times\boldsymbol{s}}{2}\,\frac{\hbar}{mc}.$
(70)

It may then seem surprising that in this limit
$\langle\boldsymbol{S}_{E\perp}\rangle$ vanishes whereas
$\langle\boldsymbol{S}_{c\perp}\rangle$ remains non-zero for a state with non-
zero transverse polarization. The reason is that the difference in the
definition of the internal AM is given by

$\langle\boldsymbol{S}_{c}\rangle-\langle\boldsymbol{S}_{E}\rangle=\left(\langle\boldsymbol{R}_{E}\rangle-\langle\boldsymbol{R}_{c}\rangle\right)\times\boldsymbol{p}=\frac{(\boldsymbol{p}\times\boldsymbol{s})\times\boldsymbol{p}}{2p^{0}(p^{0}+m)}.$
(71)

In the infinite-momentum frame, the infinitely small transverse shift is
multiplied by an infinitely large momentum, leading to a finite difference.

###  IV.2 Spin sum rules

We are now ready to derive a generic spin sum rule. By this we mean a
decomposition of the expectation value of internal AM into quark and gluon
contributions

$\langle\boldsymbol{S}_{X}\rangle=\sum_{a}\langle\boldsymbol{S}_{X,a}\rangle,\qquad
X=E,M,c,$ (72)

where

$S^{i}_{X,a}\equiv\int\mathrm{d}^{3}r\,\epsilon^{ijk}\left(r^{j}-\langle
R^{j}_{X}\rangle\right)T^{0k}_{a}(r)=J^{i}_{a}-\epsilon^{ijk}\langle
R^{j}_{X}\rangle P^{k}_{a}.$ (73)

Using our master formula (57), we find that the expectation values of the
quark and gluon contributions to the total AM are given by

$\langle\boldsymbol{J}_{a}\rangle=\left(\boldsymbol{\mathcal{R}}\times\boldsymbol{p}+\frac{\boldsymbol{s}}{2}\right)A_{a}(0)+\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)B_{a}(0).$
(74)

It follows then from Eqs. (54) and (73) that

$\langle\boldsymbol{S}_{X,a}\rangle=\left[(\boldsymbol{\mathcal{R}}-\langle\boldsymbol{R}_{X}\rangle)\times\boldsymbol{p}+\frac{\boldsymbol{s}}{2}\right]A_{a}(0)+\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)B_{a}(0).$
(75)

Finally, using the results in Table 1 we obtain the key result of this work

$\langle\boldsymbol{S}_{X,a}\rangle=\langle\boldsymbol{S}_{X}\rangle
A_{a}(0)+\langle\boldsymbol{S}_{M}\rangle B_{a}(0).$ (76)

Summing over the quark and gluon contributions, we recover the constraints
$\sum_{a}A_{a}(0)=1$ and $\sum_{a}B_{a}(0)=0$ arising from Poincaré symmetry.
We show in the following that various spin sum rules discussed in the
literature simply correspond to particular cases of this remarkable simple-
looking formula.

####  IV.2.1 Helicity sum rule

First, if we restrict ourselves to the longitudinal component, we find the
unique situation that we get a sum rule _independent_ of the choice of
position vector, i.e. a sum rule with the quark and gluon contributions given
by

$\langle\boldsymbol{S}_{X,a}\rangle\cdot\hat{\boldsymbol{p}}=\frac{\boldsymbol{s}\cdot\hat{\boldsymbol{p}}}{2}\left[A_{a}(0)+B_{a}(0)\right].$
(77)

For a longitudinally polarized nucleon, we have in particular
$\boldsymbol{s}\cdot\hat{\boldsymbol{p}}=1$. This is nothing else than the
helicity sum rule derived in Refs. Ji:1996ek ; Ji:1997pf . Since the choices
$X=E,M,c$ all give the same result, this explains why the question of the
center of the nucleon did not attract much attention in the literature.

####  IV.2.2 Canonical spin sum rule

If we choose to work with the canonical position
$\langle\boldsymbol{R}_{c}\rangle=\boldsymbol{\mathcal{R}}$, we simply have to
set $X=c$ in Eq. (76). The quark and gluon contributions to the canonical spin
are then given by

$\langle\boldsymbol{S}_{c,a}\rangle=\frac{\boldsymbol{s}}{2}\,A_{a}(0)+\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)B_{a}(0).$
(78)

This expression was derived for the first time in Ref. Leader:2011cr  by
setting the origin at the center of the wave packet (i.e. the relativistic
center of spin), and hence computing
$\langle\boldsymbol{J}_{a}\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}=\langle\boldsymbol{S}_{c,a}\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}$
from Eq. (49). Our phase-space approach generalizes in a straightforward way
the derivation to an arbitrary choice of origin, i.e. to arbitrary
$\boldsymbol{\mathcal{R}}$, thanks to Eq. (50).

####  IV.2.3 Covariant spin sum rule

If we are interested in the internal AM about the relativistic center of mass,
it suffices to set $X=M$ in Eq. (76). We then find

$\langle\boldsymbol{S}_{M,a}\rangle=\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)\left[A_{a}(0)+B_{a}(0)\right],$
(79)

This is the only case where the contributions associated with the $A_{a}$ and
$B_{a}$ form factors have the same momentum dependence. In particular, even
though $\boldsymbol{S}_{M,a}$ is not conserved, its expectation value is
proportional to $\langle\boldsymbol{S}_{M}\rangle$ and hence transforms in a
covariant way. This has to do with the fact that $\langle R^{\mu}_{M}\rangle$
transforms as a Lorentz four-vector.

Focusing on the transverse components, we get

$\langle\boldsymbol{S}_{M\perp,a}\rangle=\gamma\,\frac{\boldsymbol{s}_{\perp}}{2}\left[A_{a}(0)+B_{a}(0)\right]$
(80)

with $\gamma=p^{0}/m$. This result was derived recently in Ref. Ji:2020hii  by
computing $\langle
J^{i}_{a}\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}=\epsilon^{ijk}\langle\int\mathrm{d}^{3}r\,r^{j}T^{0k}_{a}(r)\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}$
from Eq. (49) like in Ref. Leader:2011cr , but this time discarding terms
proportional to $p^{j}$ or $p^{k}$ in
$\langle\int\mathrm{d}^{3}r\,r^{j}T^{0k}_{a}(r)\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}$.
The authors motivated the latter prescription by the fact that the Pauli-
Lubański pseudovector $\langle
W^{\mu}\rangle=\frac{1}{2}\,\epsilon^{\mu\alpha\beta\lambda}\langle
M_{\alpha\beta}\rangle p_{\lambda}$ does not receive any contribution from
terms proportional to $p^{\alpha}$ or $p^{\beta}$ in $\langle
M^{\alpha\beta}\rangle$, which were accordingly interpreted as the
“contribution coming from the center-of-mass motion” of the nucleon.
Unfortunately, we did not find in Ref. Ji:2020hii  a clear definition of the
center of mass, nor a direct calculation of the contribution from its motion.
Moreover, the prescription applies only at the level of expectation values and
requires therefore a more rigorous justification from the operator level.

Our phase-space approach combined with the discussion about the relativistic
definition of the center of the nucleon from Sec. II justifies and clarifies a
posteriori the above ad hoc prescription. Indeed, if we consider the
expectation value of spatial moments about the relativistic center of mass, we
obtain from the master formula (57)

$\displaystyle\langle\int\mathrm{d}^{3}r\left(r^{i}-\langle
R^{i}_{M}\rangle\right)T^{\mu\nu}_{a}(r)\rangle$
$\displaystyle=\left(\mathcal{R}^{i}-\langle
R^{i}_{M}\rangle-\frac{\epsilon^{0i\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{2m(p^{0}+m)}\right)\frac{p^{\mu}p^{\nu}A_{a}(0)+m^{2}g^{\mu\nu}\bar{C}_{a}(0)}{p^{0}}$
(81)
$\displaystyle\quad+(\delta^{i}_{\lambda}-\delta^{0}_{\lambda}\,\tfrac{p^{i}}{p^{0}})\,\frac{p^{\\{\mu}\epsilon^{\nu\\}\lambda\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{p^{0}m}\,\frac{A_{a}(0)+B_{a}(0)}{2}.$

According to Table 1, we can write

$\mathcal{R}^{i}-\langle
R^{i}_{M}\rangle=\frac{(\boldsymbol{p}\times\boldsymbol{s})^{i}}{2m(p^{0}+m)}=\frac{\epsilon^{0i\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{2m(p^{0}+m)}$
(82)

showing that the contribution from the motion of the center of the wave packet
relative to the center of mass compensates exactly the part associated with
the Wigner rotation. The first line in Eq. (81) then vanishes identically and
we are left with

$\langle\int\mathrm{d}^{3}r\left(r^{i}-\langle
R^{i}_{M}\rangle\right)T^{\mu\nu}_{a}(r)\rangle=(\delta^{i}_{\lambda}-\delta^{0}_{\lambda}\,\tfrac{p^{i}}{p^{0}})\,\frac{p^{\\{\mu}\epsilon^{\nu\\}\lambda\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{p^{0}m}\,\frac{A_{a}(0)+B_{a}(0)}{2}.$
(83)

In particular, for the AM we have

$\epsilon^{ijk}\langle\int\mathrm{d}^{3}r\left(r^{j}-\langle
R^{j}_{M}\rangle\right)T^{0k}_{a}(r)\rangle=\epsilon^{ijk}\left[\frac{\epsilon^{kj\alpha\beta}\mathcal{S}_{\alpha}p_{\beta}}{2m}\,\frac{A_{a}(0)+B_{a}(0)}{2}\right],$
(84)

where the expression inside the square brackets is effectively what one
obtains from applying the ad hoc prescription to
$\langle\int\mathrm{d}^{3}r\,r^{j}T^{0k}_{a}(r)\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}$,
as was done in Ref. Ji:2020hii . We note however that applying naively the ad
hoc prescription to
$\langle\boldsymbol{K}_{a}\rangle_{\boldsymbol{0},\boldsymbol{p},\boldsymbol{s}}$
would give a contribution from $\bar{C}_{a}(0)$, while the correct expression
derived from Eq. (83) does not.

####  IV.2.4 A new spin sum rule

For completeness, let us finally consider the internal AM about the
relativistic center of energy. Setting $X=E$ in Eq. (76), we get

$\langle\boldsymbol{S}_{E,a}\rangle=\frac{m}{2p^{0}}\left(\boldsymbol{s}+\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{m(p^{0}+m)}\right)A_{a}(0)+\frac{p^{0}}{2m}\left(\boldsymbol{s}-\frac{\boldsymbol{p}(\boldsymbol{p}\cdot\boldsymbol{s})}{p^{0}(p^{0}+m)}\right)B_{a}(0),$
(85)

To the best of our knowledge, this result is derived for the first time.

Since $\boldsymbol{S}_{E}=\boldsymbol{W}/P^{0}$, one may wonder whether this
new spin sum rule is related to the sum rule based on the Pauli-Lubański
pseudovector proposed in Ref. Ji:2012vj . Beside the factor of $1/P^{0}$, it
appears that the decomposition into quark and gluon contributions is in fact
different. In Ref. Ji:2012vj  the Pauli-Lubański pseudovector is decomposed as

$W^{\mu}=\sum_{a}\mathcal{W}^{\mu}_{a}$ (86)

with the quark and gluon contributions defined as999Note that
$\mathcal{W}^{\mu}_{a}$ is not the Pauli-Lubański pseudovector of the quark
subsystem since the latter would involve $P^{\mu}_{a}$ instead of $P^{\mu}$.

$\mathcal{W}^{\mu}_{a}=\frac{1}{2}\,\epsilon^{\mu\alpha\beta\lambda}(M_{a})_{\alpha\beta}P_{\lambda}.$
(87)

In other words, one has

$\frac{\boldsymbol{\mathcal{W}}_{\\!a}}{P^{0}}=\boldsymbol{J}_{a}+\frac{\boldsymbol{K}_{a}\times\boldsymbol{P}}{P^{0}}$
(88)

whereas we defined

$\boldsymbol{S}_{E,a}=\boldsymbol{J}_{a}+\frac{\langle\boldsymbol{K}\rangle\times\boldsymbol{P}_{a}}{P^{0}}$
(89)

as follows from Eqs. (73) and (64). Clearly, the quark and gluon contributions
to $\langle\boldsymbol{S}_{E}\rangle$ are defined relative to the center of
energy of the whole system and hence involve $\langle\boldsymbol{K}\rangle$.
In contrast, $\langle\boldsymbol{\mathcal{W}}_{\\!a}\rangle$ involves only
$\langle\boldsymbol{K}_{a}\rangle$ which is related to the energy dipole
moment arising from the quark or gluon subsystem Lorce:2018zpf . This explains
in particular why $\bar{C}_{a}(0)$ appears in
$\langle\boldsymbol{\mathcal{W}}_{\\!\perp,a}\rangle$ Leader:2012ar ;
Hatta:2012jm ; Harindranath:2013goa ; Leader:2013jra  and not in
$\langle\boldsymbol{S}_{E\perp,a}\rangle$.

##  V Conclusions

A spin sum rule is usually understood as a decomposition of the angular
momentum about the _center_ of the nucleon into quark and gluon contributions.
While the longitudinal spin or helicity sum rule is well established,
different transverse spin sum rules have been proposed in the literature. In
this paper we show that there is no contradiction between the published
results, since they correspond to different definitions of what constitutes
the center of the nucleon. Transverse angular momentum is always more
complicated in a relativistic theory because, unlike longitudinal angular
momentum, it commutes neither with longitudinal boosts nor momentum.

In the literature, spin sum rules are often obtained from the expectation
value of the angular momentum operator. The latter has to be treated with care
since a naive calculation using momentum eigenstates leads to ambiguous or
divergent results. Some papers also make use of the Pauli-Lubański
pseudovector but have trouble with providing a physical meaning to the
contribution coming from the boost generators, and discard terms asserted to
arise from the motion of the center of mass without a clear definition of the
latter.

In this work, we stress the importance of defining clearly the center of the
nucleon. We review the various possibilities offered in the relativistic
context (namely the relativistic centers of energy, mass and spin), and remind
in passing that boost generators provide the information about the position of
the center of energy. We then define expectation values from a phase-space
perspective. This allows us to avoid ambiguities encountered in the standard
approach, and to compute for the first time the contribution from the motion
of the center of the nucleon.

Our key result is found in Eq. (76) which, together with Table 1, gives the
generic expression for the quark and gluon contributions to the nucleon spin,
based on the three possible definitions of the center of the nucleon.
Projecting on particular spin components with particular choices for the
center of the nucleon, we reproduce several spin sum rules reported in the
literature. In particular, we show that the new sum rule obtained by Ji and
Yuan Ji:2020hii  is not in contradiction with the older sum rule derived by
Leader Leader:2011cr , but simply relies on a different definition for what is
meant by internal angular momentum or spin. Namely, if one requires spin to
generate rotations about the center of the nucleon, then one is forced to work
with the canonical spin operator, leading to the Leader sum rule. If one
requires instead spin to transform in a covariant way, then one is forced to
work with the angular momentum about the relativistic center of mass, leading
to the Ji and Yuan sum rule. We note that a third sum rule can be obtained
based on the angular momentum about the relativistic center of energy or
inertia, which has the simplest expression in terms of the Poincaré
generators.

###### Acknowledgements.

I am very grateful to Elliot Leader for many valuable comments and suggestions
about the manuscript.

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