No, there's no controversy. It's just something that comes directly out of the symplectic classification associated with the symmetry group. For Relativity, that's the Poincaré group, and in the non-relativistic limit, it's the Galilei group and its central extension: the Bargmann group. In order to have consistency with the "correspondence principle" and to have a consistent non-relativistic limit, you should actually also be doing the symplectic decomposition with the (trivial) central extension of the Poincaré group, rather than the Poincaré group itself.
This is the picture that emerges:
Tardyons have 3 pairs of symplectic coordinates - which is what underlies the Heisenberg relation - if spin 0, and a 4th pair if spin non-zero. The extra pair are associated with the coordinates of a sphere. It corresponds to the degree of freedom usually labelled by $m$ (for the axial component of angular momentum) in quantum theory.
Luxons have 3 pairs of symplectic coordinates, if they belong to the sub-class which I've termed "helical". In the helical subclass, angular momentum is helical; i.e. its axis is parallel to linear momentum; and the helicity (i.e. the component parallel to the momentum) is an invariant. As a special case, this includes "spin 0" - which actually means helicity 0. The photon is a helical luxon, and it has non-zero helicity.
The luxons that don't fall within the helical sub-class have 4 pairs of symplectic coordinates. Unlike tardyons, the extra pair does not describe spin but is associated with the coordinates of a cylinder. This larger, and more generic, sub-family of luxons, I've termed "cylindrical".
There is no consensus on what, if anything, a cylndrical luxon may correspond to.
Tachyons have 3 pairs of symplectic coordinates only if they are "spin 0". Otherwise, they have 4; and the 4th pair is associated with either a one-sheeted hyperboloid, a double cone, or two-sheeted hyperboloid.
So, there are three subfamilies in the tachyon class. Again, there is no consensus on what (if anything) they would correspond to.
All "spin 0" classes have a well-defined canonical position vector $$ and the angular momentum $$ and moving mass moment $$ have the same decomposition as in non-relativistic theory: $ = ×$ and $ = M - t$, where $$ and $M$ the ("moving") mass (in relativity, "total energy" $E = Mc^2$ is often used in place of $M$).
The parameter $t$ is arbitrary. For a free non-interacting system, all the canonical quantities are constant, and the constancy of $$ entails the constancy of $ = d/dt = /M$, which is just a back-door way of expressing The Law Of Inertia.
All of what I've described applies both relativistically and non-relativistically, except that the tachyon and luxon classes - for the non-relativstic version - converge into an unnamed class I've termed the "synchron". (Its defining property is $M = 0$, but $ ≠ $.) Like the luxons, they also have a helical sub-family - inherited as the non-relativistic limit of the helical luxon class.
For all cases, $M^2 - α P^2$ is an invariant, with the distinction between the relativistic and non-relativistic cases being drawn, respectively, by $α = 1/c^2 > 0$ versus $α = 0$. The distinction between tardyon, luxon and tachyon rests entirely in the sign of this invariant.
If $M^2 - α P^2 > 0$, then they are tardyons and are distinguished by the property that they have both a center of mass and a rest frame in which $ ≠ $. The value of $M$ in that frame is $m$ - the "rest mass" and $M^2 - α P^2 = m^2$. For the non-relativistic case $M$, itself, is the rest mass $m$ and is invariant.
If $M^2 = α P^2$ and $ ≠ $, then it's a luxon (which, in the non-relativistic case, becomes a synchron with $M = 0$). They have no rest frame, and therefore the term "rest mass" makes no sense for them. But since $M^2 - α P^2 = 0^2$, it's common practice to wing it and just say that their "rest [sic] mass" is 0. However, referring to what follows, you could treat them like "tachons with zero impulse", just as well as "tardyons with zero rest mass". They're a limit case of both classes.
In the non-relativistic limit, $Π = |P|$ is an invariant and may be considered as an "impulse", with a synchron being that which conveys an instantaneous action-at-a-distance transfer of impulse; i.e. the instantaneous version of what is traditionally called a "line of force". They have infinite speed and exist only at an instant. They don't move in time at all, but exist in space as purely spatial objects. The "time of existence" is canonical for them, and is given by $t = -·/P^2$, and is an invariant for this class.
If $M^2 = α P^2$, trivially, with $ = $ and $M = 0$, then that's the "homogeneous" class, which I'll describe further below.
If $M^2 < α P^2$, then it's a tachyon. This can only happen in the relativistic case $α > 0$. They have no rest frame, so the term "rest mass" makes no sense here, either. Instead, it's the impulse $Π^2 = P^2 - M^2/α$ that has meaning. That's the momentum square magnitude in any "synchron" frame of reference where the tachyon is instantaneous at infinite speed. In an infinite speed frame, for tachyons, $M = 0$, like the synchron.
What makes all this important is that it runs central to the issue of the spin-orbit decomposition.
For the spin non-zero classes, you need a spin-orbit decomposition in order to even be able to talk about spin. And that only happens with the tardyon class, with it being given by
$$ = × + , \hspace 1em = M - t + \frac{α×}{m + M}.$$
You can't write down anything meaningful, without $$ (and $m$).
If the latter of these relations is solved for $$, you get $ = ( + t)/M$, plus the extra contribution $-α×/(M(m+M))$ that is entirely relativistic and spin-dependent - and together this gives you the classical version of what is called the Newton-Wigner position vector.
There's nothing like this for tachyons, and there is only marginally something like this for luxons ... only in the case of the helical subclass, where you can sorta write down a position operator (notwithstanding the "no position operator" theorem in the literature you may see mis-used as a folklore theorem on its "impossibility"). But there's no spin for the helical sub-class, and no 4th pair of symplectic coordinates at all. It's just 3 pairs (given by the components of $$ and $$ themselves), and they satisfy the same complementarity relation as in the spin 0 case.
There is also the homogeneous class. They have no center of mass, and (in fact) are invariant with respect to spatial and temporal translations, but may have non-zero $$ and $$. For relativity, the vectors $_± = ± /\sqrt{-α}$ are each analogous to spin, except that they are complex, because $α > 0$. So, there are 2 pairs of symplectic coordinates; unless both vectors vanish (i.e. if $ = $ and $ = $), in which case they are also isotropic and boost-invariant (i.e. invariant with respect to change in the moving frame of reference) and have 0 symplectic coordinates. An "isotropic, boost-invariant, homogeneous" system is just another way of describing a "vacuum", and may be taken as the definition of such. So, the general case may be thought of as a "vacuum with spin". For the non-relativistic case, there may be 0, 1 or 2 sets of symplectic coordinates; the case of 0 being (again) the vacuum, the case of 1 being where $ = $ and $ = ≠ $ (with $S^2$ then being an invariant), and the case of 2 being where $ ≠ $, with $K^2$ and $·$ both being invariant. In the relativistic case, for the homogeneous class, $·$ and $K^2 - α J^2$ are the invariants, and one can distinguish 3 sub-classes based on the sign of the latter invariant, with the $K^2 - α J^2 = 0$ case being the relativistic version of the above-mentioned $ = $, $ ≠ $ non-relativistic sub-class.
If you were to work out the symplectic decomposition in detail, what you would end up finding out is that - like the Wigner classification - the vectors $ = M + ×$ and $W_0 = ·$ actually play a central role in everything. They are what, together, comprise 4 components what is known as the Pauli-Lubanski vector. There's also actually good reason to also treat $W_4 = ·$ as a 5th component, but it is not really needed here. They have an invariant: $W^2 - α {W_0}^2$ and the vector satisfies the identity $· = M W_0$, which limits its number of independent components to just 2 (plus $W_4$ as a third orphan component).
For the so-called "spin 0" family, this invariant is always 0, because $ = $ and $W_0 = 0$, for this class. It's actually from there, that you get the trivial spin-orbit decomposition: you just define $ = ( + t)/M$, where $t$ can be arbitrarily chosen, and substitute into $ = $ to solve for $$ and get $ = -×/M = ×/M = ×$.
For the helical families, the invariant is still 0, but $W_0 ≠ 0$. This can only happen in the non-relativistic case with synchrons, or in the relativistic case with luxons. In all cases, $W_0$ is an invariant with $$ being parallel to $$. The "helicity" (which is defined as the component of angular momentum along an axis parallel to $$) is just $W_0/P$, itself and is invariant.
For the tardyons, the invariant $W^2 - α {W_0}^2$ is always 0 or positive, and if it is positive, then it is $W^2 - α {W_0}^2 = m^2 S^2$: the square of the product of rest-mass and spin. In that case, the independent components of $(, W_0)$ trace out a sphere of radius $m S$.
For luxons, the invariant is 0 (for the helical subfamily) or positive (for the larger cylindrical family). In the latter case, the independent components of $(, W_0)$ trace out a cylinder of radius $\sqrt{W^2 - α {W_0}^2}$.
For tachyons, the invariant $W^2 - α {W_0}^2$ may be positive, zero or negative; and the independent components of $(, W_0)$ trace out a one-sheeted hyperboloid (if positive), a double cone (if zero) or a two-sheeted hyperboloid (if negative).
So, what corresponds to "spin" for tardyons, morphs into "cylindrical" something or another when you pass through the boundary case of luxons, into something totally different, involving hyperboloids or double-cones, by the time you get to tachyons. I have no idea what it represents or corresponds to; and I don't think any consensus has ever been arrived at in the literature, either, on what it ought to represent.
I worked much of this out in more explicit detail here
What's the physical meaning of the statement that "photons don't have positions"?
in the process of addressing the issues of what corresponds to a "spin-orbit decomposition" and "position operator" for helical luxons. So, you might consider this an addendum to that earlier reply.
