# Is conservation of statistics logically independent of spin?

If the number of fermions is $n$, we expect the quantity $(-1)^n$ to be conserved, i.e., $n$ never changes between even and odd. This is known as conservation of statistics. In the normal context of particles with the properties traditionally expected for fermions and bosons, it can be derived from conservation of angular momentum. If $n$ were to change from, say, even to odd, then the total angular momentum of the system would have to change from an integer to a half-integer.

In the context of tachyons, there's an odd twist, however. The spin-statistics theorem is based on the assumption that the field has to commute at points that are spacelike in relation to one another. This commutation relation wouldn't hold for tachyons, so they get a free pass on spin-statistics. In fact Feinberg, in the classic paper that introduced the term "tachyon," found that the most plausible possibility was that tachyons would be spin-0 fermions (Feinberg 1967).

Feinberg says on p. 1099, discussing selection rules,

[...]there are restrictions following from the conservation of statistics. I shall simply assume here that if we assign a number $+1$ for bosons and $-1$ to fermions and multiply these numbers for a multiparticle system that these products are conserved in any transition.[16]

Footnote 16 says,

See, for example, Greenberg and Messiah (Ref. 10) where this is proven, however, under assumptions that may be invalid for tachyon theories.

Given Feinberg's assumption, we have two separate conservation laws, conservation of statistics and conservation of angular momentum, and the result is that if tachyons are unique in combining integer spin with Fermi statistics, they have a special conservation law that only applies to them: if $t$ is the number of tachyons, then $(-1)^t$ is conserved. This means, for example, that you can't produce just one tachyon, and you also can't have a process that produces a tachyon plus an electron, even though in normal field theory it's perfectly OK to produce two unlike fermions.

Does conservation of statistics really have some independent logical status, or is Feinberg just making an assumption that could be false, but that might make the field theory he's constructing easier to work with or more familiar? The understanding of tachyons in QFT has progressed a lot since 1967, so I wonder if this issue is better understood today than it was then.

G. Feinberg, "Possibility of Faster-Than-light Particles". Phys. Rev. 159 no. 5 (1967) 1089. Copy available here (may be illegal, or may fall under fair use, depending on your interpretation of your country's laws).

## 1 Answer

There's no related exception for tachyons. Tachyons' statistics must be determined a priori. Most typically, tachyons have to be bosons – and under certain additional assumptions, they have to be scalar (spin-zero) bosons. They differ from massive bosons just by the fact that the mass term $m^2\phi^2/2$ has the opposite sign – opposite sign of their $m^2$.

It's also untrue that the tachyon fields don't commute at spacelike separations. They do. One may derive this statement from the canonical commutation relations.

One must be careful about the mode expansion for tachyons. One could naively talk about tachyon modes that are combinations of $\exp(ip\cdot x)$ for spacelike real momenta $k$ to obey $k^2=m^2\lt 0$. But that's not the right approach. The right approach is to keep the usual oscillating dependence on the spatial dimensions, $\exp(i\vec k \cdot \vec x)$, and choose $k^0$, the energy, in such a way that $k^2=m^2$ is obeyed. This will force us to have an imaginary $k^0$ which means that the tachyon field will be composed of the exponentially decreasing and the exponentially increasing term (as functions of time).

If the increasing term is there, it will soon win so at very late times, the tachyon field looks like an exponentially growing instability.

It is not quite possible to identify the exact vacuum in the presence of tachyons – it's some state near the configuration where the tachyon field sits near the maximum but it's unstable so there's no Gaussian-like wave functional around that point of the configuration space. But if we do define some states, it will still be true that the angular momentum is conserved and the statistics is conserved as the modulo 2 consequence of it.

• So in your opinion is the Feinberg paper just totally wrong? Wrong for 1967, or just wrong with hindsight in 2013? There's a lot of interesting material in your post, but very little that addresses the question, which is about the logical foundation of conservation of statistics. "There's no related exception for tachyons." Do you mean no exception to the spin-statistics theorem? Related to what? Is Feinberg mistaken? "One may derive this statement from the canonical commutation relations." Feinberg states that the canonical commutation relations don't and can't apply. Is his reasoning wrong? – Ben Crowell May 9 '13 at 15:45
• I think my answer speaks for itself and it's clear what the answers to your repeated questions are. It may be better to avoid repetitiveness; and to avoid writing things that you preemptively indicate to be irritated by. Otherwise canonical commutation relations always hold in any theory that may be derived by a quantization from its classical limit. Saying anything else means not to obey any rules or standards. Also, there can't be spin-1/2 tachyons because no Dirac equation can lead to $m^2\lt 0$. And yes, almost everything people said about tachyons in QFT in the 1960s was misguided. – Luboš Motl May 9 '13 at 19:06