# Spinning Tachyons

In all examples that I know, tachyons are described by scalar fields. I was wondering why you can't have a tachyon with spin 1. If this spinning tachyon were to condense to a vacuum, the vacuum wouldn't be Lorentz invariant---seems exotic but not a-priori inconsistent. Is there some stronger consistency requirement which rules out spinning tachyons? If someone could provide a reference that would be helpful too!

Here's another confusion: I was reading Wikipedia, which claims that tachyons should be spinless and obey Fermi-Dirac statistics(?). (They reference an original paper by G. Feinberg which unfortunately I am not wealthy enough to download). The claim about Fermi-Dirac statistics is baffling---isn't the Higgs field a boson? Does anyone understand what they're talking about?

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Higgs boson is not a tachyon, so what are you talking about? Also, spin-statistics theorem holds only for states with non-negative mass and also under other assumptions. As soon as any of those is violated more exotic possibilities arise. See the list of assumptions here: en.wikipedia.org/wiki/Spin-statistics_theorem#Proof – Marek Feb 21 '11 at 11:17
The Higgs is a boson, but I'm not sure how that's related to your question. The Higgs is not a faster-than-light particle. – Tim Goodman Feb 21 '11 at 11:18
The Higgs particle is not a tachyon, but appears when expanding the Higgs field around the vacuum of spontaneously broken symmetry. Around the vacuum where the Higgs field has vanishing expectation value, the full gauge symmetry is restored but the Higgs field has negative mass^2. – truebeliever1234 Feb 21 '11 at 11:57
The Higgs particle is a tachyon in the only sense that modern field theory accepts--- a particle which makes an unstable vacuum for the zero-charge state. – Ron Maimon Dec 31 '11 at 11:57
– Ben Crowell May 5 '13 at 2:13

It's not completely self-evident but it is true that in consistent theories, tachyons have to be scalar particles - much like the Higgs boson when expanded around the maximum of the potential (zero vev) - which obey, of course, Bose-Einstein statistics. (The claim about Fermi-Dirac is just wrong, or was meant to apply to Faddeev-Popov ghosts or similar fields, not physical tachyons.)

In non-interacting string theory, one sees that this conclusion is true because the ground state energy of a single-string Hilbert space is equal to $L_0=-1$ for the scalar tachyon, so any addition of spin - through the string oscillators - raises $L_0$ at least by one, bringing us to the massless or massive level (non-negative $m^2$).

$L_0=-1$ was true for the bosonic string ground state; in the case of the superstring, using the RNS formalism, the ground state has $L_0=-1/2$ but we also have antiperiodic fermions that only raise $L_0$ by $1/2$: that's still enough to show that any addition of spin - which only comes via internal oscillators - brings us to the massless or massive level, above the tachyonic interval.

The scalar character of the tachyon may also be seen in the effective field theory. Dirac or Weyl tachyons are impossible because the Dirac mass term $$-m \bar \psi \psi$$ has to be Hermitian. It implies that $m$ has to be real, which means that the particle is positively massive. A tachyonic fermion would need an imaginary $m$ but that would produce a non-Hermitian action.

The same is true for spin-one particles. Spin-one particles may only get their mass consistently by the Higgs mechanism: the relevant term arises from the covariant version of the kinetic terms for the Higgs fields: $$D_\mu \phi^\dagger D^\mu \phi$$ Again, this has to be Hermitean, and if it is Hermitean, $\phi^\dagger \phi$ that is left if the Higgs field has a nonzero vev, is automatically positively definite, which produces the usual mass term $$m^2 A_\mu A^\mu/2$$ with a positive coefficient.

Concerning a tachyonic vev, well, any vev of a tachyonic field that solves the equations of motion has to be Lorentz-breaking - because it is a non-constant function of spacetime. The spin of a tachyon would just add one more aspect to this story. But it is intuitively natural that the tachyons have to be scalars - the value of the tachyon away from the maximum of the potential measures "how much the instability has already advanced", and this quantity is naturally a scalar.

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Thank you for your answer Lubos! You claim that massive vectors can only be made to consistently interact via the Higgs mechanism (or maybe stueckelburg.) I have heard this before, but I am wondering if there is a reference I can visit to understand this myself. The claim in wikipedia about tachyons being fermions seems unbelievable.. but I was curious about where this strange idea came from. – truebeliever1234 Feb 21 '11 at 14:49
BTW, if a scalar tachyon acquires constant expectation value it does not break Lorentz symmetry. On the other hand, if a vector acquires constant expectation value, it manifestly does. This is what I was referring to in my question. – truebeliever1234 Feb 21 '11 at 14:54
Dear @truebeliever1234, a tachyon can't have a constant (in spacetime) vev because a constant vev doesn't satisfy the equations of motion, $(\nabla_\mu \nabla^\mu + m^2)T=0$. The first term with the derivatives drops for a constant $T$ and the second one doesn't, so the equation is violated. That's the point of tachyons that instead of constants, you must either have waves moving in spacelike directions, or exponentially growing/decreasing functions of time. Constants are not allowed. – Luboš Motl Feb 21 '11 at 15:15
@truebeliever Have a look at the following paper and papers that cite it. As I recall, some of what they found in this paper was later reinterpreted by Sen&Zwiebach in terms of decay of unstable D-branes. The stuff on Lorentz violation probably involves an uncontrolled approximation and as far as I know has not been backed up by later studies. Still, the idea is intersting. Spontaneous Breaking of Lorentz Symmetry in String Theory. V.Alan Kostelecky, (Indiana U.) , Stuart Samuel, (City Coll., N.Y.) . IUHET-139, CCNY-HEP-88/4, May 1988. 8pp. Published in Phys.Rev.D39:683,1989. – pho Feb 21 '11 at 21:27
I agree with Lubos too. ;) – pho Feb 22 '11 at 13:47

If the polarizations of a tachyonic vector boson are Lorentz covariant, then its norm has to be indefinite, not positive definite, or positive semidefinite as in the case of massless vector bosons. This is true despite the fact that the polarizations have to be transverse to the 4-momentum because the 4-momentum is spacelike. Negative probabilities don't make any sense.

Tachyonic scalars have to be bosons, not fermions!

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I have seen this argument, but I don't believe it holds water since the ghost polarizations only appear if you assume that the spin-1 tachyon must have a particle interpretation---specifically that the field can be expanded in Fourier modes with spacelike 4-momenta. However, in all physical applications of tachyons the 4-momentum is actually timelike, and then the mass-shell condition implies that the field grows exponentially in time. This is a signature of instability of the vacuum. Under this assumption you find that the free spin-1 tachyon does not propagate ghosts. – truebeliever1234 Feb 21 '11 at 15:42
If you perform the Fourier transform of the spatial wave vectors, then if $m^2 = - k^2$, when the magnitude of the spatial wave vector exceeds $k$, we still have spacelike 4-momenta. – QGR Feb 21 '11 at 16:00

Your question was solved by Eugene Wigner when he classified all possible representations of the Poincare group, i.e. all possible groups that can be translated in space or time, and boosted by changes of velocity in relativity.

Wigner found the invariants of the Poincare group, first the mass, and second the spin. Furthermore he found the allowed spins which depend on the character of the mass.

For m^2>0, spin = 0,1/2,1,3/2,... and Polarization may be measured from p = -s,1-s ... s

For m^2=0, spin = 0,+/-1/2,+/-1,+/-3/2 which are also are the possible measureable polarizations

When m^2<0, the only allowed representation of spin are the trivial spin=0, and a set of infinite dimensional groups. So there a no finite spin groups that describe tachyons. What the infinite dimensional groups allow you to do, I do not know, the math get somewhat hard there.

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This assumes that the result of tachyonic condensation is Lorentz invariant, so that Wigner's classification applies. – Ron Maimon Dec 31 '11 at 11:59