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Do other particles besides scalars admit tachyonic solutions? For example fermions or gauge-boson tachyons? The picture in my head is that a tachyonic scalar simply rolls off some unstable potential until it finds a stable position in field space (so the higgs field is basically a tachyon until it condenses). But I don't see a similar picture for fermions or gauge bosons (they have higher dimensional operators, but not a potential in the same sense as a scalar particle). Furthermore for fermions I believe we can just perform a chiral phase transformation to get rid of any imaginary part of the mass term.

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The second bounty is for jjcale. Other people can add answers to it, but I don't think I will be awarding any of there bounties for this question. – DJBunk Jul 27 '12 at 1:44

The case for the fermion is sneaky because the fermion field is (anticommuting) Grassman-valued. The Grassman numbers can't get 'big enough' to roll away. But more concretely speaking, supposed we tried to construct such a Lagrangian: $$ \mathcal{L}=\bar\psi(i\gamma^\mu\partial_\mu)\psi+im\bar\psi\psi $$ where the fermion mass term comes in with 'apparently' the wrong (tachyonic) sign. But as you pointed out, upon performing a chiral rotation: $\psi\rightarrow e^{i\alpha\gamma_5}\psi$ by 180 degrees 90 degrees, you can recover the standard form for the Dirac lagrangian. [this fact actually plays an interesting role chiral anomalies and in CP violation in non-abelian gauge theories]

The case for massless vector fields is another sneaky story. In this case, the vector field is endowed with a gauge symmetry, a subset of which is a shift symmetry: $$ A_\mu(x)\rightarrow A_\mu(x)+b_\mu $$ where $b_\mu$ is a constant space-time vector. So, a field that has 'rolled away' to large values carries the same energy as one with small field value. (In fact they are physically equivalent by the gauge principle).

The only case remaining now is the massive spin-1 field (the Maxwell-Proca field) with a wrong-sign mass term: $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu} F^{\mu\nu}-\frac{1}{2}m^2A_\mu A^\mu $$ In this unfortunate case, gauge symmetry is lost, and the energy is unbounded from below. So I guess this counts as being tachyonic. I should say, however, that most realistic models generate gauge boson masses via spontaneous symmetry breaking to retain the Ward identities. And in these cases, the massive vector boson always acquire a positive mass thanks to real-valued gauge coupling constants.

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A tachyonic Dirac field needs to have an imaginary value for m, not a negative one. I don't see a real connection with anomalies, other than that the mass term sign is not chirally invariant. – Ron Maimon Jul 22 '12 at 9:59
Point taken, and Lagrangian duly corrected. Thanks! The connection with anomalies is irrelevant to the original question. I added that remark for contextual interest only. – QuantumDot Jul 23 '12 at 0:19
@QuantumDot So you are saying that 1) there are no tachyonic fermions since a field redefinition removes the 'unstable' mass? and that 2) Nothing will stabilize a tachyonic massive spin 1 field ? – DJBunk Jul 25 '12 at 13:50
@DJBunk I guess my post does imply 'yes' to your question (1). But to question (2), I should clarify that as written, the massive spin-1 field has a tachyonic direction. But that is not to say that nothing will stabilize it. [Also, all analysis so far is classical. There are quantum mechanical effects that can lead to unstable modes in fluctuations of composite fields i.e. the chiral condensate in QCD]. – QuantumDot Jul 26 '12 at 0:48

For a classification of the unitary representations of the Poincare group see

Here also tachyonic representations are described : Beside the simple scalar field there are also infinite-component tachyonic fields possible corresponding to non-trivial representations of the little group SO(D−2,1) .

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I started a second bounty to give to you for the reference, there is just a waiting period before I can assign it. – DJBunk Jul 27 '12 at 1:33

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