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How to show that higher derivative theories (mostly) breaks unitarity?

Spinor field $\psi_{a_{1}...a_{n}\dot {b}_{1}..\dot {b}_{m}} $, which refer to the $\left( \frac{n}{2}, \frac{m}{2} \right)$ representation of the Lorentz group, must satisfy

$$ (\partial^{2} + M^{2})\psi_{a_{1}...a_{n}\dot {b}_{1}..\dot {b}_{m}} = 0 , \quad \partial^{\dot {b}_{1}a_{1}}\psi_{a_{1}...a_{n}\dot {b}_{1}..\dot {b}_{m}} = 0 , $$ for representing the unitary one-particle state of the Poincare group with spin $n + m$, which can be integer or half-integer, and mass $M$.

Then, these conditions can be combined in some field equation, which refers to the some Lagrangian. There is a requirement that the equation must not contain derivatives higher than second-order. If the requirement is violated, one says that it breaks the unitarity of the theory of corresponding field.

So, the question: how exactly can be showed that it breaks the unitarity? And does it always break the unitarity?

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Possible duplicates: and links therein. – Qmechanic Sep 29 '13 at 0:20
The equations your wrote are not derivable from a Lagrangian except for some particular cases! Lagrangian formulation requires more fields. When people talk about higher derivative equations violating unitarity, they mean the maximal order of time derivative as one can write something that is either nonlinear in $\dot{\phi}$ or contains space gradients of higher order. A typical example is Galileon. – John Sep 29 '13 at 6:48
@John . Okay. But how to show that the field equations with derivatives higher than second rank generally breaks the unitarity in the field theories? – user8817 Sep 30 '13 at 13:17

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