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When discussing higher dimensional operators in a theory with fermions, why do I never see anyone ever talk about the dimension five operator $\partial_\mu\bar\psi\partial^\mu\psi$?

How does the Fermion field behave when such a strange operator is its Lagrangian?

$$\mathcal{L}=\bar\psi(i\partial_\mu\gamma^\mu-m)\psi+\frac{1}{\Lambda}\partial_\mu\bar\psi\partial^\mu\psi$$

Since it is still quadratic in $\psi$, I expect this to be fairly easy to analyze. What happens when such a term is added?

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I'm not an effective field theory guy myself, but you can pass the derivative to the other field and get $\overline{\psi} \partial^2 \psi.$ But then, using on-shellness (which for some reason is accepted in that field), $\psi$ obeys the Klein-Gordon-equation, so $\partial^2 \psi = \pm m^2 \psi$. So up to off-shell stuff going on, you've written down a mass term for the fermion. –  Vibert Dec 7 '12 at 5:34
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The equation of motion for $\psi$ would be $(\Lambda^{-1}\Box - i\gamma^\mu\partial_\mu + m)\psi=0$. I believe this violates the positivity postulates of quantum mechanics. See Weinberg, Quantum Theory of Fields, Vol I, Ch 10.7. The current $\overline{\psi}\gamma^\mu\psi$ isn't conserved either. –  MarkWayne Dec 7 '12 at 6:34

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This particular extra term may be removed by a field redefinition $$\psi\to \psi' = \psi - K \cdot \gamma^\mu \partial_\mu \psi $$ for an appropriate value of $K\sim 1/\Lambda$, up to terms that are even higher dimension operators. This also modifies the mass. This field redefinition is an explicit off-shell way to realize Vibert's comment that one is just modifying the mass in which he assumed the equations of motion.

Mark Wayne is also right that (despite the equivalence, up to even higher-order terms), the propagator violates positivity of quantum mechanics (we really mean the positive norm of states with particle excitations which is linked to positive probabilities, a "must": excitations with negative norm are known as "[bad] ghosts"). If one writes down the full propagator for this (free) theory, one gets some additional poles which have a coefficient of the wrong sign. However, these pathologies occur at $p^2\sim \Lambda^2$ where we expect the theory to break down, anyway: the OP wrote the extra term as a correction in effective field theory that is meant to be used at energies $p^2\ll \Lambda^2$. This pathology with non-positivity may be fixed by new physics near $\Lambda$ and because of the equivalence, indeed, we can see that we may adjust the even higher-order terms (which may also arise from integrating out other fields and interactions) so that the full theory is exactly equivalent to the ordinary massive Dirac fermion and therefore has no ghosts.

That means that as long as we use this as an effective field theory, knowing that it may have further higher-order modification and new physics in general that kicks in near $p^2\sim \Lambda^2$, it is healthy as an effective field theory. Again, if we wanted to use it as a full theory even at energies of order $\Lambda$, it would be inconsistent.

In general, higher-order "free" terms like that – similarly $\square\phi\cdot \square\phi$ for bosons etc. – have been proposed as ways to make the propagators softer in the UV, like $1/p^4$ instead of $1/p^2$, which would improve the convergence of Feynman diagrams. The price is that if such a theory is to be taken seriously near $\Lambda$, the scale determining the size of the higher-dimensional operators, then this theory leads to new negative-norm excitations which is an inconsistency. So the addition of these new terms makes the theory break down "earlier", despite the intriguing softening of the UV problems.

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@QuantumDot The conclusion of this answer is correct. However, the problem with higher order derivatives in free actions is the unboundedness of the Hamiltonian. The problem with negative norm states only shows up when one does not quantize the theory properly (and even in this case the vacuum is not normalizable). This is a very spread mistake which does not have any consequence because in both cases the theory needs a completion at $\Lambda$. –  drake Dec 7 '12 at 8:16
    
Dear @drake, if you allow, I would disagree. In some cases, you may say that to "quantize the theory properly" means to avoid the negative norms at any cost, and convert the minus signs to unboundedness of energy. You would have this freedom if the higher-derivative term were actually the leading one. But in this case, the norms are already determined by the ordinary kinetic and mass terms and we're just adding corrections. So we're not able to change the overall sign of the norm of the excitations. –  Luboš Motl Dec 7 '12 at 8:22
    
Instead, we see new poles and we may compute the relative sign of the poles - the coefficients of the new ones and the old ones - and regardless of conventions, the relative sign is minus. So one of the poles of the propagator inevitably corresponds to wrong-sign-of-norm excitations regardless of any conventions. Quite generally, I think it's right to treat Lagrangians with the wrong sign of the kinetic terms to describe negative-norm (yes, very pathological) excitations and it is bad practice to pretend that the wrong sign of kinetic terms is "just" about the indefinite energy. –  Luboš Motl Dec 7 '12 at 8:24
    
In the usual kinetic+mass terms, it's simply appropriate to read the coefficients in the standardized way: the coefficient of the kinetic term decides about the norm of the excitations of the field; the mass term describes the mass (which may be tachyonic in some cases, for unusual signs). The field redefinitions needed to claim that the wrong sign of the kinetic term still allow all the excitations to be positive norm aren't really allowed, so a theory with physical excitations with a wrong sign of the kinetic terms is already inconsistent as a physical theory. Denying it is wrong. –  Luboš Motl Dec 7 '12 at 8:27
    
Dear LubosMotl, if one quantizes the theory correctly (annihilation operators go with $e^{-i(-|E|)t}$, with $-|E|$ the energy of the mode), one obtains a theory with positive definite states and unbounded from bellow Hamiltonian. The $i\epsilon$ terms of the new poles have the correct sign even though the $1/(p^2-m^2)$ has the wrong sign (because the energy of the mode is negative). If you look at the classical theory where everything is clear, the Hamiltonian is also unbounded from bellow. So if one quantizes the theory in the way you like, one obtains a wrong classical limit. Kind regards. –  drake Dec 7 '12 at 9:28

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