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In the book Quantum Field Theory and the Standard Model by M. D. Schwartz, the author has used negative kinetic term in most of the Lagrangians. See equation (7.91) at page 97, for example. (Focus on the first term on the right side of equal sign.) $${\cal L}=-\frac{1}{2}\phi\Box\phi-\frac{1}{2}m^{2}\phi^{2}+\frac{g}{3!}\phi^{3}.\tag{7.91}$$

Is it okay to have negative kinetic term?

I was under the impression that kinetic term must be positive. I understand that a theory should be bounded from below in terms of energy. So the cubic interaction term is also problematic. But I am concerned about the kinetic term (with the box operator) in this question.

The book is using $(+,-,-,-)$ metric. I don't have any confusion about the part.

Sorry for such a trivial question, but I am very new to field theory.

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  • $\begingroup$ I have already checked the post: physics.stackexchange.com/q/100557 But I am still confused. A precise answer directed towards this context will be much appreciated. $\endgroup$
    – baba26
    Sep 4, 2020 at 3:14
  • $\begingroup$ Don’t you consider a kinetic term to be one with $(\partial\phi/\partial t)^2$ such as $(\nabla\phi)^2$? Do you see how to get this? $\endgroup$
    – G. Smith
    Sep 4, 2020 at 3:21
  • $\begingroup$ You are right @G.Smith . This is not a kinetic term to be precise. And I didn't know how to get this. But now ( after seeing the posted answer ) I am clear about the sign and also understand how to actually get this term. Thanks. $\endgroup$
    – baba26
    Sep 8, 2020 at 19:21

3 Answers 3

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The kinetic term in that Lagrangian is not actually negative. The usual way to write the kinetic term for a scalar field is ${\cal L}_{0}=\frac{1}{2}(\partial_{\mu}\phi)(\partial_{\mu}\phi)$. However, if you use an integration by parts (allowed because the Lagrange density ${\cal L}$ will always appear in an integrated action) to move one derivative, you get an equivalent Langrangian ${\cal L}'=-\frac{1}{2}\phi(\partial^{\mu}\partial_{\mu}\phi)$, which is what appears in your question (using the box $\Box=\partial^{\mu}\partial_{\mu}$). So the negative sign comes entirely from the integration by parts.

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The term that commonly appears in Lagrangian is

$${\frac 12}\partial_\mu \phi \partial^\mu \phi$$

so based on identity

$$\partial_\mu \phi \partial^\mu \phi= \partial_\mu (\phi\partial^\mu \phi)-\phi \partial_\mu\partial^\mu\phi $$

when you integrate the Lagrangian for all spacetime on both sides of this equation,and assuming that the field is zero at infinity and applying Gauss's theorem then the first term on the right side of equation(which is essentially a quadrivergence)disappears , and so that only the kinetic term retains but now with a different appearance than it had in the original Lagrangian and appearing also a negative sign. So both ways to express the Lagrangian are equivalent. Bye

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The fundamental importance of the sign of the Lagrangian is that the variation of the action is minimal with respect to the real "trajectory" of the system, regardless of the sign of the kinetic term of the Lagrangian. For example, in the Lagrangian of the relativistic free particle, the sign is minus in order to guarantee the minimum of the action, in the real trajectory of the particle. The opposite occurs with the sign of the kinetic term in a non-relativistic free particle where, on the contrary, the kinetic term of the Lagrangian must be positive to guarantee the minimum of the action, along the real trajectory of the particle.

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