Is there a well-known Lagrangian that, writing the corresponding equation of motion, gives the Klein-Gordon equation in QFT? If so, what is it?
What is the canonical conjugate momentum? I derive the same result as in two sources separately, but with opposite sign, and I am starting to suspect that the error could be in the Lagrangian I am departing from.
Is there any difference in the answers to these two questions if you choose $(+---)$ or $(-+++)$? If so, which one?
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asked Feb 3, 2013 at 5:15
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Yes. The standard scalar field which all QFT books (e.g. Peskin & Schroeder, Zee) start with yields the KG equation. For that reason it is also called the Klein-Gordon field. The Lagrangian (density) is \begin{align} \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2. \end{align} Here the metric is $(+---)$.
By definition it is $\pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}$. This gives $\pi = \partial^0 \phi$.
It is purely convention, there is no right choice. The only difference in using a different metric will be in how we write things down - any quantities that involve contraction with the metric $\eta_{\mu \nu}$ will change by a minus sign. For example in the Lagrangian, using the metric (- + + +), the first term is changed to $-\frac{1}{2} \partial_\mu \phi \partial^\mu \phi$. But this is still equal to $\frac{1}{2}(\partial_t^2 \phi - \nabla^2 \phi)$ regardless of which metric we use.
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$\begingroup$ Thanks a lot for your answer, it has put me on the right track. Eqs (1.14) and (1.15) in the preprint of Srednicki have the key to the changed sign of question 2. I found it thank to your indications. $\endgroup$ Commented Feb 3, 2013 at 13:24
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1$\begingroup$ The answer above is unclear to me. The Lagrangian above is (1) 1/2 (ħ ∂μφ) (ħ ∂νφ) g^μν – 1/2 m^2 φ^2 If you change the sign of the metric, the first term changes sign and the second term does not change. To get the same Lagrangian, the sign of the first or second term must change. No? $\endgroup$– rjpettiCommented Sep 10, 2021 at 15:21
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$\begingroup$ @rjpetti : It seems you are right. But I think that we must not expect any respond from the answerer "nervxxx" since I feel he/she is inactive (no recent answers,questions, comments etc). $\endgroup$– VoulkosCommented Sep 10, 2021 at 17:46
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$\begingroup$ ...the same is valid for the OP "Eduardo Guerras Valera"... $\endgroup$– VoulkosCommented Sep 10, 2021 at 18:37
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1$\begingroup$ The Lagrangian is invariant up to an overall scalar, so the point about the spectrum being bounded or unbounded when you allow negative kinetic energy terms can always be avoided by a trivial re-definition, the point about not satisfying the relativistic energy-momentum relation is the non-trivial fact which fixes the overall sign of the kinetic terms with respect to the potential terms for each choice of metric. $\endgroup$ Commented Sep 14, 2021 at 8:27