# What is the Lagrangian from which the Klein-Gordon equation is derived in QFT?

1. Is there a well-known Lagrangian that, writing the corresponding eq of motion, gives the Klein-Gordon Equation in QFT? If so, what is it?

2. 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.

3. Is there any difference in the answers to that two questions if you choose (+---) or (-+++)? If so, which one?

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## 1 Answer

1. 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 (+ - - -).

2. By definition it is $\pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}$. This gives $\pi = \partial_0 \phi$.

3. 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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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. –  Eduardo Guerras Valera Feb 3 '13 at 13:24