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?
Asked
Active
Viewed 1.9k times
10
Urb
- 2,724
Eduardo Guerras Valera
- 3,805
1 Answers
19
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.
