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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.

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.

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 $\partial_t^2 \phi - \nabla^2 \phi$ regardless of which metric we use.

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.