I am following along these lecture slides. For scalar functions $f$ and $g$ we define the Klein-Gordon inner product as follows: $$ <f,g> \ = \ i \int_{\Sigma} d^{3}\mathbf{x}\ \left[ f^{\ast}(t,\mathbf{x}) \frac{\partial g(t,\mathbf{x})}{\partial t} - \frac{\partial f^{\ast}(t,\mathbf{x})}{\partial t} g(t,\mathbf{x}) \right] $$

Where we take $\Sigma$ to a 3D hypersurface of constant $t = t_{0}$. If $f$ and $g$ are the solutions to the Klein-Gordon then the above is independent of the choice of $t_0$ used to integrate it.

One can use this to show that the modes $u_{\mathbf{k}} \propto e^{-i \sqrt{\mathbf{k}^2 + m^2} t + i \mathbf{k} \cdot \mathbf{x}}$ are orthogonal to one another.

My question is; where does this inner product come from? I under stand that $i \phi^{\ast} \frac{\partial \phi}{\partial t} - i \phi^{\ast} \frac{\partial \phi}{\partial t} $ is a conserved current, but why would you put a conserved current under and integral sign to define an inner product? It seems arbitrary. What if one came up with a different inner product? What benefits does the KG inner product have?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.