# Klein-Gordon inner product

Studying the scalar field and Klein-Gordon equation in quantum field theory I came across this definition for the inner product in the space of the solutions of the K.G. equation:

$$\langle \Phi_1 | \Phi_2 \rangle = i\int \mathrm{d}\vec{x}(\Phi_1 ^* \overleftrightarrow{\partial_0}\Phi_2) = i\int \mathrm{d}\vec{x} (\Phi_1 ^* \partial_0\Phi_2 - \Phi_2 \partial_0\Phi_1^*).$$

I see that this definition should be invariant under Poincaré transformations, but I couldn't prove it.

Moreover I couldn't find the reason why such a scalar product is introduced. Aren't there other possible scalar products? Why choose this one?

• Something to think about: consider the current $J_\mu = i\Phi_1^* \partial_\mu \Phi_2 - i\Phi_2 \partial_\mu \Phi_1^*$ and maybe set $\mu = 0$... What can you say about $J_\mu$? – Vibert Jan 16 '13 at 22:46
• For a thorough treatment of that expression, consider this paper. – Frederic Brünner Jan 17 '13 at 3:29

The Klein-Gordon inner product is a natural construction for functions defined on the mass hyperboloid $k^2=m^2$, because if you write your function in momentum space, $$\phi(x)\sim \int\widetilde{\mathrm dk}\ \mathrm e^{-ikx}a(k)+\text{h.c.}$$ with $\widetilde{\mathrm dk}$ the measure on $k^2=m^2$, then the Fourier coefficients become (ref. 1, sec 3-1-2) $$a(k)=\langle\phi,\exp_k\rangle$$ where $\exp_k(x)\equiv\mathrm e^{-ikx}$.
One should point out that, even if $\langle\cdot,\cdot\rangle$ is a natural construction, the real reason we define this particular integral is that it appears in the proof of the LSZ formula for scalars (ref. 1, sec 5-1-4).
• But if you have an arbitrary inner product $(,)$ you get $(\phi(x),\exp_k(x)) = c(k)$ where the $c(k)$ are the coefficients in expansion of $\phi$ now, $i.e.$ with respect to this other arbitrary inner product $\phi(x)=\int [c(k) \exp_{k}(x) + c(k)^{\ast} \exp_{-k}(x)]$. What restricts me from upgrading $c(k)$ to creation/annihilation operators (taking $[c(k),c(p)]=...$ and so on) and quantizing the field in this way with respect to the arbitrary inner product? I guess the question boils down now to what is the defining property of $a(k)$ here? – QuantumEyedea Jul 10 '18 at 20:00