= Solution
In <Minkowski spacetime> with signature $(-+++)$, $p\cdot x=-\omega_{\mathbf p}t+\mathbf p\cdot\mathbf x$, where $\omega_{\mathbf p}=p^0=\sqrt{|\mathbf p|^2+m^2}>0$. Hence $i\partial_t\psi_{\mathbf p}=\omega_{\mathbf p}\psi_{\mathbf p}$: these are <positive-frequency solutions> relative to the inertial time translation.
On a constant-time <Cauchy hypersurface>, the <Klein-Gordon inner product> and the spatial <Fourier transform> identity give
$$
(\psi_{\mathbf p},\psi_{\mathbf q})=
\frac{\omega_{\mathbf p}+\omega_{\mathbf q}}{2\sqrt{\omega_{\mathbf p}\omega_{\mathbf q}}}
e^{i(\omega_{\mathbf p}-\omega_{\mathbf q})t}\delta^3(\mathbf p-\mathbf q)
=\delta^3(\mathbf p-\mathbf q).
$$
Also $(\overline\psi_{\mathbf p},\overline\psi_{\mathbf q})=-\delta^3(\mathbf p-\mathbf q)$ and $(\psi_{\mathbf p},\overline\psi_{\mathbf q})=0$: the latter is proportional to $(\omega_{\mathbf p}-\omega_{\mathbf q})\delta^3(\mathbf p+\mathbf q)$, which vanishes. This is the <Klein-Gordon plane-wave normalization>.
For completeness, spatial Fourier transformation of the field equation gives $\ddot{\widehat\Phi}(t,\mathbf p)+\omega_{\mathbf p}^2\widehat\Phi(t,\mathbf p)=0$. Its solutions have the two branches $e^{\mp i\omega_{\mathbf p}t}$. Selecting the positive-frequency branch leaves precisely the Fourier superpositions of $\psi_{\mathbf p}$. These plane waves form a generalized basis, normalized with the <Dirac delta distribution>, rather than individually normalizable vectors. For a normalizable packet,
$$
\Phi_+(x)=\int d^3p\,c(\mathbf p)\psi_{\mathbf p}(x),\qquad
\boxed{(\Phi_+,\Phi_+)=\int d^3p\,|c(\mathbf p)|^2>0\quad\text{if }c\ne0.}
$$
Back to article page
