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Use metric signature $(+,-,-,-)$ and units with propagation speed one. The scalar Euler-Lagrange equation givesWith the stated spatial Fourier convention, $-{\nabla }^2$ becomes $k^2$. Each transformed mode obeysHere $F$ and $V$ are the initial transformed field and velocity. Reality requires $F(-\mathbf{k})=F(\mathbf{k}{)}^{\ast }$ and the same for $V$; the Fourier convention places no extra $(2\pi {)}^3$ in these evolution formulas.

Including the interaction gives the two coupled real Klein-Gordon fields,Adding and subtracting diagonalizes them:Both squared masses are positive for $0<g<m^2$. For arbitrary initial data $F_i={\tilde{\varphi }}_i(0)$ and $V_i={\dot{\tilde{\varphi }}}_i(0)$, defineThen the general solution is ${\tilde{\varphi }}_1=(U_{+}+U_{-})/2$, ${\tilde{\varphi }}_2=(U_{+}-U_{-})/2$, with the corresponding reality conditions on initial data.

For the specified sinusoidal initial displacement, put ${\Omega }_{\pm }=\sqrt{q^2+m^2\mp g}$. Both normal fields initially have amplitude $A\sin (\mathbf{q}\cdot \mathbf{x})$ and zero time derivative. ConsequentlyWriting $\bar{\Omega }=({\Omega }_{-}+{\Omega }_{+})/2$ and $\delta \Omega =({\Omega }_{-}-{\Omega }_{+})/2$ gives ${\varphi }_1=A\sin (\mathbf{q}\cdot \mathbf{x})\cos \bar{\Omega }t\cos \delta \Omega t$ and ${\varphi }_2=A\sin (\mathbf{q}\cdot \mathbf{x})\sin \bar{\Omega }t\sin \delta \Omega t$. This is field-amplitude beating of coupled Klein-Gordon modes. The conserved energy includes the interaction term $-g{\varphi }_1{\varphi }_2$; separate uncoupled-field energies need not be conserved, and an envelope zero is not by itself exact complete energy transfer.