= Solution
Use <metric signature> $(+,-,-,-)$ and units with propagation speed one. The scalar <Euler-Lagrange equation> gives
$$
\frac{\partial\mathcal L_{\rm KG}}{\partial\phi}=-m^2\phi,\qquad
\frac{\partial\mathcal L_{\rm KG}}{\partial(\partial_\mu\phi)}=\partial^\mu\phi,
\qquad
\boxed{(\partial_t^2-\nabla^2+m^2)\phi=0.}
$$
With the stated spatial Fourier convention, $-\nabla^2$ becomes $k^2$. Each transformed mode obeys
$$
\ddot{\tilde\phi}+\omega_k^2\tilde\phi=0,\qquad
\omega_k=\sqrt{k^2+m^2},\qquad
\boxed{\tilde\phi(\mathbf k,t)=F(\mathbf k)\cos\omega_kt+
\frac{V(\mathbf k)}{\omega_k}\sin\omega_kt.}
$$
Here $F$ and $V$ are the initial transformed field and <velocity>. Reality requires $F(-\mathbf k)=F(\mathbf k)^*$ 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>,
$$
\boxed{(\Box+m^2)\phi_1-g\phi_2=0,\qquad
(\Box+m^2)\phi_2-g\phi_1=0.}
$$
Adding and subtracting diagonalizes them:
$$
(\Box+m^2-g)\phi_+=0,\qquad
(\Box+m^2+g)\phi_-=0,\qquad
\omega_\pm(k)=\sqrt{k^2+m^2\mp g}.
$$
Both squared masses are positive for $0<g<m^2$. For arbitrary <initial data> $F_i=\tilde\phi_i(0)$ and $V_i=\dot{\tilde\phi}_i(0)$, define
$$
U_\pm(\mathbf k,t)=(F_1\pm F_2)\cos\omega_\pm t+
\frac{V_1\pm V_2}{\omega_\pm}\sin\omega_\pm t.
$$
Then the general solution is $\tilde\phi_1=(U_++U_-)/2$, $\tilde\phi_2=(U_+-U_-)/2$, 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>. Consequently
$$
\boxed{\phi_1=\frac A2\sin(\mathbf q\cdot\mathbf x)
[\cos\Omega_+t+\cos\Omega_-t],\qquad
\phi_2=\frac A2\sin(\mathbf q\cdot\mathbf x)
[\cos\Omega_+t-\cos\Omega_-t].}
$$
Writing $\bar\Omega=(\Omega_-+\Omega_+)/2$ and $\delta\Omega=(\Omega_--\Omega_+)/2$ gives $\phi_1=A\sin(\mathbf q\cdot\mathbf x)\cos\bar\Omega t\cos\delta\Omega t$ and $\phi_2=A\sin(\mathbf q\cdot\mathbf x)\sin\bar\Omega t\sin\delta\Omega t$. This is <field-amplitude beating of coupled Klein-Gordon modes>. The conserved <energy> includes the interaction term $-g\phi_1\phi_2$; separate uncoupled-field <energies> need not be conserved, and an envelope zero is not by itself exact complete <energy> transfer.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1-coupled-fields.png]
{title=Coupled scalar-mode beating}
Back to article page
