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## Main Question or Discussion Point

I understand that the ansatz to $$(\Box +m^{2})\phi(\mathbf{x},t)=0$$ (where ##\Box\equiv\partial^{\mu}\partial_{\mu}=\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}##) is of the form ##\phi(\mathbf{x},t)=e^{(iE_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}##, where ##E_{\mathbf{k}}=\sqrt{\mathbf{k}^{2}+m^{2}}##, but I've seen in several lecture notes that one first represents ##\phi(\mathbf{x},t)## in terms of its spatial Fourier transform $$\phi(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{\phi}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ first and then require that at each instant in time, the Fourier modes ##\tilde{\phi}(\mathbf{k},t)## satisfy the equation $$\left(\partial^{2}_{t}+(\mathbf{k}^{2}+m^{2})\right)\tilde{\phi}(\mathbf{k},t)=0$$ (where ##\partial^{2}_{t}\equiv\frac{\partial^{2}}{\partial t^{2}})##.

Is it correct to say that although we represent ##\phi(\mathbf{x},t)## in terms of its spatial Fourier transform at a particular

If so, then the approach makes sense to me, since the general solution to ##\left(\partial^{2}_{t}+(\mathbf{k}^{2}+m^{2})\right)\tilde{\phi}(\mathbf{k},t)=0## is given by $$\tilde{\phi}(\mathbf{k},t)=N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)$$ where ##N(\mathbf{k})## is some normalisation factor to be specified.

The general solution to the Klein-Gordon equation would then be given by $$\phi(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)e^{i\mathbf{k}\cdot\mathbf{x}}=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}+b(\mathbf{-k})e^{i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}\right)$$ (assuming that ##N(-\mathbf{k})=N(\mathbf{k})##).

Have I understood this correctly or am I missing something?

Is it correct to say that although we represent ##\phi(\mathbf{x},t)## in terms of its spatial Fourier transform at a particular

*fixed*instant in time, ##t##, we require that that it should be possible to do this at every instant in time, hence requiring that the Fourier modes, ##\tilde{\phi}(\mathbf{k},t)## are*time-dependent*and satisfy the equation given above?If so, then the approach makes sense to me, since the general solution to ##\left(\partial^{2}_{t}+(\mathbf{k}^{2}+m^{2})\right)\tilde{\phi}(\mathbf{k},t)=0## is given by $$\tilde{\phi}(\mathbf{k},t)=N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)$$ where ##N(\mathbf{k})## is some normalisation factor to be specified.

The general solution to the Klein-Gordon equation would then be given by $$\phi(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)e^{i\mathbf{k}\cdot\mathbf{x}}=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}+b(\mathbf{-k})e^{i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}\right)$$ (assuming that ##N(-\mathbf{k})=N(\mathbf{k})##).

Have I understood this correctly or am I missing something?