- #1

"Don't panic!"

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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?