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In expanding the classical Klein-Gordon field in Fourier space to write it in terms of $\phi(\mathbf{p})$ instead of $\phi(\mathbf{x})$, I reached the following result. $$\int \mathrm{d}^3p\exp({i\mathbf{p}\cdot\mathbf{x}})\left[\frac{\partial^2}{\partial \mathrm{t}^2}+|\mathbf{p}|^2 + m^2\right]\phi(\mathbf{p},t) =0$$

Now, how is it concluded that $$\left[\frac{\partial^2}{\partial \mathrm{t}^2}+|\mathbf{p}|^2 + m^2\right]\phi(\mathbf{p},t) =0.$$

I suspect there is a physical rather than mathematical reasoning for this, since, at least apparently, the integrand could have the whole complex plane as its range. (Or is the assumption that $\phi(\mathbf{x})$ is real, and therefore $\phi ^*(\mathbf{p})=\phi(\mathbf{-p})$, relevant?)

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  • $\begingroup$ Are you varying the action in order to obtain the equation of motion? In that case the $\phi$ in your first equation should be $\delta \phi$ and since one considers arbitrary variations $\delta \phi$, the bracketed expression must be zero for this to always vanish. $\endgroup$
    – Danu
    Aug 16, 2014 at 18:33
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    $\begingroup$ Hint to the question (v1): Try Fourier transformation. $\endgroup$
    – Qmechanic
    Aug 16, 2014 at 18:34

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Your first equation can be compared to (in one dimension) $$ g(x)=\int dp\exp(ipx)f(p)=0 \,\,\, (*) $$ Note that this is a statement applies to all x-values. The inverse fourier transform of $g(x)$ should give $f(p)$, $$ f(p)=\frac{1}{2\pi}\int dx\exp(-ipx)g(x) $$ As $g(x)=0$ according to to $(*)$ then $f(p)=0$.

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This is essentially the statement that the Fourier transform is injective and therefore invertible. In looser language, it states that the functions $\mathbf x\mapsto\exp(i\mathbf p\cdot\mathbf x)$ are linearly independent, so any sum of them (i.e. $\int\mathrm d^3\mathbf p$) that gives zero must have identically zero coefficients.

Thus, it applies to any (nice enough) function $f$: if $$ \tilde f(\mathbf x)=\frac{1}{(2\pi)^{3/2}}\int \mathrm d^3\mathbf p f(\mathbf p)\exp(i\mathbf p\cdot\mathbf x)=0 $$ then you can conclude that $f(\mathbf p)$ itself is identically zero.

The standard way to see this is via a direct inversion. The Fourier transform is essentially its own inverse, and one can calculate the double transform to see that $$ f(\mathbf p)=\frac{1}{(2\pi)^{3/2}}\int \mathrm d^3\mathbf x \tilde f(\mathbf x)\exp(-i\mathbf p\cdot\mathbf x) $$ under appropriate niceness conditions on $f$. With that calculation in hand, it is trivial to show that if $\tilde f$ vanishes then $f$ must do so as well.

This does take some getting used to. You are correct in your feeling that the fact that an integral $\int g(p)\mathrm dp$ vanishes does not by itself imply that the integrand is zero. Instead, the principle at play is slightly subtler: the integral is of the form $\int g(x,p)\mathrm dp$, and it vanishes for each and every $x$. In the former case you have a single piece of information, which is not enough to pin down an arbitrary function of position, but in the latter case you have one condition for each real $x$, and that turns out (if the dependence of $g(x,p)$ on $p$ for different $x$s is 'different enough') to be sufficient.

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