Starting with the action for a free scalar field $$S[\phi]=\frac{1}{2}\int\;d^{4}x\left(\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-m^{2}\phi^{2}(x)\right)=\int\;d^{4}x\mathcal{L}$$ Naively, if I expand this to second-order, I get $$S[\phi+\delta\phi]=S[\phi]+\int\;d^{4}x\frac{\delta S[\phi(x)]}{\delta\phi(x)}\delta\phi(x)+\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)$$ Now, assuming that ##\phi(x)## satisfies the equations of motion (EOM), then the first-order term vanishes, however, I'm unsure how to calculate the second-order variation. So far, my attempt is $$\delta^{2}S=\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\=\int\;d^{4}x d^{4}y\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}\delta\phi(x)\delta\phi(y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(y)}\delta\phi(x)\delta(\partial_{\mu}\phi(y))+\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\mu}\phi(y))}\delta(\partial_{\mu}\phi(x))\delta(\partial_{\mu}\phi(y))\right)$$ However, I am unsure how to progress (integration by parts doesn't seem to work as nicely in this case), as naively it seems as though the only term that would survive is ##\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}##, but I've seen references stating that ##\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}## is of the form ##\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}=\Box +m^{2}##.(adsbygoogle = window.adsbygoogle || []).push({});

Any help would be much appreciated.

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# I How to compute second-order variation of an action?

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