Suppose one starts with the standard Klein-Gordon (KG) Lagrangian for a(adsbygoogle = window.adsbygoogle || []).push({}); freescalar field: $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}$$ Integrating by parts one can obtain an equivalent (i.e. gives the same equations of motion) Lagrangian $$\mathcal{L}=-\frac{1}{2}\phi\left(\Box +m^{2}\right)\phi$$ where ##\Box:=\partial_{\mu}\partial^{\mu}##.

My question is, what is the correct form for the Euler-Lagrange equations in this second case? Naively, I find that $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi +0=-m^{2}\phi-\frac{1}{2}\Box\phi=0$$ which clearly isn't correct. Should the Euler-Lagrange equation be modified to $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=0\;?$$ such that $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi

-\frac{1}{2}\Box\phi=-(\Box+m^{2})\phi =0$$

Apologies for a potentially silly question, but I'm a bit stuck on this issue.

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# I Equivalent Klein-Gordon Lagrangians and equations of motion

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