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For K-G field, the lagrangian density

$$L=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$$

In ryder quantum field theory book i saw that in next step he writes

$$L=-\frac{1}{2}\phi(\partial_\mu\partial^\mu+m^2)\phi$$

My question is how did we got to second step from first,where did the minus sign came from and how did he pulled out a phi from the derivative in first step?

Miyase
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Souvik
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    Didn't he just integrate the term by parts and dropped the surface term? – Radu Moga Jun 18 '22 at 15:47
  • Can you please show it? – Souvik Jun 18 '22 at 16:08
  • As @RaduMoga said, this is just the previous Lagrangian minus $\partial_\mu(\phi\partial^\mu\phi)$ (it's a good exercise to check this). Subtracting a total derivative doesn't change the action, or the Euler–Lagrange equation. – J.G. Jun 18 '22 at 16:43

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We can write $\partial_\mu \phi \partial^\mu \phi=\partial_\mu(\phi \partial^\mu \phi)-\phi\partial_\mu \partial^\mu \phi$ by the product rule for differentiation. The term $\partial_\mu(\phi \partial^\mu \phi)$ is a surface term that vanishes when the Langrangian density is integrated over spacetime (under certain assumptions such as the fields decaying fast enough at infinity). This essentially means we can make the replacement $\partial_\mu \phi \partial^\mu \phi \rightarrow -\phi\partial_\mu \partial^\mu \phi$ in the Lagrangian density.

Radu Moga
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  • Can you please say why is total derivative term the surface term? – Souvik Jun 18 '22 at 17:26
  • That is what it is meant by surface terms in general. See also this post https://physics.stackexchange.com/questions/370185/what-is-the-surface-term – Radu Moga Jun 18 '22 at 17:43