Evaluate the Klein-Gordon action

[auditor]

[SOLVED] Evaluate the Klein-Gordon action

I'm interested in evaluating the Klein-Gordon action in P&S, p. 287. It goes as follows
$$S_0 = \frac{1}{2} \int d^4 x \! \phi \left( - \partial^2 -m^2 \right) \phi + \left(\text{surface term} \right)$$
The surface terms drops out, that's fine. I would like to see a detailed calculation of the first part of the integral. Anyone to the rescue?

Thanks in advance!

 

[auditor]

I came to my own rescue, it seems (or rather Mark Srednicki did). For reference, the evaluation is done in his "Quantum Field Theory", p. 55. A draft version may be downloaded from professor Srecnicki's homepage. I'm not allowed to post URL's quite yet, but google his name and you'll find his homepage.

 

[denian]

The Klein-Gordon action is a mathematical expression that describes the dynamics of a scalar field, represented by the variable \phi. It is commonly used in quantum field theory to understand the behavior of particles with spin 0, such as the Higgs boson. The action is given by the integral of the Lagrangian density over space and time, and is defined as:

S = \int d^4 x \mathcal{L}

where \mathcal{L} is the Lagrangian density. In the case of the Klein-Gordon action, the Lagrangian density is given by:

\mathcal{L} = \frac{1}{2} \phi \left( - \partial^2 -m^2 \right) \phi

This expression represents the energy of the field \phi, and its dynamics are governed by the Klein-Gordon equation, which is obtained by varying the action with respect to \phi. The surface term mentioned in the question is a boundary term that arises when integrating by parts, and it does not contribute to the dynamics of the field.

To evaluate the integral, we can use the Fourier transform of the field \phi:

\phi(x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}} \left( a_p e^{i p \cdot x} + a_p^{\dagger} e^{-i p \cdot x} \right)

where a_p and a_p^{\dagger} are the creation and annihilation operators, and E_p is the energy of the field. Substituting this into the Lagrangian density, we get:

\mathcal{L} = \frac{1}{2} \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} \left( a_p e^{i p \cdot x} + a_p^{\dagger} e^{-i p \cdot x} \right) \left( - \partial^2 - m^2 \right) \left( a_p e^{i p \cdot x} + a_p^{\dagger} e^{-i p \cdot x} \right)

Expanding this out and simplifying, we get:

\mathcal{L} = \frac{1}{2} \int \frac{d^