- #1

saadhusayn

- 22

- 1

- Homework Statement
- I refer to the attached set of lecture notes,equation 4.184, P70. We expand an action ##S[\phi]## around a background field ##\varphi(x) = \phi(x) -f(x)##

- Relevant Equations
- We need to know how to Taylor expand a functional.

If we expand the action ##S[\phi]## about ##\varphi(x)##,

## S[\phi] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi (x)} f(x) + \frac{1}{2!}\int d^dx_1 d^d x_2 \frac{\delta^2 S[\varphi]}{\delta \varphi(x_1) \delta \varphi(x_2)} f(x_1) f(x_2) + \mathcal{O}(f^3)##

But according to (4.184) I am supposed to get

$$ S[\phi] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi (x)} f(x) - \frac{1}{2}\int d^dx f(x) \Delta f(x) + \mathcal{O}(f^3)$$ The operator ##\Delta## has only been introduced as the Klein Gordon operator before. So how does the quadratic term become

$$-\frac{1}{2}\int d^dx f(x) \Delta f(x)$$

Thank you.

## S[\phi] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi (x)} f(x) + \frac{1}{2!}\int d^dx_1 d^d x_2 \frac{\delta^2 S[\varphi]}{\delta \varphi(x_1) \delta \varphi(x_2)} f(x_1) f(x_2) + \mathcal{O}(f^3)##

But according to (4.184) I am supposed to get

$$ S[\phi] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi (x)} f(x) - \frac{1}{2}\int d^dx f(x) \Delta f(x) + \mathcal{O}(f^3)$$ The operator ##\Delta## has only been introduced as the Klein Gordon operator before. So how does the quadratic term become

$$-\frac{1}{2}\int d^dx f(x) \Delta f(x)$$

Thank you.