Expanding an action about a background field

[saadhusayn]

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.

 

[mark726]

Hello,

Thank you for your post. The difference between the two expressions is due to the different ways of expanding the action. The first expression expands the action in terms of the field $\varphi$, while the second expression expands the action in terms of the perturbation $f(x)$. When expanding in terms of $\varphi$, we use the functional derivatives with respect to $\varphi$, while when expanding in terms of $f(x)$, we use the functional derivatives with respect to $f(x)$.

To understand why the quadratic term becomes $-\frac{1}{2}\int d^dx f(x) \Delta f(x)$, we can look at the definition of the functional derivative:

$$\frac{\delta S[\varphi]}{\delta \varphi(x)} = \lim_{\epsilon \rightarrow 0} \frac{S[\varphi + \epsilon \delta(x-x')]-S[\varphi]}{\epsilon}$$

If we expand the action around $\varphi$, we get:

$$S[\varphi + \epsilon \delta(x-x')] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi(x)} \epsilon \delta(x-x') + \frac{1}{2!}\int d^dx_1 d^d x_2 \frac{\delta^2 S[\varphi]}{\delta \varphi(x_1) \delta \varphi(x_2)} \epsilon^2 \delta(x_1-x') \delta(x_2-x') + \mathcal{O}(\epsilon^3)$$

Plugging this into the definition of the functional derivative, we get:

$$\frac{\delta S[\varphi]}{\delta \varphi(x)} = \lim_{\epsilon \rightarrow 0} \frac{\int d^d x' \frac{\delta S[\varphi]}{\delta \varphi(x')} \epsilon \delta(x-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)} \epsilon^2 \delta(x_1-x') \delta(x_2-x') + \math