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I refer to this set of lecture notes by Hugh Osborn, equation 4.184 on p.70. We expand an action $S[\phi]$ around a background field $\varphi(x) = \phi(x) -f(x)$

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

$$ S[\phi] = S[\varphi] + \int d^d x \frac{\delta S[\varphi]}{\delta \varphi (x)} f(x) - \frac{1}{2}\int d^dx f(x) \Delta f(x) + \mathcal{O}(f^3).\tag{4.184}$$ The operator $\Delta$ has only been introduced as the Klein Gordon operator before. So how does the quadratic term become

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

Can we integrate the quadratic term by parts to get this?

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Presumably Osborn is assuming that the action $S$ is a local functional, such that the Hessian becomes on the form $$ \frac{\delta^2 S[\varphi]}{\delta \varphi(x_1) \delta \varphi(x_2)}~=~-\Delta \delta^d(x_1-x_2), $$ where $\Delta$ is differential operator, and such that each term has an even number of derivatives. Also Osborn is ignoring boundary terms.

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  • $\begingroup$ Thanks, this makes sense. In an earlier chapter, we had $\frac{\delta S[\varphi]}{\delta \varphi(x) } = -\Delta \varphi(x)$ and thus $\frac{\delta^2 S[\varphi]}{\delta \varphi(x_1) \delta \varphi(x_2) } =- \Delta_1 \delta^{(d)}(x_1 - x_2)$. The rest is just integration by parts. $\endgroup$
    – saad
    Commented Oct 6, 2019 at 14:20

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