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In the linear sigma model, the Lagrangian is given by

$ \mathcal{L} = \frac{1}{2}\sum_{i=1}^{N} \left(\partial_\mu\phi^i\right)\left(\partial^\mu\phi^i\right) +\frac{1}{2}\mu^2\sum_{i=1}^{N}\left(\phi^i\right)^2-\frac{\lambda}{4}\left(\sum_{i=1}^{N}\left(\phi^i\right)^2\right)^2. \tag{11.65} $

(for example see Peskin & Schroeder (P&S) page 349).

When perturbatively computing the effective action for this Lagrangian the derivative $ \frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} $ needs to be computed. (for instance, Eq. (11.67) in P&S):

$$ \frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} ~=~ -\partial^2\delta^{kl} +\mu^2\delta^{kl}-\lambda\left[\phi^i\phi^i\delta^{kl}+2\phi^k\phi^l\right].\tag{11.67}$$

My question: where are two delta functions?

If you don't understand why there to need them, I write full calculation:

\begin{eqnarray} \frac{\delta^{2} \mathcal{L} \left[\phi\right]}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}&=&\frac{\delta^{2}}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}\left\{ \frac{1}{2}\sum_{i=1}^{N}\left(\partial_{\mu z}\phi^{i}\left(z\right)\right)\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)+...\right\} \\&\stackrel{}{=}&\frac{\delta}{\delta\phi^{a}\left(x\right)} \left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\frac{\delta}{\delta\phi^{b}\left(y\right)}\phi^{i}\left(z\right)\right)\right)+...\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)} z\left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta^{ib}\delta\left(z-y\right)\right)\right)+...\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)} \left\{ \left(\partial^{\mu}\,_{z}\phi^{b}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+...\right\} \\&=& \left(\partial^{\mu}\,_{z}\delta^{ab}\delta\left(x-z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+... \\&=& -\delta^{ab}\left(\partial_{\mu}\,_{z}\partial^{\mu}\,_{z}\delta\left(x-z\right)\right)\delta\left(z-y\right)+... \\& \tag{1} \end{eqnarray}

You may see two delta functions there.

You may see two delta functions there. Next we assert that $x=y=z$ and we have $$-\delta^{ab}\left(\partial_{\mu}\,_{z}\partial^{\mu}\,_{z}\delta\left(0\right)\right)\delta\left(0\right)+... \tag{2} $$

illuminato
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1 Answers1

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  1. First we should focus on what we ultimately are trying to calculate, namely the second functional derivative of the action (as opposed to the e.g. the Lagrangian density) $$\frac{\delta^2 S}{\delta\phi^{\alpha} (x)\delta\phi^{\beta}(y)}.\tag{A}$$ Therefore the result (A) (which is worked out in this Phys.SE post) contains one $4$-dimensional Dirac delta distribution (rather than, say, two or zero). To conform with eq. (A), OP should include an integration over $z$ in his eq. (1).

  2. P&S are admittedly not making the above point very clear. In fact, P&S are using the notation of a 'same-spacetime' functional derivative$^1$
    $$\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots, \tag{B}$$ which does not contain Dirac delta distributions, as explained in my answer to the previous Phys.SE post. This explains why there are no Dirac delta distributions in eq. (11.67).

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$^1$ For completeness, let us mention that most of P&S's formulas can be explained by the notation of a 'same-spacetime' functional derivative (B), but eq. (11.58) is even beyond this notation. To make sense of eq. (11.58) replace all appearances of ${\cal L}_1$ on the rhs. with $S_1$.

Qmechanic
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