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 $$-\delta^{ab}\left(\partial_{\mu}\,_{z}\partial^{\mu}\,_{z}\delta\left(0\right)\right)\delta\left(0\right)+... \tag{2} $$


1 Answer 1

  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).


$^1$ 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 $S_1$.

  • $\begingroup$ I wrote your answer but I did't understand where are my delta functions δ(0). $\endgroup$
    – illuminato
    Aug 27, 2017 at 10:39
  • $\begingroup$ Integration over z doesn't need because I started with lagrangian instead of action. $\endgroup$
    – illuminato
    Aug 27, 2017 at 11:18
  • $\begingroup$ The Lagrangian also contains spatial $z$-integrations. Do you mean Lagrangian density? $\endgroup$
    – Qmechanic
    Aug 27, 2017 at 11:42
  • $\begingroup$ Yes, I mean Lagrangian density. $\endgroup$
    – illuminato
    Aug 27, 2017 at 16:39
  • $\begingroup$ Well, it should be the action. $\endgroup$
    – Qmechanic
    Aug 27, 2017 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.