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  where are delta functions?

+ 0 like - 0 dislike

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   $

(for example see Peskin & Schroeder 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 need them, I write full calculation:

\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)+...   \\&

You may see two delta functions there.

asked Aug 25, 2017 in Theoretical Physics by illuminates (15 points) [ revision history ]
edited Aug 25, 2017 by illuminates

1 Answer

+ 0 like - 0 dislike

There are no $\delta (x-z)$ nor $\delta (z-y)$; this is your error. There is $x$ everywhere instead.

answered Aug 26, 2017 by Vladimir Kalitvianski (102 points) [ no revision ]

I understand it, I wrote δ(xz) and δ(zy) for illustrative purposes. In reality are there two δ(0)?

In reality there are two factors 1.  All non trivial factors are $\delta_{ab}$, etc.

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