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I am reading the section 11.4 of Peskin and Schroeder's book (page 373), and there is a step I could not follow. To calculate the effective action of linear sigma model, the determinant of $\frac{\delta^{2}\mathcal{L}}{\delta\phi^{i}\delta\phi^{j}}$, and in peskin's book, it shows that $$\frac{\delta^2\mathcal{L}}{\delta\phi^i\delta\phi^{j}}=-\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda[(\phi_{\mathrm{cl}}^k)^2\delta^{ij} + 2\phi_{\mathrm{cl}}^i\phi_{\mathrm{cl}}^j].\tag{11.67}$$ Then, orient the coordinates to make the $\phi^{i}_{cl}$ points in the Nth direction, $$\phi^{i}_{cl}=(0,0,...,0,\phi_{cl}),\tag{11.68}$$ and the operator in the RHS of the first equation is just a KG operator $(-\partial^{2}-m^{2}_i),$ where $$m^2_i=\begin{cases} \lambda\phi^2_{cl}-\mu^2\quad acting\; on\;\eta^1,...,\eta^{N-1};\\3\lambda\phi^2_{cl}-\mu^2\quad acting\; on\;\eta^N.\end{cases}\tag{11.69}$$ I think for all $\eta^i$, the value of $m^2_i$ should equal to $3\lambda\phi^2_{cl}-\mu^2$, how the $\lambda\phi^2_{cl}-\mu^2$ is obtained?

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Simply read the diagonal N×N matrix $$ -\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda[ v^2 \delta^{ij} + 2\phi_{\mathrm{cl}}^i\phi_{\mathrm{cl}}^j], \tag{11.67}$$ where, to allay your confusion, I call $\phi_{cl}=v$, so $(\phi_{\mathrm{cl}}^k)^2=v^2$.

The non-gradient part of this diagonal matrix is, for $i,j= 1...,N-1$, $$(\mu^2-\lambda v^2, \mu^2-\lambda v^2,...,\mu^2-\lambda v^2),$$ whereas the last entry (only!) in the diagonal is $$ \mu^2-\lambda v^2-2\lambda v^2= \mu^2-3\lambda v^2. $$

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  • $\begingroup$ I think I must have misunderstood something. I think if $i=j=N-1$, as $\phi^{N-1}_{cl}=(0,0,...,\phi_{cl},0) $, the N-1 term in the matrix still should be $\mu^2-3\lambda\phi_{cl}$. Where did I go wrong? $\endgroup$
    – David Shaw
    Jun 1, 2022 at 3:49
  • $\begingroup$ 11.68 tells you the penultimate component is zero. You wrote a meaningless vector, instead! $\endgroup$ Jun 1, 2022 at 5:08
  • $\begingroup$ I think I know what is my mistake. As $\phi^i=\phi^i_{cl}+\mu^i$, I regard the $\mu^i$ as a term linear with $\phi^i$, so that $\phi^{N-1}=(0,...,\phi_{cl},0)$ but it seems does not need to be the cases. All scalar fields are oriented along the Nth direction. Thank you! $\endgroup$
    – David Shaw
    Jun 1, 2022 at 8:53
  • $\begingroup$ Indeed, (11.68) simply states that the vector $\phi^{i}_{cl}$ has components $\phi_{cl}^1=0= \phi_{cl}^2=...=\phi_{cl}^{N-1}$ , $\phi_{cl}^N=v$. $\endgroup$ Jun 1, 2022 at 10:01
  • $\begingroup$ What confuse me is in fact the superscript. As $\phi^i$ is a set of real scalar field, choosing $i=1$ seems represents that one of the real scalar fields, but it seems also represents the component of one of the scalar field. So what is the actual meaning of the $\phi^i$ here? $\endgroup$
    – David Shaw
    Jun 1, 2022 at 16:10
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Hint: Eqs. (11.67) & (11.68) yield

$$ \frac{\delta^2\mathcal{L}}{\delta\phi^i\delta\phi^{j}}=-\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda\phi_{\mathrm{cl}}^2 [\delta^{ij} + 2\delta_N^i\delta_N^j].$$

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