It's said in Chapter VI.4 of A. Zee's book Quantum Field Theory in a Nutshell, a theory defined as $L(U(x))=\frac{f^2}{4}Tr(\partial_{\mu}U^{\dagger}\cdot\partial^{\mu}U)$, can be write in the form of a non-linear $\sigma$ model (up to some order)


where $U(x)=e^{\frac{i}{f}\vec{\pi}\cdot\vec{\tau}}$ is a matrix-valued field belonging to $SU(2)$, $\vec{\pi}$ is a three components vector, $\vec{\tau}$ are Pauli matrices. Maybe it's not hard but I meet some problems to derive it.

I suppose the first step is the Taylor expansion of $U$, $U=1+\frac{i}{f}\vec{\pi}\cdot\vec{\tau}-\frac{1}{2f^2}(\vec{\pi}\cdot\vec{\tau})^2+...$, and then $\partial^{\mu}U=\frac{i}{f}\partial^{\mu}(\vec{\pi}\cdot\vec{\tau})-\frac{1}{f^2}(\vec{\pi}\cdot\vec{\tau})\partial^{\mu}(\vec{\pi}\cdot\vec{\tau})$, then


Now there are my questions,

(1) Can I write $\partial^{\mu}(\vec{\pi}\cdot\vec{\tau})=\partial^{\mu}\vec{\pi}\cdot\vec{\tau}$? Then by $\vec{\tau}^2=1$, I get


which is almost correct but differ to the wished answer by a pre-factor $\frac{1}{2}$.

(2) Suppose $\partial^{\mu}(\vec{\pi}\cdot\vec{\tau})=\partial^{\mu}\vec{\pi}\cdot\vec{\tau}$ is correct, however, if I do $\partial^{\mu}U=\frac{i}{f}U\partial^{\mu}(\vec{\pi}\cdot\vec{\tau})=\frac{i}{f}U\partial^{\mu}\vec{\pi}\cdot\vec{\tau}$ first, it seems $\partial^{\mu}U^{\dagger}\cdot\partial^{\mu}U=|\frac{i}{f}U\partial^{\mu}\vec{\pi}\cdot\vec{\tau}|^2=\frac{1}{f^2}(\partial\vec{\pi})^2$, say, only the first term of the wished answer.

I probably made something wrong somewhere, can anyone hit me?

  • 2
    $\begingroup$ I think you are somehow imagining a wrong question. There is nothing called "the" non-linear sigma model. You can define "a" non-linear sigma model by choosing whetever you like as the target Lie group in which your fields are valued in. Depending on what group you choose you get a different sigma-model. So you can always talk of the SU(2) NLSM where the fields are basically restricted to be on S^3. I would recommend that you see chapter 13, 14, 15 of this book to get a good picture of the issue, amazon.com/Quantum-Critical-Phenomena-International-Monographs/… $\endgroup$ – user6818 Nov 21 '13 at 22:45
  • $\begingroup$ Hm, google chrome is trying to tell me that the revisions for this question are in "Greek". Huh? $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 27 '13 at 11:01

First of all, the Pauli matrices are not space-time dependent, so of course you can pass the derivative right through them. Second of all, $\operatorname{Tr} [\partial(\vec{\pi}\cdot\vec{\tau})]^2 = \operatorname{Tr}\partial_\mu \pi^i \partial^\mu \pi^j \tau^i \tau^j $

Now remember $\tau^i \tau^j = i \epsilon_{ijk} \tau^k + \delta^{ij} I_{2x2}$

So compute the trace and be done already!

(It pays to write things out in full if you aren't sure what you are doing)


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