2
$\begingroup$

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)

$L=\frac{1}{2}(\partial\vec{\pi})^2+\frac{1}{2f^2}(\vec{\pi}\cdot\partial\vec{\pi})^2+...$,

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

$(\partial_{\mu}U^{\dagger})(\partial^{\mu}U)=\frac{1}{f^2}[\partial(\vec{\pi}\cdot\vec{\tau})]^2+\frac{1}{f^4}.[(\vec{\pi}\cdot\vec{\tau})\partial(\vec{\pi}\cdot\vec{\tau})]^2$.

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

$L=\frac{1}{4}(\partial\vec{\pi})^2+\frac{1}{4f^2}(\vec{\pi}\cdot\partial\vec{\pi})^2$,

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?

$\endgroup$
  • 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
3
$\begingroup$

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)

$\endgroup$

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.