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Derive non-linear $\sigma$ model from a theory of SU(2) matirx

+ 2 like - 0 dislike
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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?

This post imported from StackExchange Physics at 2014-03-17 05:58 (UCT), posted by SE-user hongchaniyi
asked Nov 21, 2013 in Theoretical Physics by hongchaniyi (15 points) [ no revision ]
retagged Mar 17, 2014
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/…

This post imported from StackExchange Physics at 2014-03-17 05:59 (UCT), posted by SE-user user6818
Hm, google chrome is trying to tell me that the revisions for this question are in "Greek". Huh?

This post imported from StackExchange Physics at 2014-03-17 05:59 (UCT), posted by SE-user Dimensio1n0

1 Answer

+ 1 like - 0 dislike

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)

This post imported from StackExchange Physics at 2014-03-17 05:59 (UCT), posted by SE-user lionelbrits
answered Nov 22, 2013 by lionelbrits (110 points) [ no revision ]

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