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Eugeniu Plamadeala edited untitled.tex
about 9 years ago
Commit id: e3d852b2bcf08911a8acd5dede5e719ffcdb24c5
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index 0c41e7b..bf43f9a 100644
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$$ -S_G = \frac{1}{4\pi K} \int d\tau dx \left[ i \partial_\tau \phi_R \partial_x \phi_R - i \partial_\tau \phi_L \partial_x \phi_L - ... \right]$$
It looks just like the perfect metal action, so I conclude that (dropping the factor of K):
$$ \rho_R(x) =
\frac{-1}{2\pi} \frac{1}{2\pi} \partial_x \phi_R $$
$$ \rho_L(x) = \frac{-1}{2\pi} \partial_x \phi_R $$
From the commutation for the non-chiral boson.