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I'm currently studying field theory and I'm having some trouble with conserved charge given in field components. If we have a complex scalar action of a field $\phi=(\phi_1,\phi_2)^T$ that is

$$S[\phi]={\int}\left[ \partial^\mu\phi^\dagger\partial_\mu\phi-m^2\phi^\dagger\phi-\frac{1}{2}\lambda(\phi^\dagger\phi)^2\right]d^4x.$$

In a global $SU(2)$-symmetry when $\phi$ belongs to $SU(2)$ two dimensional representation $$\phi\rightarrow\phi'=g\phi,\quad g=e^{i\chi_iT^i},$$

where $T^i$ is the generator chosen as $T^i=\frac{1}{2}\tau^i$ and $\tau^i$ are Pauli matrices. Now I've gotten that the conserved charges are

$$Q_i=-i\int d^3x\:\phi^\dagger \overset{\leftrightarrow}{\partial^0}(T^i\phi).$$

Now the next part says that written with the components of the field these should produce

$$Q_1=\mathrm{Im}\int d^3x\:\phi_2^\dagger \overset{\leftrightarrow}{\partial^0}\phi_1,\quad Q_2=\mathrm{Re}\int d^3x\:\phi_2^\dagger \overset{\leftrightarrow}{\partial^0}\phi_1, \quad Q_1=\frac{i}{2}\int d^3x\:(\phi_2^\dagger \overset{\leftrightarrow}{\partial^0}\phi_2-\phi_1^\dagger \overset{\leftrightarrow}{\partial^0}\phi_1)$$

Now I have no idea on how to get these. Why does the generator $T^i$ vanish for example?

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Aren't they just plugging in the Pauli matrices explicitly? – user37496 Feb 10 '14 at 17:49
I tried that yes but for some reason didn't get the same results. – Apogee Feb 10 '14 at 17:56
So perhaps so what you tried so people can help you. – NowIGetToLearnWhatAHeadIs Feb 10 '14 at 18:53

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