We commonly investigate the properties of SU(2) on the basis of SO(3). However, I want to directly calculte the infinitesimal generator of SU(2) according to the definition $$X_{i}=\frac{\partial U}{\partial \alpha_{i}}$$ from Lie group theory. But, where are the problems of the methods I used below?

First, I parameterize the SU(2) with the $(\theta, \phi, \gamma) $ like this: $$U=\begin{bmatrix} e^{i\theta}sin\phi & e^{i\gamma}cos\phi \\ -e^{-i\gamma}cos\phi & e^{-i\theta}sin\phi\end{bmatrix}$$ and the E is when $(\theta, \phi, \gamma)$= $(0,\frac{\pi}{2},0)$.

Second, I use the definition of infinitesimal generator like this: $$ I_{1}=\frac{\partial U}{\partial \theta}|_{(0,\frac{\pi}{2},0)}=i\begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}$$

$$ I_{2}=\frac{\partial U}{\partial \phi}|_{(0,\frac{\pi}{2},0)}=i\begin{bmatrix}0 & i\\ -i & 0 \end{bmatrix}$$

$$ I_{3}=\frac{\partial U}{\partial \gamma}|_{(0,\frac{\pi}{2},0)}=i\begin{bmatrix}0 & 0\\ 0 & 0 \end{bmatrix}$$ Here is the question...

Why do I get the 0 matrix? We should expect to have the Pauli Matrix. Isn't it?
Where is the problem from?


The problem is that your coordinates aren't well defined at $\theta=0$ and $\phi=\pi/2$. Note in particular that $$ U|_{(0,\frac{\pi}{2},\gamma)} = \begin{pmatrix}1&0\\0&1\end{pmatrix} $$ for any value of $\gamma$. A simpler choice is $$ \tilde{U} = \begin{pmatrix} x+iy & z+iw \\ -z+iw & x-iy \end{pmatrix}, $$ with $$ x = \sqrt{1 - y^2 - z^2 - w^2}. $$ Differentiating this you find $$ d\tilde U = i\begin{pmatrix} dy & +i\,dz + dw \\ -i\,dz + dw & -dy \end{pmatrix} - \frac{y\,dy+z\,dz+w\,dw}{\sqrt{1-y^2-z^2-w^2}}\begin{pmatrix}1&0\\0&1\end{pmatrix} $$ from which you can read off the Pauli matrices at the point $(x,y,z,w)=(1,0,0,0)$.

  • $\begingroup$ Thank you! "The parameters must be well defined" it seems that i should do this step cautiously. $\endgroup$ – Legend_Dyson May 22 '13 at 4:55

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