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Let $X_i$ be the generators of $SU(2)$ and let the parameters of the rotation be $\theta, \phi, \delta$ such that the matrix $R = e^{i(\phi X_{1} + \delta X_{2} + \theta X_{3})}$, where $R$ is an element of $SU(2)$.

So, $R = 1 + i(\phi X_{1} + \delta X_{2} + \theta X_{3}) + \frac{1}{2!}[i(\phi X_{1} + \delta X_{2} + \theta X_{3})]^2 + ...$ via Taylor expansion about the zero matrix $\textbf{0}$.

So, $-i \frac{\partial R}{\partial \phi}|_{\theta, \phi, \delta = 0} = -i\ [iX_{1} + (i \phi X_{1}+i \delta X_{2} + i \theta X_{3})(i X_{1}) + ...]|_{\theta, \phi, \delta = 0} = X_{1}$, and similarly $-i \frac{\partial R}{\partial \delta}|_{\theta, \phi, \delta = 0} = X_{2}$, and $-i \frac{\partial R}{\partial \theta}|_{\theta, \phi, \delta = 0} = X_{3}$.

Therefore, $X_{1} = -i \frac{\partial R}{\partial \phi}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \phi}\begin{pmatrix}a & b \\ -b^* & a^* \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \phi}\begin{pmatrix}e^{i\theta}cos \phi & e^{i\delta}sin \phi \\ -e^{-i\delta}sin \phi & e^{-i\theta}cos \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}-e^{i\theta}sin \phi & e^{i\delta}cos \phi \\ -e^{-i\delta}cos \phi & -e^{-i\theta}sin \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}$, which is one of the Pauli matrices.

But, $X_{2} = -i \frac{\partial R}{\partial \delta}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \delta}\begin{pmatrix}a & b \\ -b^* & a^* \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \frac{\partial}{\partial \delta}\begin{pmatrix}e^{i\theta}cos \phi & e^{i\delta}sin \phi \\ -e^{-i\delta}sin \phi & e^{-i\theta}cos \phi \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & ie^{i\delta}sin \phi \\ ie^{-i\delta}sin \phi & 0 \end{pmatrix}|_{\theta, \phi, \delta = 0} = -i \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$. But, I don't think that the zero matrix is one of the generators of the $SU(2)$ group. In fact, it is not even a Pauli matrix. I was wondering if someone can point out the error in my calculation of $X_{2}$.

nightmarish
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1 Answers1

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So I discussed this today and as it turns out there's a coordinate singularity at the identity in those co-ordinates. So the identity is given by $\theta=\phi=\delta=0$, but equally well the identity is given by $\theta=\phi=0, \delta=\text{anything}$. In fact they're singular whenever $\phi=0$, since the parameter $\delta$ is redundant.

Now we want to compute the lie algebra which is identified with the tangent space at the identity of the group, but one of our co-ordinates $\delta$ isn't good there, so differentiating with respect to that co-ordinate makes no sense, and gives the non-nonsensical result that the zero matrix is a generator, when they should be pauli matrices.

However suppose instead we let $a=x+iy$ and $b=u+iv$, we have the relation that $x^2+y^2+u^2+v^2=1$, so we can let $x=+\sqrt{1-(y^2+u^2+v^2)}$ and our matrix is:

$$\begin{pmatrix}x+iy & u+iv \\ -u+iv & x-iy \end{pmatrix},$$

and we compute derivatives with respect to $y,u,v$ evaluated at $y=u=v=0$.

Then $-i\cdot \frac{\partial}{\partial y}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}=\sigma_z$

And $-i\cdot \frac{\partial}{\partial u}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}=\sigma_y$

And finally $-i\cdot \frac{\partial}{\partial v}\Big|_{y=u=v=0}$ gives $\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}=\sigma_x$

snulty
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