I am just a beginner of group theory. I saw an $SU(2)$ example (example 4.16) in the book by Nadir Jeevanjee, An introduction to Tensors and group theory for Physicists. For $SU(2)$ elements, they satisfy:


The generic element is:

$$\begin{pmatrix} \alpha\ \ \beta\\ -\bar{\beta}\ \bar{\alpha} \end{pmatrix} ,\ \alpha,\beta\in\mathbb{C} ,\ |\alpha|^2+|\beta|^2=1. $$

The author just directly write one kind of parametrization: $$\begin{pmatrix} e^{i(\psi+\phi)/2}\cos{\frac{\theta}{2}}\ \ ie^{i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\\ ie^{-i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\ \ e^{-i(\psi+\phi)/2}\cos{\frac{\theta}{2}} \end{pmatrix}. $$

The three angles $\theta,\psi,\phi$ are Euler angles. From $|\alpha|^2+|\beta|^2=1$, I can get a 3-sphere.

If $\alpha=u+iv,\ \beta=x+iy$, I get: $u^2+v^2+x^2+y^2=1.$ I can use a parametrization:

$$ u=\cos{\phi_1}\\ v=\sin{\phi_1}\cos{\phi_2}\\ x=\sin{\phi_1}\sin{\phi_2}\cos{\phi_3}\\ y=\sin{\phi_1}\sin{\phi_2}\sin{\phi_3}. $$ But apparently this is not the case which is related to Euler angles. How can I get the parametrization by the author and relate to the Euler angles?


1 Answer 1


To get the “usual” Euleurian expression as the product $R_z(\zeta)R_y(\eta)R_z(\gamma)$, with $$ R_k(\varphi)=e^{i\varphi \sigma_k} $$ you need to set \begin{align} \zeta=\psi+\pi/4\, ,\qquad \gamma=\phi-\pi/4\, ,\qquad \eta=\theta\, . \end{align} The rest you can read off: $$ \alpha=e^{i(\psi+\phi)}\cos\theta/2 = \cos\left((\psi+\phi)/2\right)\cos(\theta/2) + i \sin\left((\psi+\phi)/2\right)\cos(\theta/2) $$ so $u=\cos\left((\psi+\phi)/2\right)\cos\theta/2$ etc.

  • $\begingroup$ Thank you so much for your help!!! $\endgroup$
    – Hsu Bill
    Jan 4, 2022 at 17:30

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