# Derivation of Cayley-Klein parameters in Goldstein

In his derivation for Cayley-Klein parameters, Goldstein introduces the matrix

$$\mathbf{Q}=\begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}$$

and says that the following unitary condition on $$\mathbf{Q}$$

$$\mathbf{Q}^\dagger\mathbf{Q}=\mathbf{1}=\mathbf{Q}\mathbf{Q}^\dagger$$

gives the following equations:

$$\alpha\alpha^*+\gamma\gamma^*=1\\ \beta\beta^*+\delta\delta^*=1\\ \alpha^*\beta+\gamma^*\delta=0$$

These equations were apparently derived from $$\mathbf{Q}^\dagger\mathbf{Q}=\mathbf{1}$$, and I understand how to arrive at them. However, if you do the same thing with $$\mathbf{Q}\mathbf{Q}^\dagger=\mathbf{1}$$ you get another equation:

$$\alpha\gamma^*+\beta\delta^*=0$$

This equation is not obviously not independent from the three equations listed above, but Goldstein seems to think it is. Is this last equation independent from the previous three or not?

## 2 Answers

Actually, $$\mathbf{Q}\mathbf{Q}^\dagger=\mathbf{1}$$ can be derived from $$\mathbf{Q}^\dagger\mathbf{Q}=\mathbf{1}$$ (a matrix commutes with its inverse), so you shouldn't gain any new information by looking at $$\mathbf{Q}\mathbf{Q}^\dagger$$.

Another way of seeing it is to actually derive the coefficients of $$\mathbf{Q}$$ from the $$3$$ equations you wrote. WLOG, it is possible to write:

\begin{align*} \alpha = \cos(\theta) e^{i \phi_{\alpha}}, \, &\beta = \sin(\theta) e^{i \phi_{\beta}}, \\ \gamma = -\sin(\theta) e^{i \phi_{\gamma}}, \, &\delta = cos(\theta) e^{i \phi_{\delta}} \end{align*}

With $$\phi_{\alpha} - \phi_{\beta} = \phi_{\gamma} - \phi_{\delta} \, [2 \pi]$$.

From this it is easy to see that the other equation you wrote ($$\alpha \gamma^* + \beta \delta^*=0$$) is verified.

So in the end, looking at the coefficients of $$\mathbf{Q}\mathbf{Q}^\dagger$$ is indeed redundant.

$$\mathbf{Q}^{\dagger}\mathbf{Q}=\left(\begin{array}\\ \alpha^* & \gamma^* \\ \beta^* & \delta ^* \end{array}\right)\left(\begin{array}\\ \alpha & \beta \\ \gamma & \delta \end{array}\right)=\left(\begin{array}\\ \alpha^*\alpha+\gamma^*\gamma & \alpha^*\beta+\gamma^*\delta \\ \beta^*\alpha+\delta^*\gamma & \beta^*\beta + \delta^*\delta \end{array}\right)=\left(\begin{array}\\ 1 & 0 \\ 0 & 1 \end{array}\right)$$

I think this is what you were after?

• This is how you arrive at those three equations, however I already knew how to do that. I was wondering whether adding $\alpha\gamma^*+\beta\delta^*=0$ (which is derived from $\mathbf{Q}\mathbf{Q}^\dagger=1$ instead of $\mathbf{Q}^\dagger\mathbf{Q}=1$) to the group of three equations I listed above would be redundant. – user109923 Dec 8 '20 at 1:55