Lets look at Euler angles, a representation of rotations

Finite rotations

🧠 Remember: the unitary transformation for a rotation about the vector angle $\vec \theta=(\theta_1,\theta_2,\theta_3)$ is

$$ \hat U_{\vec \theta}=e^{-i\vec \theta \cdot \hat L/\hbar}=e^{-i(\theta_x \hat L_x + \theta_y \hat L_y + \theta_z \hat L_z)/\hbar} $$

where $\hat L$ is the orbital angular momentum operator meaning that $[\hat L_i,\hat L_j]\ne 0$ for $i\ne j$

• The operators are none commutative meaning that we can’t do

  $$ e^{-\tfrac{i}{\hbar}\,(\theta_1 \hat{L}_x + \theta_2 \hat{L}_y + \theta_3 \hat{L}_z)}\ne e^{-\tfrac{i}{\hbar}\,\theta_1 \hat{L}_x}\,e^{-\tfrac{i}{\hbar}\,\theta_2 \hat{L}_y}\, e^{-\tfrac{i}{\hbar}\,\theta_3 \hat{L}_z} $$

  Making the mathematics impractical so instead we will try to represent it in terms of $\hat L^2$ and $\hat L_z$

🗒️ Note: while we can choose a coordinate system which makes the rotation 1 dimensional simplifying the math. This would not help as our measuring apparatus would likely not be in this arbitrary direction and so to find the eigenvalues in a meaningful direction (say $z$) we would need to worry about the commutations of $\hat L$ when we decompose.

Euler angles

In parametrisation we write a rotation about a fixed axes as three consecutive rotations

1. Rotation about the $z$-axis by an angle $\alpha$ where $0\le \alpha \le 2\pi$
2. Rotation about the $x$-axis by an angle $\beta$ where $0\le \beta \le\pi$
3. Rotation about the $z$-axis by an angle $\gamma$ where $0\le\gamma\le 2\pi$

• 🗒️ Note: yes we are only rotating about 2 axis ie $z-x-z$ but that is enough

• If we perform this sequence the wavefunction transforms as

  $$ \begin{aligned} \psi (\vec{r}) \to \psi'(\vec{r}) &= \psi \left( [R_z(\gamma) R_x(\beta) R_z(\alpha)]^{-1} \vec{r} \right)
  \\ &= \psi \left( R_z(-\alpha) R_x(-\beta) R_z(-\gamma) \vec{r} \right) = \hat{U}_{\alpha, \beta, \gamma} \psi (\vec{r}) \end{aligned} $$

• Now we write down $\hat U_{\alpha,\beta,\gamma}$

  $$ \begin{aligned} \hat{U}_{\alpha, \beta, \gamma} \psi(\vec{r}) &= \psi \big( R_z(-\alpha) R_x(-\beta) R_z(-\gamma) \vec{r} \big) \\ &= e^{-i \alpha \hat{L}_z / \hbar} \psi \big( R_x(-\beta) R_z(-\gamma) \vec{r} \big) \\ &= e^{-i \alpha \hat{L}_z / \hbar} e^{-i \beta \hat{L}_x / \hbar} \psi \big( R_z(-\gamma) \vec{r} \big) \\ &= e^{-i \alpha \hat{L}_z / \hbar} e^{-i \beta \hat{L}_x / \hbar} e^{-i \gamma \hat{L}_z / \hbar} \psi(\vec{r}) \end{aligned} $$

  🗒️ Note: this looks super similar to what we said was not allowed previously, the only difference is that here we rotate by each direction one after the other

• Which thus gives us

  $$ \boxed{\hat{U}_{\alpha, \beta, \gamma}=e^{-i \alpha \hat{L}_z / \hbar} e^{-i \beta \hat{L}_x / \hbar} e^{-i \gamma \hat{L}_z / \hbar}} $$

• Finally we can show that $\hat U _{\alpha,\beta,\gamma}$ is unitary

  $$ \hat{U}{\alpha, \beta, \gamma} \hat{U}{\alpha, \beta, \gamma}^{\dagger} = e^{-i\gamma \hat{L}_z / \hbar} e^{-i\beta \hat{L}_x / \hbar} e^{-i\alpha \hat{L}_z / \hbar} e^{i\alpha \hat{L}_z / \hbar} e^{i\beta \hat{L}_x / \hbar} e^{i\gamma \hat{L}_z / \hbar} = \hat{I} $$

Representation of rotations

💼 Case: lets consider the basis $\ket{l,m}$ such that