# Derivation of Angular Momentum Operator

I was reading a book on theoretical quantum mechanics and the authors introduced the (orbital) angular momentum operator as the operator that generates rotations around an (arbitrary) axis. To do this, they first showed that any rotational matrix corresponding to a rotation of angle theta around an axis e can be written as: $$\mathbf R\left(\mathbf e, \,\vartheta\right) = \exp\left[\vartheta\boldsymbol\Omega_e\right],$$

where $\Omega$ is a skew-symmetric matrix.

They then proceeded by examining how a unitary operator corresponding to a rotation acts on the wave function and eventually related the two using the formula:

\begin{align} \psi\left(e^{-\vartheta\boldsymbol{\Omega}_e}\mathbf x\right)&=\left(\hat{I}-\vartheta\boldsymbol\Omega_e\mathbf x\cdot\nabla+\frac{\vartheta^2}{2!}\left(\boldsymbol\Omega_e\mathbf x\cdot\nabla\right)^2\right)\psi\left(\mathbf x\right)+\cdots \\ &=\left(e^{-\vartheta(\boldsymbol\Omega_e\mathbf x)\cdot\nabla}\psi\right)\left(\mathbf x\right).\tag{27.82} \end{align}

which they asked the reader to prove. I've been trying to prove the result (by simply differentiating the left hand side of the equation with respect to theta) for quite a while now, but I'm unfortunately not really getting anywhere. Maybe it's just a simple trick that I'm missing.

Anyway, I'd be grateful for any suggestions or hints.

## 1 Answer

The thing is that checking this in Cartesian coordinates by hand can get messy really fast. If you really want to go for it, you can try spherical coordinates instead and consider just a rotation around the $z$ axis. Then the rotation $\exp \left[ - \vartheta \, \Omega_z \right]$ applied on $\mathbf{x}$, with $\left( \Omega_z \right)_{ij}=\varepsilon_{ij3}$ and $\mathbf{x}=r \left\{\sin\theta \cos \phi,\sin\theta \sin \phi,\cos\theta\right\}$, simply does $\phi \rightarrow\phi+\vartheta$. Then your statement can be written as

\begin{equation} \psi \left(r,\theta,\phi+\vartheta \right) = \exp \left[\vartheta \, \partial_\phi \right] \, \psi \left(r,\theta,\phi \right) \, , \end{equation} which is just a Taylor expansion.

Then you can just argue that this covers the general case as one can just use a coordinate system adapted to the rotation such that the rotation axis matches the $z$ axis. This relays on the fact that any rotation can be written in the angle-axis parameterization.