Angular Momentum as a Generator of Rotations in 3D

Let us consider an infinitesimal rotation $\mathbf {\alpha } \!$ directed along the axis about which the rotation takes place.We then have

$\mathbf {w} '=\mathbf {w} +\mathbf {\alpha } \times \mathbf {w} =\mathbf {w} +\delta \mathbf {w}$

The changes $\delta \mathbf {w}$ (in the radial vector $\mathbf {w} \!$ of the particle) due to such a rotation is:

$\delta \mathbf {w} =\mathbf {\alpha } \times \mathbf {w}$

so

$\psi \left(\mathbf {w} +\mathbf {\delta } \mathbf {w} \right)=\left(1+\mathbf {\alpha } \cdot \left(\mathbf {w} \times \mathbf {\nabla } \right)\right)\psi \left(\mathbf {w} \right)$

The expression

$1+\mathbf {\alpha } \cdot (\mathbf {w} \times \mathbf {\nabla } )$

is the operator of an infinitesimally small rotation. We recognize the equation

$\mathbf {w} \times \mathbf {\nabla } ={\frac {i}{\hbar }}\mathbf {L}$

Therefore, the infinitesimal rotation operator is

$\mathbf {R} _{inf}=1+{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L}$

This expression is only until the first order correction. The actual rotation operator is calculated by applying this operator N times where N goes to infinity. Doing so, we get the rotation operator for finite angle

$\mathbf {R} =e^{{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }$

In this form, we recognize that angular momentum is the generator of rotation. And we can write the equation relating the initial vector before rotation with the transformed vector as

$\mathbf {w} '=e^{{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }\mathbf {w} e^{-{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }$

This expression of the rotation operator is also valid when the rotation angle is not infinitesimal. What's more, this equation also implies that if we have a scalar instead of $\mathbf {w} \!$, it would be invariant. We can also calculate the effect of the unitary operator $e^{{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }$ on the states:

$\langle r_{0}|e^{{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }\mathbf {\hat {\mathbf {r} }} e^{-{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }=\langle r_{0}|{\hat {\mathbf {r'} }}=r_{0}'\langle r_{0}|$

$\Rightarrow \psi '(r_{0})=\langle r_{0}|\psi '\rangle =\langle r_{0}|e^{{\frac {i}{\hbar }}\mathbf {\alpha } \cdot \mathbf {L} }|\psi \rangle =\langle r_{0}'|\psi \rangle =\psi (r_{0}')$

This is the wavefunction evaluated at a rotated point.

A sample problem: the infinitesimal rotation.