Angular Momentum as a Generator of Rotations in 3D

Let us consider an infinitesimal rotation described by a vector $\mathbf{\alpha} \!$ directed along the axis about which the rotation takes place and whose magnitude is the angle of the rotation. We then have

$\mathbf{r}' = \mathbf{r} + \mathbf{\alpha} \times \mathbf{r} = \mathbf{r} + \delta\mathbf{r},$

where $\delta \mathbf{r}=\mathbf{\alpha}\times \mathbf{r}$ is the change in the position vector $\mathbf{r}\!$ of the particle due to such a rotation.

Let us now consider a function of position, $\psi(\mathbf{r}).$ Substituting in the rotated coordinate and expanding to first order in $\delta\mathbf{r},$ we obtain

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

Figure 1: Illustration of the rotation considered here.

The expression,

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

may therefore be interpreted as an operator that performs an infinitesimally small rotation of position coordinates. Noticing that

$\mathbf{r}\times\mathbf{\nabla}=\frac{i}{\hbar}\hat{\mathbf{L}},$

we may write this infinitesimal rotation operator as

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

Note that this expression only applies to infinitesimal rotations. We may construct a rotation operator for finite rotations, however, as follows. Let $\mathbf{\alpha}$ be a finite rotation. Let us imagine performing this rotation as a sequence of $N\!$ rotations by $\frac{\mathbf{\alpha}}{N},$ where $N\!$ is large. Each of these rotations may be treated as infinitesimal. The full rotation operator becomes

$\hat{R}(\mathbf{\alpha})=\left (1+\frac{i}{\hbar}\frac{\mathbf{\alpha}}{N}\cdot\hat{\mathbf{L}}\right )^N.$

If we now let $N\rightarrow\infty,$ we obtain

$\hat{R}(\mathbf{\alpha})=e^{i\mathbf{\alpha}\cdot\hat{\mathbf{L}}/\hbar}.$

In this form, we recognize that angular momentum is a generator of rotations, similarly to how linear momentum generates translations.

The transformation rule for an operator is thus

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

This expression is valid for any rotation. We see that, if the operator commutes with both position and momentum, then it will remain unchanged by a rotation.

We can also calculate the effect of the unitary operator $e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}$ on the wave function, as follows. We first determine the effect of the operator on a position eigenstate:

$\langle \mathbf{r}_0|e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}\mathbf{ \hat{\mathbf{r}}} e^{-\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}=\langle\mathbf{r}_0|\hat{\mathbf{r}}'=\mathbf{r}'_0\langle \mathbf{r}_0|$

As expected, the effect is to produce another position eigenstate, this one at the image of the rotation. The effect on the wavefunction is therefore as follows.

$\psi'(\mathbf{r}_0)=\langle \mathbf{r}_0|\psi'\rangle=\langle \mathbf{r}_0|e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}|\psi\rangle=\langle \mathbf{r}'_0|\psi\rangle=\psi(\mathbf{r}'_0)$

This is just the wave function evaluated at the rotated point.