View source for Angular Momentum as a Generator of Rotations in 3D

{{Quantum Mechanics A}}
Let us consider an infinitesimal rotation described by a vector <math> \mathbf{\alpha} \!</math> directed along the axis about which the rotation takes place and whose magnitude is the angle of the rotation.  We then have
:<math> \mathbf{r}' = \mathbf{r} + \mathbf{\alpha} \times \mathbf{r} = \mathbf{r} + \delta\mathbf{r}, </math>
 
where <math>\delta \mathbf{r}=\mathbf{\alpha}\times \mathbf{r}</math> is the change in the position vector <math>\mathbf{r}\!</math> of the particle due to such a rotation.

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

:<math>\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)</math>

[[Image:Rotation.jpg]]

'''Figure 1: Illustration of the rotation considered here.'''

The expression,

:<math>1+\mathbf{\alpha}\cdot(\mathbf{r}\times\mathbf{\nabla}),</math>

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

:<math>\mathbf{r}\times\mathbf{\nabla}=\frac{i}{\hbar}\hat{\mathbf{L}},</math>

we may write this infinitesimal rotation operator as

:<math>\hat{R}_{inf}=1+\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}.</math>

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

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

If we now let <math>N\rightarrow\infty,</math> we obtain

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

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

:<math>\hat{O}'=e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}\hat{O}e^{-\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}.</math>

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 <math>e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\hat{\mathbf{L}}}</math> on the wave function, as follows.  We first determine the effect of the operator on a position eigenstate:

:<math>\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|</math>

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.

:<math>\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)</math>

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