Angular Momentum as a Generator of Rotations in 3D

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Angular Momentum as a Generator of Rotations in 3D
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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,


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


we may write this infinitesimal rotation operator as


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


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


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.

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