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)

Image:Rotation.jpg

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.

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