# Finite rotations about an arbitrary axis in three dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.4.3.

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The operators for an infinitesimal rotation in 3-d are

 $\displaystyle U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{x}L_{x}}{\hbar}\ \ \ \ \ (1)$ $\displaystyle U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{y}L_{y}}{\hbar}\ \ \ \ \ (2)$ $\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (3)$

If we have a finite (larger than infinitesimal) rotation about one of the coordinate axes, we can create the operator by dividing up the finite rotation angle ${\theta}$ into ${N}$ small increments and take the limit as ${N\rightarrow\infty}$, just as we did with finite translations. For example, for a finite rotation about the ${x}$ axis, we have

$\displaystyle U\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=\lim_{N\rightarrow\infty}\left(I-\frac{i\theta L_{x}}{N\hbar}\right)^{N}=e^{-i\theta L_{x}/\hbar} \ \ \ \ \ (4)$

What if we have a finite rotation about some arbitrarily directed axis? Suppose we have a vector ${\mathbf{r}}$ as shown in the figure:

The vector ${\mathbf{r}}$ makes an angle ${\alpha}$ with the ${z}$ axis, and we wish to rotate ${\mathbf{r}}$ about the ${z}$ axis by an angle ${\delta\theta}$. Note that this argument is completely general, since if the axis of rotation is not the ${z}$ axis, we can rotate the entire coordinate system so that the axis of rotation is the ${z}$ axis. The generality enters through the fact that we’re keeping the angle ${\alpha}$ arbitrary.

The rotation by ${\delta\theta\hat{\mathbf{z}}\equiv\delta\boldsymbol{\theta}}$ shifts the tip of ${\mathbf{r}}$ along the circle shown by a distance ${\left(r\sin\alpha\right)\delta\theta}$ in a counterclockwise direction (looking down the ${z}$ axis). This shift is in a direction that is perpendicular to both ${\hat{\mathbf{z}}}$ and ${\mathbf{r}}$, so the little vector representing the shift in ${\mathbf{r}}$ is

$\displaystyle \delta\mathbf{r}=\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r} \ \ \ \ \ (5)$

Thus under the rotation ${\delta\boldsymbol{\theta}}$, a vector transforms as

$\displaystyle \mathbf{r}\rightarrow\mathbf{r}+\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r} \ \ \ \ \ (6)$

Just as with translations, if we rotate the coordinate system by an amount ${\delta\boldsymbol{\theta}}$, this is equivalent to rotating the wave function ${\psi\left(\mathbf{r}\right)}$ by the same angle, but in the opposite direction, so we require

$\displaystyle \psi\left(\mathbf{r}\right)\rightarrow\psi\left(\mathbf{r}-\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r}\right) \ \ \ \ \ (7)$

A first order Taylor expansion of the quantity on the RHS gives

$\displaystyle \psi\left(\mathbf{r}-\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r}\right)=\psi\left(\mathbf{r}\right)-\left(\delta\boldsymbol{\theta}\times\mathbf{r}\right)\cdot\nabla\psi \ \ \ \ \ (8)$

The operator generating this rotation will have the form (in analogy with the forms for the coordinate axes above):

$\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]=I-\frac{i\delta\theta}{\hbar}L_{\hat{\theta}} \ \ \ \ \ (9)$

where ${L_{\hat{\theta}}}$ is an angular momentum operator to be determined.

Writing out the RHS of 8, we have

 $\displaystyle \psi\left(\mathbf{r}\right)-\left(\delta\boldsymbol{\theta}\times\mathbf{r}\right)\cdot\nabla\psi$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\left(\delta\theta_{y}z-\delta\theta_{z}y\right)\frac{\partial\psi}{\partial x}+\left(\delta\theta_{x}z-\delta\theta_{z}x\right)\frac{\partial\psi}{\partial y}-\left(\delta\theta_{x}y-\delta\theta_{y}x\right)\frac{\partial\psi}{\partial z}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\delta\theta_{x}\left(y\frac{\partial\psi}{\partial z}-z\frac{\partial\psi}{\partial y}\right)-\delta\theta_{y}\left(z\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial z}\right)-\delta\theta_{z}\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\delta\boldsymbol{\theta}\cdot\frac{i}{\hbar}\mathbf{r}\times\mathbf{p}\psi\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\frac{i}{\hbar}\delta\boldsymbol{\theta}\cdot\mathbf{L}\psi\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]\psi \ \ \ \ \ (14)$

Comparing this with 9, we see that

$\displaystyle L_{\hat{\theta}}=\hat{\boldsymbol{\theta}}\cdot\mathbf{L} \ \ \ \ \ (15)$

where ${\hat{\boldsymbol{\theta}}}$ is the unit vector along the axis of rotation. Since all rotations about the same axis commute, we can use the same procedure as above to generate a finite rotation ${\boldsymbol{\theta}}$ about an arbitrary axis and get

$\displaystyle U\left[R\left(\boldsymbol{\theta}\right)\right]=e^{-i\boldsymbol{\theta}\cdot\mathbf{L}/\hbar} \ \ \ \ \ (16)$