Finite rotations about an arbitrary axis in three dimensions

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

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

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)

3 thoughts on “Finite rotations about an arbitrary axis in three dimensions

  1. Pingback: Vector operators; transformation under rotation | Physics pages

  2. Pingback: Rotation of a vector wave function | Physics pages

  3. Pingback: Total angular momentum – finite rotations | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *