# Rotations through a finite angle; use of polar coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.3.

The angluar momentum operator ${L_{z}}$ is the generator of rotations in the ${xy}$ plane. We did the derivation for infinitesimal rotations, but we can generalize this to finite rotations in a similar manner to that used for translations. The unitary transformation for an infinitesimal rotation is

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

For rotation through a finite angle ${\phi_{0}}$, we divide up the angle into ${N}$ small angles, so ${\varepsilon_{z}=\phi_{0}/N}$. Rotation through the full angle ${\phi_{0}}$ is then given by

$\displaystyle U\left[R\left(\phi_{0}\hat{\mathbf{z}}\right)\right]=\lim_{N\rightarrow\infty}\left(I-\frac{i\phi_{0}L_{z}}{N\hbar}\right)^{N}=e^{-i\phi_{0}L_{z}/\hbar} \ \ \ \ \ (2)$

The limit follows because the only non-trivial operator involved is ${L_{z}}$, so no commutation problems arise.

In rectangular coordinates, ${L_{z}}$ has the relatively non-obvious form

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle XP_{y}-YP_{x}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (4)$

so it’s not immediately clear that 2 does in fact lead to the desired rotation. Trying to calculate the exponential with ${L_{z}}$ expressed this way is not easy, given that the two terms ${x\frac{\partial}{\partial y}}$ and ${y\frac{\partial}{\partial x}}$ don’t commute.

It turns out that ${L_{z}}$ has a much simpler form in polar coordinates, and there are two ways of converting it to polar form. First, we recall the transformation equations.

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \rho\cos\phi\ \ \ \ \ (5)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle \rho\sin\phi\ \ \ \ \ (6)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}}\ \ \ \ \ (7)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \tan^{-1}\frac{y}{x} \ \ \ \ \ (8)$

From the chain rule, we can convert the derivatives:

 $\displaystyle \frac{\partial}{\partial x}$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial x}\frac{\partial}{\partial\rho}+\frac{\partial\cos\phi}{\partial x}\frac{\partial}{\partial\left(\cos\phi\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial x}\frac{\partial}{\partial\rho}-\sin\phi\frac{\partial\phi}{\partial x}\frac{\partial}{\left(-\sin\phi\right)\partial\phi}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x}{\rho}\frac{\partial}{\partial\rho}-\sin\phi\frac{-y/x^{2}}{1+y^{2}/x^{2}}\left(\frac{-1}{\sin\phi}\right)\frac{\partial}{\partial\phi}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x}{\rho}\frac{\partial}{\partial\rho}-\frac{y}{\rho^{2}}\frac{\partial}{\partial\phi} \ \ \ \ \ (12)$

Using similar methods, we get for the other derivative

 $\displaystyle \frac{\partial}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial y}\frac{\partial}{\partial\rho}+\frac{\partial\sin\phi}{\partial x}\frac{\partial}{\partial\left(\sin\phi\right)}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{y}{\rho}\frac{\partial}{\partial\rho}+\frac{x}{\rho^{2}}\frac{\partial}{\partial\phi} \ \ \ \ \ (14)$

Plugging these into 4 we have

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\left[x\left(\frac{y}{\rho}\frac{\partial}{\partial\rho}+\frac{x}{\rho^{2}}\frac{\partial}{\partial\phi}\right)-y\left(\frac{x}{\rho}\frac{\partial}{\partial\rho}-\frac{y}{\rho^{2}}\frac{\partial}{\partial\phi}\right)\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{x^{2}+y^{2}}{\rho^{2}}\frac{\partial}{\partial\phi}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (17)$

Another method of converting ${L_{z}}$ to polar coordinates is to consider the effect of ${U\left[R\right]}$ for an infinitesimal rotation ${\varepsilon_{z}}$ on a state vector expressed in polar coordinates ${\psi\left(\rho,\phi\right)}$. Shankar states that

$\displaystyle \left\langle \rho,\phi\left|U\left[R\right]\right|\psi\left(\rho,\phi\right)\right\rangle =\psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (18)$

If you don’t believe this, it can be shown using a method similar to that for the one-dimensional translation. In this case, we’re dealing with position eigenkets in polar coordinates, so we have

$\displaystyle U\left[R\right]\left|\rho,\phi\right\rangle =\left|\rho,\phi+\varepsilon_{z}\right\rangle \ \ \ \ \ (19)$

Applying this, we get

 $\displaystyle \left|\psi_{\varepsilon_{z}}\right\rangle$ $\displaystyle =$ $\displaystyle U\left[R\right]\left|\psi\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left[R\right]\int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho,\phi\right\rangle \left\langle \rho,\phi\left|\psi\right.\right\rangle \rho d\rho\;d\phi\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho,\phi+\varepsilon_{z}\right\rangle \left\langle \rho,\phi\left|\psi\right.\right\rangle \rho d\rho\;d\phi\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left|\rho^{\prime},\phi^{\prime}\right\rangle \left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime} \ \ \ \ \ (23)$

where in the last line, we used the substitution ${\phi^{\prime}=\phi+\varepsilon_{z}}$. (The substitution ${\rho^{\prime}=\rho}$ is used just to give the radial variable a different name in the integrand.) We can use the same limits of integration for ${\phi}$ and ${\phi^{\prime}}$, since we just need to ensure that the integral covers the total range of angles. It then follows that

 $\displaystyle \left\langle \rho,\phi\left|\psi_{\varepsilon_{z}}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\left\langle \rho,\phi\left|\rho^{\prime},\phi^{\prime}\right.\right\rangle \left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\delta\left(\rho-\rho^{\prime}\right)\delta\left(\phi-\phi^{\prime}\right)\left\langle \rho^{\prime},\phi^{\prime}-\varepsilon_{z}\left|\psi\right.\right\rangle \rho^{\prime}d\rho^{\prime}\;d\phi^{\prime}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (26)$

Combining this with 1 we have

$\displaystyle \left\langle \rho,\phi\left|I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right|\psi\right\rangle =\psi\left(\rho,\phi-\varepsilon_{z}\right) \ \ \ \ \ (27)$

Expanding the RHS to order ${\varepsilon_{z}}$ we have

$\displaystyle \left\langle \rho,\phi\left|I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right|\psi\right\rangle =\psi\left(\rho,\phi\right)-\varepsilon_{z}\frac{\partial\psi}{\partial\phi} \ \ \ \ \ (28)$

from which 17 follows again.

Once we have ${L_{z}}$ in this form, the exponential form of a finite rotation is easier to interpret, for we have, from 2

 $\displaystyle e^{-i\phi_{0}L_{z}/\hbar}$ $\displaystyle =$ $\displaystyle \exp\left[-\phi_{0}\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\phi_{0}\frac{\partial}{\partial\phi}+\frac{\phi_{0}^{2}}{2!}\frac{\partial^{2}}{\partial\phi^{2}}+\ldots \ \ \ \ \ (30)$

Applying this to a state function ${\psi\left(\rho,\phi\right)}$, we see that we get the Taylor series for ${\psi\left(\rho,\phi-\phi_{0}\right)}$, so the exponential does indeed represent a rotation through a finite angle.