Rotation of a vector is easy and straight forward, just apply a rotation matrix on the vector. But rotating a function may be tricky, we need to transform the coordinate one by one.

r \rightarrow r' = R\cdot r

\displaystyle f(r)  \rightarrow f(r')

When rotating spherical harmonic, the thing becomes get. We can treat the spherical harmonic as a eigen state of angular momentum operator. The problem will becomes easy.

Recall that, the rotation operator for a eigen-state is

\displaystyle R(\alpha, \theta, \phi) = e^{-i\alpha J_z} e^{-i\theta J_y} e^{-i \phi J_z},

The matrix element is the Wigner D-matrix,

D^j_{m'm} = \left<jm'|R|jm\right>

The spherical harmonic is Y_{lm}(\Omega)

Thus, the rotation of spherical harmonic is

\displaystyle R(Y_{lm}) = \sum_{m=-j}^{j} Y_{lm'} D^j_{m'm}

or

Y'_{l} = Y_{l} \cdot D_{l}(R)

Capture.PNG

The above picture is Y_{20} rotated by \theta = 45 \deg.

We can see, the spherical harmonic formed a close set under Wigner D-matrix, that combination of them are themselves.


The derivation is follow. Notice that

Y_{lm} = \langle \hat{r} | lm\rangle

\langle \hat{r}| U(R) = \langle R(\hat{r}) |

Thus,

Y_{lm}(R(\hat{r})) = \langle R(\hat{r}) |lm\rangle = \langle \hat{r} |U(R) |lm \rangle

We can use the identity

\sum_{m'} |lm'\rangle \langle lm' | = 1

\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} \langle \hat{r} | lm' \rangle \langle lm' |U(R) |lm \rangle

Than

\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} Y_{lm'}(\hat{r}) D_{m'm}^{l}(R)

or

Y'_{l} = Y_{l} \cdot D_{l}(R)


It is easy to show that

(D_{m'm}^{l}(R))^{\dagger} = D_{mm'}^{l*}(R) = D_{m'm}^l(R^{-1})

or

D_{l}^{\dagger} = D_{l}^{T*}

Note that

D_{l} ^{\dagger} \cdot D_{l} = 1 =  D_{l} \cdot D_{l} ^{\dagger}

 

Advertisements