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,

$D^j_{m'm} = \left$

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

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

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

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