How Do Spherical Harmonic Tensors Behave Under the Laplacian Operator?

[latentcorpse]

Let $C_{i_1i_2 \dots i_l}$ be a symmetric traceless tensor of rank $l$. Let $\hat{x}= \frac{x}{|x|}$ be a three dimensional unit vector on the unit sphere. Define a tangential derivative such that $\nabla_i \hat{x_j} = \delta_{ij} - \hat{x_i} \hat{x_j}$. For the spherical harmonic $Y_l(\hat{x})=C_{i_1i_2 \dots i_l} \hat{x_{i_1}} \hat{x_{i_2}} \dots \hat{x_{i_l}}$ show that

$\nabla^2 Y_l( \hat{x} ) = -l(l+1) Y_l( \hat{x})$

I'm not really getting anywhere here as I can't see how the $\nabla^2$ moves through the tensor so that i can act it on the x's.

 

[diazona]

Try this:
$$\nabla^2 Y_l(\hat{x}) = \delta^{mn}\nabla_m[\nabla_n Y_l(\hat{x})]$$
I'm not sure if you can consider the tensor coefficients $C_{i_1 i_2\cdots i_l}$ to be position-independent, i.e. to have the property
$$\nabla_m C_{i_1 i_2\cdots i_l} = 0$$

 

[fzero]

Try this:
$$\nabla^2 Y_l(\hat{x}) = \delta^{mn}\nabla_m[\nabla_n Y_l(\hat{x})]$$
I'm not sure if you can consider the tensor coefficients $C_{i_1 i_2\cdots i_l}$ to be position-independent, i.e. to have the property
$$\nabla_m C_{i_1 i_2\cdots i_l} = 0$$

They're spherical harmonics, so the $C_{i_1 i_2\cdots i_l}$ are indeed constants. The coefficients for low orders can be deduced from the table at http://en.wikipedia.org/wiki/Table_of_spherical_harmonics