Orthonormal basis for $L^2(\mathbb R^3)$

Question

It is well-known that the Hermite functions form an orthonormal basis for $L^2(\mathbb R)$.

Is there an orthonormal basis for $L^2(\mathbb R^3)$?

Thanks.

Answer 1

This is an elaboration of the comment left by L. S. the Unknown

Given any orthonormal basis $\{\phi_i\}_{i\in\mathbb N}$ of $L^2(\mathbb R)$, we can form an orthonormal basis for $L^2(\mathbb R^n)$ simply by taking products of the basis vectors like this: $$ \phi_{i_1,i_2,\ldots,i_n}(x_1,x_2,\ldots,x_n)=\phi_{i_1}(x_1)\phi_{i_2}(x_2)\cdots\phi_{i_n}(x_n). $$ The reason it works is because $$ \bigl\langle\phi_{i_1,i_2,\ldots,i_n}(x_1,x_2,\ldots,x_n),\phi_{j_1,j_2,\ldots,j_n}(x_1,x_2,\ldots,x_n)\bigr\rangle=\prod_{k=1}^n\langle \phi_{i_k},\phi_{j_k}\rangle, $$ which equals $1$ if and only if $i_k=j_k$ for all $k$, and otherwise equals $0$ - the defining property of an orthonormal basis.