We know that $B_2(H)$ (collection of Hilbert Schmidt operators on $H$) forms a Hilbert space with respect to the inner product $\left<A,B\right>=\text{tr}(B^*A)$. I was wondering if there is any explicit orthonormal basis for $B_2(H)$ ?
I can notice that $\{e_i\otimes e_j\}_{i\in I}$ forms a orthonormal set in $B_2(H)$, where $\{e_i\}_{i\in I}$ is an orthonormal basis for $H$.
And it's not only orthonormal, but complete: If $A$ is orthogonal to $e_i\otimes e_j$ for all $i,j$, then you have that $$ 0=\langle A,e_i\otimes e_j\rangle=\operatorname{Tr}(A\,e_j\otimes e_i)=\langle Ae_i,e_j\rangle. $$ From there you get that $\langle Ax,y\rangle=0$ for all $x,y$, so $A=0$, and thus your set is an orthonormal basis.