Direct sum of Hilbert spaces

Let $H$ be a separable Hilbert space with an orthonormal basis $\{e_n\}_{n =0}^{\infty}$. Consider a direct sum, $H \oplus H$. What is the orthonormal basis of $H \oplus H$ ? Is it $(e_n, e_m)_{n,m \in \mathbb{N}}$?

In finite dimensional case, say the dimension of $H$ is $n$ we have that $\dim H \oplus H = 2n$, in infinite dimensional case the dimension will be the same, that is $\aleph_0$.

Thank you for any help I try to understand the Hilbert space direct sum.

• No. Let $\{e_n\}$ be an orthonormal basis of $H$. Writing $(x,y)$ the vectors of $H\oplus H$, an orthonormal basis is given by $\{(e_n,0)\}\cup\{(0,e_n)\}$. Just the union. But for the tensor product Hilbert space $H\otimes H$ (which is the completion of the algebraic tensor product for the ad hoc inner product), an orthonormal basis is $\{e_n\otimes e_m,\,n,m\}$, which is maybe what got you confused. – Julien Jun 3 '13 at 22:33
• An easy way to see that the vectors $(e_n,e_m)$ don't constitute an orthonormal basis is to calculate the inner products and/or norms of some of them. Just a little more work gets you that they're not a basis at all, because they're linearly dependent; for example $(e_1,e_1)+(e_2,e_2)=(e_1,e_2)+(e_2,e_1)$. – Andreas Blass Jun 4 '13 at 2:45
• @julien Your comment is really an answer... – ˈjuː.zɚ79365 Jun 4 '13 at 6:15