Completeness of $L^2([0,1]) \otimes L^2([0,1])$.

Question

I'm trying to understand the Hilbert space of a quantum system of two particles in the infinite square well. The claim is that this is $L^2([0,1] × [0,1])$, which is said to be isomorphic to the tensor product $L^2([0,1]) \otimes L^2([0,1])$. The latter space is not entirely obvious to me yet. Indeed, to have any hopes of being a Hilbert space, we need an inner product. The most natural choice seems to be $$ \left<v_1 \otimes w_1, v_2 \otimes w_2 \right>_{\otimes} := \left<v_1, v_2 \right>_{L^2} \cdot \left<w_1, w_2 \right>_{L^2}. $$ Subsequently we must determine whether this space is complete w.r.t. the metric induced by the norm induced by this inner product. Supposedly, it is not, and we need to take its completion. I would like to know why. To this end, I construct the element $x: =\sum_{n=1}^∞ \frac{1}{n}e_n \otimes e_n$. As $$\Vert x \Vert = \sqrt{\sum_{n=1}^∞ \frac{1}{n^2}} = \frac{π}{\sqrt{6}}, $$ the series converges in the completion of $L^2([0,1]) \otimes L^2([0,1])$. How can I show that it does not converge within $L^2([0,1]) \otimes L^2([0,1])$?

Answer 1

You have to be careful with what you mean when you write $L^2 \otimes L^2$. There are different notions of tensor products which mean genuinely different things in this case: There is the algebraic tensor product $A\otimes_{alg} B$ which is the space of all finite sums $\sum_{i=1}^N a_i\otimes b_i$ and there are various notions of topological tensor products which are defined to be completions of the algebraic tensor product w.r.t. to various topologies. In your case the "Hilbert space tensor product" or "Hilbert-Schmidt tensor product" is defined as the completion of the algebraic tensor product w.r.t. to the norm induced by the scalar product you found. This is sometimes denoted as $\widehat{\otimes}$ to distinguish it from the algebraic tensor product.

Now it is true that $L^2([0,1]) \widehat{\otimes} L^2([0,1])$ is isometrically isomorphic to $L^2([0,1]\times[0,1])$, but it is NOT true that it is isomorphic to the algebraic tensor product because the latter is not complete.

To see this, you can show different things. One way would be to consider $L^2 \otimes_{alg} L^2$ as a subspace of $L^2([0,1]^2)$ via the isometric embedding $f\otimes g \mapsto ((x,y)\mapsto f(x)g(y))$ and show that this is a dense, non-closed subspace of $L^2([0,1]^2)$.

To see this, consider the measurable sets generated by $L^2 \otimes_{alg} L^2$ and compare them with $L^2([0,1]^2)$, i.e. looking at characteristic functions $\chi_A$ with $A\subseteq [0,1]^2$ contained both spaces. You could try to prove that the algebraic tensor product contains all $A$ in the set-algebra generated by $\{A_1\times A_2 \mid A_i\subseteq[0,1]\}$ (up to sets of measure zero of course) whereas $L^2 \widehat{\otimes} L^2$ contains all the $A$ in the corresponding sigma-algebra. So if can exhibit a subset that is contained in one, both not the other, the corresponding characteristic function proves that $L^2 \otimes_{alg} L^2$ is not closed in $L^2([0,1]^2)$.