Tensor product of wavefunctions

Question

In a system of two non-interacting particles in a one-dimensional infinite square well, we represent the eigenstate of the whole system as the tensor product of the eigenstates of the individual particles: $|n_1,n_2\rangle=|n_1\rangle \otimes |n_2\rangle$, since both kets belong to different Hilbert spaces.

However, if we solve the problem by separation of variables in the time-independent Schrödinger equation

$$ \widehat{H}_{0}\left(x_{1}, x_{2}\right)=\widehat{H}_{0}\left(x_{1}\right)+\widehat{H}_{0}\left(x_{2}\right) $$

$$\psi_{n_{1} n_{2}}\left(x_{1}, x_{2}\right)=\psi_{n_{1}}\left(x_{1}\right) \psi_{n_{2}}\left(x_{2}\right)=\frac{2}{L}\sin{\left(\frac{n_1\pi x_1}{L}\right)}\sin{\left(\frac{n_2\pi x_2}{L}\right)}$$

Why is not the tensor product used in case the approach is made through wavefunctions?

Answer 1

I'd say that it is merely for historical and cultural reasons. Mathematically there is an isomorphism $$ L^2[{\mathbb R}^3]\simeq L^2[{\mathbb R}]\otimes L^2[{\mathbb R}]\otimes L^2[{\mathbb R}], $$ but we customarily use the first form for a single particle moving in 3d.