# Tensor product of wavefunctions

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?

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.