# When is separating the total wavefunction into a space part and a spin part possible?

The total wavefunction of an electron $$\psi(\vec{r},s)$$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s}$$ where $$\phi(\vec{r})$$ is the space part and $$\zeta_{s,m_s}$$ is the spin part of the total wavefunction $$\psi(\vec{r},s)$$. In my notation, $$s=1/2, m_s=\pm 1/2$$.

Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.

Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $$\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $$\psi(\vec{r}_1,\vec{r}_2;s_1,s_2)$$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.

Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.

[any arbitrary] wavefunction of an electron $$\psi(\vec{r},s)$$ can always be written as $$\psi(\vec{r},s)=\phi(\vec{r})\zeta_{s,m_s} \tag 1$$ where $$\phi(\vec{r})$$ is the space part and $$\zeta_{s,m_s}$$ is the spin part of the total wavefunction $$\psi(\vec{r},s)$$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $$\phi_1$$ and $$\phi_2$$, and two orthogonal spin states, $$\zeta_1$$ and $$\zeta_2$$, and define $$\psi = \frac{1}{\sqrt{2}}\bigg[\phi_1\zeta_1+\phi_2\zeta_2 \bigg].$$
If the hamiltonian is separable into spatial and spin components as $$H = H_\mathrm{space}\otimes \mathbb I+ \mathbb I \otimes H_\mathrm{spin},$$ with $$H_\mathrm{space}\otimes \mathbb I$$ commuting with all spin operators and $$\mathbb I \otimes H_\mathrm{spin}$$ commuting with all space operators, then there exists an eigenbasis for $$H$$ of the separable form $$(1)$$.
To build that eigenbasis, simply diagonalize $$H_\mathrm{space}$$ and $$H_\mathrm{spin}$$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)