When can i separate spin from the wavefunction? I am currently working on a Tight-Binding model and for the derivation of nearest neighbor spin interactions I have terms like
$$
\left<x\uparrow\right|\eta\;\mathbf{l}\cdot\mathbf{s}\left|z\downarrow\right>
$$
where $\left|x\uparrow\right>$ is an atomic orbital.For the simplification it is very important that i can assume 
$$
\left|x\uparrow\right> = \left|x\right>\left|\uparrow\right>
$$
My question is:


*

*is this always possible in a solid body or do i neglect some effects with that? (like spin polarization or something?)

 A: @ChrisWhite is right I believe you don't need to worry about that since using the tensor product notation $|x s\rangle = |x \rangle  \otimes |s\rangle$ and the angular momentum and spin operators just act on their part of the Hilbert space so we can write them as $l=l\oplus1_s$ and $s=1_x \oplus s$ and they would act in the following way 
$$
l|x s\rangle = (l\oplus1_s)|x \rangle  \otimes |s\rangle = l|x \rangle  \otimes |s\rangle
$$
an viceversa for the spin. Then the scalar product is
$$
l \cdot s = \sum_i  (l_i\oplus1_s)(1_x \oplus s_i) = \sum_i l_i \oplus s_i
$$
which will act as you wanted
$$
(l \cdot s)|x s\rangle =\left( \sum_i l_i \oplus s_i\right)|x \rangle  \otimes |s\rangle = \sum_i l_i|x \rangle  \otimes s_i|s\rangle
$$
The problem with factorisation usually appears when you consider more than one electron, then by the antisymmetry requirement you will have to include certain parts of the Hilbert space of two particles which you won't be able to write as the product of two single particle states. For instance if the spatial part of the wave function is antisymmetric you would need for instance the symmetric $s_z=0$ state of the $s=1$ triplet 
$$
\frac{1}{\sqrt(2)}(|\uparrow \downarrow \rangle +| \downarrow \uparrow \rangle  )
$$
which you cannot factorise as two separate one particle spin states.
