Identical particle state with spin

I think the question already has the answer. In this case $|psi_i rangle$ is the spatial part, and $|sigma_i rangle$ is the spin part of $i^{text{th}}$ particle. These two are different properties and lie in seperate spaces. One is position living in a Hilbert space of infinite dimensions and the other is spin Hilbert space of two dimensions, representing an internal property. Combined wavefunction is the direct product of each space for just one particle.

Since there are two fermions associated with different spaces, we shall take the direct product of each of the particles. Anti-symmetrizing the wavefunction using Slater determinant yields the equation mentioned by OP.

See the below for detailed explanation.

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• The state vector $|Psi^{(n)} rangle$ of a system of $N$ particles is an element of the direct product space, $$ |Psi^{(n)} rangle = |psi{(1)} rangle otimes |psi{(2)} rangle otimes |psi{(3)} rangle otimes­ cdots otimes­ |psi{(n)} rangle $$ formed from the $N$ single-particle spaces $|psi(i) rangle$ associated with each particle.

• There are various properties (more precisely, degrees of freedom) associated to each particle, e.g. position, spin etc. and each of them lives in a different Hilbert space of certain dimension. For two properties $P_1$ and $P_2$ that are independent of each other for a particle, dimension $N_1$ and $N_2$, respectively (either or both of which may be infinite), the combined system should be represented as, $$P_{12} = P_1 otimes P_2$$ of dimension $N_1 times N_2$.

• The main point of this digression is that the state space of one spin-$1/2$ particle can be represented as direct product, $$|Psi rangle = |psi_{text{spatial}}rangle ­ otimes |sigma_{text{spin}}rangle$$ of a quantum space $|psi_{text{spatial}}rangle$ describing the particle’s spatial state (which is spanned, e.g., by an infinite set of position eigenstates $|mathbf{r}rangle$), and a two-dimensional quantum space $|sigma_{text{spin}}rangle$ describing the particle’s internal structure.

Neither answer is right in general.

• The only actual requirement is that the state be [anti]symmetric with respect to exchange. That's it.

• The first state you give, $|psirangle=frac{1}{sqrt{2}}(|psi_1rangle_{sigma_1} otimes|psi_2rangle_{sigma_2}-|psi_2rangle_{sigma_2} otimes |psi_1 rangle_{ sigma_1})$, is a special case, but it is not general. This can be thought of as a single 'configuration', or Slater determinant, and those are often good approximations in the absence of strong correlations. But there are valid states which are superpositions of multiple different Slater determinants and cannot be reduced into this form.

• The second state you give, $|psirangle=|phirangleotimes|chirangle$, generally with spatial and spin parts of definite exchange symmetry, is again a special case, in which the spatial and spin parts are factorizable. If the system's hamiltonian has no spin-orbit coupling, then you are guaranteed an eigenbasis of states of this form. But in general (i) states need not have this form, (ii) your system's hamiltonian may well have spin-orbit coupling, and thus (iii) the eigenstates won't have this form.