Bra-ket of products I was trying to solve the following problem.

(Lifted from Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund) 
I came across across a solution for the problem as follows

(Lifted from http://theochemlab.asu.edu/teaching/chm598/ch2soln.pdf) 
The 4th line in the solution doesn't seem to make sense to me. From what I understand, if we let: 
$\chi$ be an $M \times O$ matrix 
$\psi$ be an $M \times N$ matrix 
$\alpha$ (and $\beta$) be an $N \times O$ matrix 
then $\langle \psi_i^\alpha | \psi_j^\beta \rangle$ is an $N \times N$ matrix and $\langle \alpha | \beta \rangle$ is an $O \times O$ matrix. This means that we cannot multiply the 2 resulting matrices, right? Also, how was $\langle \chi_{2i-1} | \chi_{2i} \rangle$ expanded into $\langle \psi_i^\alpha | \psi_j^\beta \rangle \langle \alpha | \beta \rangle$?
 A: I will consider non-relativistic quantum mechanics in this answer. The wave function of a particle with spin can be thought of as living in the tensor product space $L^2(\mathbb R^3,\mathbb C)\otimes V$, where $V$ is the spin space (usually just $\mathbb C^2$). This tensor product vector space becomes a Hilbert space (after a possible quotient and completion in the general case) through the extension of the inner product on elementary tensors
$$(\psi\otimes\chi,\phi\otimes\eta):=(\psi,\phi)(\chi,\eta),\qquad\forall\psi,\phi\in L^2(\mathbb R^3,\mathbb C),\chi,\eta\in\mathbb C^2$$
to the whole tensor product space.
A: First off, while $\alpha$ and $\beta$ can indeed be represented by matrices (of $2 \times 1$ size, not of $N \times O$), it can not be done for spatial orbitals $\psi$, and consequently, for spin orbitals $\chi$ as well. Thus, the last paragraph of your question does not make sense.
Secondly, everything is pretty simple here, so you do not need any matrix representation. The thing you have to understand is that when you use Dirac bra-ket notation with spin orbitals you integrate over both spin and spatial coordinates of electrons, i.e.
$$
\langle \chi_{2i-1} | \chi_{2i} \rangle
=
\iint \chi_{2i-1}^{*}(\vec{x}) \chi_{2i}(\vec{x}) \mathrm{d}\vec{r} \mathrm{d}\omega \, ,
$$
where integration over $\mathbf{R}^3$ is assumed for spatial coordinate $\vec{r}$. Now if you express spin orbitals as mentioned in the question you get
$$
\langle \chi_{2i-1} | \chi_{2i} \rangle
=
\iint \phi_{i}^{\alpha *}(\vec{r}) \alpha^{*}(w) \phi_{i}^{\beta}(\vec{r}) \beta(\omega) \mathrm{d}\vec{r} \mathrm{d}\omega \, ,
$$
and the double integral clearly breaks into the following product of two singe integrals
$$
\langle \chi_{2i-1} | \chi_{2i} \rangle
=
\int_{\mathbf{R}^3} \phi_{i}^{\alpha *}(\vec{r}) \phi_{i}^{\beta}(\vec{r}) \mathrm{d}\vec{r} \int_{-1/2}^{+1/2} \alpha^{*}(w) \beta(\omega) \mathrm{d}\omega \, .
$$
The second integral in the product is zero by definition of spin functions $\alpha(w)$ and $\beta(\omega)$.
This way you prove that if two spin orbitals "come from different worlds" (one from "$\alpha$-world" and another from "$\beta$-world") they are orthonormal, since the second part of the product above is zero. If two spin orbitals "come from the same world", then the first part of the product above will be zero, since spatial parts of spin orbitals from "the same world" are orthonormal as stated ath the very beginning of the question. And again spin orbitals are orthonormal.
