Factorization of two-particle states in internal state and relative motion Taken from Subatomic Physics, by Henley and Garcia (ch. 9, p. 242)

(...)
$$ a +b \to c+ d$$ 
Symbolically, the initial state can be described as
$$ |\text{initial}\rangle = |a\rangle |b\rangle |\text{relative motion}\rangle$$
where $|a\rangle$ and $|b\rangle$ describe the internal state of the two subatomic particles and $|\text{relative motion}\rangle$ is the part of the wave function characteristic of the relative motion of $a$ and $b$.

I think my background in non-relativistic quantum mechanics is quite solid, but I've never encountered such a notation. What I wanted to know is, what is its precise meaning? Is there any assumption for such a factorization?
Assume that the only thing I know is that a two-particle state is an element of the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$, where $\mathcal{H}_i$ is the single particle space.
Thanks in advance!
 A: You have a two particle system composed of particles A and B. Each particle has a Hilbert space given by e.g. $\mathcal{H}_A= \mathcal{H}_{int,A} \otimes \mathcal{H}_{space,A}$ where the first factor describes the internal properties (e.g. spin) whilst the latter describes translational properties (the spatial wavefunction).
I think the notation you are looking at just means they have grouped the two particle state as:
$\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B=\mathcal{H}_{int,A} \otimes \mathcal{H}_{int,B}\otimes\mathcal{H}_{space,A} \otimes \mathcal{H}_{space,B}$
If the Hamiltonian only couples states in space and not their internal motions then a product state will generically start as $|int_A\rangle |int_B\rangle |space_A \rangle |space_B \rangle $ and evolve into some entangled state $|int_A\rangle |int_B \rangle |space_{A,B}\rangle$ so I guess your source has decided to write the subspace factorization as 
$$\mathcal{H}=\mathcal{H}_{int,A}\otimes\mathcal{H}_{int,B}\otimes\mathcal{H}_{space,AB}$$
They've called the last factor "relative motion" but I think it's just meant to mean the (entangled) spatial wavefunction for both particles (although it is a two body system so they might be implying one should do the usual two-body setup with reduced masses etc).
A: The "internal state" here must be a reference to the internal structure of a particle which is not itself fundamental (i.e. not a quark or electron, but could be a pion or a nucleon). Then your ${\cal H}_1$, ${\cal H}_2$ refer to internal degrees of freedom such as quark content. (With a pair of atoms it would be the state of the electrons). When you want to consider the motion of the particle as a whole, you need the position of its centre of mass, represented by a position operator $\hat{\bf x}$. For two particles each has such an operator. The ket labelled "relative motion" is in a Hilbert space describing these external motional degrees of freedom. They are in a tensor product with the other degrees of freedom in the normal way. (Just like the way that spin is in a tensor product with spatial degrees of freedom).
An analogous situation arises when we discuss an atom such as hydrogen. We are usually only interested in the internal degrees of freedom, so we introduce a wavefunction $\psi(\bf r)$ where ${\bf r} = {\bf r}_e - {\bf r}_p$ is the relative coordinate. Or we talk about a state vector $|n,l,m\rangle$, which also refers only to the motion of the electron relative to the proton, so it is called the "internal state". But if we want to consider the motion of the whole atom, for example in order to discuss its momentum when it interacts with a light wave, then we remember that it also has the centre of mass position ${\bf R} = (m_p {\bf r}_p + m_e {\bf r}_e)/(m_p+m_e)$ and an associated state of motion which could be labelled for example by the momentum $P$ of the whole atom. Then the complete state is given by $ |n,l,m\rangle \otimes |P\rangle$.
