Single electron on a two atoms chain : factorisation of hilbert space by external and internal degrees of motion I have a question on the first pages of the book "A Short Course
on Topological
Insulators" by János K. Asbóth, László Oroszlány and András Pályi
But actually we can see it here : http://theorie.physik.uni-konstanz.de/burkard/sites/default/files/ts15/TalkSSH.pdf
Presentation of the problem
We work with a 1 dimensional chain where there are two types of atoms $A$ and $B$. The unit cell is labelled by $m$. We study the motion of one electron.
We have different interaction terms : $v$ and $w$.
They work with the following SSH hamiltonian model :
$$ H=v \sum_{m=1}^N (|m,B\rangle\langle m,A| +hc)+w\sum_{m=1}^{N-1} (|m+1,A\rangle\langle m,B|+hc)$$
Where $hc$ is for the hermitic conjugate.
Thus, if we write the hamiltonian, we have something that looks like :
$$ H = \begin{bmatrix}0&v&0&0&0&0&0&0\\v&0&w&0&0&0&0&0 \\ 0&w&0&v&0&0&0&0 \\ 0&0&v&0&w&0&0&0\\ 0&0&0&w&0&v&0&0\\ 0&0&0&0&v&0&w&0 \\ 0&0&0&0&0&w&0&v \\ 0&0&0&0&0&0&v&0 \end{bmatrix}$$
And the Hilbert space can be seen as a tensorial product of :
$\mathcal{H}_{tot}=\mathcal{H}_{ext} \otimes \mathcal{H}_{int}$
Where the external degree of freedom is represented by the letter $m$, and the internal one by the fact we are on site $A$ or $B$.
Thus : $|m,\alpha\rangle = |m\rangle \otimes |\alpha \rangle$ where $\alpha=A,B$.
My question
But there is something I misunderstand here.
I agree that we can see the total Hilbert space of the problem as a tensorial product of an internal and external degrees of freedom hilbert spaces. 
But in the same time, if we consider the state $|m,A\rangle$, we would see a gaussian centered on the atom $A$ in cell $m$. And then $|A\rangle$ would be a gaussian centered in $0$ and $|m\rangle$ would "shift it" to the position $m$ right ? But if we write everything on the $x$ basis we have :
$$ \psi(x)=\psi_m(x) \psi_A(x)$$ and it should be $0$ outside of the support of $\psi_A$. And as $\psi_A$ is a gaussian centered in $0$ we would have a wavefunction that is zero everywhere if we go far enough.
Where is the mistake I do in my vision of the problem?
Isn't $|m,A\rangle$ a gaussian centered on the atom $A$ that is in the cell $m$ ? If so, what represents the kets $|m\rangle$ and $|A\rangle$ physically (what those wavefunctions look like).
 A: It is wrong to write $\psi(x)=\psi_m(x)\psi_A(x)$. The correct wave function $\psi_{mA}(x)$ that represents the state $|m,A\rangle$ should be
$$\psi_{mA}(x)=e^{-m\partial_x}\psi_{A}(x)=\left(1-m\partial_x+\frac{1}{2!}m^2\partial_x^2-\frac{1}{3!}m^3\partial_x^3+\cdots\right)\psi_{A}(x) =\psi_{A}(x-m),$$
where we have used the formula of Taylor expansion. Physically, this can be understood by noticing that $p=-\text{i}\partial_x$ is the momentum operator that generates translation, and the meaning of the state $|m,A\rangle$ is the simply the Gaussian packet $\psi_A(x)$ translated by the displacement $m$.
The tensor product in $|m,A\rangle=|m\rangle\otimes|A\rangle$ does not mean to multiply two wave functions together directly. It just means that if you consider the following linear superposition, the result can be expanded in the tensor product basis as
$$(c_m|m\rangle+c_n|n\rangle)\otimes(c_A|A\rangle+c_B|B\rangle)=c_mc_A|m,A\rangle+c_mc_B|m,B\rangle+c_nc_A|n,A\rangle+c_nc_B|n,B\rangle.$$
This is what defines a tensor product structure in the Hilbert space. Any algebraic structure satisfying such property of binlienar maps can be called a tensor product. You can see that in terms of the wave function, the following algebraic structure is indeed a tensor product
$$(c_m e^{-m\partial_x}+c_n e^{-n\partial_x})\otimes(c_A \psi_A(x)+c_B \psi_B(x))=c_mc_A \psi_A(x-m)+c_mc_B \psi_B(x-m)+c_nc_A \psi_A(x-n)+c_nc_B \psi_B(x-n).$$
In this sense, the operators $e^{-m\partial_x}$ form a set of basis of the external Hilbert space, which can be denoted as $|m\rangle$ abstractly. There is no wave function associated with $|m\rangle$, because $|m\rangle=e^{-m\partial_x}$ is actually represented as a linear operator in this case.

Well, if one insists to understand the $|m\rangle$ state as a wave function, one possible interpretation is to consider it to be a Dirac delta function located at $x=m$ (the center of the $m$th unit cell).
$$\psi_m(x)=\delta(x-m).$$
But still, the tensor product $|m\rangle\otimes|A\rangle$ does not correspond to multiplying the wave functions $\psi_m(x)$ and $\psi_A(x)$ together in a point-wise manner. It should actually be understood as a convolution of the two wave functions:
$$\psi_{mA}(x)=(\psi_{m}*\psi_A)(x)=\int\text{d}y\psi_m(y)\psi_A(x-y)=\psi_A(x-m).$$
The convolution also satisfies the algebraic properties of the tensor product and is hence a legitimate representation of the tensor product. This point of view is secretly equivalent to the above operator point of view, because, in functional analysis, the Dirac delta function (or the shifted Dirac delta function) is actually defined to be the kernel of the identity operator (or the translation operator).
A: I think your problem is due to a misunderstanding of what the tensor product state space, spanned by the basis $\{\lvert m,X \rangle\}$ where $m=1,2,3,4$ and $X=A,B$, means. The SSH model you are considering is specified by the Hamiltonian whose states are spanned by this tensor product of basis vectors. In the context of this model, the state $\lvert A \rangle$ does not necessarily have any meaning on its own. Therefore, I believe your interpretation of beginning with the state $\lvert A\rangle$ being a Gaussian centered on anything is incorrect. This state on its own has no meaning. Within this model, the states must be specified by the full tensor product.
A: OK, let's see if this is what you want: consider a general position:
\begin{align}
x = m \cdot a_0 + \tilde{x}
\end{align}
and consider the following basis function:
\begin{align}
&\psi_M(x) \propto \sum_{m = 0}^{N} \delta(m\cdot a_0) \\
\lim_{x-\infty}\quad& \psi_A(x) = 0
\end{align}
then, your wave-function can be written as:
\begin{align}
\psi(x) = \psi_M(m\cdot a_0)\otimes\psi_A(\tilde{x})
\end{align}
