Can the Sum of Two Tensor Products Be Written as a Single Tensor Product? In general, if I have $|\Psi\rangle = (|\Psi_{1_1}\rangle \otimes |\Psi_{1_2}\rangle + |\Psi_{2_1}\rangle \otimes |\Psi_{2_2}\rangle)$,  can I find $|\Psi_{3_1}\rangle$ and $|\Psi_{3_2}\rangle$, such that $|\Psi\rangle = |\Psi_{3_1}\rangle \otimes |\Psi_{3_2}\rangle$? Here $\otimes$ means tensor product and $|\Psi\rangle$ and means a vector. No assumption is made about any relationship between the $|\Psi_{i_j}\rangle$, except that they are all the same dimension and their components are complex numbers.
The motivation is the quantum double slit experiment, where the wave state, $|\Psi\rangle$, between the slits and the detector, is the sum of two interfering waves, and $|\Psi\rangle$ is still in a "pure state", which means that $|\Psi\rangle$ can also be written as a tensor product
 A: The matrix $A=\begin{pmatrix} 1& 0& 0& 0\\ 0&0& 0&0\\ 0&0&0&0\\0&0&0&1\end{pmatrix}$ is a sum of two simple tensors which is not a simple tensor itself. In particular, $A=E_{11}\otimes E_{11}+E_{22}\otimes E_{22}$ where $E_{11}=\begin{pmatrix} 1&0\\0&0\end{pmatrix}$ and $E_{22}=\begin{pmatrix} 0&0\\0&1\end{pmatrix}$. In fact, this is one of the strengths of quantum science: not all quantum states are pure states. In linear algebra terms, not all tensors are simple tensors.
A: In general the answer is no and you can clearly see why when you look at the dimensions: $\dim (V\otimes W)$ is much larger than $\dim (V\times W)$ whenever $V$ and $W$ are both of dimension greater than $1$. This tells you that in general a tensor is more than just a couple of vectors.
Tensors of the form $a\otimes b$ are called pure tensors.

Edit: As was correctly noted in the comments, the dimensional observation does not provide a proof just by itself, it is rather a convenient way to remember this fact. If $\{e_i\}_{i=1}^n$ is a basis of $V$ and $\{f_i\}_{i=1}^m$ is a basis of $W$ then a basis of $V\otimes W$ is given by $\{e_i\otimes f_j\}$ (dimension $nm$) and an element of $V\otimes W$ can be represented by an $n\times m$ matrix where the entries are coordinates in that basis. Now with this description, pure tensors are matrices of rank $1$. Except when $\dim V$ or $\dim W$ is equal to $1$, not all matrices have rank $1$. 
The subset of pure tensors is not a subspace, however, so in a way the dimensional observation might be misleading.
