I have two unit column vectors $a = (\alpha_1, \alpha_2)^T$ and $b = (\beta_1, \beta_2)^T$. Unit means that $\alpha_1^2 + \alpha_2^2 = 1$ and similarly for $b$.
If I take their tensor product, it is the vector $p = (\alpha_1\beta_1, \alpha_1\beta_2, \alpha_2\beta_1, \alpha_2\beta_2)^T$
Suppose that I know the values $p_i$ for $p = (p_1, p_2, p_3, p_4)^T$.
Is it possible to get the values for $a$ and $b$ from these? I just thought that because there are 4 values in $p$ and 4 in $a$ and $b$, it might be possible, especially given the additional information about them being unitary.
But I haven't been able to simplify the equations in a way that I'd get value for one of the unknowns ($\alpha_i$ or $\beta_i$) in terms of the knowns ($p_i$).
For instance, $p_1^2+p_3^2$ $=\alpha_1^2\beta_1^2+\alpha_2^2\beta_1^2$ $=(\alpha_1^2+\alpha_2^2)\beta_1^2$ $=\beta_1^2$. You can find similar combinations of terms that will give you the other three elements.
EDIT: as noted in a comment, this isn't strictly accurate; rather, this only works up to sign, and in fact the answer isn't uniquely obtainable because negating both input vectors gives the same output. In the case where the $\alpha$s or $\beta$s can be negative (and in particular, where entries of the tensor product are negative) it's still possible to determine whether there's a valid solution: Assign to each $\alpha_i$ the sign of the $\alpha_i\beta_1$ term of the tensor product. Then if every vector $v_j=\langle\alpha_i\beta_j:1\leq i\leq n\rangle$ has either the same sign pattern or an opposite sign pattern to $v_1$, we can assign signs to the $\beta_j$ based on which of those two it is. If it's not the case — for instance, if $p_{11}\gt 0$, $p_{12}\gt 0$, $p_{21}\gt 0$, but $p_{22}\lt 0$ — then there is no solution.