Is it possible to retrieve $|\psi_1\rangle,|\psi_2\rangle$ from their tensor product $|\psi_1\rangle\otimes|\psi_2\rangle$?

Question

Consider two quantum states$$\left| \psi_1 \right> = \alpha \left|0\right> + \beta\left|1\right>$$ and $$\left| \psi_2 \right> = \gamma \left|0\right> + \delta\left|1\right>$$ Now tensor product of two states gives $$\left| \psi \right> = \left|\psi_1\right> \otimes \left|\psi_2\right>$$

$$\left| \psi \right> = \alpha\gamma \left|00\right> + \alpha\delta\left|01\right> + \beta\gamma \left|10\right> + \beta\delta\left|11\right>$$

Is it possible to factor the state $\left| \psi \right> $ and get $\left| \psi_1 \right> $ and $\left| \psi_2 \right> $ back?

Answer 1

Yes, it is possible. You have two expressions for the state $x_0|00\rangle + x_1|01\rangle + x_2|10\rangle + x_3|11\rangle = \alpha\gamma \left|00\right> + \alpha\delta\left|01\right> + \beta\gamma \left|10\right> + \beta\delta\left|11\right>$, you just need to solve the system of equations

$$\begin{cases} \alpha\gamma = x_0 \\ \alpha\delta = x_1 \\ \beta \gamma = x_2 \\ \beta \delta = x_3 \\ \end{cases}$$

If the system of equations has a solution, you got your answer, and as a bonus you know that your quantum system is separable (i.e., the qubits are not entangled).

Answer 2

Yes, you can factor it. For example, this is what cirq.sub_state_vector does and also how the amplitude displays work in Quirk.

The basic idea is that to factor out $A$ from $A \otimes B$ you look at a non-zero part of the wavefunction where B is held constant. This tells you what $A$ has to be proportional to, and you can check that the other parts of the wavefunction are consistent with this. Then you use the ratios between those pieces to derive $B$.