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

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?

• Can you be more precise about what you mean by "undo"? You could take a partial trace on one of the systems to recover the other system. Apr 26, 2021 at 18:41
• I mean can I retreive back $\left| \psi_1 \right>$ and $\left| \psi_2 \right>$ given $\left| \psi\right>$ Apr 26, 2021 at 18:54

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
$$\begin{cases} \alpha\gamma = x_0 \\ \alpha\delta = x_1 \\ \beta \gamma = x_2 \\ \beta \delta = x_3 \\ \end{cases}$$
• +1 Might be worth mentioning that if the system of equations fails to have a solution then $x_0|00\rangle + x_1|01\rangle + x_2|10\rangle + x_3|11\rangle$ is entangled. Apr 26, 2021 at 19:42
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 $$B$$.