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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – Rammus
    Apr 26 at 18:41
  • $\begingroup$ I mean can I retreive back $\left| \psi_1 \right> $ and $\left| \psi_2 \right> $ given $\left| \psi\right> $ $\endgroup$
    – User1086
    Apr 26 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

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

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).

  • 1
    $\begingroup$ +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. $\endgroup$ Apr 26 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 $B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.