0
$\begingroup$

Consider a qubit

$$\left| \psi \right> = (\alpha_1 + i\alpha_2 ) \left|0\right> + (\beta_1 + i\beta_2 )\left|1\right>$$

Now if i pass this through a series of quantum gates or any typical quantum circuit can I know the values of $\alpha$'s &$\beta$'s. before measuring it or without measuring it?

$\endgroup$
2
  • $\begingroup$ what do you mean "before measuring it or without measuring it?" you could run your circuit in reverse to revert back to the original state. $\endgroup$ May 16, 2022 at 20:50
  • $\begingroup$ I mean is it possible for me to see the amplitude values or the probabilities of being in state zero or state one.(in complex form like original) $\endgroup$
    – User1086
    May 16, 2022 at 21:00

1 Answer 1

1
$\begingroup$

The single qubit state is $$|\psi \rangle = \alpha |0\rangle + \beta|1\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \hspace{2 cm} \alpha, \beta \in \mathbb{C}$$ and if you know what $\alpha$ and $\beta$ is ahead of time then of course you can keep track of all the matrix vector multiplications as applying a quantum gate is nothing but applying a unitary matrix, $U \in \mathbb{C}^2 \times \mathbb{C}^2 \ \ \ \textrm{s.t} \ \ \ U^\dagger U = I $, to the vector $|\psi \rangle$. So from the information of the resulting vector from the matrix-vector multiplcation, you can determine the probability of observing $|0\rangle$ and $|1\rangle$. This is what it is being done if you use something like statevector_simulator fro Qiskit.

For a single qubit, this is not a problem... but when you start getting into the 50 qubits range... storing $2^{50}$ complex values is quite daunting already and we are not talking about the cost of matrix-vector multiplication yet. With all that being said, here I am assuming a general setting. There are cases where this exact simulation can be done efficiently classically.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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