2
$\begingroup$

Disclaimer: I recently started learning quantum information.

I've been exploring creating the $|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|00\rangle + |11\rangle)$ Bell state (starting with the state $|00\rangle$).

I know we can apply the $H$ gate to the leftmost qubit, then use the same qubit as a control for the $CNOT$ gate applied on both qubits to entangle them.

Before the $CNOT$ gate is applied, the two qubits have no correlation. They are separable qubits, where one is in a superposition.

So I wonder: Is it necessary, in this case, to apply a first $CNOT$ gate to the two qubits before applying the $H$ gate and the final $CNOT$ that creates the entanglement?

(I know that using both methods will still create the $|\Phi^+\rangle$ Bell state. I just want to find out if it's necessary to create any correlation between the two qubits before performing any operations on them)

$\endgroup$

1 Answer 1

2
$\begingroup$

No, it is not necessary to apply a $CNOT$ gate before the typical construction of $|\Phi^+\rangle$, that is applying $H$ on the control qubit followed by a $CNOT$.

Indeed, non-local quantum gates like $CNOT$ (but there are many others like $CZ$ for example) are necessary to create a correlation between qubits. However, such correlation (also called entanglement) is created when the control qubit is in superposition, otherwise the result of applying the non-local gate is deterministic and no entanglement is created.

Regarding the specific example of the $|\Phi^+\rangle$ Bell state, it is easy to demonstrate (let the leftmost qubit be "qubit $0$" and the rightmost qubit be "qubit $1$"):

  1. Let $|\Phi_0\rangle = |00\rangle$ be the initial state of the system, as you mentioned.
  2. Applying $CNOT(0,1)$ now would do nothing.
  3. Applying $H$ on qubit 0 leads us to the state $|\Phi_1\rangle = \frac{|00\rangle + |10\rangle}{\sqrt{2}}$.
  4. Applying $CNOT(0,1)$ now leads us to the final state $|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$.

When the control qubit is in state $|1\rangle$ before applying a $CNOT$ an action will take place, but still the result would be deterministic due to lack of superposition.

$\endgroup$

Your Answer

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

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