Is it necessary to apply CNOT for the sole purpose of "establishing a correlation" between two qubits

Question

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)

Answer 1

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.