0
$\begingroup$

I am currently taking a first course in quantum information theory and having trouble understanding how quantum gates are possible in practice. How does a unitary quantum gate evolve the qubit without interacting with the state and entangling it?

If the qubit is system A in the state $|\phi(t_0)\rangle$, and the quantum gate is system B in the state $|\psi(t_0)\rangle$, then the composite system AB is in a pure unentangled state $|\Omega(t_0)\rangle=|\phi(t_0)\rangle \otimes |\psi(t_0)\rangle$. The state evolves unitarily by $|\Omega(t)\rangle=U(t)^{AB}|\Omega(t_0)\rangle$. We only end up with an unentangled state if the interaction Hamiltonian is very weak.

How should I think of the system AB?

$\endgroup$
1
  • $\begingroup$ I asked a related question here, in case it helps you. $\endgroup$
    – knzhou
    Feb 14, 2018 at 16:23

1 Answer 1

1
$\begingroup$

Think of it like a Controlled-NOT operation. In your example, the control qubit would be B and the target qubit would be A.

If you put the control qubit into a superposition state like $|+\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$, then you do risk the control becoming entangled with the target. But if the control qubit is in a computational basis state, either $|0\rangle$ or $|1\rangle$, then you will find that applying the CNOT operation never entangles the control with the target.

In other words, if the application of the gate B to the target system A is unconditional, then there will be no entanglement created between the two. It's only when the application of B to A depends strongly on some internal state of B, and that state is in superposition, that the two systems might become entangled.

In practical situations there may be some internal-state-of-B dependence in the application of B to A, but it should be negligible. It's like a superposed photon bouncing off the mirrors in a Mach-Zhender interferometer. By conservation of momentum the photon must push the mirrors, and you might expect this to entangle the photon with the mirror and act as a detection of the photon's path. But actually the uncertainty in the position and momentum of the mirrors is large enough that the entanglement is negligible.

$\endgroup$

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.