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?

  • $\begingroup$ I asked a related question here, in case it helps you. $\endgroup$ – knzhou Feb 14 '18 at 16:23

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.


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