Symmetry in Conditional Phase Shift Gates and Realizing CNOT through HCZH

Question

Why are conditional phase shift gates, such as CZ, symmetrical? Why do both the control and target qubit pick up a phase?

Furthermore, assuming that they are symmetrical, when using a CNOT gate as an H gate on the target qubit, a CZ gate, and another H gate on the target qubit, wouldn't the CZ gate cause the control qubit to pick up a phase?

For example, if the control qubit is an equal superposition of $|0\rangle$ and $|1\rangle$ i.e. $|+\rangle$ and then this version of the CNOT gate is implemented, wouldn't the control qubit end up with a phase?

Answer 1

A $Z$ gate's effect is "If the target is ON then negate the phase". A controlled gate's effect is "If the control is ON then apply the gate to the target". Therefore a controlled Z's effect is "If the control is ON then if the target is ON then negate the phase".

Said another way, the controlled Z's effect is "If the control and target are both ON, negate the phase". This is a symmetric effect because it conditions on both the control and the target in the same way. The fact that we distinguish between the control and the target is just an accident of history.

Thinking of the control and target as fundamentally different is particularly tricky when it comes to the CNOT gate, as you have noticed. Assuming that the control cannot be affected is a holdover from classical-style thinking that no longer applies in the quantum case.

Classically, you think of a NOT gate as toggling the target qubit. Quantumly, the NOT gate's effect can instead be thought of as negating the phase of the $|-\rangle$ state. The $X$ gate leaves $|+\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$ alone and negates $|-\rangle = \frac{1}{\sqrt{2}} (|0\rangle - |1\rangle)$, in the same way that the $Z$ gate leaves $|0\rangle$ alone and negates $|1\rangle$.

So, quantumly, a CNOT's effect is "If the control is ON then if the target is $|-\rangle$ then negate the phase". Equivalently, "If the control is ON and the target is $|-\rangle$ then negate the phase". I hope that makes it clear that "control" and "target" are not really the most appropriate names. It's more like "the Z-axis control and the X-axis control".

Answer 2

It's not that both the qubits are independently picking up a phase, it's that the two qubit state itself is picking up that phase.

$$\mathbf{CZ}|1\rangle\otimes |1\rangle = |1\rangle\otimes(-1\times |1\rangle) = -|1\rangle\otimes |1\rangle$$

So although the second qubit is the one that has the Z applied and is actually picking up the phase, that entire $|11\rangle$ state is picking up the phase. If it helps you can look at the effect on a generic state in the computational basis:

$$\mathbf{CZ}(\alpha |00\rangle + \beta |01\rangle + \gamma |10\rangle + \delta |11\rangle) = \alpha |00\rangle + \beta |01\rangle + \gamma |10\rangle - \delta |11\rangle$$

so the key action is that CZ flips the sign of a $|11\rangle$ two qubit state.

If you follow this through on your example, you should see that the sign flip does affect the state of both qubits in the $|11\rangle$ case, but then when passed through the H gates in the correct way to get your desired behavior.