Is it possible to push back an $H$ gate to a $CZ$ gate?

Question

Given the above scenario. Is it possible to "push back" the $H$ gate operation to occur before $CZ$?

Formally I am looking for some operation $CZ\cdot(U_1\otimes U_2) = H\cdot CZ$.

Answer 1

The Hadamard and the CZ gates don't commute with each other. So, it is not possible to just push it back straightforwardly. If you are still interested in obtaining a unitary U such that: $H.CZ=CZ.U$, that is possible. In particular, you can use the propagation relations $$H=\frac{X+Z}{\sqrt{2}}\;,\quad CZ_{a,b}\,X_a\,CZ_{a,b}=X_aZ_b$$ to arrive at the required unitary $U=\frac{X_1Z_2+Z_1}{\sqrt{2}}$.

Answer 2

No, it's not possible. Not without changing the CZ to a CX.

In particular, consider an X error, on the top qubit, crossing from right to left. If the Hadamard is there then the X error turns into a Z error which commutes with the CZ so it exits left with no term on the bottom qubit.

If the CZ is the rightmost operation, then the X error first crosses the CZ creating a Z kickback on the bottom qubit. This non-identity term on the bottom qubit can't be removed by further single qubit operations.

Error kickback onto other qubits is an observable property, so it can't be the case that a single-qubit-gates-then-CZ circuit correctly implements the CZ-then-H circuit.

You can also check this by brute force search. Since the circuit is clifford, you only need to iterate over all the possible Clifford operations that meet your requirements:

import stim

def find_equivalent():
    CZ: stim.Tableau = stim.Tableau.from_named_gate("CZ")
    goal: stim.Tableau = CZ.copy()
    goal.append(stim.Tableau.from_named_gate("H"), [0])

    single_qubit_cliffords = []
    for x in "XYZ":
        for z in "XYZ":
            if x != z:
                for sign1 in [+1, -1]:
                    for sign2 in [+1, -1]:
                        single_qubit_cliffords.append(
                            stim.Tableau.from_conjugated_generators(
                                xs=[sign1 * stim.PauliString(x)],
                                zs=[sign2 * stim.PauliString(z)],
                            )
                        )

    for g1 in single_qubit_cliffords:
        for g2 in single_qubit_cliffords:
            achieved = CZ * (g1 + g2)  # Note: + is direct sum (diagonal concatenation)
            if goal == achieved:
                return g1, g2

    raise ValueError("Not found")


print(find_equivalent())
```