Pauli errors and correction before non-Clifford gate

Question

If we have a circuit of Clifford gates, any Pauli gates (such as those required by error correction) can be pushed to the end of the circuit. However, when we have a non-Clifford gate like a T-gate, we must do the Pauli corrections immediately since these non-Clifford gates can turn Pauli corrections into non-Pauli corrections.

When we implement a T-gate via magic states, we see this in action. Consider this circuit (source: https://www.nature.com/articles/s41467-023-42482-1)

The S-gate is non-Pauli and therefore we must wait to know the outcome of the Z measurement reliably, implement the S-gate, and then continue with the rest of the circuit.

Now consider this implementation of the T-gate from this paper - https://arxiv.org/pdf/2405.03970.

The idea here is that the output register has a Pauli correction that depends on the bits $s$ and $m$. Why can't one continue with the rest of the circuit and push this $Z^{s\oplus m}$ gate (if it turns out to be needed) to the end of the circuit, even if there are more T-gates? We would simply collect all the $s$ and $m$ outcomes from each T-gate slowly and the only correction is a Z gate which seems like it can be pushed to the end of the circuit. What is the error in this argument?

Answer 1

If there are Hadamard gates later in the circuit, then the $Z^{s\oplus m}$ gets converted into $X^{s\oplus m}$, so it doesn't commute with later $T$ gates. You need to implement that correction before the next $T$ gate.