Trying to prove $U$ implements a logical $CCZ$ gate on the $[[8,3,2]]$ code?

Question

I know from several different papers that the gate $U=TT^{\dagger}TT^{\dagger}TT^{\dagger}TT^{\dagger}$ actually implements a logical $CCZ$ gate on the $[[8,3,2]]$ quantum code. However, I am having difficulty proving this to myself.

My understanding is that $UX_{L_{i}}U^{\dagger}$ should have the same effect on $X_{L_{i}}$ as $CCZX_{i}CCZ^{\dagger}$ has on $X_{i}$ (in order for $U$ to implement a logical $CCZ$).

However, I cannot figure this out as I know that $X_{L_{1}} = X \otimes X \otimes X \otimes X \otimes I \otimes I \otimes I \otimes I$ and $$U: X_{L_{1}} \rightarrow SX \otimes S^{\dagger}X \otimes SX \otimes S^{\dagger}X \otimes I \otimes I \otimes I \otimes I.$$

Whereas $X_{1} = X \otimes I \otimes I$ and $$CCZ: X \otimes I \otimes I \rightarrow X \otimes CZ_{2,3} = X \otimes |0\rangle\langle0| \otimes I + X \otimes |1\rangle\langle1| \otimes Z.$$

However, I do not see from the above workings how $U$ has the same effect on $X_{L_{1}}$ as $CCZ$ has on $X_{1}$?

Answer 1

You are almost there, you just lack some information about the $[[8, 3, 2]]$ code. Your goal is to prove that the $X_i$ logical operator is mapped to the $X_i C_{j,k}$ logical operator, which requires you to know how to perform such a $C_{j,k}$. Consulting this code's error correction zoo page, logical $CZ$ gates seems to have a transversal implementation, found out by this paper.

An alternating pattern of $S$ and $S^\dagger$ gates on the qubits of a face of the cube translates to a $CZ$ gate between the two logical qubits whose logical $X$ is not supported on this face. This is exactly what your computation leads you to.

It might be worth noting that $XS \otimes XS^{\dagger} \otimes XS \otimes XS^{\dagger} \otimes I \otimes I \otimes I \otimes I$ and $SX \otimes S^{\dagger}X \otimes SX \otimes S^{\dagger}X \otimes I \otimes I \otimes I \otimes I$ denotes the same logical operator because they only differ by a stabilizer: the $Z$-stabilizer over the $[1,2,3,4]$ face. This essentially tells you that $X_{1}C_{2,3} = C_{2,3}X_{1}$ which is reassuring.