Which codes have transversal $T$ gate?

Question

A previous post transversal P (phase) gate shows that codes where all stabilizer elements have weights that are multiple of 4 will have a transversal $P$ gate.

"Transversal" seems to have multiple definitions; here a gate $G$ is transversal for the code with stabilizer $S$ if it is in the normalizer of $S$ in the full unitary group : $$G \in U_n : G' S G = S\,.$$ (note $G$ need not be in the clifford group). This is shown using $$P^\dagger X P = \imath XZ\,.\\ P^\dagger Z P = Z\,.$$ At first guess I thought that codes where all stabilizer elements have weight that's a multiple of 8 would have a transversal $T$ gate. ($T^2=P$); but $$T^\dagger X T = w_8 P Z X\,,\\ T^\dagger Z T=Z\,,$$ where $w_8^8=1$. So it's not obvious if a similar argument can be used. Does anyone know how to extend the result for $P$ gate to $T$?

Answer 1

The $ 11 $ qubit $ d=3 $ error-correcting quantum code we find here

https://arxiv.org/abs/2310.17652

is probably the smallest code with transversal $ T=\begin{bmatrix} 1 & 0 \\ 0 & e^{i \pi/4}\end{bmatrix} $.

The code is permutationally invariant and has codewords $$ |\overline{0} \rangle= \sqrt{\frac{5}{16}} | D^{11}_0 \rangle + \sqrt{\frac{11}{16}} | D^{11}_8 \rangle $$ and $$ |\overline{1} \rangle = \sqrt{\frac{5}{16}} | D^{11}_{11} \rangle + \sqrt{\frac{11}{16}} | D^{11}_3 \rangle $$ where $ | D^{11}_w \rangle $ represents a normalized uniform superposition over all $ 11 $ qubit computational basis kets of weight $ w $. So $ | D^{11}_0 \rangle =|00000000000 \rangle $ while $ | D^{11}_8 \rangle $ is $ 1/\sqrt{\binom{11}{8}} $ times the uniform sum of all $ \binom{11}{8} $ many of the weight $ 8 $ basis kets. Similarly for $ | D^{11}_{11}\rangle $ and $ | D^{11}_3 \rangle $

The logical $ T $ gate for this code is implemented by $ (T^3)^{\otimes 11} $.

The code has weight enumerator $$ A=(1, 0, \frac{187}{10}, 0, \frac{4081}{60}, 0, \frac{3619}{20}, 0, \frac{1001}{4}, 0, \frac{6061}{12}, 0) $$ and dual weight enumerator $$ B=(1, 0, \frac{187}{10}, \frac{1265}{24}, \frac{4081}{60}, \frac{1771}{4}, \frac{3619}{20}, \frac{8371}{10}, \frac{1001}{4}, \frac{60643}{60}, \frac{6061}{12}, \frac{29149}{40}) $$ So one can verify that the distance is 3 by observing that $$ A_0=1=B_0 \\ A_1=0=B_1 \\ A_2=18.7=B_2 \\ A_3=0\neq \frac{1265}{24}=B_3 $$ A sanity check here is that the A weight enumerators should sum to $ 2^{n-k} $ (for a proof of this see Do the coefficients of the weight enumerator polynomial add up to $2^{n-k}$ for any $[\![n,k]\!]$ code?) where $ n=11 $ is the number of physical quibts and $ k=1 $ is the number of logical qubits. Similarly, the B weight enumerators should sum to $ 2^{n+k} $. We can check indeed that the weight enumerators I gave above have $ \sum_{i=0}^{11} A_i= 1024 $ and $ \sum_{i=0}^{11} B_i= 4096 $

Aside on weight enumerators: The weight enumerators of a code are defined by

\begin{align*} A_i &= \frac{1}{(Tr(\Pi))^2} \sum_{E \in \mathcal{E}_i} Tr(E \Pi)Tr(E^\dagger \Pi) \\ B_i &= \frac{1}{Tr(\Pi)} \sum_{E \in \mathcal{E}_i} Tr( E \Pi E^\dagger \Pi) \end{align*} Here $\Pi$ is the code projector and $\mathcal{E}_i$ are the Pauli errors with weight $i$. The code has at least distance $ d $ if and only if both $ A_d < B_d $ and $$ A_i=B_i $$ for all $ i \leq d-1 $.

To construct a code projector for an $ ((n,2,d)) $ with logical code words $ | \overline{0} \rangle $ and $ | \overline{1} \rangle $ the code projector $ \Pi $ is given by $$ \Pi= | \overline{0} \rangle \langle \overline{0} | + | \overline{1} \rangle \langle \overline{1} | $$

Answer 2

A well known way to construct codes with transversal T gates is from a triorthogonal check matrix.

Each column of the matrix is a qubit. Each row of the matrix is an $X$ stabilizer. All pairs of rows must have an even number of 1-on-qubit-in-both-rows. All triplets of rows must have an even number of 1-on-qubit-in-all-three-rows.

This won't exhaust all ways of getting a good logical T, but it's a simple way to get one.