Why does fault-tolerant transversal phase gate $P$ only work with doubly-even codes?

Question

I read in this topic that to get a fault-tolerant transversal relative phase gate of $\pi/2$ -- $P$ gate in the reference --, one has to select a CSS code which satisfies doubly-even distances.

I don't really get the reason of such a restriction. Indeed, it says verbatim:

Since the phase gate commutes with the Pauli $Z$ operator it is clear that all $g_z$ stabilizers are preserved. For any $g_x$, we have

$$ P g_x P^\dagger = i^{w(g_x)} g_x g_z $$

where $w(g_x)$ is the weight of stabilizer $g_x$, i.e. the number of non-identity factors in it. Since $C_2^\perp$ is doubly-even, we see that

$$ P g_x P^\dagger = g_x g_z $$

and thus $P g_x P^\dagger \in S$.

My observation is that even if $i^{w(g_x)} g_x g_z \notin S$, this is still equivalent, up to a global phase $i^{w(g_x)}$, to $g_x g_z \in S$.

As consequence, $g_x g_z$ can still be considered a stabilizer for $P|\varphi\rangle$. Allowing for any (self-dual) code, not only doubly-even ones.

Is my observation correct?

Answer 1

It might help to think about what the logical states actually look like. Up to normalisation, the logical 0 is $$ |0\rangle_L=\left(\prod_x(I+g_x)\right)|00\ldots 0\rangle. $$ Now, if you want a logical phase gate $P$, you need $P|0\rangle_L=|0\rangle_L$, but what you've managed to achieve is $$ P|0\rangle_L=\left(\prod_x(I+i^{w(g_x)}g_x)\right)|00\ldots 0\rangle. $$ From this, you see that the factors $i^{w(g_x)}$ are not global phases, but relative phases. The only way this output is $|0\rangle_L$ is if $i^{w(g_x)}=1$ for all $x$.

Answer 2

Claim: A CSS code $ CSS(H_X,H_Z) $ has transversal phase gate $ P $ if and only if every $ X $ type stabilizer generator $ g_x $ has one of the following two properties:

(1) $ wt(g_x) $ is congruent to 0 mod 4 (doubly even) and $$ g_z \in S_Z $$ (This case applies when $ H_X=H_Z $ and more generally for all codes where $ H_X $ is properly contained in $ H_Z $, for example the $ [[15,1,3]] $ quantum reed muller code)

or

(2) $ wt(g_x) $ is congruent to 2 mod 4 (singly even) and $$ -g_z \in S_Z $$

$ H_X $ is the classical parity check matrix corresponding to the $ X $ type stabilizer generators (which generate the $ X $ stabilizers $ S_X $) and $ H_Z $ is the classical parity check matrix corresponding to the $ Z $ type stabilizer generators (which generate the $ Z $ type stabilizers $ S_Z $).

Here $ g_z:=Hg_xH $ is just $ g_x $ with every $ X $ replaced by a $ Z $ ($H$ with no subscript is Hadamard).

$ H_X $ is the classical parity check matrix corresponding to the $ X $ type stabilizer generators (which generate the $ X $ stabilizers $ S_X $) and $ H_Z $ is the classical parity check matrix corresponding to the $ Z $ type stabilizer generators (which generate the $ Z $ type stabilizers $ S_Z $).

Proof of Claim: Suppose $ P $ is transversal for $ CSS(H_X,H_Z) $. In other words $ P^{\otimes n} $ implements a logical operation. That is equivalent to saying that $ P^{\otimes n} $ preserves the code space. Since this a stabilizer code that is equivalent to saying that $ P^{\otimes n} $ is in the normalizer $ N(S) $ of the code stabilizer $ S $. Since $ CSS(H_X,H_Z) $ is a CSS code then there exists a choice of stabilizer generators which are all either $ X $ type Pauli operators or $ Z $ type Pauli operators. $ P^{\otimes n} $ certainly normalizes all the $ Z $ type stabilizer generators, in fact it commutes with them. Thus $ P $ is transversal if and only if for every $ X $ type stabilizer generator $ g_x $ we have $$ (P^{\otimes n})g_x (P^{\otimes n})^\dagger \in S $$ As noted above $$ P Z P^\dagger=Z \; , \; P X P^\dagger= iXZ $$ so we have $$ (P^{\otimes n})g_x (P^{\otimes n})^\dagger=i^{wt(g_x)} g_x g_z $$ where $ g_z $ is a $ Z $ type Pauli operator obtained from the $ X $ type Pauli operator $ g_x $ by switching all the $ X $s to $ Z $s. Now we prove the first direction of the theorem. Suppose that $ P $ is transversal. Then $$ i^{wt(g_x)} g_x g_z \in S $$ for every $ X $ type stabilizer $ g_x $. Since $ g_x $ is already in the stabilizer that implies $$ i^{wt(g_x)} g_z \in S $$ Since elements of the stabilizer must have $ 1 $ as an eigenvalue then $ wt(g_x) $ must be even. Thus we have that either $$ -g_z \in S $$ if $ g_x $ is singly even or $$ g_z \in S $$ if $ g_x $ is doubly even.

For the reverse implication pick an arbitrary $ X $ type generator $ g_x $ then either (1) $ wt(g_x) $ is congruent to 0 mod 4 (doubly even) and $$ g_z \in S_Z $$ in which case $$ (P^{\otimes n})g_x (P^{\otimes n})^\dagger=i^{wt(g_x)} g_x g_z=g_x g_z \in S $$ or (2) $ wt(g_x) $ is congruent to 2 mod 4 (singly even) and $$ -g_z \in S_Z $$ So $$ (P^{\otimes n})g_x (P^{\otimes n})^\dagger=i^{wt(g_x)} g_x g_z=-g_x g_z \in S $$ And of course $$ (P^{\otimes n})g_z (P^{\otimes n})^\dagger= g_z \in S $$ for any $ Z $ type stabilizer generator $ g_z $. So we indeed have that $$ P^{\otimes n} \in N(S) $$

Corollary 1: If $ H_X $ is contained in $ H_Z $ and $ CSS(H_X,H_Z) $ is doubly even then $ P $ is transversal

This covers some interesting slightly less standard cases like the transversal $ P $ for the $ [[15,1,3]] $ code. Even more specifically,

Corollary 2: If $ H_X = H_Z $ and $ CSS(H_X,H_Z) $ is doubly even then $ P $ is transversal

This is a pretty classic fact and is often stated in connection with the Steane $ [[7,1,3]] $ code. This fact is stated, for example, here Transversal logical gate for Stabilizer (or at least Steane code) were it is further stated that for this special class of code transversal $ P $ must implement either logical $ P $ or logical $ P^\dagger $.

The case (2) that I discuss above is quite unusual and doesn't come up in any well known examples that I am aware of. But its not hard to construct an ad hoc example like this $ [[6,2,2]] $ code with stabilizer generators $$ XXXXXX $$ and $$ -ZZIIIII,-IIZZII,-IIIIZZ $$ For this $ [[6,2,2]] $ code I think $ P^{\otimes 6} $ implements some very strange 2 qubit gate like the negative of the controlled $ Z $ gate or something.