This question concerns about a formal definition of transversal operator. I understood that transversal operator are a group of operators which are efficient in terms of circuit depth and can be used as logical operators for stabilizer codes. However I'm not sure I got when a unitary $U$ is transversal. Or maybe it is better saying that $U$ ``admits'' a transversal implementation?

Can I state that if a unitary $U$ maps the Pauli group into itself, then $U$ is said to be transversal (or admit transversal implementation)?

  • $\begingroup$ isn't it just that a logic gate is "transversal" if you can apply it by applying the corresponding physical qubit gate to each of the physical qubits which define your logical qubit? $\endgroup$
    – Lior
    Feb 26, 2022 at 21:26
  • $\begingroup$ @Lior it is usually relaxed a little bit by saying that two physical qubits within the same logical qubit do not interact (i.e. you are not forced to have exactly the same physical gates applying on all the physical qubits within a given logical qubit). $\endgroup$ Feb 27, 2022 at 18:21

2 Answers 2


Based on informal conversations: there is no actually agreed upon definition of a transversal operator. People use it to mean different things. Typically it refers to either the operation being fast or simple or trivially-fault-tolerant.

The most conservative definition of transversal is "the logical operation is achieved by broadcasting the same operation over the physical qubits". For example, the logical T gate in the 15 qubit Steane code is performed by applying a T gate to each of the 15 physical qubits.

Sometimes transversal means specifically "the logical operation has constant depth, regardless of code distance". For example, the transversal S gate in the folded surface code uses physical gates that aren't the S gate. It uses two qubit gates across the folded halves.

And sometimes transversal is so weak it only means "the logical operation uses a lot of the physical operation". For example, there's a "transversal" CCZ in the surface code that involves $O(d)$ layers of three surface codes being drifted past each other while doing oodles of CCZs interleaved with just in time error correction.

  • $\begingroup$ Ya there are many definitions. The most conservative definition you refer to above is often called "strong transversality" for example it is called that in the paper transversal S gate in the folded surface code you mention in this answer. $\endgroup$ Mar 21 at 15:07
  • $\begingroup$ Out of curiosity, in your most conservative definition of transversality: "the logical operation is achieved by broadcasting the same operation over the physical qubits" I know you would include $ H^{\otimes 7} $ implementing logical $ H $ on the Steane code but what about $ P^{\otimes 7} $ implementing logical $ P^\dagger $ on the Steane code? We are broadcasting the same operation, $ P $, over all the physical qubits but the actual logical operation achieved is $ P^\dagger $ rather than $ P $. Would you still say $ P $ is transversal in the most conservative definition for the Steane code? $\endgroup$ Mar 21 at 15:42
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    $\begingroup$ @IanGershonTeixeira I'd say implementing it up to Pauli gates is good enough, because those can be tracked in the classical control system. Really that whole definition is already ridiculously overly strict. $\endgroup$ Mar 21 at 16:44
  • $\begingroup$ haha ok I guess because $ P^\dagger=ZP $ that's as good a reason as any. $\endgroup$ Mar 21 at 16:46

As pointed by Craig Gidney's answer, there might exist different definitions.

Another important definition consists in saying that a logical gate $G_L$ is being implemented transversally if each physical qubit composing the logical qubit on which $G_L$ has to be implemented does not interact with any other physical qubit composing this same logical qubit ( * )

In the particular case of a single qubit logical gate, for a logical qubit composed of $n$ physical qubits, it means that:

$$G_L = \otimes_{i=1}^n G_i$$

Where $G_i$ is a single qubit physical gate acting on the $i$'th physical qubit of the logical qubit. Note that the $G_i$'s can all be different.

For a logical two qubit gate between two logical qubits, a typical example of a transversal operation consists in saying that the physical qubit $i$ of the first logical qubit will only interact with the physical qubit $i$ of the second logical qubit. For instance the image below describes a transversal cNOT for the seven qubit Steane code (in this case all the physical gates are identical, but again this is not a requirement in general)

enter image description here

The interest in considering such gates is that they don't propagate errors. Typically if a given physical qubit in one logical qubit has an error, it won't propagate on two physical qubits having an error on this same logical qubit after the transversal implementation (because two physical qubits within the same logical one will not interact).

Of course, for two-qubit gates, one error might induce two errors, but they will affect two different logical qubits. It is not a problem because the danger only comes from having multiple physical errors inside the same logical qubit. Different logical qubits are associated with independent error correction procedures and you can resist single-qubit errors occuring on different logical qubits.

( * ) In the context of concatenated codes, because you have different concatenation levels, the definition I am giving should be "generalized". It is done by saying that the $i$'th qubit of a given codeblock will only interact with the $i$'th qubit of any other codeblock, but this is really the same spirit as the one I am providing in this answer. The story gets more complicated because you have different "levels" (concatenation levels) to define your logical qubits.

A ref for this. Look at the last paragraph of page 22.

  • $\begingroup$ Thanks. What about the case of two qubit transversal operators? For example, in the color code a CNOT is transversal in the sense that it’s performed by doing physical CNOTs between data qubits - in this case errors may propagate, no? $\endgroup$
    – Lior
    Feb 28, 2022 at 6:16
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    $\begingroup$ @Lior you are welcome. I edited to put further clarification for cNOTs. I hope it is better now. In short: you are correct by saying that a transversal logical cNOT will propagate errors, but those errors propagate on two different logical qubits. Because of that you are not in a trouble. You are having troubles if a given logical qubit has two errors inside (I implicitly assume that the code distance is $3$ but the same "overall" philosophy would hold for higher distances). $\endgroup$ Feb 28, 2022 at 16:02
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    $\begingroup$ Got it. Still have a long way to go before I’ll be able to claim an even basic grasp of quantum error correction… $\endgroup$
    – Lior
    Feb 28, 2022 at 22:40
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    $\begingroup$ Ok I just read your last sentence in your question. You are asking if when $U$ is Clifford (i.e. it maps the Pauli group into itself by conjugation), then $U$ admits a transversal implementation. I am afraid that it does not hold. One counter example would come from the 15 qubit code which admits a transversal $T$ gate (which is non Clifford). From Eastin-Knill theorem we cannot have a complete gateset of transversal gates. Hence there exists a Clifford gate that is not transversal for this code. Now for many codes usually all Clifford are transversal. But this is not always the case. $\endgroup$ Mar 1, 2022 at 11:02
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    $\begingroup$ Yes, I see clearly the difference now. Thanks a lot! $\endgroup$ Mar 1, 2022 at 11:38

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