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From Arthur Pesah's blog on "Computing with Quantum Codes using Transversal Gates", found here: https://arthurpesah.me/blog/2023-12-25-transversal-gates/

He gives the following examples of logical gates that are fault tolerant and non-fault tolerant.

The first gate is fault-tolerant.:

Fault tolerant gate

He gives the example that if an $X$ error occurs on the top left qubit, then when we apply the first CNOT gate, it propagates to two $X$ errors (as $CNOT: XI \rightarrow XX$):

fault tolerant gate with errors

In this case, the blog states that "suppose $d=4$ and suppose any vertical string of $X$ is an example of a minimum-weight logical $X$ operator. With only $2$ errors present before applying the gate, we can get a logical error once the gate has been applied, thus reducing the effective distance by $2$."

He then offers an example of a non-fault tolerant gate:

non-fault tolerant

Again, we look at what happens if an $X$ error occurs on the top left qubit. This time the error propagates to four $X$ errors:

non-fault tolerant errors

In this case, the blog states that "a single $X$ error can propagate into $d$ $X$ errors and create a logical error, effectively reducing distance to $1$."

The blog post define a fault tolerant logical gate for a family of codes with a growing distance to be "a gate that doesn't reduce the effective distance to O(1) when the family of codes is growing", where the effective distance is defined as being the minimum number of physical errors that create a logical error after applying the gate.


My problem is that I don't fully understand why the first gate is fault tolerant and the second is not. My understanding of fault tolerance as defined in this blog post is that the distance of the code must depend on the size of the input.

I understand that, regardless of the size of the input (say $L \times L$), errors will always propagate into $L$ errors and create a logical error after the first gate is applied. This is because, the CNOTs are acting on intersecting pairs of qubits.

However, because the first gate is acting on disjoint pairs of physical qubits, can the $2$ $X$ errors ever propagate to more than $2$ $X$ errors? I don't understand how this gate works? Why is it fault tolerant?

I don't understand why the distance has been reduced to $2$ in the first example? I don't see why the $2$ $X$ errors create a logical error? I would have thought that $XXXX$ was the logical error for this code?

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However, because the first gate is acting on disjoint pairs of physical qubits, can the 2 𝑋 errors ever propagate to more than 2 𝑋 errors?

Yes, 2 $X$ errors can propagate to 4 $X$ errors via the CNOTs.

I don't understand how this gate works? Why is it fault tolerant?

The gate halves the distance of the code. So the gate doesn't reduce the effective distance to $O(1)$ when the family of codes is growing.

I don't understand why the distance has been reduced to 2 in the first example? I don't see why the 2 𝑋 errors create a logical error? I would have thought that 𝑋𝑋𝑋𝑋 was the logical error for this code?

Consider 2 $X$ errors on the left-most column of qubits, one on the first and one on the third qubit from the top. These 2 $X$ errors will spread to a weight 4 $X$ operator on all qubits in the left-most column.

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    $\begingroup$ Ah, I think I understand. So in the gate shown in the first example, we can have a logical error ONLY if we begin with $2$ $X$ errors on the $1^{st}$ and $3^{rd}$ qubits? But the $2$ errors cannot propagate into $4$ errors if they are originally on the $1^{st}$ and $2^{nd}$ qubits because these are not connected to qubit $3$ and $4$ via CNOTs? $\endgroup$
    – am567
    Feb 14 at 16:21
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    $\begingroup$ yes that's right $\endgroup$
    – Peter-Jan
    Feb 15 at 18:07

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