# Universal gate set for the $[[15,1,3]]$ code

The $$[[15,1,3]]$$ triorthogonal code implements transversal $$T$$. Since it is a CSS code, two blocks will also have a transversal $$CNOT$$ gate. To get a universal gate set all that is required is an implementation of the Hadamard gate $$H$$.

The Hadamard gate can be implemented via gate teleportation from the state $$(I \otimes H) \frac{|00\rangle+|11\rangle}{\sqrt{2}}= \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle)$$

It is my impression that the standard proposal for a universal gate set is to use the $$[[7,1,3]]$$ Steane code to implement all the Clifford gates transversally then use magic state distillation to implement the $$T$$ gate on the Steane code.

My extremely primitive understanding of these things is that fault tolerantly preparing stabilizer states is easy but fault tolerantly preparing magic states like $$T | + \rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\pi/4}|1\rangle)$$ is extremely hard.

If this is the case then wouldn't it be way easier to prepare the stabilizer state $$(I \otimes H) \frac{|00\rangle+|11\rangle}{\sqrt{2}}= \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle)$$ (which has stabilizer generators $$XZ,ZX$$) and then use gate teleportation to use this state to implement $$H$$ fault tolerantly on the $$[[15,1,3]]$$ code and so do universal computation with the $$[[15,1,3]]$$ code instead of the Steane code.

To summarize my question: if stabilizer states are easy and magic states are hard then wouldn't it be better to do universal computation with $$[[15,1,3]]$$ and $$H$$ instead of $$[[7,1,3]]$$ and $$T$$?

• For code concatenation specifically, your rate would go down roughly as $\left(\frac{1}{2\cdot 15}\right)^\ell$ as opposed to $\left(\frac{1}{2\cdot 7}\right)^\ell$ (counting ancillas). It doesn't seem obvious if the ease of preparing $|+\rangle$ is worth the loss in rate here. However, for fault tolerance based on qLDPC codes, it seems much more favorable since a gate acting as $\bar{T}^{\otimes k}$ would additionally aid in the logical gate target problem. (Personally, this is my favored approach: $\bar{T}^{\otimes k}$ or $\bar{CCZ}^{\otimes k}$ and teleporting $H$) Commented Jun 16 at 19:06