# Does the Eastin-Knill theorem hold for repetition codes?

The Eastin-Knill theorem states that

For any nontrivial local-error-detecting quantum code, the set of transversal, logical unitary operators is not universal.

see original paper. Does this theorem hold for 1D repetition codes (e.g. a bit-flip code)?

Indeed, a bit-flip code cannot correct against local phase errors on individual physical qubits, so the code is not "local-error-detecting" in this sense. If the theorem does not apply, is there a known example of a universal and transversal gate set for the bit-flip code?

TL;DR: There is no universal gateset with transversal implementation in the repetition code. This is the case not due to Eastin-Knill theorem, whose assumptions the code fails to satisfy, but due to the fact that logical states have varying amount of entanglement between physical qubits and that transversal operators cannot change the amount of it. Consequently, transversal operators fail to provide access to the full code subspace.

## Code subspace

Consider a repetition code encoding one logical qubit into a block of $$n$$ physical qubits $$a_1a_2\dots a_n$$. Its code subspace is spanned by the logical computational basis states

$$|0\rangle_L=|0\rangle_1|0\rangle_2\dots|0\rangle_n\\ |1\rangle_L=|1\rangle_1|1\rangle_2\dots|1\rangle_n.\tag1$$

The key observation for our argument is that the computational basis $$(1)$$ consists of product states while all other logical states

\begin{align} |\psi\rangle_L&=\alpha|0\rangle_L+\beta|1\rangle_L\\ &=\alpha|0\rangle_1|0\rangle_2\dots|0\rangle_n+\beta|1\rangle_1|1\rangle_2\dots|1\rangle_n \end{align}\tag2

for non-zero $$\alpha,\beta\in\mathbb{C}$$ have varying amount of entanglement between the physical qubits $$a_1a_2\dots a_n$$.

## Transversal operators

The most general transversal logical operator on a single logical qubit takes the form

$$U_L=U_1\otimes U_2\otimes \dots \otimes U_n\tag3$$

where $$U_k$$ acts on qubit $$a_k$$. Clearly, $$U_L$$ fails to change entanglement between the physical qubits. Therefore, there is no transversal logical operator on a single logical qubit capable of creating superpositions $$(2)$$ within the code subspace.

In fact, this argument extends to logical operators acting on multiple logical qubits. Consider for example a logical operator $$V_L$$ on two logical qubits $$a_1a_2\dots a_n$$ and $$b_1b_2\dots b_n$$. Transversality means that

$$V_L=V_1\otimes V_2\otimes\dots\otimes V_n\tag4$$

where $$V_k$$ acts on qubits $$a_k$$ and $$b_k$$. Clearly, no such operator may change the amount of entanglement between the pair $$a_ib_i$$ and the pair $$a_jb_j$$ for $$i\ne j$$.

## Conclusion

Thus, logical quantum circuits consisting of transversal gates may not change the amount of entanglement between physical qubits within any logical qubit. In particular, any such circuit initialized to $$|0\rangle_L$$ cannot access states outside the logical computational basis. Therefore, there is no universal gateset which can be implemented transversally in the repetition code.

• Thanks for your detailed answer. So if I understand correctly, in order to have a universal transversal gateset, one would require a code in which all logical states have the same amount of entanglement? Jan 12 at 16:21
• Yes but to be precise we should specify the subsystems whose entanglement we are concerned with here. Universal set of transversal gates requires all logical states to have the same amount of entanglement between different physical qubits (or corresponding groups of physical qubits) in a logical qubit (or multiple logical qubits), e.g. $a_i$ with $a_j$ (or $a_ib_i$ with $a_jb_j$). By contrast, the amount of entanglement between different logical qubits, e.g. $a_1\dots a_n$ with $b_1\dots b_n$ depends on logical circuit and is not preserved by logical gates (transversal or otherwise). Jan 12 at 17:06