Rigorous proof that quantum error correction is impossible without entanglement

Question

I am looking for a paper in which it is explicitly said that if it is not possible to entangle qubits, for instance because the gates are too noisy, then quantum error-correction is impossible to implement.

Is there such a reference somewhere?

Answer 1

TL;DR: Every entanglement-free quantum error-correcting code has distance one and therefore is not useful for the purposes of practical quantum error correction.

Intuition. Prohibition on entanglement forces all logical states to be product states, localizing logical qubits on physical qubits. But then no-cloning implies that each logical qubit is encoded in exactly one physical qubit, resulting in the trivial code distance.

Proof. We will proceed by induction on the number $n$ of physical qubits in the block. Suppose that every entanglement-free quantum error-correcting code on $n$ qubits has distance one. Let $C$ be such a code on $n+1$ physical qubits. We will think of our $(n+1)$-qubit block as consisting of a qubit $q$ and an $n$-qubit block $Q$. Let $U_L$ be a logical operator. We will show that $U_L$ is supported$^1$ on one qubit. First, observe that the assumption that $C$ is entanglement-free, implies then either $C=A_q\otimes|\psi_Q\rangle$ for some linear subspace $A_q\subseteq\mathcal{H}_q$ and some state $|\psi_Q\rangle\in\mathcal{H}_Q$ or $C=|\psi_q\rangle\otimes A_Q$ for some state $|\psi_q\rangle\in\mathcal{H}_q$ and some linear subspace $A_Q\subseteq\mathcal{H}_Q$. See e.g. lemma here for proof. If $C=A_q\otimes|\psi_Q\rangle$, then $U_L$ may be chosen to act as identity on $Q$ and therefore $U_L$ is supported on $q$ only. If on the other hand $C=|\psi_q\rangle\otimes A_Q$, then $U_L$ can be chosen to act as identity on $q$ and is supported on a single qubit in $Q$ by our inductive hypothesis.$\square$

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^{$^1$ Support of an operator is the set of qubits on which it acts as non-identity.}

Answer 2

I don't think you will find such absolute binary statements. If entanglement is impossible, how do you do quantum computation in the first place? In that case, we don't need quantum computation; everything can be simulated classically, and all errors can be handled classically.

Usually, statements about error correction are given in terms of some target computation accuracy $\epsilon$ and some thresholds for the probability or rate of errors. For example, see this important result Threshold Theorem.

Even then, the Threshold theorem states that if the error rate is lower than some threshold, accurate quantum computation is feasible; it does not claim the inverse. This means we can not claim that if the error rate is above some threshold, useful results are impossible.