For quantum error correction, the necessary and sufficient condition is given in standard texts as:

$\langle \phi| E^{\dagger}_{a} E_{b} |\psi \rangle = C_{ab} \langle \phi|\psi \rangle $

$|\psi\rangle$ and $|\phi\rangle$ are codewords. $E_{a}$ and $E_b$ are arbitrary errors that can occur.

My queries would be:

  1. Why is $C_{ab}$ independent of the codewords and only dependant on the errors? Intuitively, I realize this is because otherwise, we gain some information about the codeword by detecting the error and run the risk of disturbing the superposition of the codeword. Is there a more formal mathematical explanation for the same?
  2. I realize that $C$ is a Hermitian matrix which can be diagonalized. What does it mean in terms of the codewords and the errors to diagonalize the matrix $C$?
  3. Why is it the case that the state $|\psi\rangle$ is always an eigenstate of $ E^{\dagger}_{a} E_{b}$ with $C_{ab}$ as an eigenvalue? It doesn't seem intuitive to me.

1 Answer 1


General Comment

For errors to be correctable, they must not cause us to mistake one codeword for an orthogonal one. So if two codewords are orthogonal, and they are acted upon by correctable errors, the result will remain orthogonal

$$ \langle \phi|\psi \rangle =0 \,\,\implies\,\, \langle \phi| E^{\dagger}_{a} E_{b} |\psi \rangle = 0 $$

Answer to 1

By setting $|\phi \rangle = |\psi \rangle$, your condition gives us

$$C_{ab} = \langle \psi | E^{\dagger}_{a} E_{b} |\psi \rangle.$$

The only case for which this would have no $ |\psi \rangle$ dependence is if $E^{\dagger}_{a} E_{b}=I$, which is clearly not an interesting case!

Perhaps your source assumes that a choice of an orthonormal basis for the codewords has been made, and then all else is basis dependent. Or perhaps I misunderstood the question.

Answer to 3

Even if the errors are not unitary, the operator $E^{\dagger}_{a} E_{b}$ can be decomposed as a linear superposition of unitary operators. Since the condition in your question is linear, it therefore must be as true for non-unitary errors as for unitary errors.

  • $\begingroup$ Regarding your answer to the first query, look at equation 30 of this link: arxiv.org/abs/quant-ph/0004072. The condition is stated but $C_{ab}$ depends only on errors and not the codeword. $\endgroup$
    – BlackHat18
    Jul 15, 2018 at 14:06

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