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I am trying to understand the properties of quantum error correcting codes. Consider a quantum code on a lattice, with the property that a given region $R$ is correctable (for any error localized to region $R$. This means that even if this region is thrown away, one can recover the encoded state). For a pure state $\Psi$ in the code space, let $\rho_R(\Psi)$ be its reduced density matrix in region R. Then is any of the following true? :

  1. $\rho_R(\Psi)$ is independent of $\Psi$. This would mean that there is no information about the encoded state in the region $R$, so that one can recover $\Psi$ if $R$ is removed. If this is not true, then it would mean that recovery operation somehow depends on the encoded state. This seems strange because usually one does not know what the encoded state is. So, i would really appreciate clarification on this point as well.

  2. For any two orthogonal states $\Psi, \Psi'$, $\rho_R(\Psi)=\rho_R(\Psi')$. This is analogous to the property of toric code. But i am not sure if it is true in general for codes on lattice.

Thanks!

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Property 1 is equivalent to R being a correctable region. Property 2 of course follows from property 1. It is less obvious, but property 2 also implies property 1. (You can prove this by considering different sets of orthogonal codeword states.)

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  • $\begingroup$ Thanks. I can now see how to prove equivalences between these properties. You are right that 'Property 2=> Knill-Laflamme condition' cannot be shown by working on a fixed basis in code space. Is this the correct way to argue for 'Property 2=> Knill-Laflamme'?: Take $\Psi, \Psi'$ as orthogonal pure states, and consider any linear combination of them. This linear combination with have desired reduced density matrix in $R$, if and only if $\text{Tr}_R(\Psi)$ and $\text{Tr}_R(\Psi')$ have orthogonal support (here $Tr_R$ is trace over region $R$). This will then imply Knill-Laflamme condition. $\endgroup$
    – anurag anshu
    Mar 13, 2015 at 8:21
  • $\begingroup$ I don't think considering just a single linear combination at a time is sufficient. You have to consider orthogonal pairs to use property 2. The pairs $|\psi\rangle \pm |\psi'\rangle$ and $|\psi\rangle \pm i |\psi'\rangle$ are sufficient. $\endgroup$
    – Daniel Gottesman
    Mar 16, 2015 at 15:45

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