• Members 1 post
    Oct. 28, 2021, 10:23 p.m.

    Hi! Hope everyone is doing awesome!

    Let's take 5-qubit code, for example, we have these two logical states:
    $|0_L\rangle=\frac{1}{4}[|00000\rangle+|10010\rangle+|01001\rangle+|10100\rangle+|01010\rangle-|11011\rangle-|00110\rangle-|11000\rangle$
    $\quad\quad\quad-|11101\rangle-|00011\rangle-|11110\rangle-|01111\rangle-|10001\rangle-|01100\rangle-|10111\rangle-|00101\rangle]$

    $|1_L\rangle=\frac{1}{4}[|11111\rangle+|01101\rangle+|10110\rangle+|01011\rangle+|10101\rangle+|00100\rangle-|11001\rangle-|00111\rangle$
    $\quad\quad\quad-|00010\rangle-|11100\rangle-|00001\rangle-|10000\rangle-|01110\rangle-|10011\rangle-|01000\rangle-|11010\rangle]
    $

    Stabilizers for 5-qubit code is given by:
    $\langle XZZXI, IXZZX,XIXXZZ,ZXIXZ\rangle$

    From the stabilizers and logical $X$ and $Z$, we can straightforwardly find the logical 0 and 1 from $I+S$ (identity + stabilizer). I'm wondering, how do we go the other way round, i.e., from the logical 0 and 1, is there a systematic way to work out the 4 stabilizers in this case?

    Thank you in advance!! :))

  • Members 1 post
    Nov. 7, 2021, 2:31 p.m.

    Hello. Thanks for the question! Below is my understanding.

    The 5-qubit error correcting code has four stabilizers. The choice of this set is not unique in the sense that if we replace the 2nd stabilizer by the product of the 1st and the 2nd, the new set is another choice. So, I guess your question is if we can find the set of stabilizers up to this equivalence, suppose we know the set of logical states (i.e., ground states of the stabilizer Hamiltonian).

    In principle, it is possible. The most straightforward but inefficient way is to write down all possible Pauli strings on the five qubits. Test if the two states are eigenstates of that Pauli string with eigenvalue +1. If the answer is yes, keep the stabilizer. (In the end, one needs to remove the stabilizers that are not independent.)

    I do not know a more efficient way than the rather inefficient one described in the previous paragraph. Maybe someone else knows. :P

    Fun fact: this 5 qubit error correcting code has been used to assemble a larger quantum state on a hyperbolic lattice. (See arxiv.org/pdf/1503.06237.pdf.) That is an attempt to understand some aspects of quantum gravity theory.