In quantum error correcting code, such as shor 3-qubit code, code space is spanned by the basis {|000>, |111>}. The logical X operator is XXX and the logical Z operator is IIZ. When the code state is (|000>+|111>)√2, it is an eigenstate of logical X, and When the code state is |000>, it is an eigenstate of IIZ. On the other hand, the code state (|000>+√2|111>)/√3 does not seems to be the eigenstate of any logical operator. Considering these examples, is the statement "Some code states is an eigenstate of a logical operator, and also there are code states that is not an eigenstate of any logical operator" generally true?

  • $\begingroup$ $|111\rangle$ is an eigenstate of $IIZ$, it's just a -1 eigenstate rather than +1. $\endgroup$
    – DaftWullie
    Commented Jul 9 at 8:46
  • $\begingroup$ Thank you. How about 1/√3|000>+√2/√3|111>? Is it eigenstate of any logical operator? Is the statement I gave true? $\endgroup$
    – kong
    Commented Jul 9 at 11:46

1 Answer 1


Consider a code with one logical qubit (for the sake of argument). In the same way that any single-qubit state is the +1 eigenstate of some linear combination of Pauli operators, there's a logical operator that is the eigenstate of any logical state.

For example, consider the state $$ \frac{1}{\sqrt{3}}|0_L\rangle+\sqrt{\frac23}|1_L\rangle. $$ Let's step back to a single qubit $$ |\psi\rangle=\frac{1}{\sqrt{3}}|0\rangle+\sqrt{\frac23}|1\rangle. $$ Now let's construct an orthogonal state $$ |\psi^\perp\rangle=-\frac{1}{\sqrt{3}}|1\rangle+\sqrt{\frac23}|0\rangle. $$ So, I can build a Pauli operator $$ |\psi\rangle\langle\psi|-|\psi^\perp\rangle\langle\psi^\perp|=-\frac13Z+\sqrt{\frac89}X $$ which has $|\psi\rangle$ as a +1 eigenstate.

Now, let's go back to the logical state. It is the case that it is a +1 eigenstate if $$ -\frac13Z_L+\sqrt{\frac89}X_L $$


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