# The understandings of logical operator

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?

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

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 $$-\frac13Z_L+\sqrt{\frac89}X_L$$