About a necessary condition for quantum error correcting codes

Question

I'm John and I have a question which I have been thinking about. I'm studying quantum information, especially, quantum error-correcting codes. When I learned some types of quantum codes (e.g. 5 qubits code or quantum Hamming code), I thought the following condition is also a necessary condition for quantum codes, but I have no idea to prove it. Could anyone prove it although it might be false?

Let $\mathbb{P}=\{P_1,...,P_n\}$ be a projection measurement, $\rho$ an $m$ qubits codeword and $U_i$ a unitary operator which represents an error ($i=1,...,m$:position of error). Then, the condition is $$\forall k_1,k_2\in \{1,...,n\}, \forall i\in \{1,...,m\},\mathrm{Tr}(P_{k_1}U_i\rho U_i^\dagger)\neq 0 \land \mathrm{Tr}(P_{k_2}U_i\rho U_i^\dagger)\neq 0\\ \Rightarrow \frac{P_{k_1}U_i\rho U_i^\dagger P_{k_1}}{\mathrm{Tr}(P_{k_1}U_i\rho U_i^\dagger)}=\frac{P_{k_2}U_i\rho U_i^\dagger P_{k_2}}{\mathrm{Tr}(P_{k_2}U_i\rho U_i^\dagger)}$$

Answer 1

I don't believe that this is true (at least without further qualifiers). If I'm correctly understanding what you're trying to convey, I think we can essentially simplify this statement to asking whether there can exists a state $|\psi\rangle$ and two projectors $P_1$ and $P_2$ such that $$P_1|\psi\rangle\neq P_2|\psi\rangle$$ (ignoring normalisation for simplicity). If so, your statement is false.

Now, let $|\phi\rangle$ be a state that is orthogonal to $|\psi\rangle$. Then we can choose $$ P_1=\frac{1}{2}(|\psi\rangle+|\phi\rangle)(\langle\psi|+\langle\phi|),\qquad P_2=\frac{1}{2}(|\psi\rangle-|\phi\rangle)(\langle\psi|-\langle\phi|) $$ Now if you want to start imposing specific properties of syndrome measurements for error correcting codes, that might be a different story... (even then, you might want to distinguish between non-degenerate codes and degenerate codes. Without much thought, I guess your statement is true for syndrome measurements on a non-degenerate code).