Shor 9 qubit code — how are the observables measured and eigenvalues obtained during syndrome measurement?

Question

Say we have the Shor 9 qubit code

$$|\psi_L\rangle=\tfrac{1}{\sqrt{2^3}}((|000\rangle+|111\rangle)^{\otimes3}+(|000\rangle-|111\rangle)^{\otimes3}),$$

and we have a bit flip error. My lecture notes say that error syndromes are obtained by measuring the observables:

$$Z_1Z_2,Z_2Z_3,Z_4Z_5,Z_5Z_6, Z_7Z_8,Z_8Z_9.$$

Which will give eigenvalues +1 if the two adjacent qubits in the block are the same and -1 if they are different.

I have two questions about this:

1. How does one measure the observables?

2. How are the eigenvalues obtained?

Answer 1

Note that measuring an observable is equivalent to projecting the quantum state into a particular eigenspace of the operator, and the measurement result tells you which eigenspace. So, in the case of measuring an observable on an eigenstate of that observable, you just project the state onto itself, and the outcome tells you the eigenvalue. So, (2)is equivalent to (1).

In terms of how to actually measure an observable, there's a standard technique if the observable $O$ is such that $O^2$ is the identity (such as a tensor product of Pauli matrices, as is the case here). I described this in a previous answer, but add it here for completeness: Basically, you add an ancilla (the top qubit)for each observable, perform the circuit, and then measure the ancilla in the 0/1 basis. 0 corresponds to the +1 eigenvalue of the operator $O$, while 1 corresponds to the -1 eigenvalue.

Answer 2

So say we have the encoded single qubit

$$|\psi_L\rangle=\tfrac{c_0}{\sqrt{2^3}}(|000\rangle+|111\rangle)^{\otimes3}+\tfrac{c_1}{\sqrt{2^3}}(|000\rangle-|111\rangle)^{\otimes3},$$

And then say a bit flip occurs on the 4th qubit so now we have

$$|\psi_L \rangle=\tfrac{c_0}{\sqrt{2^3}}((|000\rangle+|111\rangle)\otimes(|100\rangle+|011\rangle)\otimes(|000\rangle+|111\rangle))+\tfrac{c_1}{\sqrt{2^3}}((|000\rangle-|111\rangle)\otimes (|100\rangle-|011\rangle)\otimes (|000\rangle-|111\rangle))$$

Then we add the ancilla qubit $|0\rangle$, giving

$$|\psi_i \rangle=\tfrac{c_0|0\rangle}{\sqrt{2^3}}((|000\rangle+|111\rangle)\otimes(|100\rangle+|011\rangle)\otimes(|000\rangle+|111\rangle))+\tfrac{c_1|0\rangle}{\sqrt{2^3}}((|000\rangle-|111\rangle)\otimes (|100\rangle-|011\rangle)\otimes (|000\rangle-|111\rangle))$$

Performing the first Hadamard gate in your diagram gives

$$|\psi_i \rangle=\tfrac{c_0(|0\rangle+|1\rangle)}{\sqrt{2}\sqrt{2^3}}((|000\rangle+|111\rangle)\otimes(|100\rangle+|011\rangle)\otimes(|000\rangle+|111\rangle))+\tfrac{c_1(|0\rangle+|1\rangle)}{\sqrt{2}\sqrt{2^3}}((|000\rangle-|111\rangle)\otimes (|100\rangle-|011\rangle)\otimes (|000\rangle-|111\rangle))$$

Performing $Z_4Z_5$ gives

$$|\psi_i \rangle=\tfrac{c_0(|0\rangle+|1\rangle)}{\sqrt{2}\sqrt{2^3}}((|000\rangle+|111\rangle)\otimes(-|100\rangle+|011\rangle)\otimes(|000\rangle+|111\rangle))+\tfrac{c_1(|0\rangle+|1\rangle)}{\sqrt{2}\sqrt{2^3}}((|000\rangle-|111\rangle)\otimes (-|100\rangle-|011\rangle)\otimes (|000\rangle-|111\rangle))$$

Performing the second Hadamard gate gives

$$|\psi_i \rangle=\tfrac{c_0|0\rangle}{\sqrt{2^3}}((|000\rangle+|111\rangle)\otimes(|100\rangle+|011\rangle)\otimes(|000\rangle+|111\rangle))+\tfrac{c_1|0\rangle}{\sqrt{2^3}}((|000\rangle-|111\rangle)\otimes (|100\rangle-|011\rangle)\otimes (|000\rangle-|111\rangle))$$

Now we take a measurement in the $\{|0\rangle, |1\rangle\}$ basis. At this point I'm confused. Does it mean that we use a projective operator, say for $|1\rangle$, $P_1=|1\rangle \langle1|$, then we see what the probability of this is using $\langle\psi| P_1 |\psi\rangle$? And if the probability is 1 then the eigenvalue is -1, and if its zero the eigenvalue is +1.

Note: When I understand it fully I'll edit this so it's a complete answer.