How to obtain the set of stabilizer generator of a code, from stabilizer of its circuit tableau simulation

Question

What is the relationship between the stabilizer generator of a quantum stabilizer code, and the stabilizer generator of its circuit tableau? I notice some difference for cases.

I use Stim to simulate Clifford circuit. I prepare a Bell state, then transform it to stim.Tableau. The stabilizer of the the circuit is $[XX, ZZ]$, which is the stabilizer of Bell state. Then I create 3 qubits in zero state, its tableau show $[ZII, IZI, IIZ]$. But in the angle of stabilizer code, it is supposed to be repetition code with stablizer $[ZZI, IZZ]$. Is the stabilizer of a code supposed to be different from that of its circuit, and maybe the Bell state case just a coincidence?

Now I am having the stabilizer generator of a circuit from tableau: $[−YIYZ,−YZYI,+IIYI,+IIYZ]$. Is the output state a logical state for a quantum stabilizer code? (I use stim.Tableau.to_state_vector(), and the vector does look like an encoded state)

To be more general, if we know the stablizer of a Clifford circuit, can we know whether the output state is a logical state of a quantum code? If it is, then how to know the stabilizer generator of the code, from that of its circuit?

In addition, I observe that in my stabilizer generator from Stim, some give +1 eigenvalues for my output state, and other give -1 eigenvalues. Can we get the stabilizer generator that is all in +1 subspace?

Answer 1

A Pauli operator $P$ is said to stabilize a state $|\phi\rangle$ if $P|\phi\rangle = |\phi\rangle$.

A stabilizer $n$-qubit state is a state stabilized by $n$ independent Pauli operators (they are more, but $n$ are sufficient to define the stabilizer group). The Bell state is a stabilizer state, because it is a 2-qubit state stabilized by $XX$ and $ZZ$ (as well as $YY$).

$\{ZZI, IZZ\}$ cannot define a stabilizer state alone, you would need to add a third stabilizer generator to define a stabilizer state. To be precise, the set of quantum states stabilized by both $ZZI$ and $IZZ$ belongs to a Hilbert space of dimension 2, that is the logical codespace of the repetition code with these stabilizers.

$|000\rangle$ is just a particular quantum state of this codespace, the unique state that is stabilized at the same time by the stabilizers and by the logical $Z$ operator of the code ($ZZZ$).

Note that $\{ZZI, IZZ, ZZZ\}$ and $\{ZII, IZI, IIZ\}$ represents the same stabilizer group, so stim agrees with you that $|000\rangle$ is a logical state of the repetition code.

More generally, if you know the stabilizers of a stabilizer state, you can easily check if it is a logical state of given stabilizer code. The stabilizer of the code must be in (the linear span of) the stabilizers of the state. The exact logical state of your stabilizer state can be deduced by analyzing the remaining stabilizers of the state in terms of the logical operators of the repetition code.

A given stabilizer state is a logical state of a lot of stabilizer codes. There are at least:

$$ \sum_{k=0}^{n}\binom{n}{n-k}2^{k}k! $$

For each possible value of $k$, you need to choose $n-k$ stabilizers. For the remaining $k$ Pauli operators, you need to decide which logical operator they correspond to. You are free to choose for each of them is they correspond to a $Z$ or a $X$ logical operator (hence the factor $2^{k}$) and you are also free to change their ordering.

This is an underestimate, because you could set a phase ($1$,- $i$, $-1$, $-i$) for each of the logical operator (adding a factor $4^{k}$) and you could consider any ordered basis of the space spanned by the last $k$ operators instead of just a permutation. This is not necessary for the code stabilizers because you only care about the stabilizer space and not the individual generators.

As for your last question, (which should be in my opinion asked as a separate question), the only practical way would be to start in/construct a stabilizer state of your stabilizers, which is probably equivalent to implementing an encoding circuit for the underlying stabilizer code. Note that this is not necessary for QEC usually.