# Stabilizer for quantum error correction code

I have some very basic questions about stabilizers.

What I understood:

To describe a state $$|\psi \rangle$$ that lives in an $$n$$-qubit Hilbert space, we can either give the wavefunction (so the expression of $$|\psi\rangle$$), either give a set of commuting observable that $$|\psi\rangle$$ is an eigenvector with $$+1$$ eigenvalue.

We define a stabilizer $$M$$ of $$|\psi \rangle$$ as a tensor product of $$n$$ Pauli matrices (including the identity) that verifies $$M |\psi \rangle = |\psi\rangle$$.

And (apparently) we need $$n$$ stabilizers to fully define a state.

The things I don't understand:

1. How can a stabilizer necessarily be a product of Pauli matrices?

With $$n=1$$, I take $$|\psi \rangle = \alpha | 0 \rangle + \beta |1 \rangle$$, excepted for specific values of $$\alpha$$ and $$\beta$$, this state is only an eigenvector of $$I$$ (not of the other pauli matrices). But saying $$I$$ is the stabilizer doesn't give me which state I am working for.

1. How can we need only $$n$$ stabilizers to fully define a state?

With $$n$$ qubits we have $$2^n$$ dimensional Hilbert space. I thus expect to have $$2^n$$ stabilizers, not $$n$$ to fully describe a state.

I am looking for a simple answer. Preferably an answer based on the same materials as my question, if possible. I am really a beginner in quantum error correction.

I learned these things within a 1-hour tutorial, so I don't have references for which book I learned this from. It is what I understood (maybe badly) from the professor talking.

Here are a couple of observations which will hopefully clarify things.

1. Only some states have Pauli stabilisers.

You have correctly identified that not all states have Pauli stabilisers. An example of a state that does not have any Pauli stabilisers (apart from the identity operator of course, which we typically ignore) is $$\lvert A \rangle = \tfrac{1}{\sqrt 2}\bigl( \lvert 0 \rangle + \mathrm{e}^{i \pi /4} \lvert 1 \rangle \bigr)$$.

However, a quantum error correction code is not a single state: it is a subspace of some larger state-space. Some subspaces can be described as the set of states which are stabilised by a set of commuting operators from the Pauli group, on some number $$n$$ of qubits. That set of stabilising operators forms an abelian subgroup of the Pauli group, and is referred to as a stabiliser group. If the stabiliser group has $$r$$ generators, then the dimension of the subspace which they jointly stabilise is $$2^{n-r}$$. In the case that $$r = n$$, the space which you stabilise has dimension $$2^0 = 1$$, which means that you have identified a single state (representing in this case a somewhat degenerate example of an error correcting code, because it is 'encoding' zero qubits of information).

2. Only $$n$$ stabiliser generators are needed to characterise a state, because for each independent stabiliser (each operator for which you impose the restriction to the $$+1$$ eigenspace, which cannot be formed as a product of the other operators), you are essentially removing one qubit's worth of freedom from the range of states that you can have.

The above is of course the intuitive description. More formally — given that we are talking about commuting operators selected from the Pauli group — a more formal way of seeing this is that

• For any commuting set of $$r$$ independent Pauli operators $$\{P_1,P_2,\ldots,P_r\}$$ on $$n$$ qubits, there is a Clifford group operation $$C$$ such that $$C P_j P^\dagger = Z_j$$ for each $$1 \leqslant j \leqslant r$$, where $$Z_j$$ is the single-qubit $$Z$$ operator acting only on qubit $$j$$.

• The set of Pauli operators $$\{ Z_1, Z_2, \ldots, Z_k \}$$, as a set of stabilisers, describe a set of states in which qubit $$1$$ is fixed to the state $$\lvert 0 \rangle$$, qubit $$2$$ is also fixed to the state $$\lvert 0 \rangle$$, and so forth up to qubit $$r$$; the more of these operators there are, the more qubits there are with fixed states, and the fewer degrees of freedom there are in the states which satisfy those constraints. In particular, the subspace of states satisfying this property has dimension $$2^{n-r}$$.

• The error correcting code is related to this second set of states with single-qubit constraints by a unitary transformation, which preserves the dimension of subspaces by virtue of being invertible. Thus, the error correcting code also has dimension $$2^{n-r}$$.

Each added stabiliser generator can be said to remove one qubit of freedom from the state-space, in the sense of the formal argument above. In particular, any set of $$n$$ independent commuting Pauli stabilisers on $$n$$ qubits, will characterise some single state (though, as observed above, not all states do have such a representation).

• Some very naive question. $r$ is the number of generators of the n qubits Pauli group ? (So $r =3^n$ in general ? I am probably confused...). And I don't understand because for the $n=3$ qubit bit flip code (1 logical and 3 physical qubits), the code space has dimension $1$. It would mean that $n-r=1$ thus $r=2$..? Is $r$ the number of generator of the Pauli group or the number of stabilizers ? (I am very confused) – StarBucK Nov 19 '18 at 15:14
• Should be 4(I,X,Y,Z), not 3, you forgot the identity transformation. – XXDD Nov 20 '18 at 6:40
• I've edited my response to clarify it a bit more. The generators I was referring to was not for the Pauli group itself: I actually was referring to the short list of stabiliser operators acting as constraints. Hopefully things will be a bit clearer after my edit. – Niel de Beaudrap Nov 20 '18 at 8:29