# Why do stabilizer cut the Hilbert space into two halves?

I am trying to follow the logic of Slide 8 in this deck.

The result is that if you have $$n-k$$ stabilizers in the set of stabilizers, then the dimension of the +1 eigenspace of all the stabilizers is $$2^k$$.

The intuition in the slides is that each stabilizer has an equal number of +1 and -1 eigenvalues since each stabilizer is a Pauli operator. So indeed, it cuts the Hilbert space into two halves of equal dimension.

But why is it that if you have two different stabilizers, say $$S_1, S_2$$, they cut the space in two exactly four quarters? If one thinks of it geometrically, it's not clear why the "lines of the cut" must be orthogonal.

Take independent stabilizers $$S_1,S_2,\dots$$.

First, note that $$\mathrm{tr}(S_i)=0$$, as well as $$\mathrm{tr}(S_i S_j)=0$$, and the same for any other product of the $$S_i$$ (as those all are Pauli products with at least one non-identity).

This means that $$S_1$$ has half +1 and half -1 eigenvalues. Thus, taking its +1 eigenspace cuts the space in half.

Next, note that $$P_1=(I+S_1)/2$$ is the projector onto that space. Then, $$\mathrm{tr}(P_1S_2)=(\mathrm{tr}(S_2)+\mathrm{tr}(S_1S_2))/2=0\ .$$ That is, $$S_2$$, when restricted to the $$+1$$ eigenspace $$P_1$$ of $$S_1$$, has half +1 and half -1 eigenvalues. Projecting onto the +1 space thus again cuts the space in half.

You can continue like that.

More carefully, if you have $$n$$ qubits and $$n-k$$ linearly independent stabilisers, +1 eigenspace has dimension $$2^k$$.

The way that I like to think about this is that each stabiliser yields $$\pm 1$$ values. So, not only is there the space where all the stabilisers have $$+1$$ value, but also the spaces where the stabilisers take on the different values $$\{\pm 1\}^{n-k}$$. Now, I assert that each of these spaces is the same size, and so it must be that the space is of size $$2^n/2^{n-k}=2^k$$.

Why do I assert that? Take the space where all the stabilisers are $$+1$$, and take the set of all Pauli operators that commute with the stabilisers. This set is often denoted $$N(S)$$ for the normalizer. Now, consider some $$\sigma$$ which gives a particular syndrome (set of stabiliser values). If $$\tau\in N(S)$$ then because $$\tau$$ commutes with all the stabilisers, it is also true that $$\sigma\tau$$ has the same error syndrome as $$\sigma$$ did. In other words, for every element of $$N(S)$$, there is a counterpart in each of the different subspaces. So you can see that each of those spaces will be the same size.

• Very nice argument, thank you. Does one also need to prove here that there exists a $\sigma$ for every choice of syndrome? Jan 19 at 11:37

Here's a two-line proof: Let $$S$$ be a stabilizer group of rank $$n-k$$. The projector onto its trivial representation (i.e. the joint +1 eigenspace) is $$P = \frac{1}{|S|} \sum_{\sigma \in S} \sigma$$. But then, the dimension of this subspace is $$\mathrm{tr} P = \frac{\mathrm{tr}\mathbb{1}}{|S|} =\frac{2^n}{2^{n-k}} = 2^k$$.

Note: From a representation-theoretic point of view, the subspaces in DaftWullie's answer are exactly the isotypic components of $$S$$ for which a similar projection formula applies.