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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.

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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.

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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.

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

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