7.3 Single stabiliser states

Given $n$ independent generators of a stabiliser group $\mathcal{S}$ on a Hilbert space of $n$-qubits, we end up specifying a $1$-dimensional subspace, meaning it is spanned by a single basis vector, namely the stabiliser state. We have already talked about the single-qubit stabiliser states determined by all possible stabilisers in $\mathcal{P}_1$, namely $|0\rangle$ and $|1\rangle$ for $\langle \pm Z\rangle$, $|\pm\rangle$ for $\langle \pm X\rangle$, and $|\pm i\rangle$ for $\langle \pm Y\rangle$. We have also mentioned some of the two-qubit stabilisers states, some of which are highly entangled, such as the Bell states, and some of which are separable, such as the computational basis states (whose stabilisers groups we described by block matrices with $Z$ on the diagonal, $\mathbf{1}$ everywhere else, and signs labelling each row depending on the binary description of the state).

Here’s another two-qubit example: that of the maximally entangled state $|00\rangle+|11\rangle$. This is stabilised by $\langle XX,ZZ\rangle$, but let’s explain how we can see this. If we look first at the operator $XX$, we see that it splits the $4$-dimensional Hilbert space into two $2$-dimensional subspaces, corresponding to eigenvalues $\pm1$; by definition, it stabilises the one corresponding to eigenvalue $+1$, which is spanned by $|00\rangle+|11\rangle$ and $|01\rangle+|10\rangle$. Now the operator $ZZ$ also splits the $4$-dimensional Hilbert space into two $2$-dimensional subspaces, again corresponding to eigenvalues $\pm1$; it stabilises the one corresponding to eigenvalue $+1$, which is spanned by $|00\rangle+|11\rangle$ and $|00\rangle-|11\rangle$. Note that $|01\rangle+|10\rangle$ is in the $-1$-eigenspace of $ZZ$, even though it is in the $+1$-eigenspace of $XX$ (and vice versa for $|00\rangle-|11\rangle$). So the simultaneous $+1$-eigenspace of $XX$ and $ZZ$ is exactly the state $|00\rangle+|11\rangle$. $$\begin{aligned} |00\rangle+|11\rangle \longleftrightarrow \begin{array}{c|cc|} +&X&X \\+&Z&Z \end{array} \quad&\quad |00\rangle-|11\rangle \longleftrightarrow \begin{array}{c|cc|} -&X&X \\+&Z&Z \end{array} \\|01\rangle+|10\rangle \longleftrightarrow \begin{array}{c|cc|} +&X&X \\-&Z&Z \end{array} \quad&\quad |01\rangle-|10\rangle \longleftrightarrow \begin{array}{c|cc|} -&X&X \\-&Z&Z \end{array} \end{aligned}$$

As we have already mentioned when discussing presentations of a stabiliser group, there can be multiple different generating sets, which corresponds to the fact that there are multiple different ways of bisecting the Hilbert space. For example, the stabiliser state $|00\rangle+|11\rangle$ is completely specified by $\langle XX,ZZ\rangle$, as shown above, but also by $\langle XX,-YY\rangle$ or $\langle -YY,ZZ\rangle$. But, as we should expect, these three generating sets all generate the same group, namely $\mathcal{S}=\{\mathbf{1}\mathbf{1},XX,-YY,ZZ\}$.

How many $n$-qubit stabiliser states do we have? The answer is $$2^n\prod_{k=0}^{n-1}(2^{n-k}+1)$$ as we can show with a counting argument: we will count the number of generating sets with $n$ generators (since this is exactly the right number of generators to specify a $1$-dimensional stabiliser subspace) and then divide by the number of presentations for any given stabiliser.¹⁴⁹ There are $4^{n-1}$ choices for the first generator $G_1$ (ignoring overall sign), since it can be any $n$-fold tensor product of the four Pauli matrices, excluding the identity $\mathbf{1}\mathbf{1}\mathbf{1}\mathbf{1}$. For the second generator $G_2$, we have $(4^n/2)-2$ possibilities, since it must commute with the first generator (and we know that exactly half of the operators commute with any given operator, as shown in Exercise 7.8.3, whence $4^n/2$) and it cannot be $\mathbf{1}\mathbf{1}\mathbf{1}\mathbf{1}$ or $G_1$ (whence $-2$). Similarly, $G_3$ must commute with both $G_1$ and $G_2$, but it cannot be in the group generated by them, so there are $(4^n/4)-4$ possible choices, and so on. This means that we have $$2^n(4^n-1)\left(\frac{4^n}{2}-2\right)\left(\frac{4^n}{4}-4\right)\ldots\left(\frac{4^n}{2^{n-1}}-2^{n-1}\right)$$ possible generating sets in total. Now we need to divide by the number of presentations, but we have already calculated this in Section 7.2: it’s exactly $$(2^n-1)(2^n-2)(2^n-2^2)\ldots(2^n-2^{n-1}).$$ It is a fun algebra exercise to show that this division indeed gives the number we claimed.

This is a common technique in combinatorial arguments: first overcount, and then fix your answer by accounting for this.↩︎

As we will see, stabiliser states are ubiquitous in quantum information theory due to their versatility and relative simplicity. They play a crucial role in areas such as quantum error correction, measurement-based quantum computation, and entanglement classification.

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149. This is a common technique in combinatorial arguments: first overcount, and then fix your answer by accounting for this.↩︎