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An n-qubit state will have n generators. If I have a spin entangled state like $|0\rangle|11\rangle + |1\rangle|00\rangle$, I would say the generators are $XXX, IZZ, ZZI$.

First: Is $ZZI$ correct for the above state? I.e., can I have stabilisers that just apply a global phase?

Second: Can I have negative generators/stabilisers? E.g., for the state $|0\rangle|11\rangle$, could I have $Z(-Z)(-Z)$ as well as $ZZZ, ZII, IZZ$?

Third: If I now have, say, $-|0\rangle|111\rangle + |1\rangle|000\rangle$, how do I work out the generators or stabilisers for this? Or in general, for more difficult to do-by-inspection states?

Are there states for which there are no generators?

banercat
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1 Answers1

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A stabilizer state is a state that is stabilized by some Pauli operators (e.g. $XXX, -ZZI, i\,YIY$). We say an operator $S$ stabilizes $|\phi\rangle$ if $S|\phi\rangle = |\phi\rangle$ that is to say $|\phi\rangle$ is in the $+1$-eigenspace of $S$.

$Z$ is the only non-trivial stabilizer of $|0\rangle$, $-Z$ is the only non-trivial stabilizer of $|1\rangle$ (as $Z|1\rangle = -|1\rangle$).

$|011\rangle + |100\rangle$ is then stabilized by $XXX, IZZ$ and $-ZZI$.

The stabilizers of a stabilizer state over $n$ qubits form a subgroup of the Pauli group over $n$ qubits. This group is generated by all the possible Pauli gates on each qubit, together with some phasis from $\{1, i, -1, -i\}$: $$ \mathcal{P}_n = \left\langle\pm iI, X_{1\leq i \leq n}, Y_{1\leq i \leq n}, Z_{1\leq i \leq n}\right\rangle $$ where $Z_i$ is the Pauli operator acting as $Z$ on qubit $i$ and as $I$ otherwise.

As the stabilizers form a group, you will find more than $n$ Pauli operators that stabilize any stabilizer state, but you will find only $n$ independant non-trivial generators of this stabilizer group.

For example, $|011\rangle + |100\rangle$ is also stabilized by $-ZIZ$ and $-XYY$ as $-ZIZ = IZZ \cdot-IZZ$ and $-XYY=XXX\cdot IZZ$.

Not all quantum states are stabilizer states. Indeed, there are finitely many possible choices for the stabilizer generators, so there is a finite number of stabilizer states whereas there is an infinite number of possible quantum states.

Only states reachable through Clifford gates ($H$, $S$, $CNOT$) and Pauli measurements (starting at the $|0...0\rangle$ state) are stabilizer states. This set of quantum operations is famously non-universal.

So called "magic states" are examples of non-stabilizer states. A famous one is $|0\rangle + e^{i\frac{\pi}{4}}|1\rangle$. These states are used to implement non-Clifford gates in settings where only Clifford gates are available.

I do not know of a general procedure to work out the generators of the stabilizer group of a stabilizer state.

If you know of a Clifford circuit computing the state, you can get its defining stabilizers by computing its StabilizerTable in polynomial time (relevant paper and its Qiskit implementation). This data structure allows you to go back and forth between stabilizer formalism and quantum circuit.

AG47
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