Intuitive link between clifford group and gottesman-knill theorem

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Elements of the Pauli group are the n-Pauli matrices with $\pm 1$ or $\pm i$ on front of them. They all commute or anti-commute between them.

The Clifford group are element that preserve the n-Pauli group under conjugation.

My question:

Is there a link between the result of the Gottesman Knill theorem, and somehow the fact that if you only use gates in the Clifford group you can simplify the circuit using commutation and anti-commutation rules from the n-Pauli group.

I didn't go into the proof of this theorem, I just would like to see if there is a handwavy but intuitive way to understand it.

StarBucK

Posted 2019-12-20T11:39:16.600

Reputation: 1 004

Answers

3

Yes, absolutely. That's very much the root of the whole thing. You start with a state from a particular class: a stabilizer state, i.e. one which can be described by a bunch of projectors $(I+K)/2$, where $K$ is from the Pauli Group.

To determine how the state evolves under the action of the circuit, you only need to know how the circuit acts on the elements $K$. Since gates in the Clifford group preserve the Pauli group, that just says those tensor products of Paulis map into some other tensor product of Paulis (and, importantly, we case easily tell what that update is), and so we can easily describe the outcome of the computation.

DaftWullie

Posted 2019-12-20T11:39:16.600

Reputation: 35 722