# Shortcutting Clifford circuit computations using relations between stabilizers

## Overview

I am analyzing the output state of a Clifford circuit for various stabilizer state inputs. My circuit has midcircuit computational-basis measurements. I am curious if it is possible to save myself time on these computations by using algebraic relations between my stabilizer state inputs.

To be concrete, suppose I have $$n$$ qubit stabilizers states which differ only in one of the generators as $$\vert\psi\rangle = \langle g_1, g_2, \dots, g_n\rangle$$, $$\vert\phi\rangle = \langle g_1', g_2, \dots g_n\rangle$$, and $$\vert\chi\rangle = \langle i g_1 g_1', g_2, \dots, g_n\rangle$$. To ensure the problem is both well-defined and nontrivial, let's take $$g_1$$ and $$g_1'$$ anti commuting and $$i g_1 g_1'$$ independent of $$g_2, \dots, g_n$$. Perhaps $$g_1$$ and $$g_1'$$ are the $$X$$ and $$Z$$ paulis on the first qubit, for instance, and $$g_2, \dots g_n$$ have identity on this qubit. Given that I've calculated the action of a Clifford circuit $$C$$ on $$\vert\psi\rangle$$ and $$\vert\phi\rangle$$, is there a shortcut to computing the output of $$C$$ on $$\vert\chi\rangle$$?

## My thoughts

If $$C$$ is a unitary Clifford circuit, and I've computed $$U g_1 U^\dagger$$ and $$U g_1' U^\dagger$$, computing $$U(i g_1 g_1') U^\dagger$$ is straightforward. It is less clear to me whether there is an analogous strategy when the circuit $$C$$ has stabilizer measurements. By the principle of delayed measurement, we could push measurements to the end, but at the expense of additional qubits. As the case of $$g_1 = X$$, $$g_1' = Z$$ illustrates, the measurement properties of $$\vert \chi\rangle$$ could be quite different than either $$\vert\psi\rangle$$ or $$\vert\phi\rangle$$.

I am also thinking about using linear dependence, as I would outside the stabilizer formalism. But I am more curious about using relations "within the group theory" rather than the linear algebra.

Circuits with measurements still have "stabilizer flows" $$A \rightarrow B$$ where $$A$$ and $$B$$ are Pauli products. You can interpret $$A \rightarrow B$$ as saying "if you knew $$\langle A \rangle=c$$ where $$c \in \{-1,+1\}$$ before the operation, then you will know $$\langle B\rangle=c$$ once the operation's done". The flows of a Clifford operation $$C$$ are $$P \rightarrow C^{-1}PC$$, for all Pauli products $$P$$. But non-Clifford operations also have flows.

For example, consider the two-qubit $$X \otimes X$$ measurement $$M_{XX}$$. It has the flows $$X_1 \rightarrow X_1$$ and $$X_2 \rightarrow X_2$$. It doesn't have the flow $$Z_1 \rightarrow Z_1$$, since $$Z_1$$ anticommutes with the measurement meaning you can't know it afterwards. But it does have the flow $$Z_1 Z_2 \rightarrow Z_1 Z_2$$. And also it has a flow for the measurement result: $$X_1 X_2 \rightarrow (-1)^{m} I_1 I_2$$ where $$m$$ is the 0-or-1 measurement result. For example, if you knew $$\langle X_1 X_2\rangle=+1$$ before the measurement, then you know $$\langle (-1)^m I_1 I_2\rangle = +1$$ (i.e. $$m=\text{False}$$) afterwards.

Flows form a group. They have generators, they can be multiplied, there's an identity, etc. They can be chained and concatenated and all kinds of things. A stabilizer circuit with $$n$$ qubit inputs and $$m$$ qubit outputs will have $$n+m$$ flow generators, by the state channel duality and the fact that an $$n+m$$ qubit stabilizer state has $$n+m$$ stabilizer generators.

Note this concept has a variety of names in papers. For example, in https://arxiv.org/abs/2303.08829 it's called "pauli webs". I've also heard "pauli flows".

See Appendix A of https://arxiv.org/abs/2302.02192 for more examples of stabilizer flows for basic operations. That appendix also explains how you can test if a given operation has a given flow using a small circuit.

• Thank you for the information. In the case where there is no flow at a certain stage of the computation (meaning I'm in a stabilizer state in which one of the generators has no flow with respect to the next operation) can anything meaningful be done? Sounds like the lesson is: flows allow for the shortcuts I'm after, provided they exist for the states I'm interested in. Oct 25, 2023 at 2:07
• @Jacob Usually if there's no flow it means information is lost (e.g. anticommutes with a measurement) or you're dealing with more general effects like T gates. Oct 25, 2023 at 3:23
• I can see how this is true in the context of error correction. I have simple quantum teleportation in mind (a Clifford circuit), with stabilizer state inputs. The measurements can anticommute w/ the stabilizer, but operations controlled on the results can produce the desired teleportation. Oct 25, 2023 at 13:28
• @Jacob teleportation will give you an X->X flow and a Z->Z flow, so it doesn't anti commute with anything. Oct 25, 2023 at 18:18