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.


1 Answer 1


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.

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  • $\begingroup$ 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. $\endgroup$
    – Jacob
    Oct 25, 2023 at 2:07
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 25, 2023 at 3:23
  • $\begingroup$ 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. $\endgroup$
    – Jacob
    Oct 25, 2023 at 13:28
  • $\begingroup$ @Jacob teleportation will give you an X->X flow and a Z->Z flow, so it doesn't anti commute with anything. $\endgroup$ Oct 25, 2023 at 18:18

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