# How to reason about absorbing Pauli Product rotations into measurements?

In this image, the orange box refers to the operator $$\exp(-i P \frac{\pi}{4})$$ and the blue box refers to the Pauli Measurement $$P'$$. The idea behind these rules is to show how all the Pauli Product $$\frac{\pi}{4}$$ rotations at the end of the circuit can be absorbed into the measurements.

How would you prove these equations? In particular, I am looking for a proof that will help me reason about what happens if I measure say just the first qubit after performing a multi qubit orange box. For example, how would I absorb the orange box into the measurements in the following?

First of all, let us consider an arbitrary unitary $$U$$. If we first apply $$U$$ and then measure a POVM $$\{E_a\}$$, then this is equivalent to doing nothing and measuring the POVM $$\{U^\dagger E_a U\}$$ instead since $$\mathrm{Pr}[a|\rho] = \mathrm{tr}\left( E_aU \rho U^\dagger \right) = \mathrm{tr}\left( U^\dagger E_a U \rho \right)$$. In particular, a Pauli measurement is given by the projectors $$\frac 1 2 \left( \mathbb{I} \pm P \right)$$ and the altered measurement is described by the Hermitian operator $$O = U^\dagger P U$$.

Now note that $$U=\exp(- i\frac \pi 4 P)$$ is a Clifford unitary if $$P$$ is a Pauli operator. Thus, absorbing the action of $$U$$ into the measurement of a Pauli operator $$Q$$ will result in another Pauli measurement of $$\pm Q'=U^\dagger Q U$$. Note that the possible sign simply relabels the outcomes. Next, it is clear that $$U^\dagger Q U=Q$$ if $$P$$ and $$Q$$ commute, thus we can remove $$U$$ from the circuit. If not, a straightforward calculation shows that $$e^{i\frac \pi 4 P}Qe^{- i\frac \pi 4 P} = i PQ \equiv \pm Q'.$$

If you measure the first qubit after a $$n$$-qubit Clifford gate, than this effectively the same as doing nothing and performing a Pauli measurement on many ($$\leq n$$) qubits (since $$U$$ will generally map a Pauli operator on the first qubit to a Pauli operator supported on $$n$$ qubits).

The measurement in your final example consists of measuring four commuting Pauli operators $$P_1,\dots,P_4$$ (a syndrome measurement). The projectors of the joint measurement are given as products $$P_x = \prod_{i=1}^4 \frac 1 2 (\mathbb I + (-1)^{x_i} P_i)$$ where $$x\in \{0,1\}^4$$. To compute $$UP_x U^\dagger$$, we can thus individually transform the involved projectors.

Let us use the above reasoning sequentially:

1. $$X\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z$$ anti-commutes with $$Y\otimes\mathbb{I}\otimes\mathbb{I}\otimes\mathbb{I}$$, thus we can replace the measurement of $$-Y$$ on the first qubit by the measurement of $$-iXY\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z=Z\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z$$.
2. The Pauli operators on the second and third qubit obviously commute with the unitary, so the measurements are unchanged.
3. The last Pauli operator again anti-commutes, so we can replace it by the measurement of $$X\otimes\mathbb{I}\otimes\mathbb{I}\otimes i ZY=X\otimes\mathbb{I}\otimes\mathbb{I}\otimes X$$.

Note that we obtained an entangling measurement between the first and the fourth qubit. This is in fact a Bell measurement as the joint eigenvectors of $$X\otimes X$$ and $$Z\otimes Z$$ are the four Bell states $$\{|00\rangle \pm |11\rangle, |01\rangle \pm |10\rangle\}$$.

• Hi, I'm still trying to work through understanding your answer but just from my partial understanding so far, might you have assumed that the measurement in the example circuit was a Pauli YYYY measurement? It is in fact 4 distinct single qubit measurements, sorry if this wasn't clear. The first is -Y , the other three are Y. Mar 24, 2021 at 18:00
• @user3717194 I see, sorry for that. Anyway, you can still argue similarly, I'll update the answer. Mar 25, 2021 at 15:03

Here's a sort of simple way to do it, if you have a simulator you trust. Note that a Z-basis measurement is equivalent to a Z-controlled CNOT onto an ancilla qubit and then measuring that ancilla in the Z basis. And that a Y-basis measurement is equivalent to a Y-controlled CNOT onto an ancilla qubit and then measuring that ancilla in the Z basis.

What you can do is assert that moving the rotation before the controlled operation changes the control basis of the controlled operation:

import stim

sim1 = stim.TableauSimulator()
sim1.do(stim.Circuit("""
SQRT_X 0
CNOT 0 99  # 99 is the ancilla where we're putting the measurement result
"""))

sim2 = stim.TableauSimulator()
sim2.do(stim.Circuit("""
YCX 0 99  # Y-controlled X gate
SQRT_X 0
"""))

assert sim1.current_inverse_tableau() == sim2.current_inverse_tableau()


To be safe, you likely want to use the state channel duality to ensure the simulation works for all states instead of just one state.

import stim

sim1 = stim.TableauSimulator()
sim1.do(stim.Circuit("""
H 0
CNOT 0 98  # 98 is the state channel ancilla

SQRT_X 0
CNOT 0 99  # 99 is the ancilla where we're putting the measurement result
"""))

sim2 = stim.TableauSimulator()
sim2.do(stim.Circuit("""
H 0
CNOT 0 98  # 98 is the state channel ancilla

YCX 0 99  # Y-controlled X gate
SQRT_X 0
"""))

assert sim1.current_inverse_tableau() == sim2.current_inverse_tableau()