Derivation — C075 (phase-independent parity sum)

Setup

The protocol measures the ground-state parity $\langle Z\otimes Z\rangle$ after applying a global $\pi/2$ analysis pulse to both ions. A $\pi/2$ pulse with phase $\varphi$ rotates each qubit about the axis $\hat n(\varphi)=(\cos\varphi, \sin\varphi,0)$ in the $x$-$y$ plane: $$R(\varphi)=\exp\!\big(-i\tfrac{\pi}{4}(\cos\varphi\,\sigma_x+\sin\varphi\,\sigma_y)\big) =\cos\tfrac{\pi}{4}\,\mathbb 1-i\sin\tfrac{\pi}{4}(\cos\varphi\,\sigma_x+\sin\varphi\,\sigma_y).$$ The measured observable for analysis phase $\varphi$ is $$\Pi(\varphi)=\big(R(\varphi)\!\otimes\!R(\varphi)\big)^\dagger (Z\otimes Z)\big(R(\varphi)\!\otimes\!R(\varphi)\big).$$

Heisenberg picture for a single qubit

A $\pi/2$ rotation about $\hat n(\varphi)$ maps $Z$ into the $x$-$y$ plane: $$R(\varphi)^\dagger Z\,R(\varphi)=\sin\varphi\,\sigma_x-\cos\varphi\,\sigma_y$$ (up to the global sign convention of $R$). Denote $m_a(\varphi)=\sin\varphi\,X -\cos\varphi\,Y$ for ion $a$. Then $$\Pi(\varphi)=m_1(\varphi)\otimes m_2(\varphi) =(\sin\varphi\,X-\cos\varphi\,Y)\otimes(\sin\varphi\,X-\cos\varphi\,Y).$$ Expanding, $$\Pi(\varphi)=\sin^2\!\varphi\,(X\!\otimes\!X)+\cos^2\!\varphi\,(Y\!\otimes\!Y) -\sin\varphi\cos\varphi\,(X\!\otimes\!Y+Y\!\otimes\!X).$$

Sum of two pulses offset by $\pi/2$

Take a second pulse at $\varphi+\pi/2$: $\sin(\varphi+\pi/2)=\cos\varphi$, $\cos(\varphi+\pi/2)=-\sin\varphi$. Then $$\Pi(\varphi+\tfrac{\pi}{2})=\cos^2\!\varphi\,(X\!\otimes\!X)+\sin^2\!\varphi\,(Y\!\otimes\!Y) +\sin\varphi\cos\varphi\,(X\!\otimes\!Y+Y\!\otimes\!X).$$ Adding, the $\sin^2+\cos^2=1$ coefficients combine and the cross terms cancel: $$\Pi(\varphi)+\Pi(\varphi+\tfrac{\pi}{2}) =(\sin^2\!\varphi+\cos^2\!\varphi)(X\!\otimes\!X) +(\cos^2\!\varphi+\sin^2\!\varphi)(Y\!\otimes\!Y) =X\!\otimes\!X+Y\!\otimes\!Y.$$

This is independent of the absolute phase $\varphi$. Hence the sum of the two parity measurements gives $\langle\sigma_x\sigma_x\rangle+\langle\sigma_y\sigma_y\rangle$ regardless of the (unspecified) absolute pulse phase, which is exactly the claim.

Numerical confirmation

sympy_check.py forms $\Pi(\varphi)+\Pi(\varphi+\pi/2)$ symbolically for generic $\varphi$ and, after rewriting in exponential form, finds it equals exactly $X\!\otimes\!X+Y\!\otimes\!Y$ (the $\varphi$-dependent off-diagonal entries cancel identically). On the singlet both equal $-2$. Verified.