# The expectation of a measurement of qubit 2 after qubit 1 has been measured

In section 1.2.4 (page 13) of these lecture notes http://users.cms.caltech.edu/~vidick/teaching/fsmp/fsmp.pdf, it says

\begin{aligned}\left\langle\psi\left|X_{1}^{0} Z_{2} X_{1}^{0}\right| \psi\right\rangle+\left\langle\psi\left|X_{1}^{1} Z_{2} X_{1}^{1}\right| \psi\right\rangle &=\left\langle\psi\left|X_{1} Z_{2} X_{1}\right| \psi\right\rangle \\ &=\left\langle\psi\left|Z_{2} X_{1}^{2}\right| \psi\right\rangle \\ &=\left\langle\psi\left|Z_{2}\right| \psi\right\rangle \end{aligned}

given that $$X_{1}=X_{1}^{0}-X_{1}^{1}$$ where this is the spectral decomposition, but I don't see how this can be. How did the negative terms cancel out?

• Can you point to the section of the paper where this is shown? Commented Jun 15, 2021 at 5:53
• Hi again @epelaaez, it is in section 1.2.4 on page 13 Commented Jun 15, 2021 at 6:09

Personally, I would do the calculation a little differently. Start by writing $$\langle\psi|X^0_1Z_2X^0_1|\psi\rangle+ \langle\psi|X^1_1Z_2X^1_1|\psi\rangle= \langle\psi|X^0_1Z_2X^0_1+X^1_1Z_2X^1_1|\psi\rangle$$ Next, think about a term like $$X^0_1Z_2X^0_1$$, but expand out the tensor product. This is just $$(X^0X^0)\otimes Z$$. But since $$X^0$$ is a projector, it's just $$X^0$$. So, our full expression looks like $$\langle\psi|X^0\otimes Z+X^1\otimes Z_2|\psi\rangle=\langle\psi|(X^0+X^1)\otimes Z|\psi\rangle$$ Now, $$X^0+X^1=I$$, because it's a sum over an orthonormal basis. Hence, this is just $$\langle\psi|Z_2|\psi\rangle$$.