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?

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

1 Answer 1


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 $\langle\psi|Z_2|\psi\rangle$.

Note that I never introduced any negative terms that needed cancelling out.


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