I am reading chapter 4 of Nielsen and Chuang's QCQI book.

I cannot prove the inequality from (4.66) to (4.67) in page 195.

That inequality is the following:

$$ |\langle\psi|U^\dagger M|\Delta\rangle|+|\langle\Delta|MV|\psi\rangle| \leq \|{|\Delta\rangle}\| + \| |\Delta\rangle \|$$

$U,V$ are arbitrary unitary operators, $|\psi\rangle$ is an arbitrary state, $M$ is an POVM element, and $|\Delta\rangle = (U-V)|\psi\rangle$.

How can I prove this inequality?


1 Answer 1


From Cauchy-Schwarz inequality $|\langle u|v\rangle| \le \|u\|\|v\|$, we have

$$ |\langle\psi|U^\dagger M|\Delta\rangle| \le \|MU|\psi\rangle\|\||\Delta\rangle\|. $$

But $\|MU|\psi\rangle\| \le 1$, because $U$ is unitary and $M$ a POVM element. Therefore,

$$ |\langle\psi|U^\dagger M|\Delta\rangle| \le \||\Delta\rangle\|. $$

Similar reasoning shows that $|\langle\Delta|MV|\psi\rangle| \le \||\Delta\rangle\|$.

  • $\begingroup$ Thank you for your help $\endgroup$ Jun 19, 2021 at 4:52
  • $\begingroup$ I’m not sure how the POVM argument is valid unless the POVM element is a projective measurement element. $\endgroup$ Feb 22 at 6:43
  • $\begingroup$ @ArghyadipGhosh Have a look at the answers to this question. $\endgroup$ Feb 22 at 7:43

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