# Help proving bound on POVM measurement probabilities

I am trying to follow Nielsen and Chuang's 1 proof that the difference in measurement probabilities is bounded by the difference between two unitary operators applied to a given state.

Can someone show me how to get from Equation 4.66 to 4.67 in the proof below (see page 195 in the 10th anniversary edition):

1 Nielsen and Chuang, "Quantum Computation and Quantum Information"

I'll elaborate on the answer, since it took me some effort and I'm glad to share all the steps.

Applying the Cauchy-Schwarz inequality to $$|\langle \psi| U^\dagger M |\Delta\rangle |$$ it follows that:

$$|\langle \psi| U^\dagger M |\Delta\rangle | \leq ||\psi|| \: ||U^\dagger M \Delta ||$$ Note that angle brackets and vertical bar of the bra-ket notation were omitted to make the notation more readable.

The next step is: $$||\psi|| \: ||U^\dagger M \Delta ||\leq||\psi||\:||U^\dagger M||\: ||\Delta ||$$

This inequality derives from the definition of operator norm. If $$A$$ is a linear operator, we define the norm of $$A$$ as:

$$\|A \|_{op} = \sup \{\frac {\| Ax \|} {\| x \|} \ \forall \: x \neq 0 \}$$ Then it follows that for any $$\Delta$$ it holds $$\| U^\dagger M\Delta \|\leq\| U^\dagger M\|_{op}\|\Delta \|$$. We'll omit the $$\:_{op }$$ subscript hereinafter.

From the properties of the norm, it follows that $$||\psi|| \:\| U^\dagger M\|\:||\Delta ||\leq||\psi|| \:\| U^\dagger \|\|M\|\:||\Delta ||$$

Now remember that $$||\psi||= 1$$ and that $$\| U^\dagger\|=1$$ since $$U$$ is a unitary matrix.

To prove that $$||M||\leq 1$$ remember that $$M$$ is an element of a POVM. A POVM is a set of elements $$\{M_i\}$$ of $$n X n$$ matrices which are hermitian, positive definite and complete; note that typically they are non-projective and non-orthogonal.

From completeness it holds: $$\sum_{i=0}^{n} M_i = I$$ Since $$M_i$$ is hermitian, then it is a normal operator and any normal operator is diagonal with respect to some orthonormal basis (spectral decomposition theorem); since $$M_i$$ is hermitian and positive definite all the eigenvalues $$\lambda_j$$ are real and positive and then $$M_i=\sum_{j=0}^{m} \lambda_j |j \rangle\langle j|$$. Since $$||\sum_{i=0}^{n} M_i ||=|| I||$$ then $$||M_i||\leq1$$.

So we proved that $$|\langle \psi| U^\dagger M |\Delta\rangle | \leq||\Delta ||$$.

Applying all the previous steps to $$|\langle \psi| MV |\Delta\rangle|$$ it follows that $$|\langle \psi| MV |\Delta\rangle | \leq||\Delta ||$$.

Then we proved that: $$|\langle \psi| U^\dagger M |\Delta\rangle |+|\langle \psi| MV |\Delta\rangle| \leq \||\Delta\rangle ||+\||\Delta\rangle \|$$

I hope this can help someone else.

$$|\langle \psi| AB \phi\rangle | \leq ||\psi|| \: ||AB \phi|| \leq ||\psi||\: ||AB|| \: ||\phi|| \leq ||\psi|| \: ||A||\:||B||\: ||\phi||$$ In our case $$||\psi||=1$$ and $$||A||, ||B|| \leq 1$$. because one of $$A$$ and $$B$$ is unitary and the other is part of a POVM.