Prove $|\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\leq ||(U-V)|\psi\rangle||+||(U-V)|\psi\rangle||$

Prove that $$|\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\leq |||\Delta\rangle||+|||\Delta\rangle||$$ where $$|\psi\rangle,|\Delta\rangle=(U-V)|\psi\rangle$$ are complex column vectors, $$\langle\psi|=(|\psi\rangle)^\dagger,\langle\Delta|=(|\Delta\rangle)^\dagger$$ and $$M$$ is positive definite and $$U,V$$ are unitary matrices and $$|||\psi\rangle||=1$$

My reference says it is proved using basic linear algebra and Cauchy-Schwarz inequality.

My Attempt

Cauchy-Schwarz inequality : $$|\langle\psi|\phi\rangle|\leq |||\psi\rangle||.|||\phi\rangle||$$

Matrix Norm, $$||A||=\max_{|\psi\rangle\neq 0}\dfrac{||A|\psi\rangle||}{|||\psi\rangle||}=\sigma_{\max}$$ where $$\sigma_{\max}$$ is the largest singular value of $$A$$.

$$||U|\psi\rangle||=|||\psi\rangle||$$ for any unitary matrix $$U$$

$$M$$ is positive definite, i.e., $$M^\dagger=M$$ and $$\lambda_i\geq 0$$

$$|\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|=|\langle\psi|U^\dagger MU|\psi \rangle-|\langle\psi|U^\dagger MV|\psi \rangle+|\langle\psi|U^\dagger MV|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\\ =|\langle\psi|U^\dagger M(U-V)|\psi \rangle+\langle\psi|(U-V)^\dagger MV|\psi \rangle|=|\langle\psi|U^\dagger M|\Delta \rangle+\langle\Delta| MV|\psi \rangle| \\\leq |\langle\psi|U^\dagger M|\Delta \rangle|+|\langle\Delta| MV|\psi \rangle|$$

Now, $$|\langle\psi|U^\dagger M|\Delta \rangle|\leq ||\langle\psi|U^\dagger||.|| M|\Delta \rangle||=||\langle\psi|||.|| M|\Delta \rangle||=|||\psi\rangle||.|| M|\Delta \rangle||=|| M|\Delta \rangle||$$ $$|\langle\Delta| MV|\psi \rangle|\leq ||\langle\Delta| M||.||V|\psi \rangle||=||\langle\Delta| M||.|||\psi \rangle||=||\langle\Delta| M||=|| M|\Delta \rangle||$$

$$\implies |\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\leq ||M|\Delta\rangle||+||M|\Delta\rangle||$$

How do I proceed further to prove the required statement ?

Original Reference

• Something is wrong with the inequality as written. If the left hand side is non-zero, then multiplying $M$ by $k > 1$ maintains the positive definiteness of $k$, increases the left hand side by a factor of $k$, but keeps the right hand side the same. Commented Oct 13, 2021 at 13:17
• @BenGrossmann I haved edited the post including part from my reference. Could you please have a look ? Commented Oct 13, 2021 at 14:28
• A key point, I think, is that $\lambda_{\max}(M) \leq 1$ holds because $M$ is a POVM element. Commented Oct 13, 2021 at 14:32

1 Answer

Because $$M$$ is an element of a POVM, both $$M$$ and $$I-M$$ are positive semidefinite, which means that the eigenvalues of $$M$$ are between $$0$$ and $$1$$, which means that the eigenvalues of $$M^2$$ are between $$0$$ and $$1$$, which means that $$M^2$$ and $$I - M^2$$ are positive semidefinite, which means that \begin{align} \| M |\Delta \rangle \|^2 &= \langle \Delta|M^2 |\Delta \rangle \\ & \leq \langle \Delta |M^2 |\Delta \rangle + \langle \Delta |I - M^2 |\Delta \rangle \\ & = \langle \Delta |M^2 + (I - M^2)|\Delta\rangle = \|\Delta\|^2. \end{align} With that, you can proceed from your work to the desired conclusion.

• Positive definiteness of M only implies its eigenvalues are positive right ? Commented Oct 13, 2021 at 16:10
• Yes, and positive semidefiniteness implies that the eigenvalues are non-negative. Commented Oct 13, 2021 at 16:14
• Where does the requirement for the eigenvalues to lie in between $0$ and $1$ comes from ? What I know is that $\langle \psi|M|\psi\rangle=tr(M|\psi\rangle\langle\psi|)\in[0,1]$ Commented Oct 13, 2021 at 16:19
• The fact that $I- M$ is positive semidefinite tells you that the eigenvalues of $M$ are at most $1$. Commented Oct 13, 2021 at 17:08
• Alternatively, consider that if $|\psi\rangle$ is a unit eigenvector of $M$, then $\langle \psi |M|\psi \rangle$ is the associated eiegnvalue. Commented Oct 13, 2021 at 17:10