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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

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  • $\begingroup$ 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. $\endgroup$ Commented Oct 13, 2021 at 13:17
  • $\begingroup$ @BenGrossmann I haved edited the post including part from my reference. Could you please have a look ? $\endgroup$
    – Sooraj S
    Commented Oct 13, 2021 at 14:28
  • $\begingroup$ A key point, I think, is that $\lambda_{\max}(M) \leq 1$ holds because $M$ is a POVM element. $\endgroup$ Commented Oct 13, 2021 at 14:32

1 Answer 1

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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.

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

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