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In this question‘s answer it is mentioned that $\| M|\psi\rangle \| \leq 1$ for POVM Element $M$. I don‘t get why this is.

My thoughts so far: for the set of POVM elements $\{M_a\}$ we know that all $M_a$ are positive operators satisfying $\langle \psi | M_a | \psi \rangle \leq 1$ and that $\sum_a M_a = I$. We chose an arbitrary $M$ out of this (arbitrary) POVM set. So I tried:

  • using the sum rule $\sum_a \langle \psi | M_a | \psi \rangle = 1$ can we somehow infer $\sum_a \langle \psi | M_a^2 | \psi \rangle \leq 1$?
  • we can use the schwarz inequality to show that $\langle \psi | M_a | \psi \rangle \leq \| M_a |\psi\rangle \|$, however this does not help me since the equal sign is „the wrong way“
  • I tried throwing the Schwarz inequality at it in other ways but it didn‘t get me anywhere

Those were my ideas so far.

(Note: here $M$ is the actual POVM element, not the measurement operator. It is the definition Nielsen and Chuang uses in Box 4.1, which is where the linked question comes from)

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  • $\begingroup$ Welcome to the site! These two questions are quite different and I would suggest placing the one on positive operators in a different post. Comments are more for clarification and not intended for asking questions. Can you elaborate on your efforts to answer your question yourself? See How to ask a good question for more tips. $\endgroup$
    – Jacob
    Jun 9 at 15:03
  • $\begingroup$ @Jacob thanks for your suggestion! I will remove the part about positive operators then, since for now I am more interested in the specific case of the POVM element. $\endgroup$
    – Aemmel
    Jun 9 at 15:28

2 Answers 2

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Here's the simplest proof I could come up with. First note that by definition we have $M \leq I$ where $I$ is the identity operator. Now use $A \leq B \implies X^\dagger A X \leq X^\dagger B X$ with $A = M$, $B = I$ and $X = M^{1/2}$ to get that $$ M^2 \leq M. $$ (Another way to see this inequality is via the spectral theorem).

Then $$ \| M |\psi\rangle\|^2 = \langle \psi|M^2 |\psi \rangle \leq \langle \psi| M | \psi \rangle \leq \langle \psi| I |\psi\rangle = 1\,. $$

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  • $\begingroup$ What is the exact definition of the $\leq$ when using on operators in your case? $\endgroup$
    – Aemmel
    Jun 9 at 16:34
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    $\begingroup$ @Aemmel If $M$ is a positive semidefinite operator we may write $M \geq 0$ as shorthand. $M^2 \leq M$ means that $M - M^2$ is positive semidefinite. Essentially, since the eigenvalues of $M$ are $\leq 1$, squaring them makes them even smaller. $\endgroup$
    – Jacob
    Jun 9 at 16:45
  • $\begingroup$ In the first line, should it be multiply both sides by $M$? Also the last inequality is an equality right?(+1) $\endgroup$
    – R.W
    Jun 9 at 19:00
  • $\begingroup$ @R.W Not quite, I was using $A\leq B \implies X^\dagger A X \leq X^\dagger B X$. If you only multiply on one side then the implication is not necessarily true. For the last inequality you are correct, I'll edit. $\endgroup$
    – Rammus
    Jun 9 at 19:26
  • $\begingroup$ @Jacob Ah, I didn't know that notation. It makes a lot of sense though. Thank you for clearing that up! $\endgroup$
    – Aemmel
    Jun 9 at 19:50
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By the spectral theorem, there is an orthonormal basis $|\phi_k\rangle$ in which $M=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ with eigenvalues $\lambda_k\in[0,1]$ since $M$ is a POVM element. Let $a_k\in\mathbb{C}$ be the coefficients of $|\psi\rangle$ in this basis. Then the coefficients of $M|\psi\rangle$ are $\lambda_k a_k$. Therefore,

$$ \|M|\psi\rangle\|^2=\sum_{k=1}^n\lambda_k^2 |a_k|^2\le\lambda_{max}^2\sum_{k=1}^n |a_k|^2\le\sum_{k=1}^n |a_k|^2=\||\psi\rangle\|^2=1 $$

where $\lambda_{max}\in[0,1]$ is the largest of $M$'s eigenvalues.

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  • $\begingroup$ Thank you! Rammus was a bit quicker, so I'll accept his answer. Yours is also helpful though. $\endgroup$
    – Aemmel
    Jun 9 at 19:51

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