# Why is $\| M|\psi\rangle \| \leq 1$ for POVM $M$?

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)

• 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. Jun 9 at 15:03
• @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. Jun 9 at 15:28

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

• What is the exact definition of the $\leq$ when using on operators in your case? Jun 9 at 16:34
• @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. Jun 9 at 16:45
• In the first line, should it be multiply both sides by $M$? Also the last inequality is an equality right?(+1)
– R.W
Jun 9 at 19:00
• @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. Jun 9 at 19:26
• @Jacob Ah, I didn't know that notation. It makes a lot of sense though. Thank you for clearing that up! Jun 9 at 19:50

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.

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