Why do we have : $\|A\|=\sup\limits_{\| \psi \| = 1} \|A \psi \| = \sup\limits_{\|\phi \| = \| \psi \| = 1} | \langle \phi | A | \psi \rangle|?$

Question

I have notes where is written :

$$ \|A\|=\sup_{\| \psi \| = 1} \|A \psi \| = \sup_{\|\phi \| = \| \psi \| = 1} | \langle \phi | A | \psi \rangle|. $$

But I don't exactly know the hypothesis behind (I have "holes"), we probably supposed that $A=A^{\dagger}$ but I'm not sure.

For me, we have :

$$ \|A \psi \|=\sqrt{|\langle \psi | A^{\dagger} A | \psi \rangle|}. $$

I really don't see how we could have this other vector $| \phi \rangle$ that appears.

I have not seen the operator norm defined with two separate vectors like that, ever. I would need to see a proof that they're equivalent before I would accept it as equivalent. — Adrian Keister, Apr 26 '18 at 15:28

score 1 · Answer 1 · answered Apr 26 '18 at 15:33

There is no supplementary hypotheses, for any vector $\varphi$ you have: $$ \|\varphi\| = \sup_{\|\phi\|=1} |\langle \phi | \varphi \rangle|$$ (as $|\langle \phi | \varphi \rangle| \leq \|\phi\| \|\varphi\|$ and there is equality for $\phi=\frac{\varphi}{\|\varphi\|}$).

So: $$\|A\|=\sup_{\| \psi \| = 1} \|A \psi \| =\sup_{\psi=1}\left( \sup_{\|\phi\|=1}| \langle \phi | A | \psi \rangle| \right) =\sup_{\|\phi \| = \| \psi \| = 1} | \langle \phi | A | \psi \rangle|$$

score 0 · Answer 2 · answered Apr 26 '18 at 15:31

On one hand, $|<\phi|A|\psi>|\le \|\phi\|\|A\psi\|=\|A\psi\|$, so $\sup_{...} |<\phi|A|\psi>|\le \sup_{...} \|A\psi\|$.

On the other hand one can take $\phi=\frac{A\psi}{\|A\psi\|}$ and then $|<\phi|A|\psi>|=\frac{1}{\|A\psi\|}|<\psi|A^{+}A|\psi>|=\|A\psi\|$, so $\sup_{...} \|A\psi\|\le \sup_{...} |<\phi|A|\psi>|$

Why do we have : $\|A\|=\sup\limits_{\| \psi \| = 1} \|A \psi \| = \sup\limits_{\|\phi \| = \| \psi \| = 1} | \langle \phi | A | \psi \rangle|?$

2 Answers2