# Operator norm of creation and annihilation operators

Are the creation and the annihilation operators $a(f)$ and $a^{\dagger}(f)$ for the bosonic Fock space bounded? What is their norm? So far I did not have found any note about this in the linked Wikipedia article.

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No. Consider the one-particle sector, with states $|n\rangle$ having occupation number $n$. Then $a |n\rangle = \sqrt{n} |n-1\rangle,$ so $\|a |n\rangle \| = \sqrt n.$ This means that $a$ is not bounded. –  Vibert Jul 6 '13 at 16:28
@Vibert: Thanks for the quick answer. If you want, you can write an quick answer, so I can accept it. –  tampis Jul 6 '13 at 16:30
@Vibert I second the call for that to be an answer. There's really nothing left to say. –  Chris White Jul 6 '13 at 18:34
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## 1 Answer

OK, here you go:

No. Consider the one-particle sector, with states $|n\rangle$ having occupation number $n$. Wlog we can suppose that the states are normalized. Then $$a|n\rangle =\sqrt{n}|n−1\rangle,$$ so $$\|a|n\rangle\| = \sqrt n.$$ This means that $a$ is not bounded. The same goes for $a^\dagger$.

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