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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.

Qmechanic
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Stephan Kulla
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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. ā€“ Vibert Jul 06 '13 at 16:28
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    @Vibert: Thanks for the quick answer. If you want, you can write an quick answer, so I can accept it. ā€“ Stephan Kulla Jul 06 '13 at 16:30
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    @Vibert I second the call for that to be an answer. There's really nothing left to say. ā€“  Jul 06 '13 at 18:34

1 Answers1

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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 $$a|n\rangle =\sqrt{n}|nāˆ’1\rangle,$$ so $$\|a|n\rangle\| = \sqrt n.$$ This means that $a$ is not bounded. The same goes for $a^\dagger$.

Vibert
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