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.
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ 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. $\endgroup$– VibertCommented Jul 6, 2013 at 16:28
-
1$\begingroup$ @Vibert: Thanks for the quick answer. If you want, you can write an quick answer, so I can accept it. $\endgroup$– Stephan KullaCommented Jul 6, 2013 at 16:30
-
1$\begingroup$ @Vibert I second the call for that to be an answer. There's really nothing left to say. $\endgroup$– user10851Commented Jul 6, 2013 at 18:34
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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 $a^\dagger$.
-
$\begingroup$ Strictly speaking, the bosonic creation and annihilation operators (on the bosonic Fock space) restricted to the subspace consisting of at most $r$ bosons are in fact bounded. What you instead mean, I suppose, is that one should take the sequence $\psi_n:=\otimes^n f/\|f\|^n$ where $f$ is a single-particle vector. Then $\psi_n$ is in the domain of $a(f)$ and $\|a(f)\psi_n\|=\sqrt n \|f\|$, showing that $a(f)$ is unbounded. $\endgroup$ Commented Nov 10 at 5:12