Is the harmonic oscillator Hamiltonian an unbounded operator?

[AxiomOfChoice]

My answer would be "yes," and here's my argument: If we let
$$H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac 12 m \omega^2 x^2,$$
it is a Hermitian operator with familiar normalized eigenfunctions $\phi_n(x)$ (these are products of Hermite polynomials and gaussians) and associated eigenvalues $E_n = \hbar \omega(n + 1/2)$, $n = 0,1,2,\ldots$. I claim $H$ is unbounded because, in order to be bounded, there would need to be a constant $c\in \mathbb R$ such that $\| H \psi \| \leq c \|\psi\|$ for all $\psi \in L^2(-\infty,\infty)$, the Hilbert space of square integrable functions (i.e., the norm in question is the $L^2$ norm). No such $c$ exists, however; if we focus only on the eigenfunctions, we have
[tex] H \phi_n(x) = \hbar \omega (n + 1/2) \phi_n,<br /> [/itex]<br /> so <br /> $$\| H \phi_n(x) \| = \hbar \omega (n + 1/2) \| \phi_n(x) \| = \hbar \omega (n + 1/2),$$<br /> which can be made larger than any constant we care to pick by taking $n$ large enough. Hence $H$ is an unbounded operator.<br /> <br /> ...is this correct?[/tex]

 

[xepma]

The Hamiltonian is only bounded from below.

 

[AxiomOfChoice]

The Hamiltonian is only bounded from below.

What does "bounded from below" mean, rigorously?

 

[martinbn]

What does "bounded from below" mean, rigorously?

May be it means: the spectrum is bounded from below(left) subset of the reals?

 

[Dickfore]

What does "bounded from below" mean, rigorously?

Maybe this:

...in order to be bounded from below, there would need to be a constant $c\in \mathbb R$ such that $\| H \psi \| \geq c \|\psi\|$ for all $\psi \in L^2(-\infty,\infty)$, the Hilbert space of square integrable functions (i.e., the norm in question is the $L^2$ norm).

?

 

[dextercioby]

See this page here. It's well written, because the author(s) read good books.

http://en.wikipedia.org/wiki/Unbounded_operator