- #1
AxiomOfChoice
- 533
- 1
My answer would be "yes," and here's my argument: If we let
[tex]
H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac 12 m \omega^2 x^2,
[/tex]
it is a Hermitian operator with familiar normalized eigenfunctions [itex]\phi_n(x)[/itex] (these are products of Hermite polynomials and gaussians) and associated eigenvalues [itex]E_n = \hbar \omega(n + 1/2)[/itex], [itex]n = 0,1,2,\ldots[/itex]. I claim [itex]H[/itex] is unbounded because, in order to be bounded, there would need to be a constant [itex]c\in \mathbb R[/itex] such that [itex] \| H \psi \| \leq c \|\psi\|[/itex] for all [itex]\psi \in L^2(-\infty,\infty)[/itex], the Hilbert space of square integrable functions (i.e., the norm in question is the [itex]L^2[/itex] norm). No such [itex]c[/itex] exists, however; if we focus only on the eigenfunctions, we have
[tex]
H \phi_n(x) = \hbar \omega (n + 1/2) \phi_n,
[/itex]
so
[tex]
\| H \phi_n(x) \| = \hbar \omega (n + 1/2) \| \phi_n(x) \| = \hbar \omega (n + 1/2),
[/tex]
which can be made larger than any constant we care to pick by taking [itex]n[/itex] large enough. Hence [itex]H[/itex] is an unbounded operator.
...is this correct?
[tex]
H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac 12 m \omega^2 x^2,
[/tex]
it is a Hermitian operator with familiar normalized eigenfunctions [itex]\phi_n(x)[/itex] (these are products of Hermite polynomials and gaussians) and associated eigenvalues [itex]E_n = \hbar \omega(n + 1/2)[/itex], [itex]n = 0,1,2,\ldots[/itex]. I claim [itex]H[/itex] is unbounded because, in order to be bounded, there would need to be a constant [itex]c\in \mathbb R[/itex] such that [itex] \| H \psi \| \leq c \|\psi\|[/itex] for all [itex]\psi \in L^2(-\infty,\infty)[/itex], the Hilbert space of square integrable functions (i.e., the norm in question is the [itex]L^2[/itex] norm). No such [itex]c[/itex] exists, however; if we focus only on the eigenfunctions, we have
[tex]
H \phi_n(x) = \hbar \omega (n + 1/2) \phi_n,
[/itex]
so
[tex]
\| H \phi_n(x) \| = \hbar \omega (n + 1/2) \| \phi_n(x) \| = \hbar \omega (n + 1/2),
[/tex]
which can be made larger than any constant we care to pick by taking [itex]n[/itex] large enough. Hence [itex]H[/itex] is an unbounded operator.
...is this correct?