- #1
AxiomOfChoice
- 531
- 1
My answer would be "yes," and here's my argument: If we let
<br /> H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac 12 m \omega^2 x^2,<br />
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
<br /> H \phi_n(x) = \hbar \omega (n + 1/2) \phi_n,<br /> [/itex]<br /> so <br /> <br /> \| H \phi_n(x) \| = \hbar \omega (n + 1/2) \| \phi_n(x) \| = \hbar \omega (n + 1/2),<br /><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?
<br /> H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac 12 m \omega^2 x^2,<br />
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
<br /> H \phi_n(x) = \hbar \omega (n + 1/2) \phi_n,<br /> [/itex]<br /> so <br /> <br /> \| H \phi_n(x) \| = \hbar \omega (n + 1/2) \| \phi_n(x) \| = \hbar \omega (n + 1/2),<br /><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?
