- #1
Ravi Mohan
- 196
- 21
I am reading an intriguing article on rigged Hilbert space
http://arxiv.org/abs/quant-ph/0502053
On page 8, the author describes the need for rigged Hilbert space. For that, he considers an unbounded operator [itex]A[/itex], corresponding to some observable in space of square integrable functions [itex]\mathcal{H} [/itex], with the domain [itex]\mathcal{D}(A) [/itex]. The author states that in general, [itex] \mathcal{D}(A)[/itex] does not remain invariant under the action of [itex]A[/itex].
Now the author claims that such non-invariance makes expectation values ill-defined on the whole Hilbert space [itex]\mathcal{H}[/itex].
I am not able to understand the claim.
Let us consider [itex]\phi\in\mathcal{D}(A) [/itex]. Due to invariance, [itex]\psi=A\phi[/itex] may not belong to [itex]\mathcal{D}(A) [/itex], but it remains in [itex]\mathcal{H}[/itex]. Thus the expectation value [itex](\phi,A\phi)=(\phi,\psi)[/itex] should be well defined (or am I doing something wrong?).
http://arxiv.org/abs/quant-ph/0502053
On page 8, the author describes the need for rigged Hilbert space. For that, he considers an unbounded operator [itex]A[/itex], corresponding to some observable in space of square integrable functions [itex]\mathcal{H} [/itex], with the domain [itex]\mathcal{D}(A) [/itex]. The author states that in general, [itex] \mathcal{D}(A)[/itex] does not remain invariant under the action of [itex]A[/itex].
Now the author claims that such non-invariance makes expectation values ill-defined on the whole Hilbert space [itex]\mathcal{H}[/itex].
I am not able to understand the claim.
Let us consider [itex]\phi\in\mathcal{D}(A) [/itex]. Due to invariance, [itex]\psi=A\phi[/itex] may not belong to [itex]\mathcal{D}(A) [/itex], but it remains in [itex]\mathcal{H}[/itex]. Thus the expectation value [itex](\phi,A\phi)=(\phi,\psi)[/itex] should be well defined (or am I doing something wrong?).