bhobba said:
Yea - Rigged Hilbert spaces are a bit 'out there'.
If I hadn't studied analysis and reading that link I would have gone - ahhhhh - please someone put me out of my misery.
Mate use whatever you are comfortable with. Initially I stuck to the standard math notation and came around to the bra-ket's charms.
Thanks
Bill
Okay, I took the 15 minute whirlwind tour of rigged hilbert spaces, and I think I understand the motivation, but I'm a little confused about how it relates to Dirac notation.
First, instead of the full Hilbert space [itex]\mathcal{H}[/itex], we look at a smaller subset [itex]\Phi[/itex] that consists of the exceptionally well-behaved elements of [itex]\mathcal{H}[/itex]. These elements all have well-defined expectation values for all combinations of the observables. The example I saw of a function that was in [itex]\mathcal{H}[/itex] but not in [itex]\Phi[/itex] was [itex]\psi(x) = \dfrac{1}{x+i}[/itex]. [itex]\psi(x)[/itex] is square-integrable, so it's in [itex]\mathcal{H}[/itex], but it has no expectation value for the position operator, and so is not in [itex]\Phi[/itex].
In terms of the well-behaved functions [itex]\Phi[/itex], we can define the "bras" as linear functionals on [itex]\Phi[/itex], and the kets to be anti-linear functionals on [itex]\Phi[/itex]. Every element [itex]\phi[/itex] of [itex]\Phi[/itex] corresponds to a linear functional [itex]F_\phi[/itex] as follows:
[itex]F_\phi(\psi) = \int \phi^*(x) \psi(x) dx[/itex]
As a "bra", it would be written as [itex]F_\phi = \langle \phi |[/itex]
Similarly, every element of [itex]\Phi[/itex] corresponds to an antilinear functional
[itex]F'_\phi(\psi) = \int \psi^*(x) \phi(x) dx[/itex]
As a "ket", this would be written as [itex]F'_\phi = | \phi \rangle[/itex]
But there are additional elements that don't correspond to any element of [itex]\Phi[/itex], for example:
[itex]F(\psi) = \psi(x)[/itex]
This is the "bra" [itex]|x\rangle[/itex]
[itex]F(\psi) =[/itex] Fourier transform of [itex]\psi[/itex] evaluated at [itex]k[/itex]
This is the "bra" [itex]|k\rangle[/itex]
This all makes perfect sense to me. However, this understanding of bras and kets only justifies expressions of the form
[itex]\langle F|\phi \rangle[/itex]
and
[itex]\langle \phi|F \rangle[/itex]
where [itex]\phi[/itex] is one of the well-behaved elements of [itex]\Phi[/itex], and [itex]F[/itex] is a general functional. It doesn't seem like it justifies expressions of the form
[itex]\langle F | F' \rangle[/itex]
where both the bra and the ket are generalized functionals. For example:
[itex]\langle x | k \rangle[/itex]
and
[itex]\langle x | x' \rangle[/itex]
The operator [itex]\langle x |[/itex] as a functional only applies to well-behaved functions, in [itex]\Phi[/itex]; it doesn't apply to other functionals.
. The version of the definition that I learned in functional analysis class stuck more than the version that I learned in quantum mechanics class.