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 \mathcal{H}, we look at a smaller subset \Phi that consists of the exceptionally well-behaved elements of \mathcal{H}. These elements all have well-defined expectation values for all combinations of the observables. The example I saw of a function that was in \mathcal{H} but not in \Phi was \psi(x) = \dfrac{1}{x+i}. \psi(x) is square-integrable, so it's in \mathcal{H}, but it has no expectation value for the position operator, and so is not in \Phi.
In terms of the well-behaved functions \Phi, we can define the "bras" as linear functionals on \Phi, and the kets to be anti-linear functionals on \Phi. Every element \phi of \Phi corresponds to a linear functional F_\phi as follows:
F_\phi(\psi) = \int \phi^*(x) \psi(x) dx
As a "bra", it would be written as F_\phi = \langle \phi |
Similarly, every element of \Phi corresponds to an antilinear functional
F'_\phi(\psi) = \int \psi^*(x) \phi(x) dx
As a "ket", this would be written as F'_\phi = | \phi \rangle
But there are additional elements that don't correspond to any element of \Phi, for example:
F(\psi) = \psi(x)
This is the "bra" |x\rangle
F(\psi) = Fourier transform of \psi evaluated at k
This is the "bra" |k\rangle
This all makes perfect sense to me. However, this understanding of bras and kets only justifies expressions of the form
\langle F|\phi \rangle
and
\langle \phi|F \rangle
where \phi is one of the well-behaved elements of \Phi, and F is a general functional. It doesn't seem like it justifies expressions of the form
\langle F | F' \rangle
where both the bra and the ket are generalized functionals. For example:
\langle x | k \rangle
and
\langle x | x' \rangle
The operator \langle x | as a functional only applies to well-behaved functions, in \Phi; it doesn't apply to other functionals.