houlahound said:
Sorry not following, without wave functions there is nothing g inside those brackets.
It's worth to think about this a bit more. The general formalism without using a concrete realization of the Hilbert space can be very useful to understand general properties. It goes back to Dirac's formulation of quantum theory in 1926/27 and was made mathematically rigorous by von Neumann around the same time.
There is a one-to-one mapping between the abstract (separable) Hilbert space ##\mathcal{H}## and the wave functions, which is the realization of this separable Hilbert space as functions ##\psi:\mathbb{R}^3 \rightarrow \mathbb{C}## that are square integrable, i.e., for which the Integral ##\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2## exists, so that you can normalize this integral to 1 such that ##|\psi(\vec{x})|^2## makes sense as a probability distribution for the position of the particle.
In the abstract formalism the position and momentum of the particle are represented by self-adjoint operators in ##\mathbb{H}##, but they are not defined on the entire Hilbert space but only in a smaller space, because due to the canonical commutator relations ("Heisenberg algebra"),
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar,$$
they have no proper eigenvectors that are square integrable but only socalled generalized eigenvectors in the sense of distributions. Indeed these formal eigenstates are not normalizable but obey
$$\langle \vec{x}|\vec{x}' \rangle=\delta^{(3)}(\vec{x}-\vec{x}').$$
Further, these states are "complete" in the sense that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x}|=\hat{1}.$$
Now there is a one-to-one mapping between the abstract Hilbert-space vectors ##|\psi \rangle## and the wave function ##\psi(\vec{x})=\langle \vec{x}|\psi \rangle##. Given ##|\psi \rangle## that's how you get the wave function. If you have given the wave function, you can use the completeness relation to get back the abstract state
$$|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x}|\psi \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \psi(\vec{x}).$$
Also you get the scalar product of two vectors from the wave functions by inserting a completeness relation:
$$\langle \psi|\phi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \psi|\vec{x} \rangle \langle \vec{x}|\phi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(\vec{x}) \phi(\vec{x}).$$
There are even a few cases, where you can solve the eigenvalue problem without ever referring to wave functions by only using the commutation relations. The usual examples are: Position and momentum, angular momentum, energy spectrum of the harmonic oscillator. Also the non-relativstic hydrogen atom can be solved without wave functions (in fact it was the first way it was solved by Pauli within matrix mechanics).
