Fredrik said:
What you're describing is just the difference between the Schrödinger picture and Heisenberg picture of the Hilbert space version of QM, which was invented by von Neumann. I'm not familiar with Schrödinger's and Heisenberg's original work, but I suspect that their theories are just von Neumann's theory for specific Hilbert spaces. In Schrödinger's case, it would be L^2(\mathbb R^3), and in Heisenberg's, any finite-dimensional complex Hilbert space would do (so that vectors and operators can be represented by matrices).
(I could be completely wrong about what Heisenberg's theory was).
The Heisenberg and Schrödinger picture of time evolution are only two representations of the same quantum theory. Also the Hilbert spaces are the same. There's up to equivalence only one separable complex Hilbert space, and that's the one we are talking about in non-relativistic quantum theory for a fixed number of particles.
Let's take the quantum theory of one particle with a time-independent Hamiltonian which makes the formulae a bit easier to write than in the general case. In the Schrödinger picture by definition the state kets carry the full time dependence, i.e.,
|\psi,t \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi,0 \rangle
and (not explicitly time-dependent) observables are constant in time
\hat{O}(t)=\hat{O}(0)=\mathrm{const}.
The probability amplitudes (wave functions) are the (generalized) scalar product of a (generalized) eigenvector with the state ket:
\psi(t,o)=\langle o | \psi,t \rangle = \langle{o}| \exp(-\mathrm{i} \hat{H} t)|\psi,0 \rangle.
In the Heisenberg picture, the operators representing observables carry the total time dependence, i.e., one has
\hat{O}_{\mathrm{H}}(t)=\exp(\mathrm{i} \hat{H} t) \hat{O}(0) \exp(-\mathrm{i} \hat{H} t),[/tex]
which implies that the (generalized) eigenvectors of these operators become time dependent via
|o,t \rangle_{\mathrm{H}}=\exp(\mathrm{i} \hat{H} t) |o,0 \rangle_{\mathrm{H}}.
The state kets in the Heisenberg picture are constant in time:
|\psi,t \rangle_{\mathrm{H}}=|\psi,0 \rangle_{\mathrm{H}}=\mathrm{const}.
The wave function at time t now reads
\psi(o,t)=_{\mathrm{H}} \langle o,t|\psi,t \rangle_{\mathrm{H}} = _{\mathrm{H}} \langle o,0|\exp(-\mathrm{i} \hat{H} t)|\psi,0 \rangle_{\mathrm{H}},
where I used the self-adjointness of \hat{H}. Since the Hamiltonian is the same in both pictures and since at t=0 by definition the state kets and the (generalized) eigenvectors are the same in both pictures, this result is the same in both pictures as it must be for an observable quantity as the probility to measure a certain value of the observable O, which is given by |\psi(o,t)|^2.
Another aspect of the formulation of QT by Heisenberg/Born/Jordan and Schrödinger is the representation. Heisenberg used the harmonic oscillator as one of the most simple examples to propose his theory, and he used the discrete set of energy-eigenvectors as basis. In such a basis the wave functions are represented by an "infinite dimensional'' column vector with components
\psi_n(t)=\exp(-\mathrm{i} E_n t) \langle E_n|\psi,t \rangle
(which is of course again independent of the picture used; here I used the notation in the Schrödinger picture for convenience). The Hilbert space is thus realized as the Hilbert space of square summable series, \ell^2, with the components, \psi_n(t) as summands.
Schrödinger instead used the position representation, i.e.,
\psi(x,t)=\langle x|\psi,t \rangle.
which leads to the realization of the quantum-theoretical Hilbert space as the space of square-integrable functions, \mathrm{L}^2.
Of course, as is very clear in Dirac's representation-independent bra-ket notation, both realizations of quantum theory are equivalent.
