# Can we automatically find the Hamiltonian from knowledge for multiple wave functions?

Say we had a set of wave functions $$\psi(x,t)$$ that we new the values of for all $$x$$ and $$t=t_0..t_1$$. Say we had $$N$$ of these wavefunctions, perhaps $$N=10$$. All these wave functions start off at different starting configurations so that the time evolution of one never overlaps with the time evolution of another. i.e. they are different orbits in the state space.

Could we reconstruct a Hamiltonian from this knowledge. (This is sort of like doing a lot of experiments and trying to find a Hamiltonian that fits the experiments).

This is sort of like stochastically samplying the time evolution operator with different starting functions.

How many sets of these wave functions do you think would be adequate to reconstruct a Hamiltonian of the form $$H = \frac{\partial^2}{\partial x^2} + V(x)$$ ? Is there a minimum or maximum? Or would we need an infinite number?

• How is this substantially different from this earlier question of yours? – ACuriousMind May 26 at 15:25
• This question involves not just a single wave function (one orbit), but multiple wave functions. Hence the first one is unsolvable. But this one should be solvable. People tell me not to edit questions but add a new question. A bit like having multiple paths in a fluid stream in order to reconstruct the fluid flow instead of just one. (Or perhaps multiple examples of fluid flow in order to reconstruct the Navier Stokes equations.) – zooby May 26 at 15:26

For a completely generic quantum system with an $$N$$-dimensional Hilbert space and a time-independent Hamiltonian, you need at least $$N$$ linearly independent wavefunctions as input data. Every time-dependent state can be expressed as the sum of evolutions of the Hamiltonian eigenstates, viz. $$\lvert \psi(t)\rangle = \mathrm{e}^{\mathrm{i}Ht}\lvert \psi(0)\rangle = \sum_i \mathrm{e}^{\mathrm{i}E_i t}\psi_i\lvert E_i\rangle$$ and the Hamiltonian is fully determined by the collection $$\left\{E_i, \lvert E_i\rangle\right\}$$ (an operator is determined by its action on its eigenstates). In order to obtain the $$E_i$$ and $$\lvert E_i\rangle$$, you need enough $$\lvert\psi(t)\rangle$$ such that the linear combinations of their $$\lvert \psi(0)\rangle$$ can yield all of the $$\lvert E_i\rangle$$, which are a basis for the space of state, so you need enough $$\lvert \psi(t)\rangle$$ such that their $$\lvert \psi(0)\rangle$$ are also a basis.