The continuum eigenvalues and eigenvectors of the Schrodinger operator are the limiting low-lying eigenvalues and eigenvectors of the discrete lattice approximations. Given a Schrodinger operator

$$ H= \sum_i A_i \partial_i^2 + V(x_1,....,x_n) $$

Where V is of the appropriate class (smooth is too restrictive--- you can have delta functions too, and random potentials, but I don't know the best possible function class--- it might be any integrable potential, i.e., any potential at all EDIT: of course it can't, as the -1/r^n energy levels run away to be localized on top of the attractive spot. The correct condition on the potential is involved, but you can take it to be continuous for this discussion), you replace the x's by a square lattice of spacing $\epsilon$ and of total size L in each direction with periodic boundaries, replace the $\nabla_i$ by the lattice $\nabla_i$

$$ (H_L \psi) (x) = \sum_i {A_i\over \epsilon^2} (\psi(x_i + \epsilon) - 2\psi(x_i) + \psi(x_{i-1})) + V_L(x) \psi(x) $$

Where V_L(x) is the integral over one lattice volume of the continuum V(x) in an $\epsilon$ box centered at x, and the discrete second derivative is the difference between the forward difference and the backward difference.

Then the approximately smooth low lying eigenvectors of $H_L$ converge to the eigenvalues of H in the continuum limit, and as for the high eigenvectors, who cares, these are lattice artifacts. I am sure that it is possible to prove all this rigorously, although from the physical point of view, if it were not the case, the Schrodinger equation would be physically suspect.

You can see the convergence on a computer, if you simulate a discretized Schrodinger operator. You can prove the convergence of the discrete to continuous propagator relatively easily from the path integral. For the individual eigenvalues and eigevectors, things will be somewhat more involved. If you want a mathematical proof, I can try to sketch one.

The continuum eigenvalues and eigenvectors of the Schrodinger operator are the limiting low-lying eigenvalues and eigenvectors of the discrete lattice approximations. Given a Schrodinger operator

$$ H= \sum_i A_i \partial_i^2 + V(x_1,....,x_n) $$

Where V is of the appropriate class (smooth is too restrictive--- you can have delta functions too, and random potentials, but I don't know the best possible function class--- it might be any integrable potential, i.e., any potential at all EDIT: of course it can't, as the -1/r^n energy levels run away to be localized on top of the attractive spot. The correct condition on the potential is involved, but you can take it to be continuous for this discussion), you replace the x's by a square lattice of spacing $\epsilon$ and of total size L in each direction with periodic boundaries, replace the $\nabla_i$ by the lattice $\nabla_i$

$$ (H_L \psi) (x) = \sum_i {A_i\over \epsilon^2} (\psi(x_i + \epsilon) - 2\psi(x_i) + \psi(x_{i-1})) + V_L(x) \psi(x) $$

Where V_L(x) is the integral over one lattice volume of the continuum V(x) in an $\epsilon$ box centered at x, and the discrete second derivative is the difference between the forward difference and the backward difference.

Then the approximately smooth low lying eigenvectors of $H_L$ converge to the eigenvalues of H in the continuum limit, and as for the high eigenvectors, who cares, these are lattice artifacts. I am sure that it is possible to prove all this rigorously, although from the physical point of view, if it were not the case, the Schrodinger equation would be physically suspect.

You can see the convergence on a computer, if you simulate a discretized Schrodinger operator. You can prove the convergence of the discrete to continuous propagator relatively easily from the path integral. For the individual eigenvalues and eigevectors, things will be somewhat more involved. If you want a mathematical proof, I can try to sketch one.

EDIT: Determinant formula

If you look at the eigenvalue equation for the finite dimensional operator $H_L$,

$$det(H_L - \lambda I)$$

you find a finite degree polynomial, whose zeros are the eigenvalues of the equation in the limit $\epsilon\rightarrow 0$, $L\rightarrow\infty$.

The continuum eigenvalues and eigenvectors of the Schrodinger operator are the limiting low-lying eigenvalues and eigenvectors of the discrete lattice approximations. Given a Schrodinger operator

$$ H= \sum_i A_i \partial_i^2 + V(x_1,....,x_n) $$

Where V is of the appropriate class (smooth is too restrictive--- you can have delta functions too, and random potentials, but I don't know the best possible function class--- it might be any integrable potential, i.e., any potential at all), you replace the x's by a square lattice of spacing $\epsilon$ and of total size L in each direction with periodic boundaries, replace the $\nabla_i$ by the lattice $\nabla_i$

$$ (H_L \psi) (x) = \sum_i {A_i\over \epsilon^2} (\psi(x_i + \epsilon) - 2\psi(x_i) + \psi(x_{i-1})) + V_L(x) \psi(x) $$

Where V_L(x) is the integral over one lattice volume of the continuum V(x) in an $\epsilon$ box centered at x, and the discrete second derivative is the difference between the forward difference and the backward difference.

Then the approximately smooth low lying eigenvectors of $H_L$ converge to the eigenvalues of H in the continuum limit, and as for the high eigenvectors, who cares, these are lattice artifacts. I am sure that it is possible to prove all this rigorously, although from the physical point of view, if it were not the case, the Schrodinger equation would be physically suspect.

You can see the convergence on a computer, if you simulate a discretized Schrodinger operator. You can prove the convergence of the discrete to continuous propagator relatively easily from the path integral. For the individual eigenvalues and eigevectors, things will be somewhat more involved. If you want a mathematical proof, I can try to sketch one.

The continuum eigenvalues and eigenvectors of the Schrodinger operator are the limiting low-lying eigenvalues and eigenvectors of the discrete lattice approximations. Given a Schrodinger operator

$$ H= \sum_i A_i \partial_i^2 + V(x_1,....,x_n) $$

Where V is of the appropriate class (smooth is too restrictive--- you can have delta functions too, and random potentials, but I don't know the best possible function class--- it might be any integrable potential, i.e., any potential at all EDIT: of course it can't, as the -1/r^n energy levels run away to be localized on top of the attractive spot. The correct condition on the potential is involved, but you can take it to be continuous for this discussion), you replace the x's by a square lattice of spacing $\epsilon$ and of total size L in each direction with periodic boundaries, replace the $\nabla_i$ by the lattice $\nabla_i$

$$ (H_L \psi) (x) = \sum_i {A_i\over \epsilon^2} (\psi(x_i + \epsilon) - 2\psi(x_i) + \psi(x_{i-1})) + V_L(x) \psi(x) $$

Where V_L(x) is the integral over one lattice volume of the continuum V(x) in an $\epsilon$ box centered at x, and the discrete second derivative is the difference between the forward difference and the backward difference.

Then the approximately smooth low lying eigenvectors of $H_L$ converge to the eigenvalues of H in the continuum limit, and as for the high eigenvectors, who cares, these are lattice artifacts. I am sure that it is possible to prove all this rigorously, although from the physical point of view, if it were not the case, the Schrodinger equation would be physically suspect.

You can see the convergence on a computer, if you simulate a discretized Schrodinger operator. You can prove the convergence of the discrete to continuous propagator relatively easily from the path integral. For the individual eigenvalues and eigevectors, things will be somewhat more involved. If you want a mathematical proof, I can try to sketch one.