- #61
fog37
- 1,566
- 108
Hello.
The Hamiltonian operator ##\hat H= \frac {\hat p^2} {2m}+ \hat V(x)## is specific to the problem and physical system under study. Its specificity is tied to the potential energy ##V(x)## which varies from problem to problem. The kinetic energy operator ##\hat H= \frac {\hat p^2} {2m}## remains the same for every type of problem.
As far as finding the eigenstates of different operators for a specific problem, I would say that the eigenvalue equation of a specific operator, to find its eigenstates, is the same for all different physical problems. What changes in each problem are the applied boundary conditions. Different problems will have different energy eigenstates or momentum eigenstates or angular momentum eigenstates not because the eigenvalue equations in each problem (they always have the form ##\hat A |\Psi> = a |\Psi>## where ##\hat A## is a any Hermitian operator) are different but because the boundary conditions that are imposed are different, correct?
Only the energy eigenvalue equation ##\hat H |\Psi> = E |\Psi>## seems to naturally arise directly from Schrodinger equation once we set ##V(x)=0## in the regions of space where that is applicable. The applications of BCs will determine which type of energy eigenfunctions will spur out from this eigenvalue equation.
The eigenvalue equations for other operators don't arise the same way and are just considered as the starting point for finding the respective eigenfunctions...
The Hamiltonian operator ##\hat H= \frac {\hat p^2} {2m}+ \hat V(x)## is specific to the problem and physical system under study. Its specificity is tied to the potential energy ##V(x)## which varies from problem to problem. The kinetic energy operator ##\hat H= \frac {\hat p^2} {2m}## remains the same for every type of problem.
As far as finding the eigenstates of different operators for a specific problem, I would say that the eigenvalue equation of a specific operator, to find its eigenstates, is the same for all different physical problems. What changes in each problem are the applied boundary conditions. Different problems will have different energy eigenstates or momentum eigenstates or angular momentum eigenstates not because the eigenvalue equations in each problem (they always have the form ##\hat A |\Psi> = a |\Psi>## where ##\hat A## is a any Hermitian operator) are different but because the boundary conditions that are imposed are different, correct?
Only the energy eigenvalue equation ##\hat H |\Psi> = E |\Psi>## seems to naturally arise directly from Schrodinger equation once we set ##V(x)=0## in the regions of space where that is applicable. The applications of BCs will determine which type of energy eigenfunctions will spur out from this eigenvalue equation.
The eigenvalue equations for other operators don't arise the same way and are just considered as the starting point for finding the respective eigenfunctions...