- #1
JaredMTg
- 11
- 2
Hello,
I am taking an introductory course in quantum mechanics. One thing I am confused about is, Schrodinger's equation seems to be regarded as the "ultimate" formula which determines a particle's possible wavefunctions and energies, given a certain potential (Hamiltonian of psi = Energy eigenvalue * psi).
But then other operators, such as L^2 (angular momentum) have spherical harmonics as their eigenfunctions. If you tried to plug in the wavefunction that is a solution to the infinite-square well problem (i.e. something of the form psi = C sin sqrt(kx)) into the angular momentum operator, it would not be a solution (i.e. eigenfunction) to the L^2 operator that would return a value for L^2 multiplied by the original wavefunction.
So what if I wanted to measure the angular momentum squared of the particle in the infinite well? Would the wavefunction suddenly change to become a spherical harmonic?
In other words, how can a wavefunction be a solution to Schrodinger's equation as well as the many other operators representing real physical quantities (such as momentum) at the same time? If the measurement of the particular quantity (whether it is energy or something else like angular momentum) is what determines the solution, and therefore the wavefunction, then why is Schrodinger's equation considered any more fundamental than finding the solution of the L^2 operator?
Many thanks in advance for explaining this to me.
I am taking an introductory course in quantum mechanics. One thing I am confused about is, Schrodinger's equation seems to be regarded as the "ultimate" formula which determines a particle's possible wavefunctions and energies, given a certain potential (Hamiltonian of psi = Energy eigenvalue * psi).
But then other operators, such as L^2 (angular momentum) have spherical harmonics as their eigenfunctions. If you tried to plug in the wavefunction that is a solution to the infinite-square well problem (i.e. something of the form psi = C sin sqrt(kx)) into the angular momentum operator, it would not be a solution (i.e. eigenfunction) to the L^2 operator that would return a value for L^2 multiplied by the original wavefunction.
So what if I wanted to measure the angular momentum squared of the particle in the infinite well? Would the wavefunction suddenly change to become a spherical harmonic?
In other words, how can a wavefunction be a solution to Schrodinger's equation as well as the many other operators representing real physical quantities (such as momentum) at the same time? If the measurement of the particular quantity (whether it is energy or something else like angular momentum) is what determines the solution, and therefore the wavefunction, then why is Schrodinger's equation considered any more fundamental than finding the solution of the L^2 operator?
Many thanks in advance for explaining this to me.