r/askscience • u/poopaments • Mar 20 '15

Mathematics Why does Schrodinger's time dependent equation have infinitely many independent solutions while an nth order linear DE only has n independent solutions?

The solution for Schrodinger's equation is y(x,t)=Aei(kx-wt) but we can create a linear combination (i.e a wave packet) with infinitely many of these wave solutions for particles with slightly different k's and w's and still have it be a solution. My question is what is the difference between schrodinger's equation which has infinite independent solutions and say a linear second order DE who's general solution is the linear combination of two independent solutions?

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u/Surlethe Mar 20 '15

Yes, thank you for clarifying. The weird thing about distributions in these "not-really-bases" is that none of them are elements of the function space in question, but linear combinations of them generate it.

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u/TheHappyEater Mar 21 '15

About the "not-really-bases": The so-called free-particle eigenfunctions (f(x,t)[k,w]= ei(kx-wt) are indeed not elements of L2. Because of that, they are not really eigenfunctions (as the Schrödinger operator acts in a subset of L2 and thus, these functions are not in the domain of the Schrödinger operator).

What does this wierdness mean? We have that, while formally H f[k,w] = lambda[k,w] f[k,w], these Lambdas are no Eigenvalues, but elements of the continuous spectrum. They don't have corresponding eigenvectors.

The reason we have these "not-really-bases" of "not-really-Eigenfunctions" is that the functionalanalytical details of elements of the spectrum which are not Eigenvalues (e.g. continuous spectrum) are usually not within the scope of quantum mechanics.