I am reading David Griffiths' book on Quantum Mechanics, and he usually says that the general solution to the TDSE, given a potential V, can a DISCRETE linear combinations of the wavefunction solutions. However, in one section, he says that the linear discrete sum can be regarded as a continuous sum over continuous values of k, the wavenumber. Making a continuous transforms the discrete linear combination into an integral over all continuous values of k from negative to positive infinity, of all possible wavefunctions corresponding to the Hamiltonian provided? Why can't the linear combination be discrete instead of continuous? This is on page 46 of his book. What I don't understand, is why does he write it as an integral, and then proceed to use Fourier's trick, etc..? Why doesn't he keep it as a discrete sum, as is the usual idea with the "discrete linear combination of solutions"? Thanks. Sorry if the beginning is a bit confusing.(adsbygoogle = window.adsbygoogle || []).push({});

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# Solutions to Time-dependent Schrodinger Equation

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