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Solutions to Time-dependent Schrodinger Equation

  1. Aug 6, 2013 #1
    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.
     
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  3. Aug 6, 2013 #2

    jtbell

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    Staff: Mentor

    Bound states (infinite square well, simple harmonic oscillator, etc.) have a discrete energy or momentum spectrum. The general solution is a linear combination (weighted sum) of eigenstates.

    Unbound states (free particle, finite square well with E greater than the "top" of the well, potential barrier, etc.) have a continuous energy or momentum spectrum. The general solution is a Fourier-type integral.
     
  4. Aug 6, 2013 #3
    Ah ok. By the way, even thought I'm reading the book, I don't completely understand why bound states have discrete energy, and why unbound states have continuous energy (I'm a junior in high school, so it's understandable that I won't comprehend everything :P). Could you explain to me why that is? Thanks.
     
  5. Aug 6, 2013 #4
    The best analogy I have found is that of standing waves. Think about a big open piece of air--waves of any frequency at all can propagate through it (continuous solutions). However, if you put the air in a pipe with caps at the ends, resonance phenomena force the allowable spectrum of waves to a specific set of discrete pitches--the harmonic series. To get technical, the differential equation that describes wave propagation gets boundary conditions imposed on it, which force the solutions to take on a discrete set of forms, whereas a system with no boundary conditions can have a continuous infinity of solutions. The same thing happens in the quantum case, the only difference is that the differential equation in question is now the Schrodinger Equation.
     
  6. Aug 6, 2013 #5
    Ah ok. So because the infinite 1-D potential well has those boundary conditions (also the SHM), the general wavefunction is a discrete linear combination. But for free particles, there is no boundary solution, so its quantum states form a continuous (integral) wavefunction solution. Ok thanks! I understand it now :)
     
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