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I Delta fuction potential general solution

  1. Mar 1, 2016 #1
    Hi, in the book 'Introduction to Quantum Mechanics' by Griffiths, on page 71 in the section 'The Delta-Function Potential' he states that the general solution to time independent Schrodinger Equation is $$\psi(x) = Ae^{-\kappa x} + B e^{\kappa x}$$

    he then notes that the first term blows up as $$x \to -\infty,$$ so we must choose $$A=0.$$ Why is it that we are rejecting a wave function that goes to infinity? Is it simply because we are looking for normalizable solutions?

    Thanks.
     
  2. jcsd
  3. Mar 1, 2016 #2

    vanhees71

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    For ##x<0## you must choose ##A=0## and for ##x>0## you must have ##B=0##, if ##\kappa>0##. That's indeed, because you want to have bound states in this case, leading to normalizable wave functions. Note that there may be solutions with ##\kappa \in \mathrm{i} \mathbb{R}##, leading to scattering states with energy eigenvalues in the continuum, and these wave functions are generalized eigenfunctions that are "normalizable to a ##\delta## distribution" only.
     
  4. Mar 1, 2016 #3
    So my reasoning is correct then that we want normalizable solutions hence the requirement that the wave function is bounded?
     
  5. Mar 1, 2016 #4

    vanhees71

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    If you restrict yourself to the bound-state solutions, it's correct.
     
  6. Mar 2, 2016 #5
    Okay thanks. One quick question, do you maybe know why the delta function is said to have units 1/length?
     
  7. Mar 2, 2016 #6
    Simple explanation: in electrodynamics when you want to write (for example) charge density function for point charge in the origin. You would write that as ρ [itex] = q \delta (x) \delta (y) \delta (z) [/itex] . You will write it in that way so that ∫ [itex] \rho dV[/itex] over the whole space gives you the result [itex] q[/itex]. But ρ must have units coulomb per volume, but in your formula you have coulombs in [itex] q [/itex], so delta functions have units 1/length
     
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