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Dirac-Delta Potential (Scattering)

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Say there exists a delta potential V(x) = α(x-L). Solve the eigenvalue problem to yield the energy eigenstates.
    Say a quanton is in an initial state given by Ψ(x)=Aeikox at t=0. What is Ψ(x,t)?

    2. Relevant equations
    Schrodinger Eqn.

    3. The attempt at a solution
    I solved the first part to find the energy eigenvectors (not-normalized), after applying the boundary condition (including the discontinuity at x-L), to get:
    Ψk(x) = A(eikx + B/A(e-ikx) x<L
    Ψk(x)=A(C/A)(eikx) x>L
    B/A and C/A were obtained from the boundary conditions, and of course represent the reflection and transmission coefficients, respectively.
    Since ψ(x,t) = ∫Ψk(x)⋅φ(k)dk
    where φ(k) represents the coefficients of the wave function, obtained via Fourier Transform.
    i.e. φ(k) = ∫Ψk*(x)⋅Ψ(x)dx
    I'm having trouble, qualitatively, understanding how to compute the given integral given that the initial state has a well-defined momentum and is travelling from the left to the right; would the integral with B/A(-ikx) vanish since it represents a left moving quanton, which is orthogonal to the right moving quanton in the initial state?
     
  2. jcsd
  3. Oct 18, 2015 #2
  4. Oct 18, 2015 #3

    andrewkirk

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    The integral is an expansion in the X basis of ##\langle\psi_k|\psi\rangle##. Note that ##|\psi\rangle## is simply ##|p_0\rangle## in the P (momentum) basis, where ##p_0=\hbar k_0## (or maybe ##\frac{k_0}{\hbar}##. I'm in a rush and don't have time to check which it is). That suggests to me that the problem may be more tractable if we instead compute ##\langle\psi_k|\psi\rangle## by expanding in the P basis. I would take the case ##x>L## first, as it has only one component. What is the representation in the P basis of the ket that has representation ##x\mapsto \psi_k(x)=\alpha e^{ikx}## in the X basis?
     
  5. Oct 20, 2015 #4
    I am getting Dirac-Delta Functions for φ(k). Does this sound right?
     
  6. Oct 20, 2015 #5

    andrewkirk

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    Yes.
     
  7. Oct 20, 2015 #6
    Qualitatively, though, the initial state is a non-localized particle with well-defined energy, travelling to the right. Would this mean that the second integral (with the reflection coefficient) would be zero since that represents a particle moving in the left direction (reflected part), which would be orthogonal to right-going particle at t=0?
     
  8. Oct 20, 2015 #7

    andrewkirk

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    It wouldn't mean that, because the two integrals are not entire wave functions on their own. They are two parts of the same energy eigenket, which is (in the X basis):

    $$\left[1-\Theta(x-L)\right](\alpha e^{ik_1 x}+\beta e^{-ik_1x})+\Theta(x-L) \gamma e^{ik_2x}$$
    where ##\Theta## is the Heaviside function and ##k_2## is a function of ##k_1## and the potential step ##V##.

    We can't break off a spatial fragment of the ket and reason about that fragment being orthogonal to the initial state.
     
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