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Bound state problem

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider [itex]V(x)=-aV_{0}δ(x)[/itex]. Show that it admits a bound state of energy [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex].Are there any other bound states?Hint:solve Schrodinger's equation outside the potential for E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0.

    Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0. Calculate the change in the slope at equate it to [itex]\int_{-ε}^{ε}( \frac{d^2\psi}{dx^2}) dx[/itex]
    (where ε is infinitesimal) determined from Shrodinger's equation.

    2. Relevant equations
    Shrodinger equation in X basis : [itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E-V)\psi=0[/itex].

    3. The attempt at a solution
    First how can we prove that [itex]E=-ma^2V_{0}/2\hbar^{2}[/itex] admits a bound state, shall I substitute it in the shrodinger equation then prove that the obtained wave function goes to zero when x goes to infinity?
    For are there any other bound states?
    I tried to solve the Shrodinger's equation with E<0 and outside the potential, what I got is this :
    Let E=-A where A is a positive integer.
    [itex]\frac {d^2\psi}{dx^2}+(2m/\hbar)(E)\psi=0[/itex]
    [itex]\frac {d^2\psi}{dx^2}-(2m/\hbar)(A)\psi=0[/itex]
    [itex]\psi=Ae^{-\sqrt{2mA/\hbar^2}x}+Be^{\sqrt{2mA/\hbar^2}x}[/itex]
    [itex]Be^{\sqrt{2mA/\hbar^2}x}[/itex] blows up to zero when x goes to infinity, so B=0.
    what we finally get is [itex]\psi=Ae^{-\sqrt{2mE/\hbar^2}x}[/itex]. I reached here and I don't know how to complete.
    If anybody can help.Thanks.
     
    Last edited: Mar 2, 2016
  2. jcsd
  3. Mar 2, 2016 #2

    mfb

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    x can go both to positive and negative infinity, you have to consider both cases separately.
    Then consider x=0 separately again, with the given hint.

    Schrödinger (or Schroedinger), by the way.
     
  4. Mar 3, 2016 #3
    Let E=-U where U is a positive number.
    Take case where x<0
    the solution is[itex] \psi=Ae^{-\sqrt{2mU/\hbar^2}x}+Be^{\sqrt{2mU/\hbar^2}x}[/itex]
    as x goes to -infinity the first part will blow up so A must be zero.
    when x>0
    the solution is [itex] \psi=Ce^{-\sqrt {2mU/ \hbar^2}x}+De^{\sqrt{2mU/ \hbar^2} x } [/itex]
    as x goes to +infinity the next part will blow up so D=0.

    now for case x=0
    the solution is [itex] \psi=E e^{-\sqrt {2m (U+V)\hbar^2}x}+Fe^{\sqrt{2m(U+V)/ \hbar^2} x} [/itex]
    here when x=0 V=infinity but 0*infinity is not defined. So how we can determine the wave function here?
     
  5. Mar 3, 2016 #4

    mfb

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    The wave function has to be continuous.
    You don't need x-dependent functions to consider the case x=0 as the wave function there has a single value only.
     
  6. Mar 4, 2016 #5
    Let [itex]\psi[/itex]at x=0 be a number called it F.
    for x<0 [itex]\psi=Be^{\sqrt{2mU/\hbar^2}x}[/itex]
    for x>0[itex]\psi=Ce^{-\sqrt{2mU/\hbar^2}x}[/itex]
    As the wave function must be continuous at x=0,then we have [itex]\lim_{x\rightarrow 0^{-} } {B e^ {\sqrt {2mU/ \hbar ^ 2} x }}=[/itex] [itex]\lim_{x\rightarrow 0^{+} }{Ce^{- \sqrt {2mU/ \hbar^2} x }}= [/itex] [itex]\psi(0)=F[/itex]
    then B=C=F.
    using the normalization condition at [itex]-\infty<x<0 and 0<x<+\infty[/itex]
    we have as B=C [itex]C\int_{-\infty}^{0^-} e^{\sqrt{2mU/ \hbar^2}x}dx +C\int_{0^+}^{+\infty}e^{-\sqrt{2mU/ \hbar^2}x}dx=1[/itex]
    this will lead that C=[itex]\sqrt{2mU}/2\hbar[/itex]
    but as V=infinity at x=0 then [itex]\psi=0[/itex]at x=0,so F=0.
    and in this case U must be zero!!!as C=B=0.
    which means that there is no bound states,so what goes wrong?
     
  7. Mar 4, 2016 #6

    mfb

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    Why should the wave function be zero there?

    Everything before that line was fine.
     
  8. Mar 4, 2016 #7
    OK.
    then B=C=F=[itex]\sqrt{2mU}/2\hbar[/itex]
    and so for x<0 [itex]\psi=(\sqrt{2mU}/2\hbar )* e^{\sqrt{2mU/ \hbar^2}x}[/itex]
    and so for x>0[itex]\psi=(\sqrt{2mU}/2\hbar)* e^{-\sqrt{2mU/ \hbar^2}x}[/itex]
    and for x=0[itex]\psi=\sqrt{2mU}/2\hbar[/itex]
    But this means that all negative energies are allowed, since their is no restriction on negative energy.Is this right?
     
  9. Mar 4, 2016 #8

    mfb

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    The restriction comes from the change of the first derivative that happens at x=0.
     
  10. Mar 5, 2016 #9
    for x<0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=mU/ \hbar^2[/itex]
    for x=0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=0[/itex]
    for x>0 [itex]\left.\frac{d\psi}{dx}\right|_{x=0}=-mU/ \hbar^2[/itex]
    The change of the first derivative at x=0 is [itex]-mU/ \hbar^2[/itex]
    Now as I calculated the change of slope, I will equate to [itex]\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx[/itex]
    [itex]\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx=-2mU/ \hbar^2 e^{-\sqrt{2mU/ \hbar^2}ε} +2mU/ \hbar^2[/itex]
    but I didn't see that equating the change of slope to this integral leads to any restriction in energy?
    what is the trick here?
     
  11. Mar 5, 2016 #10

    mfb

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    There is no derivative if your function has a single point only.

    The delta potential at x=0 allows to calculate the change of the first derivative. Set it equal to ##\frac{2mU}{\hbar^2}## and you get the constraints on the energy.
     
  12. Mar 6, 2016 #11
    mmmm ok.
    I will use now the Schroedinger equation [itex]-\hbar^2 /2m\int_{-ε}^{ε}\psi' '(x)dx+\int_{-ε}^{ε}V(x)\psi(x)dx=E\int_{-ε}^{ε}\psi(x)dx[/itex]

    Then [itex]-\hbar^2 /2m[\psi'_{x>0}(0)-\psi'_{x<0}(0)]-aV_{0}\psi(0)=0 (asE\int_{-ε}^{ε}\psi(x)dx=0)[/itex]
    so [itex]-\hbar^2/2m[(-2mU/2\hbar^2)-2mU/2\hbar^2]=aV_{0}\psi(0)[/itex]
    then [itex]U=aV_{0}\sqrt{2mU}/2\hbar[/itex]
    [itex]U^2=a^2V_{0}^2mU/4\hbar^2[/itex]
    then [itex]U(U-a^2V_{0}^2m/2\hbar^2)=0[/itex]
    which means that[itex] U=0 or U=a^2V_{0}^2m/2\hbar^2[/itex]
    and so the two allowed bound states are [itex]E=0[/itex] and[itex]E= - a^2V_{0}^2m/2\hbar^2[/itex]
     
  13. Mar 6, 2016 #12

    mfb

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    U=0 leads to a contradiction elsewhere.
    Do the units match for the final result? The approach looks fine.
     
  14. Mar 6, 2016 #13
    As the hint says, I must suppose E<0 out side the potential, so there will be one bound state which is [itex]E=-a^2V_{0}^2m/2\hbar^2[/itex].I think this is right because it matches what the question wants,as
    By the way, it is [itex]E=-a^2V_{0}^2m/2\hbar^2[/itex] and not [itex]E=-a^2V_{0}m/2\hbar^2[/itex] I copied it wrong in the first post.
    shall I suppose E<0 for simplicity or things would go wrong if I have dealt with E>0?as the hint restricted me to deal with E<0 out side the potential.
     
  15. Mar 6, 2016 #14

    mfb

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    E>0 cannot be a bound state. There are states with positive energy, but they are not bound.
     
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