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Delta function potential; Schrodinger Equation

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    a0cmrk.png

    Consider the TISE for a particle of mass m moving along the x-axis and interacting an attractive delta function potential at origin:



    Part(a): What is the difference between a bound state particle and a free particle?

    Part(b): Show ##\psi _{(x)} = exp (-|k|x)## is a solution of TISE, given conditions:

    Part(c): Explain why symmetric solution is of form:

    Part(d): Use boundary conditions to show the following:

    Part(e): Investigate this solution graphically

    Part(f): Find the significance of this result

    2. Relevant equations



    3. The attempt at a solution

    Part(a)

    A bound particle as energy E<0 while a continuum eigenfunction gives energy E>0.

    Part (b)

    TISE is given by:

    [tex]\hat {H}\psi = E\psi[/tex]

    Using first boundary condition provided we get

    [tex] 2k = \frac{\lambda}{a}[/tex]

    Substituting ##\psi = e^{-k|x|}## into TISE and using what we have just found:

    [tex]-\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 - \frac{\hbar ^2 \lambda}{2ma}\delta_{(x)} = E[/tex]

    I'm not sure why the delta function goes to zero here?

    [tex]E = -\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 [/tex]

    Part (c)

    For |x|>a, V = 0:
    [tex]-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi}{\partial x^2} = E\psi[/tex]

    Solving this equation gives an exponential ##\psi = exp (-k|x|)##.

    For -a<x<a, the particle is bounded and symmetric about origin so ##\psi \propto exp(kx) + exp(-kx)## which is ##\propto cosh(kx) = A cosh(kx)##.

    Part (d)

    Boundary Conditions:

    At x = a, ##\psi_{|x|<a} = \psi_{|x|>a}##.
    Thus we get:

    [tex] e^{-ka} = A cosh (ka)[/tex]

    Second boundary condition:
    At x = a, ##\frac{\partial \psi}{\partial x}_{left} = - \frac{\partial \psi}{\partial x}_{right}##

    The delta function gives opposite gradients about a.

    [tex]e^{-ka} = A sinh (ka) [/tex]

    Using boundary condition earlier,

    [tex] tanh (ka) = 1[/tex]
    [tex] tanh(ka) = 2 - 1 [/tex]

    Can I simply substitute in ##2k = \frac{\lambda}{a}## here?

    [tex] tanh(ka) = \frac{2ka}{ka} - 1 [/tex]
    [tex] tanh (ka) = \frac{\lambda}{y} - 1 [/tex]

    Part (e)

    i71miw.png

    We can see that ##y_0 < \lambda##.

    Since ##tanh(y_0) < 1##, thus ##y_0 > \frac{\lambda}{2}##.

    Since for x≠a both delta functions are positive, it implies that the second term corresponding to the two delta functions ##\delta_{(x-a)}## and ##\delta_{(x+a)}## in the schrodinger equation are >0. Thus the total energy subtract that is less than ##-\frac{\hbar ^2\lambda^2}{8ma^2}##.

    Part(f)

    Two delta function corresponds to electron being attracted to the two protons in nucleus (which is more massive so can treat as stationary). This is a more stable (favourable) configuration, so

    [tex]H + H^+ \rightarrow H_2^+[/tex]
     
  2. jcsd
  3. Feb 19, 2014 #2

    vela

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    You're verifying the solution for x>0 and x<0. At x=0, there's a cusp, so ##\psi''## isn't defined, so the Schrodinger equation doesn't really apply.

    This isn't correct. Integrate the Schrodinger equation between ##x=a-\varepsilon## and ##x=a+\varepsilon## and take the limit as ##\varepsilon \to 0^+##. This will yield the proper boundary condition on the derivatives.

     
  4. Feb 19, 2014 #3
    So the equation##exp(-k|x|)## here only applies when x≠0. We can't find out what the solution is when x = 0? Is it because due to the nature of the delta function it is a sum of infinite series at x = 0?


    But graphically you can see that it is symmetric about 0. You can also see that at x-ε the gradient is opposite that of at x+ε.
     
  5. Feb 19, 2014 #4

    vela

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    At x=0, the lefthand side of the Schrodinger equation isn't defined, so it doesn't really tell you anything about ##\psi##. The continuity requirement, however, allows you to fill in the hole at x=0.

    It may look that way in the picture when you have a single delta function, but this time you have two. You no longer have the symmetry you had before. You need to justify the boundary condition mathematically.
     
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