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Delta Function Potential Barrier

  1. Jul 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Background: The problem is to find the uncertainty relationship for the wave equation for a delta function potential barrier where ##V(x)=\alpha\delta(x)##.

    Check the uncertainty principle for the wave function in Equation 2.129 Hint: Calculating ##\left< p^2 \right> ## is tricky, because the derivative of ##\psi## has a step discontinuity at x=0. Use the result in Problem 2.24(b). Partial answer: ##\left< p^2 \right>=(\frac{m\alpha}{\hbar})^2 ##.

    2. Relevant equations

    2.129 is ##\psi (x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m\alpha |x|/\hbar^2}##; ##E=-\frac{m\alpha ^2}{2\hbar^2}##.

    2.24(b) Let ##\theta (x)## be the step function:

    ##\theta (x)=\left\{ 1,\quad if\quad x\quad >\quad 0.\quad 0,\quad if\quad x\quad <\quad 0. \right\} ##

    (In the rare case where it actually matters, we define ##\theta(0)## to be 1/2.) Show that ##\frac{d\theta}{dx}=\delta(x)##.

    3. The attempt at a solution
    I understand it up until the last step, and I even think I understand why it didn't work. I was able to integrate in steps for <x>,<x^2>, and <p>, but this does not work for <p^2>. I understand it has something to do with the step discontinuity that arises, but I guess I am having a hard time understanding why I had to integrate through the step on this one, but not have to include the point x=0 on the rest of them (I get the correct answer as well). The integrand is smooth for the integral determining <x>, and <x^2>. <p> has a kink, so <P^2> is discontinuous. In calculus up to this point I have always handled this by integrating over the smooth parts and adding the the sums of all the smooth parts. I am having trouble visualizing why I can't just assume the area under the curves of all the pieces are the only ones relevant, but that this point at x=0 adds so much info even though it is an infinitesimal contributor. I guess that is some of the weirdness that comes from making an infinitesimal point have an infinite height.

    Thanks,
    Chris
     
  2. jcsd
  3. Jul 10, 2014 #2

    George Jones

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    So, it seems that you you have found ##d \psi/dx##. What is ##d^2 \psi/dx^2##?
     
  4. Jul 10, 2014 #3
    I did, but the second one is wrong. I am working on it now, but integrating over the point x=0.

    Chris
     
  5. Jul 10, 2014 #4

    George Jones

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    Before you integrate, you should make sure that your expression for ##d^2 \psi/dx^2## is correct. What did you get for ##d^2 \psi/dx^2##?
     
  6. Jul 11, 2014 #5
    Yes, it finally worked. Spent some time on it. It actually did work by integrating over the discontinuity separately and using ##\epsilon## limits. So that answers my question. A point can't be ignored if it blows up. It actually contributes twice what the normal part of the integral does. It is funny -- they tell you not to integrate or differentiate over these sort of things in your calculus class.

    Thanks,
    Chris
     
  7. Jul 11, 2014 #6

    George Jones

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    I am sorry, but it is still not clear to me that you know how to take the second derivative, as it is not completely straightforward (especially if you haven't seen these "tricks" before), and you resist writing down what you have.

    Let
    $$
    \psi\left(x\right) = e^{-\left| x \right|} = \begin{cases}
    e^{-x}; & x \geq 0 \\
    e^x; & x < 0
    \end{cases}.
    $$

    ##\psi## is continuous, and its derivative is
    $$
    \frac{d\psi}{dx}\left(x\right) = \begin{cases}
    -e^{-x}; & x > 0 \\
    e^x; & x < 0
    \end{cases},
    $$

    which, as you noted, has a step discontinuity. Instead of writing ##d\psi/dx## as 2 lines, the Heavisde step function ##\theta\left(x\right)## can be used to write as one line as follows.

    First, the top line, ##-e^{-x}## for positive ##x##. Since ##\theta\left(x\right)## is zero for negative ##x##, the top line is the same as ##-e^{-x}\theta\left(x\right)## for all ##x##.

    Next, the bottom line, ##e^x## for negative ##x##. Since ##\theta\left(-x\right)## is zero for positive ##x##, the bottom line is the same as ##e^x\theta\left(-x\right)## for all ##x##.

    Combining these gives
    $$
    \frac{d\psi}{dx}\left(x\right) = -e^{-x}\theta\left(x\right) + e^x\theta\left(-x\right)
    $$

    for all ##x##.

    Now, use the product rule and ##\frac{d\theta}{dx}\left(x\right) = \delta\left(x\right)## to take a second derivative. What do you get?
     
  8. Jul 11, 2014 #7
    Yes, this is the method I used. For <p^2> I calculated ##-(m\alpha/\hbar)^2## for every point other than the origin. Using epsilon for integral limits for the case where ##\epsilon \rightarrow 0##, I was able to integrate over the delta (the origin), and the quantity of that evaluation was ##2(m\alpha/\hbar)^2## adding these together I get: ##(m\alpha/\hbar)^2##, and this is the answer that matches the book. The rest is easy, but you are correct. This is very tricky. Now I must move on to hurt my brain on problem 2.27 (Griffiths) where there is a double delta potential. I went for it, and got the correct ##\psi## by an educated guess, but my analysis of the allowed energy states were bunged bad. This one seems really nasty. I feel like I need to go lift some more math weights and come back later.

    Thanks,
    Chris
     
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