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Variational method particle in box approx.

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    use the variational method to approximate the ground state energy of the particle in a one-dimentional box using the normalized trial wavefunction ∅(x)=Nx[itex]^{k}[/itex](a-x)[itex]^{k}[/itex] where k is the parameter. Demonstrate why we choose the positive number rather than the negative value. By what absolute percentage does your value differ from the true one, (.125h[itex]^{2}[/itex]/m[itex]_{e}[/itex]a^2? It is also stated that
    [itex]\oint[/itex]psi*Hpsi dx=([itex]\bar{h}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex])(4k[itex]^{2}[/itex]+k)/(2k-1).


    2. Relevant equations

    I'm pretty sure you have to use this one [itex]\oint[/itex]psi*Hpsi dx/[itex]\oint[/itex]psi*psi dx


    3. The attempt at a solution

    The first thing I tried to do is normalize the trial wavefunction which didn't get very far as I couldn't figure out how to do the integral. I also can't figure out to get either of the integrals to work because of the k powers. That is the main thing I'm blocked on. The only other thing I can think of is just to take an approximation of just the first two k values but the given integral with the incorporated hamiltonian has k included so it's wanting me to approx over all k. I need help getting to the next step. Thanks!
     
  2. jcsd
  3. Oct 7, 2011 #2

    vela

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    The integral you've been given was evaluated using the normalized wave function, so I don't think you're expected to manually crank out the integrals.

    In case you're curious, this is what Mathematica found:
    [tex]\int \phi^*(x)\phi(x)\,dx = N^2\frac{\sqrt{\pi}a^{4k+1}k\,\Gamma(2k)}{16^k\, \Gamma(2k+3/2)}[/tex]
     
    Last edited: Oct 7, 2011
  4. Oct 9, 2011 #3
    Could I assume that the integral on bottom is already normalized? what could I do then? could I just solve for k values?
     
  5. Oct 9, 2011 #4

    vela

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    Yes, you just need to find the value of k now.
     
  6. Oct 9, 2011 #5
    would I take the derivative dE/dk and then solve for k?
     
  7. Oct 9, 2011 #6

    vela

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    Yes. That variational method will always overestimate the energy, so by finding the value of k that minimizes the estimate, you know you're getting the best approximation.
     
  8. Oct 9, 2011 #7
    I'm having problems deriving and solving for k. When I take the derivative, I get

    dE/dk=([STRIKE]h[/STRIKE]/(m[itex]_{e}[/itex]a[itex]^{2}[/itex])(8k[itex]^{2}[/itex]-8k-1)/(1-2k)[itex]^{2}[/itex]=0

    and I'm having problems solving for it. am I doing something wrong?
     
  9. Oct 9, 2011 #8

    vela

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    Looks good so far. Where are you getting stuck?
     
  10. Oct 9, 2011 #9
    the derivative doesn't seem to be solveable because there's no way to move anything over to the other side. and I know that I'll get multiple values of k when I solve.
     
  11. Oct 9, 2011 #10

    vela

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    Remember a fraction is only equal to 0 only if the numerator is 0. That's probably exactly what you're getting, but you're mistakenly thinking it's not working out.
     
  12. Oct 9, 2011 #11
    Okay, but if the value of k is 0, then that doesn't go with the problem. In the problem it states that there will be a positive and negative value for the parameter and it asks why the positive number is better. And if k is 0 then the approximation won't work when you plug back into E[itex]_{min}[/itex]
     
  13. Oct 9, 2011 #12

    vela

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    I didn't say k=0. I said the numerator had to be 0. What values of k cause the numerator to vanish?
     
  14. Oct 9, 2011 #13
    ok. sorry, dumb moment.
    so I solved for k=-.11237 and k=1.11237.

    then I plugged the positive k value into E and got 3.133([STRIKE]h[/STRIKE][itex]^{2}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex]) or 19.68(h[itex]^{2}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex]) which isn't even in the ballpark of the true value.
     
  15. Oct 9, 2011 #14
    and I tried the negative number and it's even worse. Did I do something wrong?
     
  16. Oct 9, 2011 #15
    is there anyway you could work through the problem and double check my work? I've worked through it twice now and still haven't gotten an answer that looks right
     
  17. Oct 9, 2011 #16

    vela

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    Your values for k are correct, but your energies aren't. For the positive value of k, I get
    [tex]E = \frac{4.9495 \hbar^2}{a^2 m}[/tex]while the actual ground state energy is[tex]E_0 = \frac{4.9348 \hbar^2}{a^2 m}[/tex]That looks pretty good actually.
     
  18. Oct 9, 2011 #17
    how did you get that? it was given in the problem that the true initial value is

    0.125h[itex]^{2}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex]

    or .0199[STRIKE]h[/STRIKE][itex]^{2}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex]

    did you plug the k value into E[itex]_{min}[/itex]=([STRIKE]h[/STRIKE][itex]^{2}[/itex]/m[itex]_{e}[/itex]a[itex]^{2}[/itex])(4k[itex]^{2}[/itex]+k)/(2k-1)
     
  19. Oct 9, 2011 #18

    vela

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    I used[tex]E=\frac{\hbar^2\pi^2}{2ma^2} = \left(\frac{\pi^2}{2}\right)\frac{\hbar^2}{ma^2} \cong 4.9348 \frac{\hbar^2}{ma^2}[/tex]
    I think you just converted to [itex]\hbar[/itex] incorrectly. You should have [itex]h=2\pi\hbar[/itex], so
    [tex]E_n = \frac{n^2h^2}{8ma^2} = \frac{n^2(2\pi)^2\hbar^2}{8ma^2} = \frac{n^2\pi^2\hbar^2}{2ma^2}[/tex]
     
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