Variational method particle in box approx.

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Homework Statement



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).


Homework Equations



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


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!
 

Answers and Replies

  • #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]
 
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  • #3
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Could I assume that the integral on bottom is already normalized? what could I do then? could I just solve for k values?
 
  • #4
vela
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Yes, you just need to find the value of k now.
 
  • #5
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would I take the derivative dE/dk and then solve for k?
 
  • #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.
 
  • #7
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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?
 
  • #8
vela
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Looks good so far. Where are you getting stuck?
 
  • #9
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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.
 
  • #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.
 
  • #11
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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]
 
  • #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?
 
  • #13
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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.
 
  • #14
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and I tried the negative number and it's even worse. Did I do something wrong?
 
  • #15
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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
 
  • #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.
 
  • #17
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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)
 
  • #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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