Variational method particle in box approx.

[josecuervo]

Homework Statement

use the variational method to approximate the ground state energy of the particle in a one-dimensional box using the normalized trial wavefunction ∅(x)=Nx^{k}(a-x)^{k} 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^{2}/m_{e}a^2? It is also stated that
\ointpsi*Hpsi dx=(\bar{h}/m_{e}a^{2})(4k^{2}+k)/(2k-1).

Homework Equations

I'm pretty sure you have to use this one \ointpsi*Hpsi dx/\ointpsi*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!

 

[vela]

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:
\int \phi^*(x)\phi(x)\,dx = N^2\frac{\sqrt{\pi}a^{4k+1}k\,\Gamma(2k)}{16^k\, \Gamma(2k+3/2)}

 

[josecuervo]

Could I assume that the integral on bottom is already normalized? what could I do then? could I just solve for k values?

 

[vela]

Yes, you just need to find the value of k now.

 

[josecuervo]

would I take the derivative dE/dk and then solve for k?

 

[vela]

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.

 

[josecuervo]

I'm having problems deriving and solving for k. When I take the derivative, I get

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

and I'm having problems solving for it. am I doing something wrong?

 

[vela]

Looks good so far. Where are you getting stuck?

 

[josecuervo]

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.

 

[vela]

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.

 

[josecuervo]

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_{min}

 

[vela]

I didn't say k=0. I said the numerator had to be 0. What values of k cause the numerator to vanish?

 

[josecuervo]

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]^{2}/m_{e}a^{2}) or 19.68(h^{2}/m_{e}a^{2}) which isn't even in the ballpark of the true value.

 

[josecuervo]

and I tried the negative number and it's even worse. Did I do something wrong?

 

[josecuervo]

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

 

[vela]

Your values for k are correct, but your energies aren't. For the positive value of k, I get
E = \frac{4.9495 \hbar^2}{a^2 m}while the actual ground state energy isE_0 = \frac{4.9348 \hbar^2}{a^2 m}That looks pretty good actually.

 

[josecuervo]

how did you get that? it was given in the problem that the true initial value is

0.125h^{2}/m_{e}a^{2}

or .0199[STRIKE]h[/STRIKE]^{2}/m_{e}a^{2}

did you plug the k value into E_{min}=([STRIKE]h[/STRIKE]^{2}/m_{e}a^{2})(4k^{2}+k)/(2k-1)

 

[vela]

I usedE=\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}
I think you just converted to \hbar incorrectly. You should have h=2\pi\hbar, so
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}