- #1
Clever-Name
- 380
- 1
Homework Statement
We're modelling an ammonia maser with a double infinite square well defined by:
[tex]
V(x) =
\begin{cases}
V_{0} & |x| < b - \frac{a}{2}\\
0 & b-\frac{a}{2} < |x| < b+\frac{a}{2}\\
\infty & |x| \geq b + \frac{a}{2}
\end{cases}
[/tex]
I have had no trouble with the assignment up until I'm asked to solve for the transcendental equations for energy and solve for the lowest odd and even state using MATLAB
Homework Equations
We are told to use the following form for the wavefunction as defined below:
Even states:
[tex]
\Psi(x)_{even} =
\begin{cases}
Acosh(\kappa x) & 0 < x < b-\frac{a}{2}\\
Bsin(k(b + \frac{a}{2} - x)) & b-\frac{a}{2} < x < b+\frac{a}{2}
\end{cases}
[/tex]
and for the Odd states:
[tex]
\Psi(x)_{odd} =
\begin{cases}
Asinh(\kappa x) & 0 < x < b-\frac{a}{2}\\
Bsin(k(b + \frac{a}{2} - x)) & b-\frac{a}{2} < x < b+\frac{a}{2}
\end{cases}
[/tex]
The Attempt at a Solution
Since the well is symmetric we can ignore the left-side classically allowed region, the function will just be defined as plus or minus [itex]Bsin(k(b + \frac{a}{2} - x))[/itex] depending on whether we're looking at an even or odd state.
Now, applying the boundary conditions at x = b-a/2 gives me the following equations:
edit - Found one mistake, negative signs, but that still doesn't fix my 2nd problem.
Even:
[tex]
\sqrt{\frac{V_{0} - E}{E}}tan(ka) = -coth(\kappa(b-\frac{a}{2})
[/tex]
Odd:
[tex]
\sqrt{\frac{V_{0} - E}{E}}tan(ka) = -tanh(\kappa(b-\frac{a}{2}))
[/tex]
with
[tex]
k = \sqrt{\frac{2mE}{\hbar^2}}
[/tex]
and
[tex]
\kappa = \sqrt{\frac{2m(V_{0}-E)}{\hbar^2}}
[/tex]
and I should note that [itex] V_{0} < E [/itex]
Typing out all of my work would have been a huge pain, if these don't look right let me know and I can type out what I did.
Now, my first question is: Do these equations for the energy look right?
If they are right my second question will be how the heck do I solve for the energies?! I'm using the fzero function, with a given initial guess of ~0.00661eV, in MATLAB and am getting ~0.041eV for the even state but 3.7e-16 for the odd state. They should be almost identical!
If anyone wants to tinker around with it and try and test my answers we are given:
[itex] V_{0} = 0.130 eV [/itex]
[itex] b = 0.070 nm [/itex]
[itex] a = 0.112 nm [/itex]
and to get the energy guess we are told
[itex] k \approx \frac{\pi}{a} [/itex]
Last edited: