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I have a finite square well like the one on the picture below:

I have done some calculations on it and got a transcendental equation for even solutions which is like this:

$$ \boxed{\dfrac{\mathcal{K}}{\mathcal{L}} = \tan \left(\mathcal{L \dfrac{d}{2}}\right)} $$

where:

\begin{align} &\mathcal{K}\equiv \sqrt{\frac{2m (W - W_p)}{\hbar^2}} &&\mathcal{L} \equiv \sqrt{\frac{2mW}{\hbar^2}} \end{align}

I have been looking in Griffith's book where he says that transcendental equation can be solved graphically so i did try to write it down so i could draw it (i inserted $\mathcal K$ and $\mathcal L$):

$$ \sqrt{1 - \dfrac{W_p}{W}} = \tan\left( \frac{\sqrt{2mW}}{\hbar} \frac{d}{2} \right) $$

I noticed that i don't even know what am i looking for. Is it the total energy $W$? If so i must know potential $W_p$ outside the well, width $d$, and a mass of a particle $m$. But still how could i graphically solve this?


EDIT:

I set all the constants to be $W_p=d=m=\hbar=1$ and ploted a graph for an even solutions:

enter image description here

I allso plotted a graph for the odd solutions which for which transcendental equation is ($\mathcal K$ and $\mathcal L$ stay the same as before):

$$ \boxed{-\dfrac{\mathcal{L}}{\mathcal{k}} = \tan \left(\mathcal{L \dfrac{d}{2}}\right)} $$

$$ -\sqrt{\frac{1}{1 - W_p/W}} = \tan\left( \frac{\sqrt{2mW}}{\hbar} \frac{d}{2} \right) $$

enter image description here

Could anyone please confirm that i got the right graph for transcendental equation?

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1 Answer

up vote 1 down vote accepted

Draw a coordinate system where $W$ (the particle energy) is on the horizontal axis. Then, do a plot of the left hand side of the equation. Then, the right hand side. The $W$ coordinate of the point where those two curves are crossing is the equation's solution; because there, the left hand and the right hand side of the equation are equal.

Of course, you need to know all the constants to draw the curves.

Here's a plot with all constants set to 1. The vertical lines don't really exist, it's just an artistical imagining by Mathematica.

enter image description here

  • Blue: left hand side

  • Red: right hand side

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I already did try to plot this in gnuplot and the program had problems drawing simple tan(sqrt(x)). –  71GA Apr 1 '13 at 18:07
    
Thank you for the graph i don't have Mathematica and am struggling with a gnuplot. What do intersections represent in the graph you drawed? Are those possible energy states in a finite well? –  71GA Apr 1 '13 at 18:35
    
Yes, exactly. The $x$ coordinate of the intersection is the possible energy - not the vertical red lines, they are just a plot error. –  Rafael Reiter Apr 1 '13 at 19:56
    
Could you please allso check my edit and tell me if my graph for odd solutions is fine? –  71GA Apr 1 '13 at 20:23
    
I'm not doing your homework for you :) But by the looks of it, it should be fine. –  Rafael Reiter Apr 1 '13 at 20:28
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