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:

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


\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?


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?


1 Answer 1


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

  • $\begingroup$ I already did try to plot this in gnuplot and the program had problems drawing simple tan(sqrt(x)). $\endgroup$
    – 71GA
    Apr 1, 2013 at 18:07
  • $\begingroup$ 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? $\endgroup$
    – 71GA
    Apr 1, 2013 at 18:35
  • $\begingroup$ Yes, exactly. The $x$ coordinate of the intersection is the possible energy - not the vertical red lines, they are just a plot error. $\endgroup$
    – zonksoft
    Apr 1, 2013 at 19:56
  • $\begingroup$ Could you please allso check my edit and tell me if my graph for odd solutions is fine? $\endgroup$
    – 71GA
    Apr 1, 2013 at 20:23
  • $\begingroup$ I'm not doing your homework for you :) But by the looks of it, it should be fine. $\endgroup$
    – zonksoft
    Apr 1, 2013 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.