It is about a particle with mass $m$ in a potential $V(x)$:

Three constant potential

I want to do a graphical determination(at first only the symmetrical case) of the energy eigenvalues. I will show you my previous work:

Since the independent Schrödinger Equation is $\varphi''+k^2(x)\varphi=0$ with $k^2(x)=\frac{2m}{\hbar^2 }\left[E-V(q)\right]$ there is following general solution for the wave-function $$\varphi(x) = \begin{cases} 0 &\mbox{if } |x| > b \\ \alpha_+e^{ikx}+\alpha_- e^{-ikx} & \mbox{if } -b < x < -a \\ \beta_+e^{\kappa x} + \beta_-e^{-\kappa x} & \mbox{if } -a < x < a \\ \gamma_+ e^{ikx}+\gamma_- e^{-ikx} & \mbox{if } a < x < b \end{cases}$$

(i.e. if $k^2 > 0 $ there is an oscillating pattern, which we have in zone I and III (classically allowed) and if $k^2 < 0$ an exponential behavior, which we only have in zone II(classically not-allowed))

We can simplify a few things for the symmetric solution: $\varphi(x) = \varphi(-x)$

So I only have to determine

$$\frac{\varphi_3}{\varphi_3'}=\frac{\varphi_2}{\varphi_2'}, (1)$$

because this is the condition in this case, where the so-called Wronski Determinant is zero. In addition to we have some more conditions for combining the several functions in each Section I,II and II: $$\varphi_1(-b)=\varphi_3(b)=0$$ $$\varphi_1(-a)=\varphi_2(-a)$$ $$\varphi_2(a)=\varphi_3(a))$$

And of course the symmetrical charasteristic $\varphi_n(x) = \varphi(-x)$. To shorten this a here is the condition Eq. (1): $$\kappa \tanh(\kappa a) = k\cot[k(a-b)](2)$$

Let me compare this to another short example please: Imagine we have such potential: http://imgur.com/a/2PWWu" And the Wronski Determinant would be zero here(symmetrical case), when $\kappa = k\tan(ka)$. By multiplying $a$ with this equation and by let being $\eta = \kappa\cdot a$ and $\xi = k\cdot a$ you will get $$\eta = \xi \tan \xi,$$ which brings you to following graphically solution of the energy eigenvalues of this potential well(ignore the dotted $cot$-function, it would be for the antisymmetrical case, the $tan$-function is for the symmetrical case):

enter image description here

So my question is now, how can I make my Eq. (2) only dependent from one value like in the example I showed above? Can someone give me a tip please, because I am clueless atm.


You potential is even so we expect on general grounds that solutions will divide into even and odds. I will discuss only the even case. The candidate wavefunction is \begin{align} \psi(x)&=\left\{ {\renewcommand{\arraystretch}{1.25}\begin{array}{ll} B\sin(k(b+x))&\hbox{if }-b\le x\le a\, ,\\ A\left(e^{\kappa x}+e^{-\kappa x}\right)&\hbox{if }-a< x < a\, ,\\ B\sin(k(b-x))&\hbox{if }a\le x\le b\, .\end{array}}\right. \end{align}

It is certainly even under the transformation $x\to -x$, and it satisfies the boundary condition that $\psi(\pm b)=0$ because the wavefunction outside the well must be $0$. This is just a generalization of your form where it is simpler to deal with boundary conditions, and where I've specialized to the even solutions.

This form assumes $E<V_0$. As always, define $$ \kappa =\displaystyle{\sqrt{2m(V_0-E)\over \hbar^2}}\, ,\qquad k=\displaystyle{\sqrt{2m E\over\hbar^2}}\, . $$

The continuity of $\psi$ and $d\psi/dx$ at $x=a$ gives two equations: \begin{align} A\left(e^{\kappa a}+e^{-\kappa a}\right)&=B\sin(k(b-a))\, ,\tag{1}\\ A\kappa\,\left(e^{\kappa a}-e^{-\kappa a}\right)&=-B\,k\,\cos(k(b-a))\, .\tag{2} \end{align} The continuity of $\psi$ and its derivative at $x=-a$ gives the same two equations.

Dividing (1) by (2) so as to eliminate $A$ and $B$, we obtain the trancendental equation: $$ \frac{1}{\kappa}\displaystyle\left({e^{\kappa a}+e^{-\kappa a}\over e^{\kappa a}-e^{-\kappa a}}\right)=-\frac{1}{k}\tan(k(b-a))\, . $$

To bring this to a solvable form, we set $$ \xi=\sqrt{2mV_0\,a^2\over \hbar^2}\, ,\qquad\qquad z=\frac{E}{V_0}\, , $$ and (if I did my substitutions right)obtain the transcendental equation $$ -\displaystyle{\frac{(e^{\xi\sqrt{1-z}}+e^{-\xi\sqrt{1-z}})} {(e^{\xi\sqrt{1-z}}-e^{-\xi\sqrt{1-z}})\sqrt{1-z}}}= \displaystyle{\frac{\tan(\xi\sqrt{z}(1-b/a))}{\sqrt{z}}}\, . $$ Once you have obtained a numerical value for $\xi$ specific to you problem you can solve graphically for $z$ in the usual way.

The case of the odd solutions proceeds in the same general manner.

  • $\begingroup$ Thanks for your answer. But can't I just draw it like this: wolframalpha.com/input/?i=cot+x+and+tanh+(x) ? But I guess that would be wrong, because I have two variables a and b. And here I only have x. $\endgroup$ – physics May 7 '17 at 18:20
  • $\begingroup$ you need to have the ratio $a/b$ somewhere... and you need to write $\xi$ or $x$ in terms of $a$ or $b$.... $\endgroup$ – ZeroTheHero May 7 '17 at 18:27
  • $\begingroup$ "Once you have obtained a numerical value for ξ specific to you problem you can solve graphically for z in the usual way." --> So you meant, that I only can solve my problem graphically for z, when I have specific values for $V_0$, $m$, $a$ and $b$ ? $\endgroup$ – physics May 7 '17 at 18:46
  • $\begingroup$ I understand how you got to this transcendental equation (the substituted one), but I do not rlly get how to form that to f(z) and then to draw it. I understand the way how I did it with the normal "potential well" above in my first post, because there was it clear. I've only used $\xi = k\cdot a$ etc. and so I got $\nu = \xi tan \xi$ for drawing it. $\endgroup$ – physics May 7 '17 at 18:51
  • $\begingroup$ the twist with this problem is that solutions depend on the ratio of $b/a$ and the height of the bump in addition to the usual dependence on $\xi$. Moreover you also have $k$ and $\kappa$ since you have two types of behaviours in your potential. Thus, it's not as "simple" as the usual finite well. I don't know that you can get away with "simple" curves such as $\xi \tan \xi$ but if you can I'd like to know. $\endgroup$ – ZeroTheHero May 8 '17 at 2:09

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