0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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