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I'm trying to get the fundamental state of an electron in a potential, as in:

$$V(X)=e|x|$$

Where $e$ is a constant. To start with I want to solve it with $e=1$, then where $e$ is big enough that it transforms the potential to something like the delta. And how i solve the problem of having a discontiunity of the derivative in zero?(i mean in the computer)

I have a conceptual problem, If I try to do it with a computer (with a method of matrix diagonalization) the energy is dependent on the big of the $x$ array. Some has a clue to give?

Another thing is that if $e>>1$, it will be like a delta function and inside can't be a bound state with a fundamental energy. Or am I wrong?

mmazz
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    Hello mazzeta and welcome to Physics Stack Exchange! As it currently stands, your question looks like an artificially constructed exercise, or some homework. We prefer that if you ask a question like that, you provide what you did to try to solve the problem, and ask a more conceptual question, as we are not a homework help service. Best of luck with your question! – auden Jul 22 '17 at 22:00
  • Sorry, only i want is a clue to keep going. I edit my post. – mmazz Jul 22 '17 at 22:10
  • Out of curiosity, why do you think it'd look like a Dirac Delta if $e\gg 1$? Couldn't you always just rescale your variable $x \rightarrow e x$? So the physics should remain the same, except for some rescalings. All $e$ does is change the global "amplitude" of $V(x)$, not its form. Or am I missing something? – Philip Jul 22 '17 at 23:06
  • Thats right. But its an electron with an e=10^19ev, i supose that is so big that i need to think like a dirac. – mmazz Jul 22 '17 at 23:37
  • Dirac could express himself clearly. Given the delta potential bound state, unrelated to your problem, what is your statement on its bound state? – Cosmas Zachos Jul 23 '17 at 00:34
  • Due diligence first. – Cosmas Zachos Jul 23 '17 at 00:38

2 Answers2

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If you interest is in solving the time independent Schroedinger equation for a potential $V~=~V_0|$ $$ -\frac{\hbar}{2m}\frac{\partial\psi}{\partial x}~+~(E~-~V_0|x|)\psi~=~0, $$ the solution is $$ \psi(x)~=~\psi_0(Ai(g^{3/2}x)~+~c~Bi(g^{3/2}x), $$ for $Ai$ and $Bi$ Airy functions. Here $g$ is all the constants in the Schrodinger equation, $g~=~2m(E~-~V_0)/\hbar$ The function $Bi(x)$ diverges so that one is not considered.

Lawrence B. Crowell
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You can already get an idea of what the ground state wave function looks like.

  1. Because the potential energy function is an even function, the ground state will also be an even function.
  2. You know that you will have decaying functions in the "forbidden" regions where $V\gt E$, but since your potential is linear rather than just constant, the decay will go "faster" than simple exponential.
  3. At the point where $E=V$, the curvature will change from negative in the allowed region to positive in the forbidden region.
  4. Because this is a bound system, the ground state wave function will have only one maximum (if we choose $\psi>0$) and no zero crossings.
  5. Because the potential is "V-shaped", which is sharper than the SHO potential, you know that the "hump" in the classically allowed region will drop more sharply than a "Bell Curve" / normal distribution would.

Finally, as you increase the value of $\bf{e}$, you should expect the wave function to get narrower, higher, and steeper; however, it will $\bf{not}$ produce the same result as a Dirac potential would, because $V(x) = e|x|$ does not turn into $V(x)=e\delta(x)$ as $\bf{e}\rightarrow \infty$

Apollonius
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