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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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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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