# Schrodinger equation for a weird potential

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1. May 7, 2016

### mixo1234

Hello everyone,

I have this weirdo potential for homework

U(x) = \frac{U_1}{ \left( 1+e^{x/a}\right)^2 } - \frac{U_2}{ \left( 1+e^{x/a}\right)}

where U1,U2 and "a" are positive

and I need to find the energies for the bound states and also the wave functions for them. I got stuck on one part for like 2 or 3 hours, tried tons of ways, but still nothing. So I'll post the most relevant way by far and I'm hoping to get some help from you. By the way, If anyone knows the name of this potential or some kind of a book (I prefer English, Russian or German, but any other language would do :D ) where I can get some help, please tell me. I looked up in Landau, Flugge and Greiner but couldnt find anything that could help...

So here it goes:

The Schrodinger equation looks like this:

-\frac{\hbar^2}{2m}\frac{d^2\Psi(x)}{dx^2} + \left(\frac{U_1}{ \left( 1+e^{x/a}\right)^2 } - \frac{U_2}{ \left( 1+e^{x/a}\right)}\right)\Psi(x)=E\Psi(x),

For the Bound states we have E=-|E|. Using these following notations:
$$w=1+e^{x/a},$$
$$\frac{2ma^2U_1}{\hbar^2} = \alpha^2,$$
$$\frac{2ma^2U_2}{\hbar^2} = \beta^2,$$
and
$$\frac{2ma^2|E|}{\hbar^2} = k^2$$
I get

(w-1)^2 \Psi''(w) + (w-1) \Psi'(w) -\left(\frac{\alpha^2}{w^2} - \frac{\beta^2}{w} + k^2 \right) \Psi(w)

and $$1<w< \infty$$.

When $$w \to \infty$$, then the equation is

(w-1)^2 \Psi''(w) + (w-1) \Psi'(w) - k^2 \Psi(w)=0

This is Eulers differential equation and the solution is $$\Psi(w) = (w-1)^{-k}$$.

So here is the problem... When
$$w \to 1$$
then I can clearly say that

\frac{\alpha^2}{w^2} - \frac{\beta^2}{w} = \alpha^2 - \beta^2
,
but I have no idea what should I do about the first and the second derivative. Any ideas?

Last edited: May 7, 2016
2. May 8, 2016

### Megaquark

If w is going to 1, then you don't have to worry about the derivatives because the coefficients both go to zero.

3. May 8, 2016

### mixo1234

Yes that would be true if we where guaranteed that both derivatives are not infinitely large.. or are we :D I have no idea what should i do at this point

4. May 8, 2016

### Megaquark

I don't know how you'd get an infinitely large derivative at a point without the wavefunction being asymptotically large. Ψ*Ψ has to be a positive number less than 1. Don't know if you could normalize a wavefunction with an infinite derivative at a point.