Solve Modified Poschl-Teller Homework Statement

[dingo_d]

Homework Statement

I have modified Poschl-Teller potential:

$$V(x)=\frac{-U_0}{\cosh^2(\alpha x)}$$. I need to expand that into Taylor and find the ground and first energy state, when keeping only quadratic terms in the potential.

Homework Equations

Schrödinger eq obviously...

The Attempt at a Solution

So I have expanded the potential, and taken only quadratic term the thing looks like:

$$V(x)=-U_0(1-\alpha^2x^2)$$

Putting that into Schrödinger and into Mathematica with the substitution that $$k^2=\frac{2mE}{\hbar^2}$$, after giving the Mathematica to solve that one (looks quite nasty) I get this:

$$\left\{\left\{\psi(x)= c_1 D_{-\frac{-k^2 \hbar ^2+\sqrt{2} \sqrt{m} \sqrt{U_0} \alpha \hbar -2 m U_0}{2 \sqrt{2} \sqrt{m}<br /> \sqrt{U_0} \alpha \hbar }}\left(\frac{2^{3/4} \sqrt[4]{m} \sqrt[4]{U_0} x \sqrt{\alpha }}{\sqrt{\hbar }}\right)+c_2<br /> D_{\frac{-\sqrt{2} k^2 \hbar ^2-2 \sqrt{m} \sqrt{U_0} \alpha \hbar -2 \sqrt{2} m U_0}{4 \sqrt{m} \sqrt{U_0} \alpha \hbar<br /> }}\left(\frac{i 2^{3/4} \sqrt[4]{m} \sqrt[4]{U_0} x \sqrt{\alpha }}{\sqrt{\hbar }}\right)\right\}\right\}$$

Where $$D_\nu(x)$$ is parabolic cylindric function. (Eeeek! XD)

Now, the problem is finding the energy. Is there any 'easy' way to find it? Because I'd have to solve this with Frobenius or series solution, and find out where to 'cut' the recursion so that it won't diverge.

Thanks :)EDIT: Never mind, you can delete this thread, I figured it out... It's just harmonic oscillator duh! ...

 

[dextercioby]

Threads are not getting deleted. So yes, the approximate potential is that of a linear oscillator whose solution for energy spectrum can be found in at least 100 different books.

About the parabolic cylindric function part, perhaps you put in the ODE the initial sech^2 x, because, if you had put the quadratic approximation, Mathematica would have returned you a solution in terms of Hermite polynomials, I guess...

 

[dingo_d]

Actually when I put the harmonic oscillator equation in Mathematica it also gave Parabolic cylindric functions as a solution (kinda weird). Now, I found out that for non negative whole number the solutions (parab. cylind. func.) turn to Hermite polynomial, so that's ok I guess.

Mathematica probably gave the most general solution.

 

[dextercioby]

Interesting. Because the confluent hypergeometric functions are the ones comprising as particular cases most of the known special functions which appear in connection with mathematical physics. But if Mathematica returns the parabl.cylind.functions, then it must be using some non-obvious algorithm.

Anyway, topic closed.