- #1
dingo_d
- 211
- 0
Homework Statement
I have modified Poschl-Teller potential:
[tex]V(x)=\frac{-U_0}{\cosh^2(\alpha x)}[/tex]. 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
Schrodinger eq obviously...
The Attempt at a Solution
So I have expanded the potential, and taken only quadratic term the thing looks like:
[tex]V(x)=-U_0(1-\alpha^2x^2)[/tex]
Putting that into Schrodinger and into Mathematica with the substitution that [tex]k^2=\frac{2mE}{\hbar^2}[/tex], after giving the Mathematica to solve that one (looks quite nasty) I get this:
[tex]\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}
\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
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
}}\left(\frac{i 2^{3/4} \sqrt[4]{m} \sqrt[4]{U_0} x \sqrt{\alpha }}{\sqrt{\hbar }}\right)\right\}\right\}[/tex]
Where [tex]D_\nu(x)[/tex] 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! ...
Last edited: