# Solving Schrodinger DE using Frobenius

1. May 3, 2012

### claret_n_blue

My DE is

$$\frac{h^2}{2m} \frac{d^2\psi}{dx^2} + \left(E - \frac{Ae^{-ax}}{x} \right) \psi = 0$$

where h, m, A < 0 and a and E are constants. I need to construct the following series solution (using the larger root of my indicial equation):

$$\psi(x) = a_0 \left[x + \frac{Am}{h^2}x^2 + \frac{1}{6} \left( \frac{2A^2m^2}{h^4} - \frac{2Em}{h^2} - \frac{2Aam}{h^2}\right)x^3+... \right]$$

at x = 0.

So far, I've managed to do a load of work and I've got:

$$\left( \frac{h^2}{2m}(k^2 - k)b_0\right)x^{k - 2} + \left(\frac{h^2}{2m}(k^2 +k) b_0 - Ab_1 \right)x^{k -1} + \sum_{\lambda = 2}^{\infty} \left[ \frac{h^2}{2m}(\lambda + k - 1)(\lambda + k)b_{\lambda} -A b_{\lambda - 1} + (E + Aa)b_{\lambda - 2} \right] x^{\lambda +k - 2} = 0$$

Which I'm pretty sure is correct, however I am stuck as to how to solve it from here. What do I do? I think the larger root of my indicial equation is 1. So subbing that into my recurrence relation gives me:

$$\sum_{\lambda = 0}^{\infty} \left[ \frac{h^2}{2m}(\lambda + k - 1)(\lambda + k)b_{\lambda + 2} -A b_{\lambda +1} + (E - Aa)b_{\lambda} \right] x^{\lambda +k} = 0$$

$$k = 1 : \frac{h^2}{2m}(\lambda + (1) - 1)(\lambda + (1))b_{\lambda + 2} -A b_{\lambda +1} + (E - Aa)b_{\lambda } = 0$$

$$= \frac{h^2}{2m}(\lambda)(\lambda + 1)b_{\lambda+2} -A b_{\lambda + 1} + (E - Aa)b_{\lambda } = 0$$

Is that bit correct? What do I do from there. I can't solve it with the three "unknowns" in $$b_{\lambda}, b_{\lambda + 1}$$ and $$b_{\lambda + 2}$$