# Solutions of the quantum SHO

1. Aug 12, 2007

### ehrenfest

The even solutions of an SHO are:

$$h^+(y) = \sum_{s = 0}^{\infty}a_s y^{2s}$$

where a is given by the recursion

$$a_{s+1} = a_s \left( \frac{4s + 1 - \epsilon}{2(s+1)(2s+1)} \right)$$

The solutions are square integrable iff

a_n = 0 so that the polynomial is finite.

What I do not understand is why my book (Robinett) says

h_0(y) = 1 and not 0 when a_0 = 0 for n = 0?

2. Aug 12, 2007

### dextercioby

Who's "n" and what does it have to do with what you wrote ?

3. Aug 12, 2007

### ehrenfest

So, there should be one solution for every n because

a_n = 0 --> epsilon = (4n +1).

So, I guess n is the degree (maybe +/- 1 of the polynomial). It is also the energy level.