# A question regarding the Power series method of solving the QHO

1. Mar 18, 2013

### klawlor419

I am working through the Griffiths QM text and I am getting caught up on some the process he uses to derive the wave functions and energy levels for the QHO, via Frobenius/Power series method.

I understand that the Schrodinger equation get recast in to a summation form over the coefficients of the power series expansion. I understand that each of the terms in the summation, which represents the original Schrodinger equation for the QHO, must be equal to zero. This is the replacement for the differential equation. Once this is realized the recursion formula for all of the coefficients of the wave function is produced.

Griffiths breaks this down into odd and even functions for the values of the index j. Discusses the divergence of the series, which was similar to asymptotic expansion he introduced at the beginning of the section. He then explains that the way to make the series convergent is by requiring the series to terminate at j=n. This imposes the condition on E that solves for the energy levels of the QHO.

What I am confused by is why, when I set n=0, I must set the first odd coefficient to zero. Does it have to due with the physicality of the wave function? On the same note, I am confused with larger values of n and this same sort of thing. Lets say that I choose the n=2 state, so that the series will be cutoff at n=2, if I choose the first odd-coefficient to be zero wouldn't I loose some portion of the wave function that could be important? How do we know that if n is even the odd terms must disappear and vice-versa?

I realize that the wave functions for the QHO should be alternatively odd and even functions. i am missing something. Thanks ahead of time for whatever help is offered.

2. Mar 18, 2013

### Bill_K

You get a recursion formula that relates coefficients cm and cm+2, which means it connects all the odd powers to each other, and all the even powers to each other. The first two coefficients c0 and c1 are completely arbitrary. So schematically you get a solution

y = c0 x (..even power series..) + c1 x (..odd power series..)

Now pick a value of E that makes the even power series terminate. However with this value, the odd power series does NOT terminate, so you must set c1 = 0.

Or... you can pick a value of E that makes the odd power series terminate. Then the even series will not terminate and you must set c0 = 0.

3. Mar 18, 2013

### klawlor419

OK I understand now. I was almost there in my understanding but slightly missing something. Thanks for the help!