1. May 6, 2007

### neelakash

1. The problem statement, all variables and given/known data

It is possibly not a homework problem.However,to do a homework problem,I require this:
Boas writes the effect of Ladder operators on y_n that satisfies
y"_n-x^2y_n=-(2n+1)y_n,n=0,1,2,3...

(D-x)(D+x)y_n=-2ny_n
(D+x)(D-x)y_n=-2(n+1)y_n

Then,she proved y_(m-1)=(D+x)y_m and the other raising operator eqn.

So far there is no problem...

Now she says,if n=0,we find a solution of (D-x)(D+x)y_n=-2ny_n by requiring (D+x)y_0=0

My question is if n=0,we have (D-x)(D+x)y_0=0.
Does that mean (D+x)y_0=0 necessarily?

2. Relevant equations

3. The attempt at a solution

treating (d+x)y_0=t,I saw that we have (D-x)t=0 or,t=c exp[x^2/2]
So,they are treating c=0?...why?

2. May 15, 2007

### maverick280857

The last point (if you take c = 0 or t = 0) you get

$$\frac{dy}{dx} + xy = 0$$

which gives

$$y(x) = Ae^{-x^{2}/2}$$

But what you're asking is the converse:

for n = 0 do we always have

$$(D-x)(D+x)y = y \implies (D+x)y = 0$$