Raising Operator (Harmonic Oscillator)

[heardie]

This is (another!) question I cannot solve
The ground state wavefunction for the harmonic oscillator can be written as
$$\chi _0 = \left( {\frac{\alpha } {\pi }} \right)^{\frac{1} {4}} \exp \left( {\frac{{ - \alpha x^2 }} {2}} \right)$$

where $$\alpha = \sqrt {\frac{{mk}} {{\hbar ^2 }}}$$

We are then asked to apply the rasing operator:
d/dy-y

where $$y = \sqrt \alpha x$$
First of all, does this mean replace all alphas, and x's with y's? (Is it even possible to get rid of all x's? I end up with a y/x in there), and then apply
d(chi)/dy-y*chi
Is that what the operator is doing? I don't see the significane of the answer. If I apply the lowering operator (d/dy + y) I still get 'an' answer, when I figure I should get something to tell me I can't go lower then the groud state (a zero perhaps, or a mathematical impossibility like division by zero)

Anyway...if anyone can shed some light on this it would eb much appreciated!

Edit: I can't figure out division on this board! I swear to god I am doing it right...I will post my operator down here...if someone can point out the latex error I'd love to know:
$$\[ \frac{d} {{dy}} - y \]$$

 

[heardie]

I swear my equations are printing in the wrong order above. For reference:
Raising operator - d/dy - y
where y = sqrt(alpha)*x

Oh and I get
$$-1/2\,{\frac {{e^{-1/2\,\alpha\,{x}^{2}}} \left( -1+4\,\alpha\,{x}^{2} \right) }{x\sqrt [4]{\alpha\,\pi }}}$$

Is this the first excited state? Pretty sure I would have made an error somewhere there!

 

[heardie]

Dont worry. Completly missed something here. All makes sesnse now. Can I delete this thread somehow?