- #1
danny271828
- 34
- 0
Homework Statement
Find the eigenfunctions and eigenvalues for the operator:
a = x + [tex]\frac{d}{dx}[/tex]
2. The attempt at a solution
a = x + [tex]\frac{d}{dx}[/tex]
a[tex]\Psi[/tex] = [tex]\lambda[/tex][tex]\Psi[/tex]
x[tex]\Psi[/tex] + [tex]\frac{d\Psi}{dx}[/tex] = [tex]\lambda[/tex][tex]\Psi[/tex]
x + [tex]\frac{1}{\Psi} [/tex][tex]\frac{d\Psi}{dx}[/tex] = [tex]\lambda[/tex]
x + [tex]\frac{d}{dx}[/tex] ln([tex]\Psi[/tex])= [tex]\lambda[/tex]
[tex]\frac{1}{2}[/tex]x[tex]^{2}[/tex] + ln([tex]\Psi[/tex]) = [tex]\lambda[/tex]x +c
[tex]\Psi[/tex] = e^(-[tex]\frac{1}{2}[/tex]x[tex]^{2}[/tex]+[tex]\lambda[/tex]x+c)
[tex]\Psi[/tex] = e^(-[tex]\frac{1}{2}[/tex]x[tex]^{2}[/tex]+[tex]\lambda[/tex])[tex]\Psi[/tex](0)
Not sure from here... I think I plug into initial equation?
a[tex]\Psi[/tex] = [tex]\lambda[/tex][tex]\Psi[/tex]
substituting and using chain rule I obtain...
x[tex]\Psi[/tex] + ([tex]\lambda[/tex]-x)[tex]\Psi[/tex] = [tex]\lambda[/tex][tex]\Psi[/tex]
so x + ([tex]\lambda[/tex] - x) = [tex]\lambda[/tex]
so [tex]\lambda[/tex] = [tex]\lambda[/tex] nope I guess I don't do that... good check though, so I guess this is the correct way to solve for eigenfunction... Can someone help me with finding [tex]\lambda[/tex]?