# Linear operator

1. Sep 23, 2006

### lokofer

let be the linear operator: (Hermitian ??)

$$L = -i(x\frac{d}{dx}+1/2)$$

then the "eigenfunctions" are $$y_{n} (x)=Ax^{i\lambda _{n} -1/2$$

then my question is how would we get the energies imposing boundary conditions? (for example y(0)=Y(L)=0 wher L is a positive integer )....:rofl: :rofl:

2. Sep 25, 2006

### HallsofIvy

Staff Emeritus
Energies? That is certainly not a Hamiltonian. In any case, since it is a first order operator, you can't satisfy two boundary conditions, only one.