So from my understanding, as long as $E>0$ you will have scattering states and these scattering states will always result in an imaginary $\psi$, but bound states can also have an imaginary $\psi$? Is this correct and or is there a better way of looking at this maybe more conceptually?
I noticed in a footnote in Griffiths Intro to QM 2nd ed. in the finite square well section he mentions that he wrote $\psi$ in terms of sin/cos instead of exponentials because we know the solutions will be even or odd because the potential is symmetric. But he also does it for the infinite square well when it's not really symmetric (at least with respect to the y-axis) and doesn't do it for the delta potential. How come he doesn't do it for the delta potential, is it not really symmetric? and why does he do it for the finite square well? What is the benefit of "exploiting the even/odd solutions" and what does he mean by that?
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$\begingroup$ Usually Schrödinger Hamiltonians have only continuous spectrum for $E>0$ with associated scattering states but there are exceptions (Anderson localisation in random systems). Why should the eigenfunction ψ be imaginary in this case? One can always multiply with a constant phase factor rendering it real (in fact states are described by unit rays rather than vectors). As to your second point, such choices are usually made for convenience. $\endgroup$– UrgjeCommented Oct 14, 2014 at 10:13
1 Answer
I also don't quite understand what you mean by "imaginary $\psi$", but let me give you some general, more mathematically correct view.
In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we also usually require that the Hamiltonian be bounded from below, which implies that the spectrum, i.e. the energies in the system have a minimum value, a ground state.
Now, the spectrum is divided into three parts (by Radon-Nikodym), two of which will be very interesting to us: First, there is the point spectrum, the eigenvalues. This spectrum corresponds to real eigenvalues which have eigenvectors, i.e. normalizable states that take on this eigenvalue. Then, there is the absolutely continuous spectrum. These are NOT eigenvalues. They do NOT have an eigenfunction in the Hilbert space, instead you can only find an approximate eigenfunction or a "generalized eigenfunction", such as the Fourier waves, which cannot be normalized. The third part, the singularly continuous spectrum is somewhat different but we won't go into that here - often, you try to prove that for what you want to do, it doesn't exist anyway.
The crucial point is that the RAGE theorem tells you what your physical intuition already told you before: The point spectrum (the eigenvalues) corresponds to bound states, the absolutely continuous spectrum corresponds to scattering states. This has nothing to do with $E>0$, although if you consider a potential that is zero at infinity, this is precisely what you expect: For energies greater than the potential, the "wave" should "escape", i.e. scatter. It's also not true that the bound states cannot occur at $E>0$, this is also entirely up to the nature of the potential and how it behaves at infinity.
The second part of your question is already answered in the comment by Urgje: If you know that your solution must have a certain symmetry, exploiting this usually makes your calculations simpler. Try doing the same calculations without exploiting the symmetry, you'll end up at the same result and just take more time.
