# Protocol for solving time independent Schrodinger equation

Just a short question about the protocol for solving the time-independent Schrodinger equation for different potentials and the reasons for accepting and rejecting solutions.

Take for example the inverted step potential $$V(x) = -\alpha \delta(x).$$ For bound state $E < 0$, we check for normalizable (physically realizable solutions) solutions and reject the others. For the scattering case $E > 0$, none of the solutions to the time independent Schrodinger equation are normalizable so we just work with the reflection coefficient and transmission coefficient.

But for the free particle $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int^{+\infty}_{-\infty}\phi(k)e^{i(kx-\frac{h k^2}{2m}t)}dk$$ we only have scattered states as solutions for the time independent Schrodinger equation, but we can form a general solution above which could be normalizable for appropriate choice of $\phi(k)$.

My queries are basically:

• Why do we reject the non-normalizable solutions of bound states if as in the case of the free particle, the general solution could still be normalizable?
• Also, could we form a normalizable general solution for the scattered states $(E >0)$ of the inverted delta function as we can for the free-particle?

Thanks.

## 1 Answer

1. We reject non-normalizable bound states because they are unphysical. A bound state is (by one definition) simply an eigenstate belonging to the discrete spectrum of the Hamiltonian. The discrete part of the spectrum always has associated eigenstates within the Hilbert space (that is, within the space of normalizable wave functions). The non-normalizable "solutions" have nothing to do with the Hamiltonian operator on our Hilbert space of wavefunction, they just belong to a same-looking differential operator on a larger function space.

2. On the other hand, a scattering state belongs to the continuous part of the spectrum - the physical picture being that a scattered state can have any energy as long as it is above the threshhold for being "captured". The rigourous formulation of where these "states" actually live (since they are not normalizable they are not, formally speaking, states - for one, they have no probability density associated to them) requires the notion of rigged Hilbert spaces. And, as a general fact, indeed you can form wavepackets $$\psi(x,t) = \int \mathrm{e}^{-\mathrm{i}Et}g(E)\psi_E(x)\mathrm{d}E$$ for compactly supported functions $g(E)$ which will be normalizable solutions to the time-dependent Schrödinger equation. For more on this, see also this question.

• Thanks for you response. Why is preference given to form a normalizable wave packet for the free particle and not the other scattered states such as the delta potential scattered state case. I have two recommended QM books (Griffiths and Zettili) which both form a normalizable wave packet for the free particle but not for other cases of scattered states (where they just anlayze reflection and transmission coefficients) such as the inverted delta potential as I described above. Do you know why this is? – user100411 Jun 17 '16 at 11:41
• @JohnDoe: The scattered non-normalizable state is easy to analyse for its behaviour. Due to linearity of the integral, it's straightforward (but potentially tedious) to deduce the behaviour of wavepackets, and for wavepackets where $g(E)$ is narrowly peaking around one $E_0$, the behaviour of the scattered state with $E_0$ is a good approximation, anyway. – ACuriousMind Jun 17 '16 at 11:43