# Bound states and parity for a arbitrary potentials

If we are given an arbitrary potential, and we are asked to find bound states and parity, what would be usual strategy to do that?

Let's we have a potential given: $$-\frac{A}{y^2+a^2} -\frac{A}{(y-r)^2+a^2}$$

$$a$$ and $$r$$ are constants here. How to find the bound states and tell the parity? Here $$A$$ would have a range of values that we will get bound states.

The ground state energy would be negative.

My understanding of this kind of problem is:

Usually the energy has to be negative $$E<0$$. Therefore getting the eigen value equation would give a good idea what are the conditions of A that could make the potential bound states.

• What equation do you know that the system's states satisfy? – catalogue_number Aug 6 '19 at 6:07
• I have updated that in my question. – user193422 Aug 6 '19 at 6:10

If you want the states themselves, It's essentialy a question about solving PDEs - you're meant to solve the Schrödinger equation for that potential, i.e. $$\frac{-\hbar^2}{2m}\nabla^2 \psi + \left[\frac{-A}{y^2+a^2}+\frac{-A}{(y-r)^2+a^2} \right]\psi = E \psi$$ That particular choice of potential doesn't look very solvable analytically though. If you actually wanted the wave functions themselves, you'd need to use some numerical solver.
However, we can still extract useful information - such as "Is there a ground state", and "What is the parity of a given solution". First of all, the variational principle states that for any wavefunction $$|\phi\rangle$$, $$\langle \phi | H | \phi \rangle \ge E_0$$, where $$E_0$$ is the ground state energy. So if you find any old trial state $$\phi$$ with $$\langle \phi | H | \phi \rangle <0$$, you have a rigorous bound on the $$E_0$$ and a proof of the existence of a bound state.
For parity, suppose now that $$\psi$$ is an exact solution to the above PDE. Then, suppose we invert the parity: Remap $$\vec{r} \mapsto -\vec{r}$$. Then we end up with $$\frac{-\hbar^2}{2m}\nabla^2 \psi + \left[\frac{-A}{(-y)^2+a^2}+\frac{-A}{(-y+r)^2+a^2} \right]\psi = E \psi$$ Clearly, there is no change - i.e. any solution of the equation remains a solution under parity change. We can't really say any more about the parity of specific solutions without first finding them though.
• I do understand the parity operations. One question, what kind of trial wave function would give the ground state energy ? The potential seems like symmetric. DO you think A gaussian trial wave function like $\psi = A e^{-br^2}$ would work? – user193422 Aug 6 '19 at 6:49