Arriving at the Quantum Mechanial Potential From The Energy Eigenvalues

Question

In Quantum Mechanics, we know that given a potential we can solve the eigen value problem to find out the energy eigen values and eigen functions. Now suppose in an experiment we have information only about the energy levels of a system and not the potential. Can we do the reverse and find out the nature of the Potential of the System uniquely?

Answer 1

No. There are isospectral potentials that have the same eigenvalues, at least in the one- and two-dimensional cases. You can generate them with a "Darboux transformation". Here's how it works in one-dimension:

Start with the Schrodinger equation

$$-\frac{\hbar^2}{2m}\Psi''+V\Psi=E\Psi\tag{1}.$$

We'd like to find another potential $\tilde{V}$ with wavefunctions $\tilde{\Psi}$ that satisfy

$$-\frac{\hbar^2}{2m}\tilde{\Psi}''+\tilde{V}\tilde{\Psi}=E\tilde{\Psi}\tag{2}$$

where $E$ is the same in both equations.

Make the ansatz

$$\tilde{\Psi}=\Psi'+A\Psi\tag{3}$$

where $A$ is a function of $x$ to be determined.

Substitute (3) into (2), and reduce the $\Psi''$ and $\Psi'''$ that appear using (1). The result is an equation in which only $\Psi$ and $\Psi'$ appear:

$$\left[\frac{\hbar^2}{m}A'+(V-\tilde{V})\right]\Psi'+\left[\frac{\hbar^2}{2m}A''+(V-\tilde{V})\right]\Psi=0\tag{4}.$$

We can satisfy this equation by choosing $A$ and $\tilde{V}$ to make the coefficient of both terms zero:

$$\frac{\hbar^2}{m}A'+(V-\tilde{V})=0\tag{5a}$$ $$\frac{\hbar^2}{2m}A''+V'+A(V-\tilde{V})=0\tag{5b}.$$

To solve these, first eliminate $V-\tilde{V}$ to get

$$\frac{\hbar^2}{2m}(A''-2AA')+V'=0\tag{6}.$$

The left side is a total derivative, so we can integrate it to get

$$\frac{\hbar^2}{2m}(A'-A^2)+V=C\tag{7}$$

where $C$ is a constant of integration.

This is a Riccati equation that can be linearized by setting

$$A=-(\log{\zeta})'\tag{8}$$

where $\zeta$ is an unknown function of $x$ to get

$$-\frac{\hbar^2}{2m}\zeta''+V\zeta=C\zeta\tag{9}$$.

This is simply Schrodinger equation (1) with potential $V$ and energy $C$!

So, take any eigenstate $\zeta$ of potential $V$, satisfying (1). Then, if $\Psi$ is an eigenstate of $V$ with energy $E$,

$$\tilde{\Psi}=\Psi'-(\log{\zeta})'\Psi\tag{10}$$

is an eigenstate of a different potential,

$$\tilde{V}=V-\frac{\hbar^2}{m}(\log{\zeta})''\tag{11},$$

with exactly the same energy $E$. You can get a whole set of isospectral potentials by taking $\zeta$ in the Darboux transformation to be different eigenstates.

This explanation was adapted from a paper which also addresses the two-dimensional case.