# Can I determine the potential term in the Schrödinger equation based on the eigenvalues? [duplicate]

Let's imagine I knew a certain system could be described by a one-dimensional Schroedinger equation. I know the mass/momentum term, but not the shape of the potential. Further for some reason I know all energy eigenvalues. Could I determine the corresponding potential?

How does this generalize to more than one dimensions?

• – Javier Mar 13 '17 at 23:59
• Gutierrez 2014 . For applications to 3d Quarkonium reflection less symmetric potentials see my comment to the answer below. – Cosmas Zachos Mar 14 '17 at 0:09

There are distinct potentials that have the same eigenvalues but one. They are called isospectral. Supersymmetric quantum mechanics is based on this. For instance, the infinite well potential $$V_1(x)=\left\{\begin{array}{cc} 0 & \quad\hbox{if }\ 0\le x \le L \\ \infty &\hbox{otherwise}\, ,\end{array}\right.$$ shares all eigenvalues with $$V_2(x)=\frac{\hbar^2\pi^2}{2mL^2}\left(2\csc^2(\pi x/L)-1\right)$$ except the ground state eigenvalue $E_0^{1}=\hbar^2\pi^2/(2mL^2)$. Indeed, more generally $E_k^{1}=E_{k-1}^2$ for $k=1,2,\ldots$, i.e. the first excited state of $V_1(x)$ is the ground state of $V_2(x)$. (The eigenfunctions are quite different for the two potentials.)