Knowing the potential, we can find the spectrum of the Schrödinger operator. The converse question is: Knowing the spectrum, can we reconstruct the potential? As an example, a harmonic potential has an equally spaced spectrum. But is the converse true?

This is, of course, similar to the 'hearing the shape of the drum' problem, which has a negative answer. But we also should notice that in classical mechanics, if the potential is symmetric, we can recover it from the oscillation period as a function of the energy of the particle. This is due to ingenious work by Abel.

  • $\begingroup$ No. Consider an $N$-dimensional vector space with an operator $A$ on that space. Given a basis of eigenvectors, one knows that $A$ is diagonal w.r.t. this basis, but the elements along the diagonal (the eigenvalues) are undetermined. $\endgroup$ – Ultima Aug 26 '15 at 9:40
  • $\begingroup$ In your last sentence, I assume you're referencing the Abel transform? $\endgroup$ – Danu Aug 26 '15 at 9:44
  • $\begingroup$ @Ultima Perhps expand that comment into an answer? $\endgroup$ – Danu Aug 26 '15 at 9:45
  • $\begingroup$ This bears some resemblence with questions appearing in DFT, such as the first Hohenberg-Kohn theorem which states that the potential is uniquely determined by the ground state density. $\endgroup$ – Norbert Schuch Aug 26 '15 at 10:10
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    $\begingroup$ It should be noted that the best you can expect is to determine the potential up to symmetries, i.e., displacement and reflection. $\endgroup$ – Norbert Schuch Aug 26 '15 at 10:19

The answer is no, I am afraid. As you may well know, the self-adjoint Laplace operator $-\Delta$ on $L^2(\mathbb{R})$ has purely absolutely continuous spectrum $\mathbb{R}^+$.

Now let $V\in L^{\infty}(\mathbb{R},\mathbb{R}^+)$ be an arbitrary bounded positive function. Then $-\Delta_x +V(x)$, where $V$ acts as a multiplicative operator is self-adjoint and has spectrum $\mathbb{R}^+$.

  • $\begingroup$ Would anything change if one restricts to potentials with only discrete spectrum? $\endgroup$ – Norbert Schuch Aug 26 '15 at 11:38
  • $\begingroup$ @NorbertSchuch not as long as the operator is bounded (and positive). You may have embedded eigenvalues in the continuous spectrum (not sure with in 1D, but for example $-\Delta_x -\Delta_y +y^2$ has the harmonic eigenvalues (of $y$, roughly speaking) embedded in the continuous spectrum, but the overall spectrum remains $\mathbb{R}^+$ nevertheless). $\endgroup$ – yuggib Aug 26 '15 at 11:50
  • $\begingroup$ I was thinking of unbounded potentials which only have point spectrum, such as e.g. the harmonic oscillator. -- But the post linked above by Danu has an example of a potential which reproduces an equally spaced spectrum but is not harmonic, so the answer is indeed still no. $\endgroup$ – Norbert Schuch Aug 26 '15 at 11:57
  • $\begingroup$ @NorbertSchuch With the harmonic example you see that $-\Delta+V$ yields a purely discrete spectrum if $(-\Delta +V -i\lambda)^{-1}$, $\lambda\in\mathbb{R}$, (the resolvent) is compact. The operator $x^2$ by itself (harmonic potential) has not discrete spectrum, but purely continuous (it is the position operator squared). Anyways, you can reproduce also a discrete spectrum with a different operators as you already noted ;-) $\endgroup$ – yuggib Aug 26 '15 at 11:59

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