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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.

Danu
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Jiang-min Zhang
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  • 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. – Ultima Aug 26 '15 at 09:40
  • In your last sentence, I assume you're referencing the Abel transform? – Danu Aug 26 '15 at 09:44
  • @Ultima Perhps expand that comment into an answer? – Danu Aug 26 '15 at 09:45
  • 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. – Norbert Schuch Aug 26 '15 at 10:10
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    It should be noted that the best you can expect is to determine the potential up to symmetries, i.e., displacement and reflection. – Norbert Schuch Aug 26 '15 at 10:19
  • @Danu I do not know the terminology. But yes, it is some integral transform. – Jiang-min Zhang Aug 26 '15 at 10:52
  • I would tend to say "yes" because the Shrodinger equation managed to get the Balmer series which was an experimental observation first. https://en.wikipedia.org/wiki/Theoretical_and_experimental_justification_for_the_Schr%C3%B6dinger_equation#Hydrogen_atom . At least it precludes a "no" – anna v Aug 26 '15 at 11:16
  • @annav the answer is no...see my answer below ;-) – yuggib Aug 26 '15 at 11:53
  • @yuggib that is mathematics predicting about all data. I am pointing out that data led to the initial mathematics vindication of the QM model with the hydrogen atom. Your "no" is qualified by the existence of this opposite way. So it may not be a mathematical theorem, but the possibility existed, fortunately for physics. – anna v Aug 26 '15 at 13:56
  • @annav but I do not think this was exactly a "reverse problem", but more a theoretical confirmation of an experimental evidence ;-) – yuggib Aug 26 '15 at 16:04
  • @yuggib They were looking for a mathematical model that would explain/have_as_solution the Balmer series rigorously,without the assumptions of the Bohr model. in my view this is reverse. – anna v Aug 26 '15 at 16:50

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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 $\mathbb{R}^+$.

yuggib
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  • Would anything change if one restricts to potentials with only discrete spectrum? – Norbert Schuch Aug 26 '15 at 11:38
  • @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). – yuggib Aug 26 '15 at 11:50
  • 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. – Norbert Schuch Aug 26 '15 at 11:57
  • @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 ;-) – yuggib Aug 26 '15 at 11:59