Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A typical problem in quantum mechanics is to calculate the spectrum that corresponds to a given potential.

  1. Is there a one to one correspondence between the potential and its spectrum?
  2. If the answer to the previous question is yes, then given the spectrum, is there a systematic way to calculate the corresponding potential?
share|improve this question

4 Answers 4

up vote 25 down vote accepted

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". The following extensive exposition by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this information is not complete even for systems as simple as quantum mechanics along a finite interval.

share|improve this answer
Do you know an example of two potentials with the same energy spectrum? –  Mark Eichenlaub Aug 13 '11 at 15:41
@David: Everything is simpler nowadays: you do not know the whole spectrum but your potential (or Theory Of your Everything, whatever) describes everything without fail. –  Vladimir Kalitvianski Aug 13 '11 at 16:37
@Mark, please see the general method described here: redshift.vif.com/JournalFiles/V09NO3PDF/V09N3LOP.pdf –  David Bar Moshe Aug 13 '11 at 17:05
@David Bar Moshe Very interesting article, it is the first time I hear the terminology "isospectral potential". But I am also very concerned about the quality of the journal. Apeiron, anybody?!! –  Revo Aug 13 '11 at 19:56

The Harmonic oscillator has the same spectrum as a weaker harmonic oscillator with a hard wall at x=0.

LATER EDIT: I see that I have to be more explicit--- the potentials

  • $V(x)= 2x^2 - 2$
  • $(x>0)$ $V(x)= x^2 - 3$ and $(x<0)$ $V(x)= \infty$

have the exact same spectrum.

share|improve this answer
But in that case you get only half the spectrum of the usual harmonic oscillator, don't you? –  Revo Aug 13 '11 at 20:10
@Revo is correct. E.g., the energy spectrum of the full harmonic oscillator is (1,3,5,7,...) all times $\hbar \omega_{full}/2$ and for the half harmonic oscillator is (3,7,11,15,...) times $\hbar \omega_{half}/2$. Just by changing $\omega_{half}$ by something independent of $n$, you can't get the terms to match up (e.g., for $n=0$; you'd have to let $3\omega_{full}=\omega_{half}$, for $n=1$ let $(7/3) \omega_{full}=\omega_{half}$, for $n=2$ let $(11/5)\omega_{full}=\omega_{half}$). –  dr jimbob Aug 13 '11 at 20:51
@dr jimbob True, since the full harmonic oscillator solutions oscillate between even and odd, half the spectrum will not satisfy the boundary conditions at the origin of the harmonic oscillator potential with a hard wall at x=0, namely the wave function must vanish there. –  Revo Aug 13 '11 at 21:09
@Everybody: you can add a constant to a potential. It's still a potential. What is this discussion? –  Ron Maimon Aug 14 '11 at 1:36
@Marty--- the question is not physics, it is mathematics--- can you find Schrodinger operators with identical spectrum. This is a trivial example. More trivial examples are any potential and its translates and reflections, and a less trivial example is any potential and its supersymmetric conjugate, when the ground state doesn't break supersymmetry. –  Ron Maimon Aug 14 '11 at 1:53

In this answer we will only consider the leading semi-classical approximation of a $1$-dimensional problem with Hamiltonian

$$ H(x,p) ~=~ \frac{p^2}{2m}+ \Phi(x), $$

where $\Phi$ is a potential. Semi-classically, the number of states $N(E)$ below energy-level $E$ is given by the area of phase space that is classically accessible, divided by Planck's constant $h$,

$$ N(E) ~\approx~ \iint_{H(x,p)\leq E} \frac{dx~dp}{h}. \qquad (1)$$

[Here we ignore the Maslov index, also known as the metaplectic correction, which e.g. yields the zero-point energy in the simple harmonic oscillator(SHO) spectrum.] Let

$$ V_0~:=~ \inf_{x\in\mathbb{R}} ~\Phi(x) $$

be the infimum of the potential energy. Let

$$\ell(V)~:=~\lambda(\{x\in\mathbb{R} \mid \Phi(x) \leq V\}) $$

be the length of the classically accessible position region at potential energy-level $V$. [Technically, the length $\ell(V)$ is the Lebesgue measure $\lambda$ of the preimage

$$\Phi^{-1}(]-\infty,V])~:=~ \{x\in\mathbb{R} \mid \Phi(x) \leq V\},$$

which does not necessarily have to be a connected interval.]

Example 1: If the potential $\Phi(x)=\Phi(-x)$ is an even function and is strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=2\Phi^{-1}(V)$ is twice the positive inverse branch of $\Phi$.

Example 2: If the potential has a hard wall $\Phi(x)=+\infty$ for $x<0$, and is strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=\Phi^{-1}(V)$ is the positive inverse branch of $\Phi$.

Example 3: If the potential $\Phi(x)$ is strongly monotonically decreasing for $x\leq0$ and strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=\Phi_{+}^{-1}(V)-\Phi_{-}^{-1}(V)$ is the difference of the two inverse branch of $\Phi$.

In Example 1 and 2, if we would be able to determine the accessible length function $\ell(V)$, then we would also be able to generate the corresponding potential $\Phi(x)$ as OP asks.

The main claim is that we can reconstruct the accessible length $\ell(V)$ from $N(E)$, and vice-versa. $$N(E) ~\approx ~\frac{\sqrt{2m}}{h} \int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}},\qquad (2) $$ $$ \ell(V) ~\approx ~\hbar\sqrt{\frac{2}{m}} \frac{d}{dV}\int_{V_{0}}^V \frac{N(E)~dE}{\sqrt{V-E}}.\qquad (3) $$

[The $\approx$ signs are to remind us of the semi-classical approximation (1) we made. The formulas can be written in terms of fractional derivatives as Jose Garcia points out in his answer.]

Proof of eq.(2):

$$ h ~N(E) ~\stackrel{(1)}{\approx}~ 2\int_0^{\sqrt{2m(E-V_0)}} \left. \ell(V) \right|_{V=E-\frac{p^2}{2m}}~dp$$ $$~\stackrel{V=E-\frac{p^2}{2m}}{=}~2\int_{V_0}^E \frac{\ell(V)~dV}{v}~=~\sqrt{2m}\int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}}, $$

because $dV~=~ - v~dp$ with speed $v~:=~\frac{p}{m}~=~\sqrt{\frac{2(E-V)}{m}}$.

Proof of eq.(3): Notice that

$$ \int_{V^{\prime}}^V \frac{dE}{\sqrt{(V-E)(E-V^{\prime})}} ~\stackrel{E=V \sin^2\theta + V^{\prime} \cos^2\theta }{=}~ 2 \int_0^{\frac{\pi}{2}} d\theta ~=~ \pi.\qquad (4) $$


$$\frac{h}{\sqrt{2m}}\int_{V_0}^V \frac{N(E)~dE}{\sqrt{V-E}} ~\stackrel{(2)}{\approx}~ \int_{V_0}^{V}\frac{dE}{\sqrt{V-E}}\int_{V_0}^{E} \frac{\ell(V^{\prime})~dV^{\prime}}{\sqrt{E-V^{\prime}}} $$ $$~\stackrel{{\rm Fubini}}{=}~\int_{V_0}^V \ell(V^{\prime})~dV^{\prime}\int_{V^{\prime}}^V \frac{dE}{\sqrt{(V-E)(E-V^{\prime})}} ~\stackrel{(4)}{=}~ \pi \int_{V_0}^V \ell(V^{\prime})~dV^{\prime},\qquad (5)$$

where we rely on Fubini's Theorem to change the order of integrations. Finally, differentiation wrt. $V$ on both sides of eq. (5) yields eq. (3).

share|improve this answer
This is repeating Jose's answer, again without explicitly stating that you are assuming that V(x)=V(-x) or a hard wall at negative x or some other way to remove the arbitrariness due to reflection symmetry. –  Ron Maimon Oct 30 '11 at 1:23
No, I have divided the reconstruction(v1) of the potential $\Phi(x)$ into two parts: i) Firstly, calculating the accessible length function $\ell(V)$ from $N(E)$ via formula (3). It is remarkable that this is always possible, and this is the main claim. ii) Secondly, calculating the potential $\Phi(x)$ from $\ell(V)$, which is only possible under additional assumptions. The even case $\Phi(x)=\Phi(-x)$ is explicitly mentioned as an example. –  Qmechanic Oct 30 '11 at 11:51
You are right, +1, I read it too quickly. It is an interesting answer, thanks. –  Ron Maimon Oct 30 '11 at 20:27
another question.. once we have constructed the potential from $ N(E) $ can we prove that the Hamiltonian $ H=p^{2} + V(x) $ will satisfy the Gutzwiller Trace formula ? . I mean if Gutzwiller trace formula and the semiclassical reconstruction of this potential are seemingly related or not :) –  Jose Javier Garcia Apr 9 '12 at 10:08
another comment, although $ N(E) $ is always postive, and if we conside even potential $ V(x)=V(-x) $, at least semiclassically is it possible to prove that $ V(x) $ will be positive if we know that $ l(V) $ is positive ??, thanks. –  Jose Javier Garcia Oct 7 '12 at 8:18

yes , at least for one dimension you can obtaine the inverse problem

let be N(E) the eigenvalue staircase then in the WKB approximation the inverse of the potential is given by

$ V^{-1} (x) = 2 \sqrt \pi \frac{d^{1/2}}{dx^{1/2}}N(x) $

so we can obtain the inverse (and hence the potential from the eignevalue staircase)

share|improve this answer
This would only work for a potential which is smooth and obeys $V(x)=V(-x)$. Then N(E) determines the potential to WKB order. –  Ron Maimon Oct 1 '11 at 16:12
this works also whenever the potential is infinite for x <0 (infinite wall at x=0) you can check for the 'bouncer' V=x or similar –  Jose Javier Garcia Oct 2 '11 at 8:34
Could you give any references Pls? –  Revo Oct 7 '11 at 11:57
you can search 'Wu and Sprung potential' Riemann Hypothesis as an inverse problem is treated there :) –  Jose Javier Garcia Oct 9 '11 at 12:40
i believe that if the length function $ l(x) $ is increasing you can get the potential... for other cases you can always reflect the function 'l' trough the line $ y=x$ i used this spectral problem to find a suitable hamiltonian for the Riemann Hypothesis in the WKB case –  Jose Javier Garcia Nov 10 '11 at 12:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.