# In quantum mechanics, given certain energy spectrum can one generate the corresponding potential?

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?
• Feb 7, 2019 at 12:00

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". An extensive exposition by Beals and Greiner (Anal. Appl. 7, 131 (2009); eprint) 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.

For additional details, see Apeiron 9 no. 3, 20 (2002), or also Phys. Rev. A 40, 6185 (1989), Phys. Rev. A 82, 022121 (2010), or Phys. Rev. A 55, 2580 (1997).

For a more experimental view, you can actually have particle-in-a-box problems with differently-shaped boxes in two dimensions that have the same spectra; this follows directly from the Gordon-Webb isospectral drums (Am. Sci. 84 no. 1, 46 (1996); jstor), and it was implemented by the Manoharan lab in Stanford (Science 319, 782 (2008); arXiv:0803.2328), to striking effect: 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.

• But in that case you get only half the spectrum of the usual harmonic oscillator, don't you?
– Revo
Aug 13, 2011 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 $(11/5)\omega_{full}=\omega_{half}$). Aug 13, 2011 at 20:51
• @Everybody: you can add a constant to a potential. It's still a potential. What is this discussion? Aug 14, 2011 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. Aug 14, 2011 at 1:53
• In other words, Ron Maimon's answer is essentially saying, that if we compare the full SHO $H=p^2/2m +m(x\omega)^2/2$ to the half SHO $H=p^2/2m + m(x\omega/2)^2/2 -(\hbar/2)(\omega/2)$, $x\geq 0$, with half frequency $\omega/2$, and with zero-point energy $(\hbar/2)(\omega/2)$ subtracted, then the two spectra are identical $E=\hbar(\omega/2)[(2n+1)+1/2]-(\hbar/2)(\omega/2)=\hbar\omega(n+1/2)$. Oct 5, 2011 at 14:40

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 $$h$$,

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

[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.] Assuming that the Hamiltonian operator is bounded from below, we get that

$$N(V_0)~=~0.$$

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. $$N(E) ~\approx ~\frac{\sqrt{2m}}{h} \int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}},\tag{2}$$ $$\ell(V) ~\approx ~\hbar\sqrt{\frac{2}{m}} \frac{d}{dV}\int_{V_{0}}^V \frac{N(E)~dE}{\sqrt{V-E}}.\tag{3}$$

[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):

\begin{align} 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\cr ~\stackrel{V=E-\frac{p^2}{2m}}{=}&~2\int_{V_0}^E \frac{\ell(V)~dV}{v}\cr ~=~~~&\sqrt{2m}\int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}},\end{align}

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

Proof of eq. (3): Notice that

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

Then

\begin{align}\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}}} \cr ~\stackrel{{\rm Fubini}}{=}&~\int_{V_0}^V \ell(V^{\prime})~dV^{\prime}\int_{V^{\prime}}^V \frac{dE}{\sqrt{(V-E)(E-V^{\prime})}} \cr ~\stackrel{(4)}{=}~& \pi \int_{V_0}^V \ell(V^{\prime})~dV^{\prime},\end{align}\tag{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). $$\Box$$

Let us for completeness try to integrate by parts (IBP).

\begin{align}\frac{1}{\hbar}\sqrt{\frac{m}{2}}\int_{V_0}^V \ell(V^{\prime})~dV^{\prime} ~\stackrel{(3)}{\approx}~& \int_{V_0}^V \frac{N(E)~dE}{\sqrt{V-E}} \cr ~\stackrel{\text{IBP}}{=}~& 2\int_{V_0}^V \!dE~N^{\prime}(E)\sqrt{V-E}, \end{align} \tag{6}

so that

\begin{align} \frac{1}{\hbar}\sqrt{\frac{m}{2}}\ell(V) ~\stackrel{(3)}{\approx}~& \frac{d}{dV}\int_{V_{0}}^V \frac{N(E)~dE}{\sqrt{V-E}}\cr ~\stackrel{(6)}{=}~&\int_{V_0}^V \frac{N^{\prime}(E)~dE}{\sqrt{V-E}}\cr ~\stackrel{\text{IBP}}{=}~& 2N^{\prime}(V_0)\sqrt{V-V_0} + 2\int_{V_0}^V \!dE~N^{\prime\prime}(E)\sqrt{V-E}.\tag{7} \end{align}

Similarly,

\begin{align} \frac{h}{\sqrt{2m}}N(E) ~\stackrel{(2)}{\approx}~& \int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}} \cr ~\stackrel{\text{IBP}}{=}~& 2\ell(V_0)\sqrt{E-V_0} + 2\int_{V_0}^E\!dV~\ell^{\prime}(V)\sqrt{E-V},\tag{8} \end{align}

so that

$$\frac{h}{\sqrt{2m}}N^{\prime}(E) ~\stackrel{(8)}{\approx}~\frac{\ell(V_0)}{\sqrt{E-V_0}} + \int_{V_0}^E\!dV~\frac{\ell^{\prime}(V)}{\sqrt{E-V}}. \tag{9}$$

• 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 :) Apr 9, 2012 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. Oct 7, 2012 at 8:18
• Calculations for later: $\quad T(E)=h N^{\prime}(E)$. $\quad N(V_0)=0$. $\quad hN(E)=\int_{V_0}^E \!dE^{\prime} ~T(E^{\prime})$. Related: physics.stackexchange.com/a/389454/2451 Jun 30, 2015 at 21:23
• Calculations for later: $\ell(V_0)=0$ unless the potential has a flat bottom. Jun 30, 2015 at 21:30

As all the answers are oriented in the theoretical side let me remind everybody of the Balmer series, an experimental observation fitted with a mathematical series, which was first modeled with the Bohr model of the hydrogen atom, and then was the cornerstone of building quantum mechanics, as it came out from the Schrodinger equation.

So in this sense a given/measured energy spectrum matched a specific potential in the Schrodinger equation.

1) Is there a one to one correspondence between the potential and its spectrum?

In the historical example above , yes. I have not seen a second potential replacing the hydrogen potential in modelling.

2) If the answer to the previous question is yes, then given the spectrum, is there a systematic way to calculate the corresponding potential

The historical example is an educated trial and error method, after all the classical potential between charges was known. It seems to point to this: use the corresponding for the generator of the spectrum classical potential to start with. A spectrum is a physical observation coming from specific atoms/molecules/ensembles . So I would replace "calculate" with "find". Physics is about finding mathematical models that fit the observables, imo.

Within a theoretical framework the "in general no" is the answer since there are infinities of possible solutions. Were we really lucky to have discovered quantum mechanics , if the probability of finding the potential that fits a spectrum is so small? IMO no, the physicists were using their physics background to find a good theoretical model that would fit the observations ( Balmer series) and use their classical knowledge of potentials as a boundary on the infinity of solutions. It is a similar logical process as using boundary conditions to reduce the infinity of solutions from the differential equations to one that fits the data.

I'll somewhat expound the solution by qmechanic, and give two examples. The solution is closely related to Landau's method for period inversion, see Mechanics Sec. 12.

The quantum number can be approximately computed as \begin{align} (n + \delta) \, 2 \pi \hbar &= \oint p \, dq \\ &= \oint \sqrt{2\,m\, [E - V(q)]} \, dq \\ &= 4 \int_0^{r_\max} \sqrt{2\,m\, [E - V(q)]} \, dq, \tag{1} \end{align} where $\delta$ is number between $0$ and $1$ (i.e., $1/2$), and $\hbar$ is the Planck constant. The integral $\oint$ means “go there and come back”. In the last step, we have assumed $V(q)$ is centrosymmetric and $V(r_\max) = E$. Differentiating (1) with respect to $E$, we get \begin{align} 2 \pi \hbar \, \frac{\partial n} {\partial E} &= \int_0^{r_\max} \frac{ \sqrt{8 m} \, dq } { \sqrt{E - V(q)} }, \tag{1} \end{align} Multiplying this by $1/\sqrt{2 \, m \, (\alpha - E)}$ and integrating over $E$, we get \begin{align} 2 \pi \hbar \, \int_0^\alpha \frac{\partial n} {\partial E} \frac{dE}{\sqrt{2 \, m \, (\alpha - E)}} &= \int_0^{r_\max} 2 \, dq \int_{V(q)}^\alpha \frac{dE}{\sqrt{\alpha - E}{\sqrt{E - V(q)}}} \\ &= 2 \, \pi \, r_\max. \end{align} Now since $V(r_\max) = \alpha$, the potential $V(r)$ is solved via the inverse function as \begin{align} \hbar \, \int_0^\alpha \frac{\partial n} {\partial E} \frac{dE}{\sqrt{2 \, m \, (\alpha - E)}} &= V^{-1}(\alpha), \end{align} where the inverse function is taken for the $r > 0$ branch.

## Example 1

For the harmonic oscillator, $E_n = \hbar \, \omega (n + 1/2)$, then $\partial n/\partial E = (\hbar \, \omega)^{-1}$, and \begin{align} \hbar \, \int_0^\alpha \frac{\partial n} {\partial E} \frac{dE}{\sqrt{2 \, m \, (\alpha - E)}} &= \sqrt{\frac{2 \, \alpha}{m \, \omega^2}} = r. \end{align} which means the potential is $V(r) = \alpha = \frac{1}{2} m \, \omega^2 \, r^2$, which is correct.

## Example 2

If $E_n = (\hbar \pi n)^2/(2 m L^2)$, then $\partial n/\partial E = L/(\hbar \, \pi) \sqrt{m/(2\,E)}$, and \begin{align} \hbar \, \int_0^\alpha \frac{\partial n} {\partial E} \frac{dE}{\sqrt{2 \, m \, (\alpha - E)}} &= \frac{L}{2\,\pi} \, \int_0^\alpha \frac{dE}{\sqrt{E \, (\alpha - E)}} \\ &= \frac{L}{2} = r. \end{align} This means that the inverse function $V^{-1}(\alpha)$ is a constant $L/2$ no matter the value of $\alpha$ (even as $\alpha$ grows to $+\infty$). In other words, the $V(r)$ grows steeply to infinity at $r = L/2$. Since we assume that the potential is symmetric, $V(r)$ also grows to infinity at $r = -L/2$. This means that the potential has two symmetric infinite hard walls at $r = \pm L/2$. This is the case of a particle in a box of size $L$. Thanks to Kyle-Kanos to point this out.

• Isn't the potential $V(\hat{x})=\frac12m\omega^2\hat{x}^2$ for the QHO? Also, how do you get the potential for the particle-in-a-box from $L/2=r$? Oct 12, 2015 at 19:35
• Sorry, I missed $r^2$ in Example 1. Thank you for pointing this out. For example 2, $V^{-1}(\alpha)$ is a constant, which means it is a horizontal line when we plot $r$ against $V$, then when we plot $V$ against $r$, we get a vertical line. Also remember that we assume that $V$ is centrosymmetric, so the mirror potential line also exists at $r = -L/2$. This is exactly the potential shape of the particle in a box, with box size $L$.
– hbp
Oct 12, 2015 at 19:44

Yes, at least for one dimension you can obtain the solution to the inverse problem as follows:

let $$N(E)$$ be 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).

• 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. Oct 1, 2011 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 Oct 2, 2011 at 8:34
• Could you give any references Pls?
– Revo
Oct 7, 2011 at 11:57
• you can search 'Wu and Sprung potential' Riemann Hypothesis as an inverse problem is treated there :) Oct 9, 2011 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 Nov 10, 2011 at 12:14

As a addition to the reply by @Jose Javier Garcia and in the spirit of the reply by @anna v, in molecular spectroscopy, anharmonic potentials are often approximately described by the Morse Potential, but this is insufficient in most cases and then the Rydberg-Klein-Rees (RKR) numerical method is very widely used to obtain the potential energy profile. This method is based on the semiclassical WKB method. (See Hirst, 'Potential Energy Surfaces; Molecular Structure and Reaction Dynamics' and articles on Wikipedia).