# Is the harmonic oscillator potential unique in having equally spaced discrete energy levels?

I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.

Any ideas? If not, is there a proof for its uniqueness?

If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems.

• That's a very good question. I guess it boils down to the more general question of operators in Hilbert spaces which have identical spectrum? I am looking forward to someone with more math knowledge who can give us the answer. – CuriousOne Aug 27 '14 at 5:47
• Does this qualify for your purposes? jetp.ac.ru/cgi-bin/dn/e_075_03_0446.pdf – CuriousOne Aug 27 '14 at 7:05
• You might want to pose the question a little more precisely. Are we talking about only one dimension, or possibly 2 or 3? What kind of boundary conditions are allowed? Can we have, e.g., particles with spin? If we can have particles with spin, then one could fine-tune a magnetic field so that each multiplet spreads out by an amount chosen so as to make the spectrum equally spaced. – user4552 Oct 7 '14 at 18:48
• What about $H=\omega L_z$? The spectrum of $L_z$ in the Hilbert space spanned by the irrep $J$ is $-m ,-m+1,\ldots, m-1,m$. Or do you insist on infinite-dimensional ladder of eigenstates? – ZeroTheHero May 12 '18 at 17:52
• Do Landau levels count as a quantum harmonic oscillator potential? – thermomagnetic condensed boson May 12 '18 at 19:12

According to Inverse spectral theory as influenced by Barry Simon Fritz Gesztesy (Submitted on 2 Feb 2010) page 4:

A particularly interesting open problem in inverse spectral theory concerns the characterization of the isospectral class of potentials $V$ with purely discrete spectra (e.g., the harmonic oscillator $V(x) = x^2$ ).

• Is that problem much more general though? Because there are many potentials with discrete spectra that are not evenly spaced (e.g. $V=x^{4}, x^{8}$). – KF Gauss May 18 '18 at 17:21
• Or by discrete spectra do they really mean evenly spaced? – KF Gauss May 18 '18 at 17:21
• @user157879 I misunderstood this. Isospectral potentials are constructed in The Spectral Class of the Quantum-Mechanical Harmonic Oscillator, McKean and Trubowitz 1982 (as I now think I understand it). The "open problem" is to find all of them. There is a math.SE member who has published in this area. – Keith McClary May 19 '18 at 3:36
• great, so it really is the case that this is an open problem. Could you update your answer with that link and/or possibly ping the math.SE member who has published in this area? – KF Gauss May 19 '18 at 4:46
• @user157879 No, it seems that isospectral potentials are proven to exist and explicit examples have been constructed. The open problem is to characterise ALL of them. If you ask the first part of your question on math.SE, CR might respond, but I think the answer would be as helpful as the McKean and Trubowitz paper. – Keith McClary May 19 '18 at 5:15

I believe my answer here to the question "How does one determine ladder operators systematically?" gives at least a partial answer to your question. It is a partial answer because I assume a little more than your bare question, but then, as we see by looking carefully at yuggib's answer observes you can obviously write down a hamiltonian with equally spaced eigenvalues and then characterise the whole family of such hamiltonians. It becomes clear that you need to talk about more than simply the Hamiltonian to answer your question: we need to define other observables and how they behave to define something akin to a "potential".

Let's look at yuggib's answer. You can obviously write down a hamiltonian with equally spaced eigenvalues. Then, as you counter, in your comment:

"Does this really qualify as a proof? Sure you can make operators with any set of discrete eigenvalues, but how do you know that they are not all just equivalent to each other and the harmonic potential? Also, I should say that I really mean a potential, because obviously you can write a Hamiltonian as a matrix and give it equally spaced eigenvalues, but how could you know what potential gave rise to it?"

For this question, I choose energy eigenstates bounded from below. Otherwise, you could get arbitrarily negative energy states and there would be no quantum ground state. This may or may not be more than what you want to assume, but I think it is physically reasonable. As I said, I'm giving a partial answer. So now our index set $I$ becomes in fact the set of semi positive integers $\mathbb{N}$. So, in the notation of yuggib's answer, choose an orthonormal basis $\left\{Y_j\right\}_{j=0}^\infty$ for our assumed separable (this is yet another assumption we must bring to bear) Hilbert space of quantum states with $P_j Y_k = \delta_{j\,k} Y_k$, where $\delta$ is clearly the Kronecker delta and $P_j$ are the projection operators onto the basis vectors. Then, the most general Hamiltonian with equispaced eigenvalues is as written in yuggib's answer with:

$$\lambda_i = E_0 + \sigma(i) \Delta$$

where $E_0$ is the ground state energy, $\Delta$ the energy spacing and $\sigma:\mathbb{N}\to\mathbb{N}$ a bijection between $\mathbb{N}$ and itself. So there are an infinite number of Hamiltonians with equispaced energy levels. All members of each family of such Hamiltonians defined by the family's ground state energy $E_0$ and spacing $\Delta$ are unitarily equivelent to one another: two members $\hat{H}_1,\,\hat{H}_2$ are equivalent by $\hat{H}_1 = U\,\hat{H}_2\,U^\dagger$ with $U$ some unitary operator.

So now, how to work this into something like a "potential"? My solution is then to abstractly define position and momentum observables $\hat{X},\,\hat{P}$ and we make our final three assumptions:

1. They fulfill the canonical commutation relationship $\hat{X}\,\hat{P} - \hat{P}\,\hat{X}=i\,\hbar\,\mathrm{id}$;

2. Measurements by these observables vary sinusoidally with time;

3. Our observables are Hermitian operators.

So now we Writing a general quantum state as:

$$\psi = \sum\limits_{j\in\mathbb{N}} \psi_j e^{-i\,\left(j\,\omega_0+\frac{E_0}{\hbar}\right)\,t}$$

so that the mean of a general observable $\hat{A}$ is:

$$\left<\left.\psi\right|\right.\hat{A}\left.\left|\psi\right.\right> = \sum\limits_{j=0}^\infty a_{j,j}|\psi_j|^2 + 2\,{\rm Re}\left(\sum\limits_{j=0}^\infty\sum\limits_{k=j+1}^\infty a_{j,k} \psi_j \psi_k^* \exp(i\,\omega_0\,(k-j)\,t)\,\right)$$

we can readily see that observables with sinusoidally varying measurement means must have two symmetrically placed, complex conjugate off-leading-diagonal diagonal stripes. Moreover the displacement from the leading diagonal must be the same for both $\hat{X},\,\hat{P}$ if they are to fulfill the CCR. The simplest case is when the two stripes are immediately above and below the leading diagonal. In this case, the means of the observables will vary like $\cos(\omega_0\, t + \phi_0)+const.$: if the two stripes are displaced $N$ steps either side of the leading diagonal, then we have variation varies like $\cos(N\,\omega_0\, t + \phi_0)+const.$. The case with the stripes displaced $N$ from the leading diagonal yield analyses that are essentially the same as the below as discussed in my other answer: they essentially pertain to a quantum oscillator with $N$ times the energy spacing of the one we talk about here.

So now, without loss of generality, in our assumed basis we can write the Hermitian observables as:

$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2}}\left(\tilde{X} + \tilde{X}^\dagger\right) \\ \hat{P} = \sqrt{\frac{\hbar}{2}}\left(\tilde{P} + \tilde{P}^\dagger\right)\end{array}$$

where:

$$\tilde{X} = \left(\begin{array}{ccccc}0&x_1&0&0&\cdots\\0&0&x_2&0&\cdots\\0&0&0&x_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)\quad\tilde{P}=\left(\begin{array}{ccccc}0&p_1&0&0&\cdots\\0&0&p_2&0&\cdots\\0&0&0&p_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$

are both arbitrary lone-striped upper triangular matrices. You'll need to look at my other answer that I referenced above for full details, but by writing down the CCR we find that:

$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2\,m\,\omega_0\,\cos\chi}}\left(a^\dagger\,e^{-i\,\xi} + a\,e^{i\,\xi}\right) \\ \hat{P} = i\,\sqrt{\frac{\hbar\,m\,\omega_0}{2\,\cos\chi}}\left(a^\dagger\,e^{-i\,(\xi+\chi)} - a\,e^{i\,(\xi+\chi)}\right) \end{array}$$

where we have defined the arbitrary complex constant $\alpha = -i\,m\,\omega_0\,e^{i\,\chi}$ by writing it in terms of a second real positive magnitude $m$ with the dimensions of mass and an arbitrary phase factor $\chi$ and where we have also defined:

$$a = \left(\begin{array}{ccccc}0&\sqrt{1}&0&0&\cdots\\0&0&\sqrt{2}&0&\cdots\\0&0&0&\sqrt{3}&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$

and its Hermitian conjugate as the wonted ladder operators.

One can quite straighforwardly show that our derived $\hat{X},\,\hat{P}$ must have continuous spectrums. So, if we change our co-ordinates so that we work in position co-ordinates, i.e. where $\hat{X}$ becomes the diagonal (multiplication) operator $\hat{X}f(x) = x f(x)$ for $f(x)\in \mathcal{H} = \mathbf{L}^2(\mathbb{R}^3)$, then we can argue as I do in my answer here that there is needfully such a co-ordinate system wherein $\hat{X}f(x) = x f(x)$ and $\hat{P} f(x) = -i\,\hbar\,\nabla f(x)$. So now the Hamiltonian in these co-ordinates is:

$$\begin{array}{lcl}\hat{H} &=& \hbar\,\omega_0 \left(a^\dagger\,a + \frac{1}{2}I\right) + \left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\\ &=& \frac{1}{2\,m\,\cos\chi} \hat{P}^2 + \frac{1}{2\,\cos\chi}\,m\,\omega_0^2\,\hat{X}^2 - \frac{\omega_0\,\tan\chi}{2}(\hat{X}\hat{P} + \hat{P}\hat{X}) +\left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\end{array}$$

so that the Schrödinger equation in these co-ordinates is:

$$i\hbar\partial_t \psi = -\frac{1}{2\,m\,\cos\chi} \nabla^2 \psi + \frac{1}{2\,\cos\chi} m\,\omega_0^2 |\vec{x}|^2 \psi + i\,\hbar\,\omega_0\,\tan\chi\,\vec{x}\cdot\nabla \psi + \left(E_0 +i\,\frac{\tan\chi\,\hbar\,\omega_0}{2} - \frac{\hbar\,\omega_0}{2}\right)\psi$$

and more "traditional" Schrödinger equation is recovered when $\chi = 0$ and $E_0 = \hbar\,\omega_0/2$:

$$i\hbar\partial_t \psi = -\frac{1}{2\,m} \nabla^2 \psi + \frac{1}{2} m\,\omega_0^2 |\vec{x}|^2 \psi$$

and we see that we must have a quadratic potential. So, in summary, let's list the assumptions that lead to this conclusion:

1. Separable quantum state space;
2. Equispaced energy levels bounded from below;
3. The existence of Hermitian position and momentum observables fulfilling the CCR and
4. Measurements from the observables vary sinusoidally with time.

I have not yet analysed the case where we relax the assumption 4. From the above, we see that equispaced energy levels implies periodic variations of measurements with time, and it seems to me that this relaxation would likely yield a much more general potential in position co-ordinates.

• Thanks for the partial answer. The assumption of sinusoidally varying observables is the critical assumption though. In the PDF mentioned in the comments of the original post, a non-quadratic polynomial potential is given with equally spaced eigenvalues, so clearly it can be done. Maybe if 4 was relaxed, a family of polynomials, or functions in general, could be discovered. – KF Gauss Dec 3 '14 at 15:32
• @user157879 But harmonically-varying-in-time eigenstates is a necessary consequence of assuming separable solutions to the time-dependent SE, so I think to relax that would require a time-varying potential. – pwf May 15 '18 at 18:02
• @pwf the answer requires sinusoidal observables, which is completely independent of the eigenstates (which are actually stationary). – KF Gauss May 15 '18 at 19:03

Very interesting problem! I will propose it to my students.

However the answer is positive up to isomorphisms of Hilbert spaces if assuming that every eigenspace has dimension $$1$$.

I stress that the result below holds true even if the initial Hamiltonian $$H$$ is not of Schroedinger form.

Proposition. Let $$H$$ be an (unbounded) self-adjoint operator over the Hilbert space $$\cal H$$ such that for a real constant $$\omega \neq 0$$ $$\sigma(H)= \sigma_p(H) = \{\omega n \:|\: n=0,1,2,\ldots\}\tag{1}$$ and every eigenspace has dimension $$1$$. Under these hypotheses there is a unitary map $$U: {\cal H} \to L^2({\mathbb R}, dx)$$ such that, for some uniquely fixed real numbers $$\alpha, \beta$$ $$U H U^{-1} = \alpha H_0 + \beta I\tag{2}$$ where $$H_0$$ is the standard (self-adjoint) Hamiltonian of the harmonic oscillator.

Proof. Let $$\psi_n$$ be the unit eigenvector of the eigenvalue $$\omega n$$ defined up to a phase. The finite span $$D$$ of the $$\psi_n$$ is dense, since they define the spectral measure of $$H$$ from (1).

We can define the standard ladder operators over the common domain $$D$$ $$a \psi_n = \sqrt{n}\psi_{n-1}$$ (with $$\psi_{-1}:=0$$) and $$a^\dagger \psi_n = \sqrt{n+1}\psi_{n+1}\:.$$ Evidently,$$D$$ turns out to be invariant under $$a$$ and $$a^\dagger$$.

Next define, over the dense domain $$D$$, the symmetric (not yet selfadjoint) operators $$A = \frac{1}{\sqrt{2}}(a+a^\dagger)$$ and $$B=\frac{i}{\sqrt{2}}(a-a^\dagger)$$.

We have $$A^2+B^2 = (\omega H + \frac{1}{2}I)|_D$$.

Here is the crucial observation, since the $$\psi_n$$ are a Hilbert basis of eigenvectors of $$H$$, $$D$$ is a core for $$H$$ and also for $$A^2+B^2+ I$$ since those vectors are analytic vectors for $$H$$ and thus also for trivially related symmetric operator $$A^2+B^2+I^2$$ By a known theorem by Nelson  $$A^2+B^2+I^2$$ is essentially self-adjoint over $$D$$.

A famous theorem by Nelson (Nelson theorem about unitary representations ) assures that there exist a strongly-continuous unitary representation of the unique connected simply-connected Lie group whose Lie-algebra is generated by $$A,B,I|_D$$. It easy to see that this Lie algebra is nothing but that of Heisenberg group.

Finally, Nelson's theorem also establishes that the unique self-adjoint extensions of $$A$$, $$B$$ and $$I|_D$$ are the self-adjoint generators of the three fundamental one-parameter groups of Heisenberg group.

In other words, we have found that $$exp(-iuA)$$ and $$exp(-ivB)$$ have the same commutation relations of $$exp(-iuX)$$ and $$exp(-ivP)$$

Here Stone-von Neumann theorem enters the game (in the generalized version due to Mackay, without requiring the irreducibility of the representation).

There exist a unitary operator $$U : {\cal H} \to \bigoplus_{k\in K}L^2(\mathbb{R}, dx)$$ such that $$U(\omega H + \frac{1}{2}I)U^{-1} = \bigoplus_{k\in K} H_0$$ so that $$H = \frac{1}{\omega} U^{-1}\left(\bigoplus_{k\in K} H_0- \frac{1}{2}I\right) U$$

It is clear that if $$K$$ contains more than one element, $$\omega H + \frac{1}{2}I$$ would have more than one eigenvector for a given eigenvalue contrarily to our hypotheses. So that $$K$$ includes only one element and we can write $$H = U^{-1} \left(\frac{1}{\omega}H_0- \frac{1}{2\omega}I\right) U\:.$$ QED

 Nelson, E.: Analytical vectors, Annals of Mathematics, 70, 572-615 (1969).

• I wanted to write the answer using the same idea and saw you've already done it. Nice answer – heaven-of-intensity May 17 '18 at 18:29
• Doesn't this contradict the explicit construction of such a potential here? jetp.ac.ru/cgi-bin/dn/e_075_03_0446.pdf – KF Gauss May 17 '18 at 18:58
• I do not know. Perhaps that Hamiltonian is unitarily equivalent to the harmonic one. This is still permitted by my answer. – Valter Moretti May 17 '18 at 19:14
• Is it possible to have two potentials of different polynomial order be unitarily equivalent? This would be very interesting, as you may be able to generate the entire family by unitary operations – KF Gauss May 17 '18 at 23:21
• Nonetheless, that means the statement "However the answer is positive up to isomorphisms of Hilbert spaces if assuming that every eigenspace has dimension 1. " does not really answer the real question – KF Gauss May 18 '18 at 2:09

There are, in general, infinitely many operators with equally spaced eigenvalues. Suppose a self-adjoint operator $A$ has a purely discrete spectrum (i.e. it is either compact or with compact resolvent) and denote by $\{\lambda_i\}_{i\in\mathbb{I}}$ its real eigenvalues ($I\subseteq \mathbb{N}$): then by the spectral theorem it can be written (on its domain of definition) as $$A=\sum_{i\in I} \lambda_i P_i$$ where $P_i$ is the orthogonal projector on the eigensubspace corresponding to $\lambda_i$.

Now this equality, read from right to left defines the operator $A$, choosing the eigenvalues and the (mutually disjoint) orthogonal projections. So playing with projections and eigenvalues you will define different operators, with equally spaced eigenvalues if you want. However it may be then necessary to prove the resulting operator is self-adjoint on a suitable domain (if the operator is unbounded).

• Does this really qualify as a proof? Sure you can make operators with any set of discrete eigenvalues, but how do you know that they are not all just equivalent to each other and the harmonic potential? Also, I should say that I really mean a potential, because obviously you can write a Hamiltonian as a matrix and give it equally spaced eigenvalues, but how could you know what potential gave rise to it? – KF Gauss Aug 27 '14 at 12:24
• two operators to be equal must have the same domain and the same action. In this setting, that means the projection and the eigenvalues must be exactly the same (as well as the domain) for the operators to be equal. – yuggib Aug 27 '14 at 12:44
• also, writing an operator as a formal sum of two non-commuting parts (as in $-\Delta + V(x)$) may not be the best way to analyze its spectrum. More general operators can be written, however if they have purely discrete spectrum they would all admit a spectral decomposition as above. – yuggib Aug 27 '14 at 12:49
• My question is specifically about potentials. Two operators can have trivially different decompositions by merely changing parameters in spectrum, such as frequency in the SHO case. But to me, they are part of the same equivalence class. Can we construct a potential that has an integer spaced spectrum that is not a second order polynomial in the position operator? – KF Gauss Aug 27 '14 at 16:54

It's been a while since I have done Quantum Mechanics, however your question reminded me of Supersymmetric Quantum Theory. In simpler words it is a theorem that states that for every potential ($V(x)$) there exists a supersymmetic (SUSY) partner potential ($\tilde{V}(x)$) that has the same functional form for the energy levels and differ in in there energies by a shift. Quantum Mechanics by Schwabl has a section devoted to this (19) and actually have the Harmonic Oscillator as an example (19.2.3). However the answer was not satisfying because the SUSY potentials they got to be for it were... $$V(x) = \frac{1}{2}\omega^2 x^2 -\frac{1}{2}\omega$$ $$\tilde{V}(x) = \frac{1}{2}\omega^2 x^2 +\frac{1}{2}\omega$$ Which are essentailly equivalent because adding constants to a Hamiltonian only shifts the energy. So according to SUSY QM the Harmonic Oscillator can only be mapped to a Harmonic Oscillator with a constant energy shift. Don't know if I completely answered your question, but I hope it helps!

The harmonic potential is not the only potential with evenly spaced energy levels.

Consider a potential $V(X)$ that is $\frac{1}{2} m \omega_1 x^2 - \frac{1}{2} \hbar \omega_1$ for $x > 0$ and $\frac{1}{2} m \omega_2 x^2 - \frac{1}{2}\hbar \omega_2$ for $x < 0$. If $\omega_2$ is a square-free multiple of $\omega_1$ then the energy levels will be spaced with levels $2 \hbar \omega_1$. The reason for this is that we can use the even $\omega_1$ energy eigenstates for the $x > 0$ part of the function and the corresponding $\omega_2$ energy eigenstates for the $x < 0$ part. At $x = 0$ we can set the functions to be equal, and the first derivative is $0$ there as well. We can not use the odd solutions because the first derivatives of the odd solutions will never be equal at $x = 0$ assuming that $\omega_2/ \omega_1$ is square-free. This is because the wave function is given by a Hermite polynomial $H_n(\sqrt{\frac{m \omega}{\hbar}} x)$ times the Gaussian curve, and for the odd solutions, the $x$ term in $H_n$ is always an integer.

In any case, let us now restrict our attention to even potentials $V(x)$, which. If $V(x)$ is even, then the harmonic potential is the only potential that has evenly spaced energy levels.

If the energy levels are evenly spaced with spacing $\Delta E$, then the time evolution operator

$$U(t) := e^{i \hat H t / \hbar}$$

will satisfy

$$U(t + \tfrac{2 \pi}{ \omega}) = e^{i \theta} U(t)$$

for all $t$, where $\omega = \Delta E/\hbar$ and $\theta$ is a physically irrelevant phase that can be set to $0$ by adding a constant to our Hamiltonian (such as setting the ground state energy to $0$).

This means that under time evolution, all state will return to their initial state every period $\frac{2 \pi}{\omega}$. Also note that by Erenfests theorem, the classical position expectation value will follow the classical equation of motion, so that too will return to its initial state after the same period.

I will now prove that the harmonic potential is the only symmetric 1D potential for which all solutions are periodic with the same period. Therefore, it is the only potential with evenly spaced energy levels.

First note that $V(x)$ must have only one minimum and no maximums. If there were a maximum, then there would exist "slow roll" solutions, rolling off of the hill, that can take arbitrarily long times to leave the hill, such that not all solutions would be periodic with the same period. Therefore, $V(x)$ must reach a minimum (WLOG at $x=0$) and increase as $|x|$ gets larger.

From the conservation of energy

$$\frac{1}{2} m v^2 = E - V(x)$$ $$v = \sqrt{\frac{2}{m}(E - V(x))}$$ $$v = \frac{dx}{dt}$$ $$\frac{dx}{\sqrt{\frac{2}{m}(E - V(x))}} = dt$$ $$\int \frac{dx}{\sqrt{\frac{2}{m}(E - V(x))}} = \int dt$$

We can consider the interval from the maximum value of $x$ that the particle reaches, $V^{-1}(E)$, to $0$.

$$\int_0^{V^{-1}(E)} \frac{dx}{\sqrt{\frac{2}{m}(E - V(x))}} = T$$

We need the right hand side to be independent of $E$ if all solutions are to have the same period. (From here on out we will suppress constants like $2$ and $m$, only wanting the integral to be independent of $E$.)

First we change variables to integrate with respect to $V$:

$$\int_0^{E} \frac{1}{V'}\frac{dV}{\sqrt{E - V}}$$

where $V' = \frac{dV}{dx}$.

Because $V$ only has one minimum, we can express $V'$ as some function $f(V)$.

$$\int_0^{E} \frac{1}{f(V)}\frac{dV}{\sqrt{E - V}}$$

In order for this integral to be independent of $E$, it must be dimensionless. The only way to accomplish this is for $f(V) \propto \sqrt{V}$. Indeed, the quantity

$$\int_0^{E} \frac{1}{\sqrt{V}}\frac{dV}{\sqrt{E - V}}$$

can also be explicitly evaluated to be independent of $E$. (It is $\pi$.)

This means that $V \propto x^2$, which is the harmonic potential. This is what we wanted to show.

• Thanks for the answer, but it ultimately rests on what you believe, rather than any strong physical proof/argument. – KF Gauss May 18 '18 at 17:19
• Well there is just one step missing: that the only potential that where all classical solutions are periodic with the same period is the harmonic potential. It might follow from some sort of argument with Fourier analysis. – user1379857 May 18 '18 at 17:22
• user157879: I have edited the answer to address your point – user1379857 May 19 '18 at 2:34
• The fundamental problem is that your conclusion is actually incorrect as some of the other comments have pointed out. There is at least one explicit potential with equal spacing shown here: jetp.ac.ru/cgi-bin/dn/e_075_03_0446.pdf but the rest of the potentials are not known and that seems to be an open problem projecteuclid.org/euclid.cmp/1103920654 – KF Gauss May 19 '18 at 4:44
• I believe the mistake here is to rely on the classical behavior of potential, which does not restrict the quantum solution properly – KF Gauss May 21 '18 at 2:03