# Do asymptotically similar potentials yield similar energy levels asymptotically?

Let there be given two Hamiltonians

$$H_1~=~ p^{2}+f(x) \qquad \mathrm{and} \qquad H_2~=~ p^{2}+g(x).$$

Let's suppose that for big big $x$, the potentials are asymptotically similar in the sense that the quotient

$$\frac{f(x)}{g(x)}~\to~1 \qquad \mathrm{for}\qquad x \to \infty.$$

Then if we quantize these Hamiltonians,

$$\hat{H}_1y(x)~=~E_{n}y(x) \qquad \mathrm{and} \qquad \hat{H}_2y(x)~=~B_{n}y(x),$$

since the potentials $f(x)$ and $g(x)$ are asymptotically close to each other, does it mean that for big quantum number $n$ $$\frac{E_{n}}{B_{n}}~\to~1 \qquad \mathrm{for}\qquad n \to \infty,$$

at least in one dimension?

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Here 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 $\Phi(x)<0$ is the potential function. Let us for simplicity assume that the potential $\Phi(x)=\Phi(-x)$ is an even function and strongly monotonically increasing for $x\geq 0$ with limit $\lim_{|x|\to \infty}\Phi(x)=0$. Let $E_n<0$ denote the energy of the $n$'th bound state, ordered increasingly $E_1 <E_2<E_3<\ldots$, so that $E_1$ denotes the ground state energy. There is also a positive continuous unbounded spectrum $E\geq0$, which we shall not discuss further.

We shall here construct a counterexample of two potentials $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that the quotient of potentials satisfies the limit $$\lim_{|x|\to \infty} \frac{\Phi^{\prime\prime}(x)}{\Phi^{\prime}(x)} ~=~1, \qquad (1)$$ but where the corresponding quotient of bound state energies satisfies $$\lim_{n\to \infty} \frac{E_n^{\prime\prime}}{E_n^{\prime}} ~\neq~1. \qquad (2)$$

The idea is to seek for a potential $\Phi$ that would generate a bound state spectrum of the form

$$E_n= -R e^{-\mu n},\qquad (3)$$

where $R>0$ is a Rydberg-like constant of dimension energy, and $\mu>0$ is a dimensionless positive constant. [The Hydrogen atom is for comparison $E_n= -R/n^2$.] Thus the number of states $N(E)$ below energy-level $E$ should roughly satisfy

$$E~\approx~ -R e^{-\mu N(E)} \qquad \Leftrightarrow \qquad N(E)~\approx~\frac{1}{\mu}\ln(-\frac{E}{R}).$$

This answer provides a semi-classical inversion formula for the potential $\Phi$ that we will use. The length $\ell(V)$ of the classically accessible region of the potential well at potential energy-level $V$ becomes

$$2\Phi^{-1}(V)~=~\ell(V) ~\approx ~\hbar\sqrt{\frac{2}{m}} \frac{d}{dV}\int_{V_{0}}^V \frac{N(E)~dE}{\sqrt{V-E}}$$ $$~\approx~\frac{\hbar}{\mu}\sqrt{\frac{2}{m}} \left(\frac{2\arctan\sqrt{\frac{V_0}{V}-1}}{\sqrt{-V}} - \frac{\ln(-\frac{V_0}{R})}{\sqrt{V-V_0}}\right). \qquad (4)$$

One may check that the accessible length function $\ell(V)$ is a monotonically increasing function for $V\in[V_1,0[$ for some choice of the constants $V_0$ and $V_1$ with $V_0<V_1<0$. Asymptotically, such potential $\Phi(x)$ behaves as an inverse square potential $-C/x^2$ for $|x|\to\infty$, where $C>0$ is a positive constant.

We now construct the potential functions $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that they are given by formula (4) in the outer region $x\geq x_1$, where $x_1:=\ell(V_1)/2>0$; and arbitrarily monotonically increasing in the inner region $0\leq x\leq x_1$. This implies that the quotient of potentials satisfies

$$\frac{\Phi^{\prime\prime}(x)}{\Phi^{\prime}(x)} ~=~1 \qquad {\rm for}\qquad |x|\geq x_1,$$

so that condition (1) is satisfied.

For large enough states $n\geq n_1$, after the inner potential well is filled, the spectrum $E_n$ eventually becomes exponentially a la (3). We now choose the inner potentials $\Phi^{\prime}(x)$ and $\Phi^{\prime\prime}(x)$ such that there fits one more bound state into the profile $\Phi^{\prime\prime}(x)$ than $\Phi^{\prime}(x)$ for $|x|\leq x_1$. Then the labeling of states $E_{n+1}^{\prime\prime}\approx E_n^{\prime}$ would be off by one for $n\geq n_1$, yielding the inequality (2),

$$\frac{E_n^{\prime\prime}}{E_n^{\prime}} ~\approx~e^{\mu} ~\neq~1. \qquad {\rm for}\qquad n\geq n_1.$$

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This is clever, but it's not a counterexample. Your answer has the energy accumulate to zero, which means there are unbound states of any positive energy, which obviously violates the OP's intention that all the states should be bound. Further, you are choosing the energy accumulation point to be zero. If you have a potential where the energy levels accumulate to E, you can always add a constant to the energy and make the ratio of the accumulation point be anything at all. I thought of something like this, but to get a real example, you need the bound $E_n$ go like $e^n$, which cannot happen. –  Ron Maimon Oct 29 '11 at 17:31
I'm of course fully aware of the continuous unbounded spectrum. I interpreted the question(v1) as asking about the discrete bounded part of the spectrum only (in particularly because OP labels his states by $n$, which traditionally implicitly implies that $n$ is an integer), and I have constructed a counterexample in that sense. –  Qmechanic Oct 29 '11 at 19:29
You are right, but this answer violates the spirit of the question, because the ratio is only because you chose the accumulation point to be at zero. Accumulating levels are not in the spirit of the question, but ok, it's still clever. –  Ron Maimon Oct 29 '11 at 21:09
Yes, it does, providing that both potentials are confining (a finite motion). For large $E_n$ you may use the quasi-classic approximation, for example, the quantization condition of Bohr-Sommerfeld, and make sure the spectrum is asymptotically the same.