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I am having some problems with a finite, shifted quantum harmonic oscillator potential, and the theorem that states:

Any attractive potential in one dimension must have at least one bound state.

Let's consider the following potential:

$$ \ \begin{cases} V_o(\frac{x}{a}-1)(\frac{x}{a}+1) & -a \leq x\leq a \\ 0 \ \ \ \ \ otherwise \end{cases} \ $$

I tried doing it by scaling the Hamiltonian:

$$ \hat{H}=\sqrt{\frac{2V_o}{m}}\cdot\frac{1}{a}\left(\frac{\hat{P}}{2}+\frac{\hat{X}}{2}\right) - V_o $$

So I suppose it should give shifted Eigenenergies of the Harmonic Oscillator :

$$ E_n =\sqrt{\frac{2V_o}{m}}\cdot\frac{1}{a} \left(\frac{1}{2}+n\right) $$ And for the bounded states: $$ E_n < 0 \iff \frac{ma^2V_o}{2} > \left(n+\frac{1}{2}\right)^2 $$

Which, for specified values of $V_o,m$ and $a$ doesn't allow a bounded ground state. But according to the theorem, one should exist. What is the problem here, or with my way of solving it?

Brendan Darrer
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Kubrik
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    The potential is not harmonic outside of interval $[-a,a]$, so it is unlikely that it gives the shifter energies of harmonic oscillator, though under some conditions they may be rather close. I suggest drawing the potential first - it may clarify some things. – Roger V. Mar 08 '23 at 11:39
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    Here is a proof of the theorem: https://physics.stackexchange.com/q/143630/2451 – Qmechanic Mar 08 '23 at 13:55

1 Answers1

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  1. The Hamiltonian of your problem is given by $$\begin{align} H &= \frac{P^2}{2m}+ V_0\left(\frac{X}{a} +1\right)\left(\frac{X}{a}-1\right) \theta(X+a) \theta(a-X) \\ &=\frac{P^2}{2m}+\frac{V_0}{a^2}\left(X^2 -a^2\right)\theta(X+a)\theta(a-X),\end{align}$$ describing neither a harmonic oscillator nor being identical to the (obviously wrong) expression for $\hat{H}$ given in your second equation.

  2. If you wish to establish a relation between your system and a harmonic oscillator, you could proceed as follows: as the inequality $x^2-a^2 \ge 0$ holds for $|x|\ge a$, the Hamiltonian $H$ and the operator $$H_{\rm HO} =\frac{P^2}{2m} + \frac{V_0}{a^2}(X^2-a^2)= \frac{P^2}{2m}+\frac{V_0}{a^2} X^2-V_0$$ obey the operator inequality $H \le H_{\rm HO}$. $H_{\rm HO}$ describes a harmonic oscillator with angular frequency $\omega =\frac{1}{a}\sqrt{2 V_0/m}$ and the pure point spectrum $\hbar \omega (n+1/2)-V_0$ (with $n=0, 1, 2, \ldots$). In addition, $H$ fulfills the (trivial) inequality $-V_0 \le H$ and we conclude that the two operator inequalities $$-V_0 \le H \le H_{\rm HO}$$ imply$^\ast$ that the ground-state energy $E_0$ of the Hamiltonian $H$ is bounded from below and above by $$-V_0 \le E_0 \le \frac{\hbar \omega}{2}-V_0, \qquad \omega=\frac{1}{a}\sqrt{\frac{2 V_0}{m}}.$$ In the case of large values of $\omega$ with $\hbar \omega /2 \ge V_0$, the upper bound can be improved to $E_0 \le 0$.

  3. The spectrum of $H$ consists of a point spectrum with eigenvalues $-V_0 \le E_r \le 0$ and a continuous spectrum $[0, \infty)$. There is no reason why the well-known theorem about the existence of at least one bound state in the case of an attractive potential in one spatial dimension should be violated in this specific case.

$^\ast$ $H \le H_{\rm HO}$ means that $\langle \psi | H \psi \rangle \le \langle \psi | H_{\rm HO} \psi \rangle$ for all $\psi \in D(H) \cap D(H_{\rm HO})= D(H_{\rm HO})$. $\inf\limits_{||\psi ||=1} \langle \psi | H \psi \rangle \le \inf\limits_{||\psi||=1} \langle \psi |H_{\rm HO} \psi \rangle$ implies the inequality $E_0 \le \hbar \omega/2 -V_0$ for the ground-state energies of $H$ and $H_{\rm HO}$. Analogously, $ -V_0 \langle \psi |\psi \rangle \le \langle \psi | H \psi \rangle$ for all $\psi \in D(H)$, implies $-V_0 \le E_0$.

Hyperon
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  • In what sense do inequalities between the Hamiltonians hold? Are they inequalities for means? for eigenenergies? – Roger V. Mar 08 '23 at 13:54
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    @Roger Vadim Apart from some subtleties concerning the domains of unbounded operators, $H \le H^\prime$ means $\langle \psi |H \psi\rangle \le \langle \psi | H^\prime \psi \rangle$ for "all" vectors $\psi$. Taking the infimum on both sides with the restriction $\langle \psi | \psi \rangle =1$, yields the inequality $E_0 \le E_0^\prime$ for the corresponding ground-state energies. – Hyperon Mar 08 '23 at 15:48
  • +1 Perhaps worth adding it as a comment/remark to your answer. – Roger V. Mar 08 '23 at 15:59
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    @Roger Vadim Thanks for the suggestion. Added now in a footnote. – Hyperon Mar 08 '23 at 16:58