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Every purely attractive potential in 1D has at least one bound state

Every purely attractive one-dimensional potential well, no matter how shallow, has at least one bound state.

More precisely, this statement can be formulated as follows. Consider a one-dimensional non-relativistic quantum particle of mass \(m\) inside a purely attractive well potential \(V(x)\): the function \(V(x)\) is such that

  • \(V(x) < 0\) for all \(x\), and
  • \(V(x) \to 0\) as \(x \to \pm \infty\).

This system has at least one bound state. That is, the ground state energy \(E_{\text{ground}}\) of the Hamiltonian \[H = -\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} + V(x)\] is strictly negative: \[E_{\text{ground}} < 0 \,.\]

Proof: [D. ter Haar, Selected Problems in Quantum Mechanics (Academic, New York, 1964).] According to the variational principle, \[E_{\text{ground}} \le \langle \psi | H | \psi \rangle = \int_{-\infty}^{\infty} dx \, \psi^*(x) H \psi(x)\] for any normalized state \(\psi(x)\). Hence, in order to prove the negativity of \(E_{\text{ground}}\), it is sufficient to find a normalized function \(\psi(x)\) for which \(\langle \psi | H | \psi \rangle < 0\). To this end, we consider a trial wave function \[\psi_{\alpha}(x) = \left( \frac{2 \alpha}{\pi} \right)^{1/4} e^{-\alpha x^2}\] with \(\alpha > 0\). The wave function is normalized: \[\langle \psi_{\alpha} | \psi_{\alpha} \rangle = \int_{-\infty}^{\infty} dx \, |\psi_{\alpha}(x)|^2 = 1 \,.\] Then, we have \[\begin{align} E(\alpha) &\equiv \langle \psi_{\alpha} | H | \psi_{\alpha} \rangle \\ &= \int_{-\infty}^{\infty} dx \, \psi_{\alpha}^*(x) \left( -\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} + V(x) \right) \psi_{\alpha}(x) \\ &= \frac{\hbar^2}{2 m} \int_{-\infty}^{\infty} dx \, \left| \frac{d \psi_{\alpha}(x)}{d x} \right|^2 + \int_{-\infty}^{\infty} dx \, V(x) |\psi_{\alpha}(x)|^2 \,, \end{align}\] and, substituting \(\psi_{\alpha}(x) = (2 \alpha / \pi)^{1/4} e^{-\alpha x^2}\), we obtain \[\begin{align} E(\alpha) &= \frac{\hbar^2}{2 m} \frac{(2 \alpha)^{5/2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} dx \, x^2 e^{-2 \alpha x^2} + \sqrt{\frac{2 \alpha}{\pi}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \alpha x^2} \\ &= \frac{\hbar^2 \alpha}{2 m} + \sqrt{\frac{2 \alpha}{\pi}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \alpha x^2} \,. \end{align}\] The energy \(E(\alpha)\) reaches its minimum at \(\alpha = \beta\), where \(\beta\) satisfies \(E'(\beta) = 0\), or \[\frac{\hbar^2}{2 m} + \frac{1}{\sqrt{2 \pi \beta}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \beta x^2} - 2 \sqrt{\frac{2 \beta}{\pi}} \int_{-\infty}^{\infty} dx \, V(x) x^2 e^{-2 \beta x^2} = 0 \,.\] Multiplying this equation by \(\beta\), we get \[\frac{\hbar^2 \beta}{2 m} + \sqrt{\frac{\beta}{2 \pi}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \beta x^2} - \frac{(2 \beta)^{3/2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} dx \, V(x) x^2 e^{-2 \beta x^2} = 0 \,,\] or \[\begin{align} \frac{\hbar^2 \beta}{2 m} + &\sqrt{\frac{2 \beta}{\pi}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \beta x^2} \\ &= \sqrt{\frac{\beta}{2 \pi}} \int_{-\infty}^{\infty} dx \, V(x) e^{-2 \beta x^2} + \frac{(2 \beta)^{3/2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} dx \, V(x) x^2 e^{-2 \beta x^2} \,. \end{align}\] Noticing that the left-hand side of the last equation is nothing but \(E(\beta)\), we obtain \[E(\beta) = \sqrt{\frac{\beta}{2 \pi}} \int_{-\infty}^{\infty} dx \, V(x) (1 + 4 \beta x^2) e^{-2 \beta x^2} \,.\] Since the integrand is the last expression is strictly negative, we find that \(E(\beta) < 0 \,.\) This means that there exists a Gaussian state \(| \psi_{\beta} \rangle\) such that \[\langle \psi_{\beta} | H | \psi_{\beta} \rangle < 0 \,,\] which in turn implies \(E_{\text{ground}} < 0\).