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 $$E_{\text{ground}} < 0$$.