I want to prove the following:
Show that for any one dimensional, time independent potential $U(x)$, where $\lim\limits_{x \rightarrow \infty}{U(x)} = 0$ and $\int_{-\infty}^{\infty}U(x) < 0$ there always exists at least one bound state with energy $E < 0$.
Hint: Approximate the expectation value of the energy $E(\alpha)$ in the state $\psi(x, \alpha) = \sqrt{\alpha} \exp(-\alpha|x|)$ for small $\alpha > 0$ and $\alpha \in \mathbb{R}$.
My question is if my proof below is correct and sound; I tried to also derive the approach given in the hint:
I argue that if $\int_{-\infty}^{\infty}U(x) < 0$ then there must exist an interval $[x_1, x_2]$ where $U(x \in [x_1, x_2]) < 0$. If I can now show that a bound state with negative energy exists even if this interval goes to zero, then we can deduct that such a state will always exist.
We reduce the interval $[x_1, x_2]$ to a point $x^{'}$; the potential at that point is then reduced to a delta distribution $U(x^{'}) = -C \delta(x - x^{'})$.
This leads to the following stationary Schrödinger Equation:
$$\frac {\partial^2} {\partial x^2} \psi(x) + \frac {2m} {\hbar^2} (C\delta(x - x^{'}) + E) \psi(x) = 0 $$
We know from a previous assignment that the normalized solution for the bound case is
$$\psi(x) = \sqrt{\alpha} \exp{(-\alpha |x|)}$$
where $\alpha = \sqrt{2mE} / \hbar$ and $\alpha \in \mathbb{C}$.
Let $\alpha = i \sqrt{2m|E|} = i \kappa$. Then the expectation value $\langle E \rangle = \langle \psi(x)| \hat{H} | \psi(x) \rangle$ with $\hat{H} = \frac{\hat{p}}{2m} + U(x)$ can be calculated as:
\begin{equation} \begin{split} \langle E \rangle & = \kappa^2 \int_{-\infty}^{\infty} \text{dx} \exp{(i \kappa |x|)} \left ( \frac{-\hbar^2}{2m} \frac {\partial^2} {\partial x^2} + U(x) \right) \exp{(- i \kappa |x|)} \\ & = \frac{\hbar^2 \kappa^4}{2m} \int_{-\infty}^{\infty} \text{dx} \exp{( i \kappa |x|)} \exp{(- i \kappa |x|)} + \kappa^2 \int_{-\infty}^{\infty} \text{dx} \exp{( i \kappa |x|)} \, U(x) \, \exp{(- i \kappa|x|)}. \\ \end{split} \end{equation}
The first integral must equal $1$ because bound states are normalizable. Since we're looking for bound states it's sufficient to consider small energies and thus $\alpha \approx 0$. Therefore, the first term vanishes. For the second integral we notice that the exponentials can now be approximated as $\exp(\pm i \kappa |x|) \approx \pm i \kappa |x|$ to first order. We retrieve:
\begin{equation} \begin{split} \langle E \rangle & = \kappa^2 \int_{-\infty}^{\infty} \text{dx} \exp{( i \kappa |x|)} \, U(x) \, \exp{(- i \kappa|x|)} \\ & \approx \kappa^4 \int_{-\infty}^{\infty} \text{dx} (i |x|) \, U(x) \, (- i |x|) \\ & = \kappa^4 \int_{-\infty}^{\infty} \text{dx} |x|^2 \, U(x). \end{split} \end{equation}
Now we remember that $\int_{-\infty}^{\infty}U(x) < 0$. Since $\alpha$ and $|x|^2$ are always positive and thus don't change the sign of the integral it follows that
$$\langle E \rangle < 0.$$
Finally, because every state $\psi(x)$ can be decomposed into eigenstates of the Hamiltonian as $\psi(x) = \sum_n c_n \psi_n(x)$ there must exist at least one bound energy eigenstate with eigenvalue $E < 0$.
