Does Every Attractive Potential in One Dimension Have a Bound State?

[Dahaka14]

Homework Statement

Exercise 5.2.2 (b.)
Prove the following theorem: Every attractive potential in one dimension has at least one bound state. Hint: Since $$V$$ is attractive, if we define $$V(\infty)=0$$, it follows that $$V(x)=-|V(x)|$$ for all $$x$$. To show that there exists a bound state with $$E<0$$, consider
$$\psi_{\alpha}(x)=\left(\frac{\alpha}{\pi}\right)^{1/4}\text{e}^{-\alpha x^{2}/2}$$
and calculate

$$E(\alpha)=<\psi_{\alpha}|H|\psi_{\alpha}>,$$ $$H=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}-|V(x)|$$.

Show that $$E(\alpha)$$ can be made negative by suitable choice of $$\alpha$$. The desired result follows from the application of the theorem approved above.

Homework Equations

The Attempt at a Solution

I evaluated the expectation value using the given wave function and special Hamiltonian and received a simpler equation of $$E=\frac{\alpha\hbar^{2}}{4m}-\int_{-\infty}^{\infty}|V(x)|\psi_{\alpha}^{2}dx$$. I have no idea where to go from here.

 

[gabbagabbahey]

Hmmm... You might try using the mean value theorem for integration.