8.3: Analytic example of variational method - Binding of the deuteron

Say we want to solve the problem of a particle in a potential $V (r) = −Ae^{−r/a}$. This is a model for the binding energy of a deuteron due to the strong nuclear force, with A = 32 MeV and a = 2.2 fm. The strong nuclear force does not exactly have the form $V (r) = −Ae^{−r/a}$, unlike the Coulomb interaction we don’t know what the exact form should be, but $V (r) = −Ae^{−r/a}$ is a reasonable model.

The potential is spherically symmetric, most attractive at $r = 0$ and falls rapidly to zero at large $r$, so we choose a trial wavefunction which does the same, say $\phi = ce^{−\alpha r/2a}$. This has only one dimensionless variational parameter, $\alpha$. The value of c follows from normalisation $\int c^2 e^{−\alpha r/a} 4\pi r^2 dr = 1$; which gives $c^2 = \alpha^3 /8\pi a^3$. (The $4\pi r^2$ comes from the problem being three dimensional).

According to the variational principle, our best estimate for the ground state using this trial function comes from minimising $\langle \phi |\hat{H}|\phi \rangle$ with respect to $\alpha$.

$$\langle\phi|H| \phi\rangle /\langle\phi \mid \phi\rangle = \frac{-\hbar^{2}}{2 m} \int_{0}^{\infty} c^{2} \left(\mathrm{e}^{-\alpha r / 2 a} \nabla^{2} \mathrm{e}^{-\alpha r / 2 a}\right) 4 \pi r^{2} d r-A \int_{0}^{\infty} c^{2} \exp [-(\alpha+1) r / a] 4 \pi r^{2} d r = \frac{\hbar^{2} \alpha^{2}}{8 m a^{2}} - A\left(\frac{\alpha}{\alpha+1}\right)^{3} \nonumber$$

From this we find the minimum for $E(\alpha )$ at $\alpha_0$

$$\frac{dE}{d\alpha} = \frac{\hbar^2\alpha}{4ma^2} − 3A \left( \frac{\alpha^2}{(\alpha + 1)^4} \right) = 0 \quad \Rightarrow \quad \frac{(\alpha 0 + 1)^4}{\alpha_0} = 12Ama^2 /\hbar^2 \nonumber$$

Solving for $\alpha_0$ gives $\alpha_0 = 1.34$, and substituting back into $\langle \phi |H|\phi \rangle$ gives $E_0 = −2.14 MeV$.

This is fairly close to the exact solution for this potential, which can be obtained analytically as a Bessel function of $\sqrt{8mA} (a/\hbar )e^{−r/2a}$ if you manage to spot that change of variables! The exact solution gives $E_0 = −2.245 MeV$.