# 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 $$E_0 = −2.245 MeV$$.

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