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

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    28790
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    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 \(E_0 = −2.245 MeV \).


    This page titled 8.3: Analytic example of variational method - Binding of the deuteron is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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