8.3: Analytic example of variational method - Binding of the deuteron
- Page ID
- 28790
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(E_0 = −2.245 MeV \).