10.16: Variation Method for a Particle in a Finite 3D Spherical Potential Well
- Page ID
- 136254
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This problem deals with a particle of unit mass in a finite spherical potential well of radius 2 ao and well height 2 Eh. The trial wave function is given below.
\[ \psi (r, \beta ) := ( \frac{2 \beta}{ \pi})^{ \frac{3}{4}} exp(- \beta r^2) \nonumber \]
\[ T = - \frac{1}{2r} \frac{d^2}{dr^2} (r \blacksquare ) \nonumber \]
\( V(r) := if [(r \leq 2), 0, 2]\)
a. Demonstrate that the wave function is normalized.
\[ \int_{0}^{ \infty} \psi (r, \beta )^2 4 \pi r^2 dr |_{simplify}^{assume,~ \beta > 0} \rightarrow 1 \nonumber \]
b. Evaluate the variational integral.
\[ E( \beta ) := \int_{0}^{ \infty} \psi (r, \beta ) [- \frac{1}{2r} \frac{d^2}{dr^2} (r \psi (r, \beta ))] 4 \pi r^2 dr ... |_{simplify}^{assume,~ \beta > 0} + \int_{2}^{ \infty} 2 \psi (r, \beta )^2 4 \pi r^2 dr \nonumber \]
\[ E( \beta ) := \frac{1}{2} \frac{3 \pi^{ \frac{1}{2}} \beta + 4 \pi ^{ \frac{1}{2}} + 16 exp(-8 \beta) 2^{ \frac{1}{2}} \beta^{ \frac{1}{2}}-4 \pi ^{\frac{1}{2}} erf((2) 2^{ \frac{1}{2}} \beta^{ \frac{1}{2}}) }{ \pi ^{ \frac{1}{2}}} \nonumber \]
c. Minimize the energy with respect to the variational parameter \( \beta\).
\( \beta\) := 5 \( \beta\) := Minimize (E, \beta ) \( \beta\) = 0.381 \( E ( \beta ) = 0.786\)
d. Calculate the average value of r.
\( \int_{0}^{ \infty} r \psi (r, \beta )^2 4 \pi r^2 dr = 1.293\)
e. Calculate the kinetic and potential energy.
Potential energy:
\( \int_{2}^{ \infty} r \psi (r, \beta )^2 4 \pi r^2 dr = 0.215\)
Kinetic energy:
\( E( \beta ) - 0.215 = 0.571\)
f. Calculate the probability that the particle is in the barrier.
\( 1 - \int_{0}^{2} \psi (r, \beta )^2 4 \pi r^2 dr = 0.107\)
g. Plot the wavefunction on the same graph as the potential energy.
