10.12: Variation Method for a Particle in a Semi-Infinite Potential Well
- Page ID
- 136124
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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 the variational approach to the particle in the semi-infinite potential well.
Kinetic energy operator: \( - \frac{1}{2} \frac{d^2}{dx^2} \blacksquare\)
Integral: \( \int_{0}^{ \infty} \blacksquare dx\)
Potential energy: \( V(x) := if[( x \leq 2), 0 , 2]\)
Trial wave function: \( \Phi (x, \beta ) := 2 \beta ^{ \frac{3}{2}}~x~exp(- \beta x)\)
If the trial wave function is not normalized, normalize it.
\( \int_{0}^{ \infty} \Phi (x, \beta )^{2} dx~ assume,~ \beta > 0 \rightarrow 1\)
Evaluate the variational energy integral.
\( E( \beta ) := \int_{0}^{ \infty} \Phi (x, \beta ) (- \frac{1}{2}) \frac{d^2}{dx^2} \Phi (x, \beta ) dx ... |_{simplify}^{assume,~ \beta >0} \rightarrow \frac{1}{2} \beta^{2} + 16 \beta^{2} e^{-4 \beta} + 8 \beta e^{-4 \beta} + 2 e^{- 4 \beta} + \int_{2}^{ \infty} 2 \Phi (x, \beta )^{2} dx\)
Minimize the energy with respect to \( \beta\):
\( \beta\) := .3 \( \beta\) := Minimize \((E, \beta )\) \( \beta = 1.053\) \( E( \beta ) = 0.972\)
Display optimized trial wave function and potential energy:
Calculate average position and most probable position of the particle:
\( \int_{0}^{ \infty} x \Phi (x, \beta )^{2} dx = 1.425\)
\( \frac{d}{dx} \Phi (x, \beta ) = 0 |_{solve,~x}^{float,~3} \rightarrow \frac{1}{ \beta} = 0.95\)
Calculate the probability of the particle in the barrier.
\( \int_{2}^{ \infty} \Phi (x, \beta )^{2} dx = 20.891%\)
Calculate the potential energy, and the kinetic energy.
\( V := \int_{2}^{ \infty} 2 \Phi (x, \beta )^{2} dx\) \(V = 0.418\)
\( T := E ( \beta ) - V\) \( T = 0.554\)