10.6: Variation Method for a Particle in a Symmetric 1D Potential Well
- Page ID
- 135912
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definite potential energy: \( V(x) := if[(x \geq -1) \cdot (x \leq 1), 0, \sqrt{ |x| - 1}]\)
Display potential energy:
Choose trial wave function: \( \Psi (x, \beta ) := ( \frac{ 2 \cdot \beta}{ \pi} )^{ \frac{1}{4}} \cdot exp (- \beta \cdot x^{2})\)
Evaluate the variational integral:
\( E ( \beta ) := \int_{- \infty}^{ \infty} \Psi (x, \beta) \cdot - \frac{1}{2} \cdot \frac{d^{2}}{dx^{2}} \Psi (x, \beta ) dx + \int_{- \infty}^{ \infty} V(x) \cdot \Psi (x, \beta )^{2} dx\)
Minimize the energy integral with respect to the variational parameter, \( \beta\).
β := .2 β := Minimize(E, β) β := 0.363 E(β) = 0.313
Display wave function in the potential well.
Calculate the probability that the particle is in the potential barrier.
\( 2 \cdot \int_{1}^{ \infty} \Psi(x, \beta )^{2} dx = 0.228\)
Define quantum mechanical tunneling.
Tunneling occurs when a quon (a quantum mchanical particle) has probability of being in a nonclassical region. In other words, a region in which the total energy is less than the potential energy.
Calculate the probability that tunneling is occurring.
\( |x| - 1 = 0.313^{2} |_{solve,~x}^{float,~4} \rightarrow {\begin{pmatrix}
1.098 \\
-1.098
\end{pmatrix}}\)
\( 2 \cdot \int_{1.098}^{ \infty} \Psi (x, \beta )^{2} dx = 0.186\)
Calculate the kinetic and potential energy contributions to the total energy.
Kinetic energy:
\( \int_{- \infty}^{ \infty} \Psi (x, \beta ) \cdot - \frac{1}{2} \cdot \frac{d^{2}}{dx^{2}} \Psi (x, \beta ) dx = 0.182\)
Potential energy:
\( \int_{- \infty}^{ \infty} V(x) \cdot \Psi (x, \beta )^{2} dx = 0.131\)