Skip to main content
Chemistry LibreTexts

10.6: Variation Method for a Particle in a Symmetric 1D Potential Well

  • Page ID
    135912
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Definite potential energy: \( V(x) := if[(x \geq -1) \cdot (x \leq 1), 0, \sqrt{ |x| - 1}]\)

    Display potential energy:

    Screen Shot 2019-02-08 at 11.43.19 AM.png

    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.

    Screen Shot 2019-02-08 at 11.43.25 AM.png

    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:

    \( \int_{- \infty}^{ \infty} V(x) \cdot \Psi (x, \beta )^{2} dx = 0.131\)


    This page titled 10.6: Variation Method for a Particle in a Symmetric 1D Potential Well is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.