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Chemistry LibreTexts

429: Variation Method for a Particle in a Box with an Internal Barrier

  • Page ID
    136125
  • \( \psi _{1} (x) = \sqrt{2} \sin{ \pi x}\)

    \( \psi _{2} (x) = \sqrt{2} \sin{3 \pi x}\)

    \( V(x) = if [(x \geq .45) (x \leq .55), 3,0]\)

    Plot trial wavefunctions and potential energy.

    Screen Shot 2019-02-11 at 1.55.32 PM.png

    Evaluate matrix elements for \(100 E_h\) internal barrier:

    \( S_{11} = \int_{0}^{1} \psi _{1} (x)^{2} dx\) \( S_{11} = 1\)

    \( S_{12} = \int_{0}^{1} \psi _{1} (x) \psi _{2} (x) dx\) \( S_{12} = 0\)

    \( S_{22} = \int_{0}^{1} \psi _{2} (x)^{2} dx\) \( S_{22} = 1\)

    \[ \begin{align} H_{11} &= \int_{0}^{1} \psi _{1} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{1} (x) dx + \int_{.45}^{.55} \psi _{1} (x)~100~ \psi _{1} (x) dx\\[4pt] & \approx 24.7711\end{align}\]

    \[\begin{align} H_{12} &= \int_{0}^{1} \psi _{1} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{.45}^{.55} \psi _{1} (x)~100~ \psi _{2} (x) dx \\[4pt] & \approx -19.1912\end{align}\]

    \[\begin{align} H_{22} &= \int_{0}^{1} \psi _{2} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{.45}^{.55} \psi _{2} (x)~100~ \psi _{2} (x) dx \\[4pt] &\approx 62.9972 \end{align}\]

    Solve the secular equations and normalization constraint for the energy and coefficients.

    Seed values for energy and coefficient: E = 5 c1 = .5 c2 = .5

    Given

    \( (H_{11} - E S_{11})c_{1} + (H_{12} - E S_{12})c_{2} = 0\)

    \( (H_{12} - E S_{12})c_{1} + (H_{22} - E S_{22})c_{2} = 0\)

    \( c_{1}^{2} S_{11} + 2 c_{1} c_{2} S_{12} + c_{2}^{2} S_{22} = 1\)

    \({\begin{pmatrix}
    E \\
    c_{1} \\
    c_{2}
    \end{pmatrix}} = Find(E, c_{1}, c_{2})\)

    \({\begin{pmatrix}
    E \\
    c_{1} \\
    c_{2}
    \end{pmatrix}} = {\begin{pmatrix} 16.7989 \\
    0.9235\\
    0.3836
    \end{pmatrix}}\)

    Plot variational results:

    \( \Phi (x) = c_{1} \Phi _{1} (x) + c_{2} \psi_{2} (x)\)

    Screen Shot 2019-02-11 at 2.02.23 PM.png

    Calculate the probability the particle is in the barrier:

    \( \int_{0.45}^{0.55} \Phi (x)^{2} dx = 0.0605\)

    Calculate potential and kinetic energy:

    \( V = 100 \int_{0.45}^{0.55} \Phi (x)^{2} dx\) \( V = 6.0541\)

    \( T = E - V\)

    \( T = 10.7448\)