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10.33: Variational Method for the Feshbach Potential

  • Page ID
    136983
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    Define potential energy: V0 = 2.5 d = 0.5 \(V(x) = V_o \tanh \left( \dfrac{x}{d}\right)^2\)

    Display potential energy:

    Screen Shot 2019-02-15 at 12.00.50 PM.png

    Choose Gaussian trial wavefunction:

    \[ \psi (x, \beta ) = \left( \frac{2 \beta}{ \pi} \right) ^{ \frac{1}{4}} exp ( - \beta x^2) \nonumber \]

    Demonstrate that the trial wavefunction is normalized.

    \[ \int_{- \infty}^{ \infty} \psi (x, \beta )^2 dx~~~assume,~ \beta > 0 \rightarrow 1 \nonumber \]

    Evaluate the variational integral.

    \[ E( \beta ) = \int_{ - \infty}^{ \infty} \psi (x, \beta ) \frac{-1}{2} \frac{d^2}{dx^2} \psi (x, \beta ) dx + \int_{- \infty}^{ \infty} V(x) \psi (x, \beta )^2 dx \nonumber \]

    Minimize the energy integral with respect to the variational parameter, \( \beta\).

    \( \beta\) = 1 \( \beta\) = Minimize (E, \( \beta\)) \( \beta\) = 0.913 E( \( \beta\)) = 1.484

    Calculate the % error given that numerical integration of Schrödingerʹs equation (see next tutorial) yields E = 1.44949 Eh.

    \[ \frac{E( \beta ) - 1.44949}{1.44949} \times 100 = 2.36 \nonumber \]

    Display wavefunction in the potential well.

    Screen Shot 2019-02-15 at 12.05.40 PM.png

    Calculate the probability that tunneling is occurring.

    \[ V(x) = 1.484 |_{float,~3}^{solve,~x} \rightarrow {\begin{pmatrix} -1.511 \\ 0 .511 \end{pmatrix}} \nonumber \]

    \[ 2 \int_{0.511}^{ \infty} \psi (x, \beta )^2 dx = 0.329 \nonumber \]


    This page titled 10.33: Variational Method for the Feshbach Potential 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.