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Chemistry LibreTexts

458: Variation Method Using the Wigner Function - The Feshbach Potential

  • Page ID
    137725
  • Define potential energy:

    \[ V_0 = 2.5~~~ d = 0.5~~~ V(x) = V_0 tanh \left( \frac{x}{2} \right) ^2\]

    Display potential energy:

    Screen Shot 2019-02-18 at 11.35.04 AM.png

    Choose trial wave function:

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

    Calculate the Wigner distribution function:

    \[ W(x, p, \beta ) = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \psi \left(x + \frac{s}{2}, \beta \right) exp(isp) \psi (\left( x- \frac{s}{2}, \beta \right) ds~ \bigg|_{assume,~ \beta > 0}^{simplify} \rightarrow \frac{1}{ \pi} e^{ \frac{-1}{2} \frac{4 \beta ^2 x^2 + p^2}{ \beta}}\]

    Evaluate the variational integral:

    \[ E( \beta ) = \int_{- \infty}^{ \infty} \int_{- \infty}^{ \infty} W(x, p, \beta ) \left( \frac{p^2}{2} + V(x) \right)dx~dp\]

    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 and display the coordinate distribution function:

    \[ Px(x, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dp\]

    Screen Shot 2019-02-18 at 11.34.59 AM.png

    Classical turning point:

    \[ V_0 tanh \left( \frac{x}{2} \right)^2 = 1.484 \bigg|_{float,~3}^{solve,~x} \rightarrow \begin{pmatrix}
    -.511\\
    .511
    \end{pmatrix}\]

    Probability that tunneling is occurring:

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

    Calculate and display the momentum distribution function:

    \[ Pp(p, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dx\]

    Screen Shot 2019-02-18 at 11.34.54 AM.png

    Display the Wigner distribution function:

    N = 60 i = 0 .. N xi = \( -3 + \frac{6i}{N}\) j = 0 .. N pj = \( -5 + \frac{10j}{N}\) Wigneri, j = W( xi, pj, \(\beta\))

    Screen Shot 2019-02-18 at 11.34.49 AM.png