10.42: Variation Method Using the Wigner Function - The Feshbach Potential
- Page ID
- 137725
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
Define potential energy:
\[ V_0 = 2.5~~~ d = 0.5~~~ V(x) = V_0 tanh \left( \frac{x}{2} \right) ^2 \nonumber \]
Display potential energy:
Choose trial wave function:
\[ \psi (x, \beta ) = \left( \frac{2 \beta}{ \pi} \right)^{ \frac{1}{4}} exp( - \beta x^2) \nonumber \]
Calculate the Wigner distribution function:
\[ W(x, p, \beta ) = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \psi \left(x + \frac{s}{2}, \beta \right) exp(isp) \psi (\left( x- \frac{s}{2}, \beta \right) ds~ \bigg|_{assume,~ \beta > 0}^{simplify} \rightarrow \frac{1}{ \pi} e^{ \frac{-1}{2} \frac{4 \beta ^2 x^2 + p^2}{ \beta}} \nonumber \]
Evaluate the variational integral:
\[ E( \beta ) = \int_{- \infty}^{ \infty} \int_{- \infty}^{ \infty} W(x, p, \beta ) \left( \frac{p^2}{2} + V(x) \right)dx~dp \nonumber \]
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 and display the coordinate distribution function:
\[ Px(x, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dp \nonumber \]
Classical turning point:
\[ V_0 tanh \left( \frac{x}{2} \right)^2 = 1.484 \bigg|_{float,~3}^{solve,~x} \rightarrow \begin{pmatrix}
-.511\\
.511
\end{pmatrix} \nonumber \]
Probability that tunneling is occurring:
\[ 2 \int_{0.511}^{ \infty} Px (x, \beta ) dx = 0.329 \nonumber \]
Calculate and display the momentum distribution function:
\[ Pp(p, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dx \nonumber \]
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, \(\beta\))