10.39: 455. Variation Method Using the Wigner Function- V(x) = |x|
- Page ID
- 137721
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \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:
\[ V(x) = |x| \nonumber \]
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:
\[ E( \beta ) = \int_{- \infty}^{ \infty} \int_{- \infty}^{ \infty} W(x, p, \beta ) \left( \frac{p^2}{2} + V(x) \right)dx~dp~ \bigg|_{simplify}^{assume,~ \beta > 0} \rightarrow \frac{1}{2} \frac{ \beta ^{ \frac{3}{2}} \pi ^{ \frac{1}{2}} + 2 ^{ \frac{1}{2}}}{ \pi ^{ \frac{1}{2}} \beta ^{ \frac{1}{2}}} \nonumber \]
Minimize the energy integral with respect to the variational parameter, \( \beta\).
\( \beta\) = 1 \( \beta\) = Minimize (E, \( \beta\)) \( \beta\) = 0.542 E( \(\beta\)) = 0.813
Calculate and display the coordinate distribution function:
\[ Px(x, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dp \nonumber \]
Classical turning points: +/- 0.813
Probability that tunneling is occurring:
\[ 2 \int_{0.813}^{ \infty} Px (x, \beta ) dx = 0.231 \nonumber \]
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, \(\beta\))