10.11: Linear Variational Method for a Particle in a Slanted 1D Box
- Page ID
- 136123
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Trial wavefunctions:
- \( \psi_{1} (x) = \sqrt{2} \sin( \pi x)\)
- \( \psi_{2} (x) = \sqrt{105} x (1-x)^{2}\)
Plot trial wavefunctions and potential energy. x = 0, .005 ... 1
Evaluate matrix elements:
\[\begin{align} S_{11} &= \int_{0}^{1} \psi _{1} (x)^{2} dx \\[4pt] &= 1 \end{align} \nonumber \]
\[\begin{align} S_{12} &= \int_{0}^{1} \psi _{1} (x) \psi _{2} (x) dx\\[4pt] &= 0.9347 \end{align} \nonumber \]
\[\begin{align} S_{22} &= \int_{0}^{1} \psi _{2} (x)^{2} dx \\[4pt] &= 1 \end{align} \nonumber \]
\( H_{11} = \int_{0}^{1} \psi _{1} (x) (- \frac{1}{2}) \frac{d^{2}}{dx^{2}} \psi _{1} (x) dx + \int_{0}^{1} \psi _{1} (x)~x~ \psi _{1} (x) dx\) \( H_{11} = 5.4348\)
\( H_{12} = \int_{0}^{1} \psi _{1} (x) (- \frac{1}{2}) \frac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{0}^{1} \psi _{1} (x)~x~ \psi _{2} (x) dx\) \( H_{12} = 5.0163\)
\( H_{22} = \int_{0}^{1} \psi _{2} (x) (- \frac{1}{2}) \frac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{0}^{1} \psi _{2} (x)~x~ \psi _{2} (x) dx\) \( H_{22} = 7.375\)
Solve the secular equations and normalization constraint for the energy and coefficients.
Seed values for energy and coefficients: E = 5 c1 = .5 c2 = .5
Given
\( (H_{11} - E S_{11})c_{1} + (H_{12} - E S_{12})c_{2} = 0\)
\( (H_{12} - E S_{12})c_{1} + (H_{22} - E S_{22})c_{2} = 0\)
\( c_{1}^{2} S_{11} + 2 c_{1} c_{2} S_{12} + c_{2}^{2} S_{22} = 1\)
\({\begin{pmatrix}
E \\
c_{1} \\
c_{2}
\end{pmatrix}} = Find(E, c_{1}, c_{2})\)
\({\begin{pmatrix}
E \\
c_{1} \\
c_{2}
\end{pmatrix}} = {\begin{pmatrix} 5.4328 \\
0.971\\
0.031
\end{pmatrix}}\)
Compare variational ground state to PIB ground state:
Calculate average position of the particle in the box:
\[ \int_{0}^{1} x \Phi (x)^{2} dx = 0.496 \nonumber \]
Calculate the probability that the particle is in the left half of the box:
\[ \int_{0}^{0.5} \Phi (x)^{2} dx = 0.5088 \nonumber \]