10.19: Trigonometric Trial Wave Function for the 3D Harmonic Potential Well
- Page ID
- 136258
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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 wave function: \( \Psi (r, \beta) := \sqrt{ \frac{3 \beta^3}{ \pi^3}} sech( \beta r)\)
Integral: \( \int_{0}^{ \infty} \blacksquare 4 \pi r^2 dr\)
Kinetic energy operator: \( T = \frac{-1}{2r} \frac{d^2}{dr^2} (r \blacksquare )\)
Potential energy operatory: \( V = \frac{1}{2} k r^2\)
a. Demonstrate the wave function is normalized.
\( \int_{0}^{ \infty} \Psi (r, \beta )^2 4 \pi r^2 dr |_{simplify}^{assume,~ \beta > 0} \rightarrow 1\)
b. Evaluate the variational integral.
\( E( \beta ) := \int_{0}^{ \infty} \Psi (r, \beta ) [ \frac{-1}{2r} \frac{d^2}{dr^2} (r \Psi (r, \beta ))] 4 \pi r^2 dr + \int_{0}^{ \infty} \Psi (r, \beta ) \frac{1}{2} r^2 \Psi (r, \beta ) 4 \pi r^2 dr\)
c. Minimize the energy with respect to the variational parameter \( \beta\).
\( \beta\) := 1 \( \beta\) := Minimize (E, \( \beta\)) \( \beta\) = 1.471 E( \(\beta \)) = 1.597
d. The exact ground state energy for the 3D harmonic oscillator is 1.5 Eh. Calculate the percent error.
\( \frac{E( \beta ) - 1.5}{1.5} = 6.488\)%
e. Compare the optimized trial wave function with the exact solution by plotting the radial distribution functions.
\( \Phi (r) := ( \frac{1}{ \pi})^{ \frac{3}{4}} exp( \frac{r^2}{2})\)
h. Calculate the overlap integral between the trial wave function and the exact wave function.
\( \int_{0}^{ \infty} \Psi (r, \beta ) \Phi (r) 4 \pi r^2 dr = 0.989\)
i. Calculate the probability that tunneling is occurring.
Classical turning point:
\( 1.597 = \frac{1}{2} r^2 |_{float,~3}^{solve,~r} \rightarrow {\begin{pmatrix}
-1.79 \\
1.79
\end{pmatrix}}\)
Tunneling probability:
\( \int_{1.79}^{ \infty} \Psi (r, \beta )^2 4 \pi r^2 dr = 12.598\)%