9.22: Numerical Solutions for a Modified Harmonic Potential
- Page ID
- 137745
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This tutorial deals with the following potential function:
\[ V(x, d) = \bigg| _{ \infty ~otherwise}^{ \frac{1}{2} k(x-d)^2~if~x \geq 0 + d \leq 0} \nonumber \]
If d = 0 we have the harmonic oscillator on the half-line with eigenvalues 1.5, 3.5, 5.5, ... for k = \( \mu\) = 1. For large values of d we have the full harmonic oscillator problem displaced in the x-direction by d with eigenvalues 0.5, 1.5, 2.5, ... for k = \( \mu\) = 1. For small to intermediate values of d the potential can be used to model the interaction of an atom or molecule with a surface.
Integration limit: xmax = 10
Effective mass: \( \mu\) = 1
Force constant: k = 1
Potential energy minimum: d = 5
Potential energy:
\[ V(x,d) = \frac{k}{2} (x-d)^2 \nonumber \]
Integration algorithm:
Given
\[ \nonumber \]
Normalize wavefunction:
\[ \psi (x) = \frac{ \psi (x)}{ \sqrt{ \int_{0}^{x_{max}} \psi (x)^2 dx}} \nonumber \]
Energy guess: E = 0.5
Calculate average position:
\[ X_{avg} = \int_{0}^{x_{max}} \psi (x) x \psi (x) dx = 5 \nonumber \]
Calculate potential and kinetic energy:
\[ V_{avg} = \int_{0}^{x_{max}} \psi (x) V(x,d) \psi (x) dx = 0.25 \nonumber \]
\[ T_{avg} = E - V_{avg} = 0.25 \nonumber \]
Exercises:
- For d = 0, k = \( \mu\) = 1 confirm that the first three energy eigenvalues are 1.5, 3.5 and 5.5 Eh. Start with xmax = 5, but be prepared to adjust to larger values if necessary. xmax is effectively infinity.
- For d = 5, k = \( \mu\) = 1 confirm that the first three energy eigenvalues are 0.5, 1.5 and 2.5 Eh. Start with xmax = 10, but be prepared to adjust to larger values if necessary.
- Determine and compare the virial theorem for the exercises above.
- Calculate the probability that tunneling is occurring for the ground state for the first two exercises. (Answers: 0.112, 0.157)