10.17: Variation Method for the Harmonic Oscillator
- Page ID
- 136255
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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 exercise deals with a variational treatment for the ground state of the simple harmonic oscillator which is, of course, an exactly soluble quantum mechanical problem.
The energy operator for a harmonic oscillator with unit effective mass and force constant is:
\( H = \frac{-1}{2} \frac{d^2}{dx^2} \blacksquare + \frac{x^2}{2} \blacksquare\)
The following trial wavefunction is selected:
\[ \psi (x, \beta ) = \frac{1}{1 + \beta x^2} \nonumber \]
The variational energy integral is evaluated (because of the symmetry of the problem it is only necessary to integrate from 0 to ∞, rather than from ‐∞ to ∞:
\( E( \beta ) = \dfrac{ \int_{0}^{ \infty} \psi (x, \beta ) \frac{-1}{2} \frac{d^2}{dx^2} \psi (x, \beta ) dx + \int_{0}^{ \infty} \psi (x, \beta ) \frac{x^2}{2} \psi (x, \beta ) dx}{ \int_{0}^{ \infty} \psi (x, \beta )^2 dx} |^{assume,~ \beta > 0}_{simplify} \rightarrow \frac{1}{4} \frac{ \beta ^2 + 2}{ \beta}\)
The energy integral is minimized with respect to the variational parameter:
\( \beta \) := 1 \( \beta \) := Minimize (E, \( \beta\)) \( \beta\) = 1.414 E( \(\beta\)) = 0.707
The % error is calculated given that the exact result is \(0.50 E_h\).
\( \frac{E( \beta ) - 0.5}{0.5} = 41.421\)%
The optimized trial wavefunction is compared with the SHO ground-state eigenfunction.
Now a second trial function is chosen:
\( \psi (x, \beta ) := \frac{1}{(1 + \beta x^2)^2}\)
Evaluate the variational energy integral:
\( E( \beta ) := \frac{ \int_{0}^{ \infty} \psi (x, \beta ) \frac{-1}{2} \frac{d^2}{dx^2} \psi (x, \beta ) dx + \int_{0}^{ \infty} \psi (x, \beta ) \frac{x^2}{2} \psi (x, \beta ) dx}{ \int_{0}^{ \infty} \psi (x, \beta )^2 dx} |^{assume,~ \beta > 0}_{simplify} \rightarrow \frac{1}{10} \frac{7 \beta ^2 + 1}{ \beta}\)
Minimize the energy integral with respect to the variational parameter:
\( \beta \) := 1 \( \beta \) := Minimize (E, \( \beta\)) \( \beta\) = 0.378 E( \(\beta\)) = 0.529
Calculate the % error given that the exact result is 0.50 Eh.
\( \frac{E( \beta ) - 0.5}{0.5} = 5.83\)%
The optimized trial wavefunction is compared with the SHO ground-state eigenfunction.
Suggestion: Continue this exercise with the following trial wavefunction and interpret the improved agreement with the exact solution.
\( \psi (x, \beta ) - \frac{1}{(1 + \beta x^2)^n}\)
where n is an integer greater than 2.