10.10: Variation Method for a Particle in a Gravitational Field
- Page ID
- 136122
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The particle of unit mass in a gravitational field for which g = 1 has the energy operator shown below.
\( - \frac{1}{2} \frac{d^2}{dz^2} \blacksquare + z \blacksquare\)
\( \int_{0}^{ \infty} \blacksquare dz\)
The following trial wave function for this problem is:
\( \Phi ( \alpha, z) := 2 ( \frac{2 \alpha}{ \pi ^{ \frac{1}{3}}})^{ \frac{3}{4}} z~exp ( - \alpha z^{2})\)
Determine whether or not the wave function is normalized.
\( \int_{0}^{ \infty} \Psi ( \alpha , z)^{2} dz |_{simplify}^{assume,~ \alpha > 0} \rightarrow 1\)
Evaluate the variational energy integral.
\( E ( \alpha ) := \int_{0}^{ \infty} \Psi ( \alpha , z) - \frac{1}{2} \frac{d^2}{dz^2} \Phi ( \alpha , z) dz ... |_{simplify}^{assume,~ \alpha >0} \rightarrow \frac{1}{2 \pi ^{ \frac{1}{2}}} \frac{3 \pi ^{ \frac{1}{2}} \alpha ^{2} + 2 (2)^{ \frac{1}{2}} \alpha ^{ \frac{1}{2}}}{ \alpha} + \int_{0}^{ \infty} z \Psi ( \alpha , z)^{2} dz\)
Minimize the energy with respect to the variational parameter \( \alpha\) and report its optimum value and the ground‐state energy.
\( \alpha := 1\) \( \alpha := Minimize(E, \alpha )\) \( \alpha = 0.4136\) \( E( \alpha ) = 1.8611\) \( E_{exact} := 1.8558\)
Plot the wave function with the distance of the particle from the surface on the vertical axis.
Find that distance below which there is a 90% probability of finding the particle.
\( \alpha := 1\)
Given \( \int_{0}^{a} \Psi ( \alpha , z)^{2} dz = .90\)
Find (a) = 1.9440
Find the most probable value of the position of the particle from the surface.
\( \frac{d}{dz} \Psi (0.4136, z) = 0 |_{float,~3}^{solve,~z} \rightarrow {\begin{pmatrix}
-1.10 \\
1.10
\end{pmatrix}}\)
Calculate the probability that the particle will be found below the most probable distance from the surface.
\( \int_{0}^{1.10} \Psi ( \alpha , z)^{2} dz = 0.4279\)
Calculate the probability that tunneling is occurring: \( \int_{1.861}^{ \infty} \Psi ( \alpha , z)^{2} dz = 0.1256\)
Kinetic energy: \( \int_{0}^{ \infty} \Psi ( \alpha , z) - \frac{1}{2} \frac{d^2}{dz^2} \Psi ( \alpha , z) dz = 0.6204\)
Potential energy: \( \int_{0}^{ \infty} z \Psi ( \alpha , z)^{2} dz = 1.2407\)
What is the apparent virial theorem for this system: \( E = 3T = \frac{3}{2} V\)