# 10.14: Variation Method for a Particle in a 1D Ice Cream Cone

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Define potential energy: V(x) := |x|

Display potential energy:

Choose trial wave function: $$\Psi (x, \beta ) := \sqrt{ \frac{ \beta}{2}} sech( \beta x)$$

$E ( \beta ) := \int_{- \infty}^{ \infty} \Psi (x, \beta ) \frac{1}{2} \frac{d^2}{dx^2} \Psi (x, \beta ) dx + \int_{- \infty}^{ \infty} V(x) \Psi (x, \beta )^{2} dx \nonumber$

Minimize the energy integral with respect to the variational parameter, $$\beta$$.

$$\beta$$ := 2 $$\beta$$ := Minimize( E, $$\beta$$) $$\beta$$ = 1.276 E( $$\beta$$) = 0.815

Display wave function in the potential well.

Calculate the probability that the particle is in the potential barrier.

$2 \int_{0}^{ \infty} \Psi (x, \beta )^2 dx = 1 \nonumber$

Define quantum mechanical tunneling.

Tunneling occurs when a quon (a quantum mechanical particle) has probability of being in a nonclassical region. In other words, a region in which the total energy is less than the potential energy.

Calculate the probability that tunneling is occurring.

$|x| = 0.815 |_{float,~4}^{solve,~x} \rightarrow {\begin{pmatrix} 0.8150 \\ -0.8150 \end{pmatrix}} \nonumber$

$2 \int_{0.815}^{ \infty} \Psi (x, \beta )^2 dx = 0.222 \nonumber$

Calculate the kinetic and potential energy contributions to the total energy.

Kinetic energy:

$\int_{- \infty}^{ \infty} \Psi (x, \beta ) -( \frac{1}{2}) \frac{d^2}{dx^2} \Psi (x, \beta ) dx = 0.272 \nonumber$

Potential energy:

$\int_{- \infty}^{ \infty} V(x) \Psi (x, \beta )^{2} dx = 0.543 \nonumber$

This page titled 10.14: Variation Method for a Particle in a 1D Ice Cream Cone is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.