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

  • Page ID
    136251
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    Define potential energy: V(x) := |x|

    Display potential energy:

    Screen Shot 2019-02-12 at 11.52.28 AM.png

    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.

    Screen Shot 2019-02-12 at 11.52.33 AM.png

    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.