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Chemistry LibreTexts

437: Variation Calculation on the 1D Hydrogen Atom Using a Trigonometric Trial Wave Function

  • Page ID
    136970
  • The energy operator for this problem is:

    \[ \frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare\]

    The trial wave function:

    \[ \Psi ( \alpha , x) := \frac{ \sqrt{12 \alpha ^3}}{ \pi} (x) sech( \alpha , x)\]

    Evaluate the variational energy integral.

    \[ E( \alpha ) := \int_{0}^{ \infty} \Psi ( \alpha , x) - \frac{1}{2} \frac{d^2}{dx^2} \Psi ( \alpha , x) dx + \int_{0}^{ \infty} \frac{-1}{x} \Psi ( \alpha , x)^2 dx |_{simplify}^{assume,~ \alpha > 0} \rightarrow \frac{1}{6} \alpha \frac{12 \alpha _ \alpha \pi^2 - 72 ln (2)}{ \pi ^2}\]

    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, \( \alpha\)) \( \alpha\) = 1.1410 E( \( \alpha\)) = -0.4808

    The exact ground-state energy for the hydrogen atom is -.5 Eh. Calculate the percent error.

    \[ | \frac{-.5 - E( \alpha )}{-.5}| = 3.8401\]

    Plot the optimized trial wave function and the exact solution, \( \Phi (x) := 2 (x) exp (-x)\).

    Screen Shot 2019-02-13 at 12.43.24 PM.png

    Find the distance from the nucleus within which there is a 95% probability of finding the electron.

    \( \alpha\) := 1. Given:

    \[ \int_{0}^{a} \Psi ( \alpha , x)^2 dx = 0.95\]

    Find (a) = 2.8754

    Find the most probable value of the position of the electron from the nucleus.

    \[ \alpha := 1.1410~~~~ \frac{d}{dx}|\frac{ \sqrt{12 \alpha ^3}}{ \pi} (x) sech( \alpha , x)| = 0~|_{float,~3}^{solve,~x} \rightarrow 1.05\]

    Calculate the probability that the electron will be found between the nucleus and the most probable distance from the nucleus.

    \[ \int_{0}^{1.05} \Psi ( \alpha , x)^2 dx = 0.3464\]

    Break the energy down into kinetic and potential energy contributions. Is the virial theorem obeyed?

    \[ T := \int_{0}^{ \infty} \Psi ( \alpha , x) \frac{-1}{2} \frac{d^2}{dx^2} \Psi ( \alpha , x) dx~~~T = 0.4808\]

    \[ V := \int_{0}^{ \infty} \frac{-1}{x} \Psi ( \alpha , x)^2 dx~~~ V = -0.9616\]

    \[ | \frac{V}{T}| = 2.0000\]

    Use the exact result to discuss the weakness of this trial function.

    Eexact := -0.5

    Using the virial theorem we know: Texact := 0.500 Vexact := -1.00

    Calculate the difference between the variational results and the exact calculation:

    E( \( \alpha\)) - Eexact = 0.0192

    T - Texact = -0.0192

    V - Vexact = 0.0384

    The variational wave function yields a lower kinetic energy, but at the expense of a potential energy that is twice as unfavorable as the kinetic energy result is favorable.

    Calculate the probability that tunneling is occurring.

    Classical turning point:

    \[ E( \alpha ) = \frac{-1}{x} |_{float,~3}^{solve,~x}\]

    Tunneling probability:

    \[ \int_{2.08}^{ \infty} \Psi ( \alpha , x)^2 dx = 0.1783\]

    Screen Shot 2019-02-13 at 12.43.29 PM.png