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Chemistry LibreTexts

438: Variation Calculation on the 1D Hydrogen Atom Using a Gaussian Trial Wavefunction

  • Page ID
    136971
  • The energy operator for this problem is:

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

    The trial wave function is:

    \[ \psi (x, \alpha ) = 2 ( \frac{2 \alpha}{ \pi^{ \frac{1}{3}}}) (x) exp( - \alpha x^2)\]

    Evaluate the energy integral.

    \[ E( \alpha ) = \int_{x}^{ \alpha} - \frac{1}{2} \frac{d^2}{dx^2} \psi (x, \alpha ) dx + \int_{0}^{ \infty} \frac{-1}{x} \psi (x, \alpha )^2 dx |_{simplify}^{assume,~ \alpha > 0} \rightarrow \frac{-1}{2 \pi^{ \frac{1}{2}}} [ (-3) \pi^{ \frac{1}{2}} \alpha + (4) 2^{ \frac{1}{2}} \alpha ^{ \frac{1}{2}}]\]

    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, \( \alpha\)) \( \alpha\) = 0.2829 E ( \(\alpha\)) = -0.4244

    The exact ground state energy for the hydrogen atom is -0.5 Eh. Calculate the percent error.

    \[ \left| \dfrac{-0.5 - E( \alpha )}{-0.5} \right| = 15.1174\]

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

    Screen Shot 2019-02-13 at 1.22.06 PM.png

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

    \( \alpha\) = 1. Given:

    \[ \int_{0}^{a} \psi (x, \alpha )^2 dx = .95\]

    Find (a) = 2.6277

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

    \[ \alpha = 0.2829 \frac{d}{dx} | \psi (x, \alpha )| = 0 |_{float,~3}^{solve,~x} \rightarrow {\begin{pmatrix}
    -1.33 \\
    1.33
    \end{pmatrix}}\]

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

    \[ \int_{0}^{1.33} \psi ( \alpha , x)^2 dx = 0.3584\]

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

    \[ \begin{align*} T &= \int_{0}^{ \infty} \psi (x, \alpha ) \frac{1}{2} \frac{d^2}{dx^2} \psi (x, \alpha ) dx \\[4pt] &= 0.4244\end{align*}\]

    \[ \begin{align*} V &= \int_{0}^{ \infty} \frac{-1}{x} \psi (x, \alpha )^2 dx \\[4pt] &= -0.8488 \end{align*}\]

    \[ | \frac{V}{T}| = 2.00\]

    Calculate the probability that tunneling is occurring.

    Classical turning point:

    \[ E( \alpha ) = \frac{-1}{x}|_{float,~3}^{solve,~x} \rightarrow 2.36\]

    Tunneling probability:

    \[ \int_{2.36}^{ \infty} \psi (x, \alpha )^2 dx = 0.0978\]

    Screen Shot 2019-02-13 at 1.22.01 PM.png