So, in class we talked about how the Schrödinger equation,

$latex displaystyle (1) , , , , i hbar psi_{t} = frac{hbar^{2}}{2m}triangle psi + V psi$

defines a wave function, $latex psi : mathbb{R}^{n+1} to mathbb{C}$, such that the quantum mechanical particle subjected to the potential field, $latex V:mathbb{R}^{n} to mathbb{R}$, has an associated probability density defined by $latex psi psi^{*}: mathbb{R}^{n+1}to mathbb{R}^{+}_{0}$. For a hydrogen atom $latex V(r) = -e^{2}/4pi epsilon_{0} r$ where $latex e$ is the charge of the nucleus and $latex epsilon_{0}$ is the permittivity of the vacuum between the atomic core and the electron. Using techniques from partial and ordinary differential equations, it can be shown that the ground state, $latex 1s^1$ orbital, wave function is given by

$latex displaystyle (2) , , , , psi(r) = frac{1}{sqrt{pi a_{0}^{3}}} e^{-r/a_{0}}$

where $latex a_{0}approx 5.27 times 10^{-11} m$ is the Bohr radius. There are other excited…