The Expectation Value of \bgroup\color{black}$v_r$\egroup in the Ground State

For \bgroup\color{black}$\ell=0$\egroup, there is no angular dependence to the wavefunction so no velocity except in the radial direction. So it makes sense to compute the radial component of the velocity which is the full velocity.

We can find the term for \bgroup\color{black}${p_r^2\over 2m}$\egroup in the radial equation.

\langle \psi_{100}\vert(v_r)^2\vert\psi_{100}\rangle&=&\int\li...
...\right) \\
&=&{\hbar^2\over m^2}\left({Z\over a_0}\right)^2 \\

Since \bgroup\color{black}$a_0={\hbar\over\alpha mc}$\egroup, we get

\begin{displaymath}\bgroup\color{black} \langle \psi_{100}\vert(v_r)^2\vert\psi_{100}\rangle=Z^2\alpha^2 c^2 \egroup\end{displaymath}

For \bgroup\color{black}$Z=1$\egroup, the RMS velocity is \bgroup\color{black}$\alpha c$\egroup or

\begin{displaymath}\bgroup\color{black} \beta=\alpha={1\over 137} \egroup\end{displaymath}

We can compute the expected value of the kinetic energy.

\begin{displaymath}\bgroup\color{black}K.E.={1\over 2}m v^2={\hbar^2\over 2m}{Z^2\over a_0^2}={1\over 2}Z^2\alpha^2mc^2=-E_{100} \egroup\end{displaymath}

This is what we expect from the Virial theorem.

Jim Branson 2013-04-22