Expectation Values in Hydrogen States

An electron in the Coulomb field of a proton is in the state described by the wave function ${1\over 6}(4\psi_{100}+3\psi_{211}-i\psi_{210}+\sqrt{10}\psi_{21-1})$. Find the expected value of the Energy, $L^2$, $L_z$, and $L_y$.

First check the normalization.

$$\bgroup\color{black} {\vert 4\vert^2+\vert 3\vert^2+\vert-i\vert^2+\vert\sqrt{10}\vert^2\over 36}={36\over 36}=1 \egroup$$

The terms are eigenstates of $E$, $L^2$, and $L_z$, so we can easily compute expectation values of those operators.

$$\begin{eqnarray*} E_n&=&-{1\over 2}\alpha^2\mu c^2{1\over n^2} \\ \langle E\ran... ...a^2\mu c^2{21\over 36}=-{1\over 2}\alpha^2\mu c^2{7\over 12} \\ \end{eqnarray*}$$

Similarly, we can just square probability amplitudes to compute the expectation value of $L^2$. The eigenvalues are $\ell(\ell+1)\hbar^2$.

$$\bgroup\color{black} \langle L^2\rangle=\hbar^2{16 (0)+9(2)+1(2)+10(2)\over 36}={10\over 9}\hbar^2 \egroup$$

The Eigenvalues of $L_z$ are $m\hbar$.

$$\bgroup\color{black} \langle L_z\rangle=\hbar{16 (0)+9(1)+1(0)+10(-1)\over 36}={-1\over 36}\hbar \egroup$$

Computing the expectation value of $L_y$ is harder because the states are not eigenstates of $L_y$. We must write $L_y=(L_+ - L_-)/2i$ and compute.

$$\begin{eqnarray*} \langle L_y\rangle&=&{1\over 72i} \langle 4\psi_{100}+3\psi_{... ...qrt{2}\hbar\over 72i} ={(2\sqrt{5}-3\sqrt{2})\hbar\over 36} \\ \end{eqnarray*}$$