Question 4 of 7: Superposition State of Hydrogen — Probabilities and Expectation Values

Question 4: Superposition State of Hydrogen — Probabilities and Expectation Values (20 marks: (a) 7, (b) 7, (c) 6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\Psi(0)=\frac{1}{\sqrt{14}}\left[2\psi_{100}-3\psi_{200}+\psi_{322}\right]$, a superposition of three orthonormal hydrogen eigenstates with coefficients $c_1=2/\sqrt{14}$ ($n{=}1,l{=}0,m{=}0$), $c_2=-3/\sqrt{14}$ ($n{=}2,l{=}0,m{=}0$), $c_3=1/\sqrt{14}$ ($n{=}3,l{=}2,m{=}2$).

Find. $P(100)$, $P(200)$, $P(322)$, $P(\text{other})$; $\langle E\rangle$; $\langle L^2\rangle$ and $\langle L_z\rangle$.

Approach. Because the three basis states are orthonormal energy (and angular-momentum) eigenstates, each measurement probability is simply $|c_n|^2$, and every expectation value is the corresponding $|c_n|^2$-weighted sum of eigenvalues.

1. Part (a) — measurement probabilities. For a superposition of orthonormal eigenstates, the Born rule gives $P(\text{state }i)=|c_i|^2$: $$ P(100)=\left(\frac{2}{\sqrt{14}}\right)^2=\frac{4}{14}=\frac{2}{7},\qquad P(200)=\left(\frac{3}{\sqrt{14}}\right)^2=\frac{9}{14},\qquad P(322)=\left(\frac{1}{\sqrt{14}}\right)^2=\frac{1}{14}. $$ Since $\Psi$ is normalized ($|c_1|^2+|c_2|^2+|c_3|^2=(4+9+1)/14=1$), no amplitude remains for any other eigenstate: $$ \boxed{P(\text{other})=1-\left(\frac{2}{7}+\frac{9}{14}+\frac{1}{14}\right)=0.} $$
2. Part (b) — expectation value of energy. The hydrogen energy levels are $E_n=E_1/n^2$ with $E_1=-13.6$ eV. Only $n$ (not $l,m$) enters the energy, so $$ \langle E\rangle = |c_1|^2E_1+|c_2|^2E_2+|c_3|^2E_3 = \frac{4}{14}(-13.6)+\frac{9}{14}\left(\frac{-13.6}{4}\right)+\frac{1}{14}\left(\frac{-13.6}{9}\right)\ \text{eV}. $$ $$ \boxed{\langle E\rangle \approx -6.18\ \text{eV}} $$ (between $E_2=-3.4$ eV and $E_1=-13.6$ eV, as expected since the $n=2$ term dominates the mixture).
3. Part (c) — expectation values of $L^2$ and $L_z$. $\psi_{100}$ and $\psi_{200}$ both have $l=0$ (so $L^2=L_z=0$ for those terms); only $\psi_{322}$ ($l=2,m=2$) contributes, with eigenvalues $L^2=l(l+1)\hbar^2=6\hbar^2$ and $L_z=m\hbar=2\hbar$: $$ \langle L^2\rangle = |c_3|^2\cdot6\hbar^2 = \frac{1}{14}(6\hbar^2) $$ $$ \boxed{\langle L^2\rangle = \frac{3\hbar^2}{7}}\qquad\qquad \langle L_z\rangle = |c_3|^2\cdot2\hbar=\frac{1}{14}(2\hbar) \qquad \boxed{\langle L_z\rangle = \frac{\hbar}{7}} $$