Question 6.4: a) Find (r^2) for all four of the degenerate n=2 states of a...

From Equation 6.47 [\left\langle \begin{matrix} \acute{n}\acute{\ell}\acute{m} \begin{vmatrix} \hat{f} \\ \end{vmatrix} n\ell m \\ \end{matrix} \right\rangle =\delta _{\ell \acute{\ell}}\delta _{m \acute{m}}\left\langle \begin{matrix}\acute{n}\ell \begin{Vmatrix} f \\ \end{Vmatrix} n\ell \\ \end{matrix} \right\rangle ] we have, for the states with \ell=1 , the following equality:

\left\langle \begin{matrix} 211\begin{vmatrix} r^ {2} \\ \end{vmatrix} 211 \\ \end{matrix} \right \rangle =\left\langle \begin{matrix} 210\begin {vmatrix} r^{2}\\ \end{vmatrix} 210 \\ \end {matrix} \right \rangle=\left\langle \begin{matrix} 21-1\begin {vmatrix} r^{2}\\ \end{vmatrix} 21-1 \\ \end{matrix} \right\rangle=\left\langle \begin {matrix}21 \begin {Vmatrix} r^{2}\\ \end{Vmatrix} 21 \\ \end{matrix} \right\rangle

To calculate the reduced matrix element we simply pick any one of these expectation values:

\left\langle \begin{matrix}21 \begin {Vmatrix} r^{2}\\ \end{Vmatrix} 21 \\ \end {matrix} \right\rangle =\left\langle \begin {matrix} 210\begin{vmatrix} r^{2}\\ \end {vmatrix} 210 \\ \end{matrix} \right\rangle =\int{r^{2}\begin{vmatrix} \psi_{210}(r)\\ \end{vmatrix}^{2} }d^{3}r=\int_{0}^{\infty }{r^{4}\begin{vmatrix} R_{21}(r)\\ \end{vmatrix} } ^{2}dr\int\begin{vmatrix} Y^{0}_{1}(\theta ,\phi )\\ \end{vmatrix}^{2} d\Omega .

The spherical harmonics are normalized (Equation 4.31 [\int_{0}^{\infty }{\begin{vmatrix} R \\ \end{vmatrix} ^{2}r^{2}} dr=1 and \int_{0}^{\pi}\int_{0}^{2\pi}{} {\begin{vmatrix} Y\\ \end{vmatrix}^{2}\sin\theta } d\theta d\phi =1 ]), so the angular integral is 1, and the radial functions are R_{n\ell} listed in Table 4.7, giving

\left\langle \begin{matrix}21 \begin{Vmatrix} r^{2}\\ \end{Vmatrix} 21 \\ \end{matrix} \right\rangle =\int_{0}^{\infty }r^{4}\frac{1}{24 a^{3}} \frac{r^{2}}{a^{2}}e^{-r/a}dr=30a^{2}

That determines three of the expectation values. The final expectation value is

\left\langle \begin{matrix}20 \begin{Vmatrix} r^{2}\\ \end{Vmatrix} 20 \\ \end{matrix} \right \rangle=\left\langle \begin{matrix} 200\begin {vmatrix} r^{2}\\ \end{vmatrix} 200 \\ \end{matrix} \right\rangle =\int{r^{2}\begin{vmatrix} \psi_{210}(r)\\ \end{vmatrix}^{2} }d^{3}r

=\int_{0}^{\infty }{r^{4}\begin{vmatrix} R_{20}(r)\\ \end{vmatrix} } ^{2}dr\int \begin {vmatrix} Y^{0}_{0}(\theta ,\phi )\\ \end{vmatrix} ^{2} d\Omega .

= \int_{0}^{\infty }r^{4}\frac{1}{2a^{3}} \left (1- \frac{1}{2}\frac{r}{a} \right)^{2}e^{-r/a}dr= 42a^{2}

Summarizing:

\left\langle \begin{matrix} 200\begin{vmatrix} r^{2}\\ \end{vmatrix} 200 \\ \end{matrix} \right\rangle =42a^{2},\left.\begin{matrix}\left\langle \begin {matrix} 211\begin{vmatrix} r^{2}\\ \end{vmatrix} 211 \\ \end{matrix} \right\rangle \\ \left\langle \begin {matrix} 210\begin{vmatrix} r^{2}\\ \end{vmatrix} 210 \\ \end{matrix} \right\rangle \\ \left\langle \begin {matrix} 21-1\begin{vmatrix} r^{2}\\ \end{vmatrix} 21-1 \\ \end{matrix} \right\rangle \end{matrix} \right\} =30a^{2}    (6.48)

(b) Find the expectation value of for an electron in the superposition state

|\psi〉=\frac{1}{\sqrt{2}}(|200〉-i|211〉)

We can expand the expectation value as

〈\psi| r^{2}|\psi〉 = \frac{1}{2 }( 〈 200|+i 〈 211|)r^{2}(| 200〉-i| 211〉)=\frac{1}{2 }(〈 200| r^{2}| 200〉 +i〈 211|r^{2}| 200〉-i〈 200|r^{2}| 211〉+〈 211|r^{2}| 211〉)

From Equation 6.47 we see that two of these matrix elements vanish, and

\left\langle \begin{matrix}\psi\begin{vmatrix} r^{2} \\ \end{vmatrix} \psi\\ \end{matrix} \right \rangle= \frac{1}{2} \left(\left\langle \begin {matrix} 20\begin {Vmatrix} r^{2} \\ \end {Vmatrix} 20\\\end {matrix} \right\rangle +\left \langle \begin {matrix} 21\begin {Vmatrix} r^{2} \\ \end {Vmatrix} 21\\\end {matrix} \right\rangle \right) =36a^{2}      (6.49)