(a) Construct the wave function for hydrogen in the state n =4 , l = 3 , m = 3 . Express your answer as a function of the spherical coordinates r, θ, and ϕ.
(b) Find the expectation value of r in this state. (As always, look up any nontrivial integrals.)
(c) If you could somehow measure the observable L_{x}^{2}+L_{y}^{2} on an atom in this state, what value (or values) could you get, and what is the probability of each?
(a) From Tables 4.3 and 4.7,
| Table 4.3: The first few spherical harmonics Y_{\ell}^{m}(\theta, \phi) | |
| Y_{2}^{+2}=\left(\frac{15}{32 \pi}\right)^{1 / 2} \sin ^{2} \theta e^{\pm 2 i \phi} | Y_{0}^{0}=\left(\frac{1}{4 \pi}\right)^{1 / 2} |
| Y_{3}^{0}=\left(\frac{7}{16 \pi}\right)^{1 / 2}\left(5 \cos ^{3} \theta-3 \cos \theta\right) | Y_{1}^{0}=\left(\frac{3}{4 \pi}\right)^{1 / 2} \cos \theta |
| Y_{3}^{\pm 1}=\mp\left(\frac{21}{64 \pi}\right)^{1 / 2} \sin \theta\left(5 \cos ^{2} \theta-1\right) e^{\pm i \phi} | Y_{1}^{\pm 1}=\mp\left(\frac{3}{8 \pi}\right)^{1 / 2} \sin \theta e^{\pm i \phi} |
| Y_{3}^{\pm 2}=\left(\frac{105}{32 \pi}\right)^{1 / 2} \sin ^{2} \theta \cos \theta e^{\pm 2 i \phi} | Y_{2}^{0}=\left(\frac{5}{16 \pi}\right)^{1 / 2}\left(3 \cos ^{2} \theta-1\right) |
| Y_{3}^{+3}=\mp\left(\frac{35}{64 \pi}\right)^{1 / 2} \sin ^{3} \theta e^{\pm 3 i \phi} | Y_{2}^{\pm 1}=\mp\left(\frac{15}{8 \pi}\right)^{1 / 2} \sin \theta \cos \theta e^{\pm i \phi} |
| Table 4.7: The first few radial wave functions for hydrogen, R_{n \ell}(r) |
| R_{10}=2 a^{-3 / 2} \exp (-r / a) |
| R_{20}=\frac{1}{\sqrt{2}} a^{-3 / 2}\left(1-\frac{1}{2} \frac{r}{a}\right) \exp (-r / 2 a) .
R_{21}=\frac{1}{2 \sqrt{6}} a^{-3 / 2}\left(\frac{r}{a}\right) \exp (-r / 2 a) . |
| R_{30}=\frac{2}{3 \sqrt{3}} a^{-3 / 2}\left(1-\frac{2}{3} \frac{r}{a}+\frac{2}{27}\left(\frac{r}{a}\right)^{2}\right) \exp (-r / 3 a) .
R_{31}=\frac{8}{27 \sqrt{6}} a^{-3 / 2}\left(1-\frac{1}{6} \frac{r}{a}\right)\left(\frac{r}{a}\right) \exp (-r / 3 a) . R_{32}=\frac{4}{81 \sqrt{30}} a^{-3 / 2}\left(\frac{r}{a}\right)^{2} \exp (-r / 3 a) . |
| R_{40}=\frac{1}{4} a^{-3 / 2}\left(1-\frac{3}{4} \frac{r}{a}+\frac{1}{8}\left(\frac{r}{a}\right)^{2}-\frac{1}{192}\left(\frac{r}{a}\right)^{3}\right) \exp (-r / 4 a) .
R_{41}=\frac{5}{16 \sqrt{15}} a^{-3 / 2}\left(1-\frac{1}{4} \frac{r}{a}+\frac{1}{80}\left(\frac{r}{a}\right)^{2}\right)\left(\frac{r}{a}\right) \exp (-r / 4 a) . R_{42}=\frac{1}{64 \sqrt{5}} a^{-3 / 2}\left(1-\frac{1}{12} \frac{r}{a}\right)\left(\frac{r}{a}\right)^{2} \exp (-r / 4 a) . R_{43}=\frac{1}{768 \sqrt{35}} a^{-3 / 2}\left(\frac{r}{a}\right)^{3} \exp (-r / 4 a) . |
(a)
\psi_{433}=R_{43} Y_{3}^{3}=\frac{1}{768 \sqrt{35}} \frac{1}{a^{3 / 2}}\left(\frac{r}{a}\right)^{3} e^{-r / 4 a}\left(-\sqrt{\frac{35}{64 \pi}} \sin ^{3} \theta e^{3 i \phi}\right)= -\frac{1}{6144 \sqrt{\pi} a^{9 / 2}} r^{3} e^{-r / 4 a} \sin ^{3} \theta e^{3 i \phi} .
(b)
\langle r\rangle=\int r|\psi|^{2} d^{3} r =\frac{1}{(6144)^{2} \pi a^{9}} \int r\left(r^{6} e^{-r / 2 a} \sin ^{6} \theta\right) r^{2} \sin \theta d r d \theta d \phi .
=\frac{1}{(6144)^{2} \pi a^{9}} \int_{0}^{\infty} r^{9} e^{-r / 2 a} d r \int_{0}^{\pi} \sin ^{7} \theta d \theta \int_{0}^{2 \pi} d \phi .
=\frac{1}{(6144)^{2} \pi a^{9}}\left[9 !(2 a)^{10}\right]\left(2 \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}\right)(2 \pi)=18 a .
(c) Using Eq. 4.133: L_{x}^{2}+L_{y}^{2}=L^{2}-L_{z}^{2}=3(4) \hbar^{2}-(3 \hbar)^{2}= 3 \hbar^{2}. with probability 1.
H \psi=E \psi, \quad L^{2} \psi=\hbar^{2} \ell(\ell+1) \psi, \quad L_{z} \psi=\hbar m \psi . (4.133).