Question 4.53: (a) Construct the wave function for hydrogen in the state n ...

(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 Lx2+Ly2 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?

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) From Tables 4.3 and 4.7,

Table 4.3: The first few spherical harmonics Ym(θ,ϕ) Y_{\ell}^{m}(\theta, \phi)
Y2+2=(1532π)1/2sin2θe±2iϕ Y_{2}^{+2}=\left(\frac{15}{32 \pi}\right)^{1 / 2} \sin ^{2} \theta e^{\pm 2 i \phi} Y00=(14π)1/2 Y_{0}^{0}=\left(\frac{1}{4 \pi}\right)^{1 / 2}
Y30=(716π)1/2(5cos3θ3cosθ) Y_{3}^{0}=\left(\frac{7}{16 \pi}\right)^{1 / 2}\left(5 \cos ^{3} \theta-3 \cos \theta\right) Y10=(34π)1/2cosθ Y_{1}^{0}=\left(\frac{3}{4 \pi}\right)^{1 / 2} \cos \theta
Y3±1=(2164π)1/2sinθ(5cos2θ1)e±iϕ 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} Y1±1=(38π)1/2sinθe±iϕ Y_{1}^{\pm 1}=\mp\left(\frac{3}{8 \pi}\right)^{1 / 2} \sin \theta e^{\pm i \phi}
Y3±2=(10532π)1/2sin2θcosθe±2iϕ Y_{3}^{\pm 2}=\left(\frac{105}{32 \pi}\right)^{1 / 2} \sin ^{2} \theta \cos \theta e^{\pm 2 i \phi} Y20=(516π)1/2(3cos2θ1) Y_{2}^{0}=\left(\frac{5}{16 \pi}\right)^{1 / 2}\left(3 \cos ^{2} \theta-1\right)
Y3+3=(3564π)1/2sin3θe±3iϕ Y_{3}^{+3}=\mp\left(\frac{35}{64 \pi}\right)^{1 / 2} \sin ^{3} \theta e^{\pm 3 i \phi} Y2±1=(158π)1/2sinθcosθe±iϕ 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,Rn(r) R_{n \ell}(r)
R10=2a3/2exp(r/a) R_{10}=2 a^{-3 / 2} \exp (-r / a)
R20=12a3/2(112ra)exp(r/2a) R_{20}=\frac{1}{\sqrt{2}} a^{-3 / 2}\left(1-\frac{1}{2} \frac{r}{a}\right) \exp (-r / 2 a) .

R21=126a3/2(ra)exp(r/2a) R_{21}=\frac{1}{2 \sqrt{6}} a^{-3 / 2}\left(\frac{r}{a}\right) \exp (-r / 2 a) .

R30=233a3/2(123ra+227(ra)2)exp(r/3a) 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) .

R31=8276a3/2(116ra)(ra)exp(r/3a) 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) .

R32=48130a3/2(ra)2exp(r/3a) R_{32}=\frac{4}{81 \sqrt{30}} a^{-3 / 2}\left(\frac{r}{a}\right)^{2} \exp (-r / 3 a) .

R40=14a3/2(134ra+18(ra)21192(ra)3)exp(r/4a) 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) .

R41=51615a3/2(114ra+180(ra)2)(ra)exp(r/4a) 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) .

R42=1645a3/2(1112ra)(ra)2exp(r/4a) 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) .

R43=176835a3/2(ra)3exp(r/4a) R_{43}=\frac{1}{768 \sqrt{35}} a^{-3 / 2}\left(\frac{r}{a}\right)^{3} \exp (-r / 4 a) .

(a)

ψ433=R43Y33=1768351a3/2(ra)3er/4a(3564πsin3θe3iϕ)=16144πa9/2r3er/4asin3θe3iϕ \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)

r=rψ2d3r=1(6144)2πa9r(r6er/2asin6θ)r2sinθdrdθdϕ \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 .

=1(6144)2πa90r9er/2adr0πsin7θdθ02πdϕ =\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 .

=1(6144)2πa9[9!(2a)10](2246357)(2π)=18a =\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: Lx2+Ly2=L2Lz2=3(4)2(3)2=32 L_{x}^{2}+L_{y}^{2}=L^{2}-L_{z}^{2}=3(4) \hbar^{2}-(3 \hbar)^{2}= 3 \hbar^{2}.  with probability 1.

Hψ=Eψ,L2ψ=2(+1)ψ,Lzψ=mψ H \psi=E \psi, \quad L^{2} \psi=\hbar^{2} \ell(\ell+1) \psi, \quad L_{z} \psi=\hbar m \psi .          (4.133).

Related Answered Questions