Question 5.15: (a) Calculate 〈 ( 1/|r1 -r2| ) 〉 for the state ψ0 (Equatio...

(a) Calculate \left\langle\left(1 /\left| r _{1}- r _{2}\right|\right)\right\rangle for the state ψ_0 (Equation 5.41). Hint: Do the d^{3} r _{2} integral first, using spherical coordinates, and setting the polar axis along r_1 , so that

\psi_{0}\left( r _{1}, r _{2}\right)=\psi_{100}\left( r _{1}\right) \psi_{100}\left( r _{2}\right)=\frac{8}{\pi a^{3}} e^{-2\left(r_{1}+r_{2}\right) / a}           (5.41).

\left| r _{1}- r _{2}\right|=\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta_{2}} .

The θ_2 integral is easy, but be careful to take the positive root. You’ll have to break the r_2 integral into two pieces, one ranging from 0 to r_1 ,the other from r_{1} \text { to } \infty Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it to E_0   (Equation 5.42) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

E_{0}=8(-13.6 eV )=-109 eV                 (5.42).

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(a)

\left\langle\frac{1}{\left|r_{1}-r_{2}\right|}\right\rangle=\left(\frac{8}{\pi a^{3}}\right)^{2} \int \underbrace{\left[\int \frac{e^{-4\left(r_{1}+r_{2}\right) / a}}{\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta_{2}}} d^{3} r _{2}\right]}_{\blacklozenge} d^{3} r _{1} .

\blacklozenge=2 \pi \int_{0}^{\infty} e^{-4\left(r_{1}+r_{2}\right) / a} \underbrace{\left[\int_{0}^{\pi} \frac{\sin \theta_{2}}{\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta_{2}}} d \theta_{2}\right]}_{\star} r_{2}^{2} d r_{2} .

\star=\left.\frac{1}{r_{1} r_{2}} \sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta_{2}}\right|_{0} ^{\pi}=\frac{1}{r_{1} r_{2}}\left[\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2}}-\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2}}\right] .

=\frac{1}{r_{1} r_{2}}\left[\left(r_{1}+r_{2}\right)-\left|r_{1}-r_{2}\right|\right]=\left\{\begin{array}{l} 2 / r_{1}\left(r_{2}<r_{1}\right) \\ 2 / r_{2}\left(r_{2}>r_{1}\right) \end{array}\right. .

\blacklozenge=4 \pi e^{-4 r_{1} / a}\left[\frac{1}{r_{1}} \int_{0}^{r_{1}} r_{2}^{2} e^{-4 r_{2} / a} d r_{2}+\int_{r_{1}}^{\infty} r_{2} e^{-4 r_{2} / a} d r_{2}\right] .

\frac{1}{r_{1}} \int_{0}^{r_{1}} r_{2}^{2} e^{-4 r_{2} / a} d r_{2}=\left.\frac{1}{r_{1}}\left[-\frac{a}{4} r_{2}^{2} e^{-4 r_{2} / a}+\frac{a}{2}\left(\frac{a}{4}\right)^{2} e^{-4 r_{2} / a}\left(-\frac{4 r_{2}}{a}-1\right)\right]\right|_{0} ^{r_{1}} .

=-\frac{a}{4 r_{1}}\left[r_{1}^{2} e^{-4 r_{1} / a}+\frac{a r_{1}}{2} e^{-4 r_{1} / a}+\frac{a^{2}}{8} e^{-4 r_{1} / a}-\frac{a^{2}}{8}\right] .

\int_{r_{1}}^{\infty} r_{2} e^{-4 r_{2} / a} d r_{2}=\left.\left(\frac{a}{4}\right)^{2} e^{-4 r_{2} / a}\left(-\frac{4 r_{2}}{a}-1\right)\right|_{r_{1}} ^{\infty}=\frac{a r_{1}}{4} e^{-4 r_{1} / a}+\frac{a^{2}}{16} e^{-4 r_{1} / a} .

\blacklozenge=4 \pi\left\{\frac{a^{3}}{32 r_{1}} e^{-4 r_{1} / a}+\left[-\frac{a r_{1}}{4}-\frac{a^{2}}{8}-\frac{a^{3}}{32 r_{1}}+\frac{a r_{1}}{4}+\frac{a^{2}}{16}\right] e^{-8 r_{1} / a}\right\} .

=\frac{\pi a^{2}}{8}\left\{\frac{a}{r_{1}} e^{-4 r_{1} / a}-\left(2+\frac{a}{r_{1}}\right) e^{-8 r_{1} / a}\right\} .

\left\langle\frac{1}{\left|r_{1}-r_{2}\right|}\right\rangle=\frac{8}{\pi a^{4}} \cdot 4 \pi \int_{0}^{\infty}\left[\frac{a}{r_{1}} e^{-4 r_{1} / a}-\left(2+\frac{a}{r_{1}}\right) e^{-8 r_{1} / a}\right] r_{1}^{2} d r_{1} .

=\frac{32}{a^{4}}\left\{a \int_{0}^{\infty} r_{1} e^{-4 r_{1} / a} d r_{1}-2 \int_{0}^{\infty} r_{1}^{2} e^{-8 r_{1} / a} d r_{1}-a \int_{0}^{\infty} r_{1} e^{-8 r_{1} / a} d r_{1}\right\}.

=\frac{32}{a^{4}}\left\{a \cdot\left(\frac{a}{4}\right)^{2}-2 \cdot 2\left(\frac{a}{8}\right)^{3}-a \cdot\left(\frac{a}{8}\right)^{2}\right\}=\frac{32}{a}\left(\frac{1}{16}-\frac{1}{128}-\frac{1}{64}\right)=\frac{5}{4 a} .

(b)

V_{e e} \approx \frac{e^{2}}{4 \pi \epsilon_{0}}\left\langle\frac{1}{\left|r_{1}-r_{2}\right|}\right\rangle=\frac{5}{4} \frac{e^{2}}{4 \pi \epsilon_{0}} \frac{1}{a} =\frac{5}{4} \frac{m}{\hbar^{2}}\left(\frac{e^{2}}{4 \pi \epsilon_{0}}\right)^{2}=\frac{5}{2}\left(-E_{1}\right)=\frac{5}{2}(13.6 eV )=34 eV .

E_{0}+V_{e e}=(-109+34) eV =-75 eV , which is pretty close to the experimental value (-79 eV).

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