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5.2 Erwartungswerte


$\displaystyle \langle$r$\displaystyle \rangle_{nlm}^{}$ = $\displaystyle \int_{0}^{\infty}$| Rnl(r)|2r3dr = a$\scriptstyle \mu$$\displaystyle {n^2 \over Z}$$\displaystyle \left[\vphantom{1 + {1\over 2} \left( 1 - {l(l+1) \over n^2}\right) }\right.$1 + $\displaystyle {1\over 2}$$\displaystyle \left(\vphantom{ 1 - {l(l+1) \over n^2}}\right.$1 - $\displaystyle {l(l+1) \over n^2}$ $\displaystyle \left.\vphantom{ 1 - {l(l+1) \over n^2}}\right)$ $\displaystyle \left.\vphantom{1 + {1\over 2} \left( 1 - {l(l+1) \over n^2}\right) }\right]$     (117)
$\displaystyle \left<\vphantom{ {1\over r} }\right.$$\displaystyle {1\over r}$ $\displaystyle \left.\vphantom{ {1\over r} }\right>_{nlm}^{}$ = $\displaystyle {Z \over a_{\mu} n^2}$     (118)
$\displaystyle \left<\vphantom{ {1\over r^2} }\right.$$\displaystyle {1\over r^2}$ $\displaystyle \left.\vphantom{ {1\over r^2} }\right>_{nlm}^{}$ = $\displaystyle {Z^2 \over a_{\mu}^2 n^3 (l+{1\over 2})}$     (119)
$\displaystyle \left<\vphantom{ {1\over r^3} }\right.$$\displaystyle {1\over r^3}$ $\displaystyle \left.\vphantom{ {1\over r^3} }\right>_{nlm}^{}$ = $\displaystyle {Z^3 \over a_{\mu}^3 n^3 l(l+{1\over 2})(l+1)}$     (120)


Alexander Wagner
2000-04-15