MOMENTO ANGULAR ORBITAL EM COORDENADAS ESFÉRICAS

MOMENTO ANGULAR ORBITAL EM COORDENADAS ESFÉRICAS

Notas de aula de Mecânica Quântica I – Marcus A.M. de Aguiar – 17/11/99

A transformação de coordenadas cartezianas para esféricas é dada por

x = rsin$\displaystyle \theta$cos$\displaystyle \phi$
y = rsin$\displaystyle \theta$sin$\displaystyle \phi$
z = rcos$\displaystyle \theta$
(1)

e a transformação inversa é

r = $\displaystyle \sqrt{x^2 + y^2+z^2}$
tan$\displaystyle \theta$ = $\displaystyle \sqrt{x^2 + y^2}$/z
tan$\displaystyle \phi$ = y/x
(2)

Em coordenadas cartezianas os operadores de momento angular são dados por:

Lx = – i$\displaystyle \hbar$(y$\displaystyle \partial$/$\displaystyle \partial$z – z$\displaystyle \partial$/$\displaystyle \partial$y)
Ly = – i$\displaystyle \hbar$(z$\displaystyle \partial$/$\displaystyle \partial$x – x$\displaystyle \partial$/$\displaystyle \partial$z)
Lz = – i$\displaystyle \hbar$(x$\displaystyle \partial$/$\displaystyle \partial$y – y$\displaystyle \partial$/$\displaystyle \partial$x)
(3)

Para ilustrar o calculo desses operadores em coordenadas esféricas faremos o caso Lz. As derivadas $ \partial$/$ \partial$x e $ \partial$/$ \partial$y são escritas como:

$\displaystyle {\frac{\partial}{\partial x}}$ = $\displaystyle {\frac{\partial r}{\partial x}}$$\displaystyle {\frac{\partial}{\partial r}}$ + $\displaystyle {\frac{\partial \theta}{\partial x}}$$\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{\partial \phi}{\partial x}}$$\displaystyle {\frac{\partial}{\partial \phi}}$

e

$\displaystyle {\frac{\partial}{\partial y}}$ = $\displaystyle {\frac{\partial r}{\partial y}}$$\displaystyle {\frac{\partial}{\partial r}}$ + $\displaystyle {\frac{\partial \theta}{\partial y}}$$\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{\partial \phi}{\partial y}}$$\displaystyle {\frac{\partial}{\partial \phi}}$

As derivadas das coordenadas esféricas em relação às cartezianas pode ser obtida derivandos implicitamente ambos os lados das equações 2 em relação á x, y ou z:

$\displaystyle {\frac{\partial r}{\partial x}}$ = $\displaystyle {\frac{x}{r}}$ = sin$\displaystyle \theta$cos$\displaystyle \phi$
$\displaystyle {\frac{1}{\cos^2{\theta}}}$$\displaystyle {\frac{\partial \theta}{\partial x}}$ = $\displaystyle {\frac{x}{\sqrt{x^2+y^2} z}}$ = $\displaystyle {\frac{\cos{\phi}}{r \cos{\theta}}}$         ou        $\displaystyle {\frac{\partial \theta}{\partial x}}$ = $\displaystyle {\frac{\cos{\theta} \cos{\phi}}{r}}$
$\displaystyle {\frac{1}{\cos^2{\phi}}}$$\displaystyle {\frac{\partial \phi}{\partial x}}$ = – $\displaystyle {\frac{y}{x^2}}$ = – $\displaystyle {\frac{\sin{\phi}}{r \sin{\theta} \cos^2{\phi}}}$         ou        $\displaystyle {\frac{\partial \phi}{\partial x}}$ = – $\displaystyle {\frac{\sin{\phi}}{r \sin{\theta}}}$

Fazendo o mesmo processo com as derivadas em relação à y obtemos

$\displaystyle {\frac{\partial r}{\partial y}}$ = sin$\displaystyle \theta$sin$\displaystyle \phi$
$\displaystyle {\frac{\partial \theta}{\partial y}}$ = $\displaystyle {\frac{\cos{\theta} \sin{\phi}}{r}}$
$\displaystyle {\frac{\partial \phi}{\partial y}}$ = $\displaystyle {\frac{\cos{\phi}}{r \sin{\theta}}}$

Voltando à equação 3 para Lz temos:

x$\displaystyle {\frac{\partial}{\partial y}}$ = rsin$\displaystyle \theta$cos$\displaystyle \phi$$\displaystyle \left(\vphantom{ \sin{\theta} \sin{ \phi} \frac{\partial}{\partia... ...a} + \frac{\cos{\phi}}{r \sin{\theta}} \frac{\partial}{\partial \phi} }\right.$sin$\displaystyle \theta$sin$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial r}}$ + $\displaystyle {\frac{\cos{\theta} \sin{\phi}}{r}}$$\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{\cos{\phi}}{r \sin{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$ $\displaystyle \left.\vphantom{ \sin{\theta} \sin{ \phi} \frac{\partial}{\partia... ...a} + \frac{\cos{\phi}}{r \sin{\theta}} \frac{\partial}{\partial \phi} }\right)$ (4)

e

y$\displaystyle {\frac{\partial}{\partial x}}$ = rsin$\displaystyle \theta$sin$\displaystyle \phi$$\displaystyle \left(\vphantom{ \sin{\theta} \cos{ \phi} \frac{\partial}{\partia... ...a} - \frac{\sin{\phi}}{r \sin{\theta}} \frac{\partial}{\partial \phi} }\right.$sin$\displaystyle \theta$cos$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial r}}$ + $\displaystyle {\frac{\cos{\theta} \cos{\phi}}{r}}$$\displaystyle {\frac{\partial}{\partial \theta}}$ – $\displaystyle {\frac{\sin{\phi}}{r \sin{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$ $\displaystyle \left.\vphantom{ \sin{\theta} \cos{ \phi} \frac{\partial}{\partia... ...a} - \frac{\sin{\phi}}{r \sin{\theta}} \frac{\partial}{\partial \phi} }\right)$ (5)

Subtraindo 5 de 4 e multiplicando por – i$ \hbar$ obtemos Lz. Os dois primeiros termos de cada equação se cancelam e os últimos termos se somam:

Lz = – i$\displaystyle \hbar$(cos2$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \phi}}$ + sin2$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \phi}}$) = – i$\displaystyle \hbar$$\displaystyle {\frac{\partial}{\partial \phi}}$ (6)

O cálculo de Lx e Ly é completamente análogo e resulta em

Lx = i$\displaystyle \hbar$(sin$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{\cos{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$) (7)
Ly = i$\displaystyle \hbar$(- cos$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{\sin{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$) (8)

Os operadores L+ e L– podem ser calculados também:

L+ = $\displaystyle \hbar$ei$\scriptstyle \phi$($\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{i}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$) (9)
L– = (L+)$\scriptstyle \dagger$ = $\displaystyle \hbar$e-i$\scriptstyle \phi$(- $\displaystyle {\frac{\partial}{\partial \theta}}$ + $\displaystyle {\frac{i}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$) (10)

Fica como exercício provar essas equações, principalmente a última relação para L–.

Finalmente calculamos L2. Começamos com Lx2:

Lx2 = –$\displaystyle \hbar^{2}_{}$(sin$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle {\frac{\cos{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$)(sin$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle {\frac{\cos{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$)
= – $\displaystyle \hbar^{2}_{}$$\displaystyle \left(\vphantom{ \displaystyle{ \sin^2{\phi}\frac{\partial^2}{\p... ...\frac{\cos^2{\phi}}{\tan{\theta}} \frac{\partial}{\partial \theta} - } }\right.$sin2$\displaystyle \phi$$\displaystyle {\frac{\partial^2}{\partial \theta^2}}$–$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan^2{\theta}\cos^2{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$+2$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial^2}{\partial \phi \partial \theta}}$+$\displaystyle {\frac{\cos^2{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \theta}}$–$\displaystyle \left.\vphantom{ \displaystyle{ \sin^2{\phi}\frac{\partial^2}{\p... ...\frac{\cos^2{\phi}}{\tan{\theta}} \frac{\partial}{\partial \theta} - } }\right.$
$\displaystyle \left.\vphantom{ \displaystyle{ \frac{\sin{\phi}\cos{\phi}}{\tan^... ...frac{\cos^2{\phi}}{\tan^2{\theta}} \frac{\partial^2}{\partial \phi^2}} }\right.$$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan^2{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$+$\displaystyle {\frac{\cos^2{\phi}}{\tan^2{\theta}}}$$\displaystyle {\frac{\partial^2}{\partial \phi^2}}$$\displaystyle \left.\vphantom{ \displaystyle{ \frac{\sin{\phi}\cos{\phi}}{\tan^... ...frac{\cos^2{\phi}}{\tan^2{\theta}} \frac{\partial^2}{\partial \phi^2}} }\right)$

Da mesma forma obtemos:

Ly2 = –$\displaystyle \hbar^{2}_{}$(-cos$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle {\frac{\sin{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$)(-cos$\displaystyle \phi$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle {\frac{\sin{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$)
= – $\displaystyle \hbar^{2}_{}$$\displaystyle \left(\vphantom{ \displaystyle{ \cos^2{\phi}\frac{\partial^2}{\pa... ...\frac{\sin^2{\phi}}{\tan{\theta}} \frac{\partial}{\partial \theta} +} }\right.$cos2$\displaystyle \phi$$\displaystyle {\frac{\partial^2}{\partial \theta^2}}$+$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan^2{\theta}\cos^2{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$-2$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial^2}{\partial \phi \partial \theta}}$+$\displaystyle {\frac{\sin^2{\phi}}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle \left.\vphantom{ \displaystyle{ \cos^2{\phi}\frac{\partial^2}{\pa... ...\frac{\sin^2{\phi}}{\tan{\theta}} \frac{\partial}{\partial \theta} +} }\right.$
$\displaystyle \left.\vphantom{ \displaystyle{ \frac{\sin{\phi}\cos{\phi}}{\tan^... ...frac{\cos^2{\phi}}{\tan^2{\theta}} \frac{\partial^2}{\partial \phi^2}} }\right.$$\displaystyle {\frac{\sin{\phi}\cos{\phi}}{\tan^2{\theta}}}$$\displaystyle {\frac{\partial}{\partial \phi}}$+$\displaystyle {\frac{\sin^2{\phi}}{\tan^2{\theta}}}$$\displaystyle {\frac{\partial^2}{\partial \phi^2}}$$\displaystyle \left.\vphantom{ \displaystyle{ \frac{\sin{\phi}\cos{\phi}}{\tan^... ...frac{\cos^2{\phi}}{\tan^2{\theta}} \frac{\partial^2}{\partial \phi^2}} }\right)$

e

Lz2 = –$\displaystyle \hbar^{2}_{}$$\displaystyle {\frac{\partial^2}{\partial \phi^2}}$

Somando tudo obtemos finalmente

L2 = – $\displaystyle \hbar^{2}_{}$$\displaystyle \left(\vphantom{ \displaystyle{-\frac{\partial^2}{\partial \theta... ... \theta} + \frac{1}{\sin^2{\theta}}\frac{\partial^2}{\partial \phi^2}} }\right.$–$\displaystyle {\frac{\partial^2}{\partial \theta^2}}$+$\displaystyle {\frac{1}{\tan{\theta}}}$$\displaystyle {\frac{\partial}{\partial \theta}}$+$\displaystyle {\frac{1}{\sin^2{\theta}}}$$\displaystyle {\frac{\partial^2}{\partial \phi^2}}$$\displaystyle \left.\vphantom{ \displaystyle{-\frac{\partial^2}{\partial \theta... ... \theta} + \frac{1}{\sin^2{\theta}}\frac{\partial^2}{\partial \phi^2}} }\right)$

 

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