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11.3 Solutions independent of $ \theta$ and $ \varphi$

Initially we shall just restrict ourselves to those cases where the wave function is independent of $ \theta$ and $ \varphi$, i.e.,

$\displaystyle \phi$(r,$\displaystyle \theta$,$\displaystyle \varphi$) = R(r). (11.8)

In that case the Schrödinger equation becomes (why?)

- $\displaystyle {\frac{\hbar^2}{2m}}$$\displaystyle {\frac{1}{r^2}}$$\displaystyle {\frac{\partial}{\partial r}}$$\displaystyle \left(\vphantom{r^2 \frac{\partial}{\partial r}R(r)}\right.$r2$\displaystyle {\frac{\partial}{\partial r}}$R(r)$\displaystyle \left.\vphantom{r^2 \frac{\partial}{\partial r}R(r)}\right)$ + V(r)R(r) = ER(r). (11.9)

One often simplifies life even further by substituting u(r)/r = R(r), and multiplying the equation by r at the same time,

- $\displaystyle {\frac{\hbar^2}{2m}}$$\displaystyle {\frac{\partial^2}{\partial r^2}}$u(r) + V(r)u(r) = Eu(r). (11.10)

Of course we shall need to normalise solutions of this type. Even though the solution are independent of $ \theta$ and $ \varphi$, we shall have to integrate over these variables. Here a geometric picture comes in handy. For each value of r, the allowed values of x range over the surface of a sphere of radius r. The area of such a sphere is 4$ \pi$r2. Thus the integration over r,$ \theta$,$ \varphi$ can be reduced to

$\displaystyle \int_{\textrm{all\ space}}^{}$f (r) dxdydz = $\displaystyle \int_{0}^{\infty}$f (r) 4$\displaystyle \pi$r2dr. (11.11)

Especially, the normalisation condition translates to

$\displaystyle \int_{0}^{\infty}$| R(r)|2 4$\displaystyle \pi$r2dr = $\displaystyle \int_{0}^{\infty}$| u(r)|2 4$\displaystyle \pi$dr = 1 (11.12)


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Next: 11.4 The hydrogen atom Up: 11. 3D Schrödinger equation Previous: 11.2 Spherical coordinates

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk