Dimensionless Radial Equation Hydrogen Atom

[Summer95]

Homework Statement

Show that in terms of the dimensionless variable $\xi$ the radial equation becomes $\frac{\mathrm{d}^{2} u}{\mathrm{d} \xi^{2}}=(\frac{l(l+1)}{\xi^{2}}-\frac{2}{\xi}-K)u$

Homework Equations

$u(r)\equiv rR(r)$
$\xi \equiv \sqrt{2\mu U_{0}}\frac{r}{\hbar}$ dimensionless variable
$K\equiv \frac{E}{U_{0}}$
Radial equation: $-\frac{\hbar^{2}}{2\mu}\frac{1}{r^{2}}\frac{\mathrm{d} }{\mathrm{d} r}(r^{2}\frac{\mathrm{d} }{\mathrm{d} r}R)+\frac{\hbar^{2}l(l+1)}{2\mu r^{2}}R+U(r)R=ER$
and $U(r)$ is the coulomb potential

The Attempt at a Solution

The radial equation becomes:
$-\frac{U_{0}}{\xi^{2}}\frac{\mathrm{d} }{\mathrm{d} r}(r^{2}\frac{\mathrm{d} }{\mathrm{d} r}\frac{u}{r})+\frac{U_{0}l(l+1)}{\xi^{2}}\frac{u}{r}+U(r)\frac{u}{r}=E\frac{u}{r}$
and after taking the derivatives and some canceling:
$-\frac{r}{\xi^{2}}\frac{\mathrm{d}^{2}}{\mathrm{d}r^{2}}u+\frac{l(l+1)}{\xi^{2}}\frac{u}{r}+\frac{U(r)}{U_{0}}\frac{u}{r}=K\frac{u}{r}$
and so
$\frac{r^{2}}{\xi^{2}}\frac{\mathrm{d}^{2} }{\mathrm{d} r^{2}}u=(-K+\frac{l(l+1)}{\xi^{2}}-\frac{e^{2}}{4\pi \varepsilon _{0}rU_{0}})u$

so two of the terms on the right are fine. I then substituted the ground state of hydrogen in for $U_{0}$ in the last term, and the only way that comes out correctly is if $\mu$ (the effective mass) $=16m$. Is this the case? I don't really understand how $\mu$ is defined. and I don't understand how to integrate with respect to $\xi$ in the first term. Thank you so much in advance!

edit: actually, not sure where I got the extra factor of 4 in there but $\mu=m$ what is the point of defining $\mu$?

 

[DrClaude]

μ is the reduced mass:
$$
\mu = \frac{m_1 m_2}{m_1+m_2}
$$
It appears when one separates the center-of-mass motion (which depends on the total mass) from the relative motion (which depends on the reduced mass) for a system of two particles.

For a hydrogen atom, since the mass of the proton is 3 orders of magnitude bigger, one gets
$$
\mu = \frac{m_e m_p}{m_e+m_p} \approx \frac{m_e m_p}{m_p} = m_e
$$