For the hydrogen atom we have a Coulomb force exerted by the proton forcing the electron to orbit around it. Since the proton is 1837 heavier than the electron, we can ignore the reverse action. The potential is thus
|
V (r) = − {{e}^{2}\over
4π{ϵ}_{0}r}.
| (11.13) |
If we substitute this in the Schrödinger equation for u(r), we find
|
−{{ℏ}^{2}\over
2m} {{∂}^{2}\over
∂{r}^{2}}u(r) − {{e}^{2}\over
4π{ϵ}_{0}r}u(r) = Eu(r).
| (11.14) |
The way to attack this problem is once again to combine physical quantities to set the scale of length, and see what emerges. From a dimensional analysis we find that the length scale is set by the Bohr radius {a}_{0},
|
{a}_{0} = {4π{ϵ}_{0}{ℏ}^{2}\over
m{e}^{2}} = 0.53 × 1{0}^{−10}\ \textrm{m}.
| (11.15) |
The scale of energy is set by these same parameters to be
|
{{e}^{2}\over
4π{ϵ}_{0}{a}_{0}} = 2\ \textrm{Ry},
| (11.16) |
and one Ry (Rydberg) is 13.6\ \textrm{eV}. Solutions can be found by a complicated argument similar to the one for the Harmonic oscillator, but (without proof) we have
|
{E}_{n} = −{1\over
2}\left ( {{e}^{2}\over
4π{ϵ}_{0}{a}_{0}}\right ) {1\over {
n}^{2}} = −13.6 {1\over {
n}^{2}}\ \textrm{eV}.
| (11.17) |
and
|
{R}_{n} = {e}^{−r∕(n{a}_{0})}({c}_{
0} + {c}_{1}r + \mathop{\mathop{…}} + {c}_{n−1}{r}^{n−1})
| (11.18) |
The explicit, and normalised, forms of a few of these states are
Remember these are normalised to
|
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{∞}{R}_{
n}{(r)}^{∗}{R}_{
m}(r)\kern 1.66702pt dr = {δ}_{nm}.
| (11.21) |
Notice that there are solution that do depend on θ and φ as well, and that we have not looked at such solutions here!