11.4: The hydrogen atom
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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)=−e24πϵ0r
If we substitute this in the Schrödinger equation for u(r), we find
−ℏ22m∂2∂r2u(r)−e24πϵ0ru(r)=Eu(r).
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 a0,
a0=4πϵ0ℏ2me2=0.53×10−10 m
The scale of energy is set by these same parameters to be
e24πϵ0a0=2Ry
and one Ry (Rydberg) is 13.6 eV . Solutions can be found by a complicated argument similar to the one for the Harmonic oscillator, but (without proof) we have
En=−12◻e24πϵ0a0⌈1n2=−13.61n2eV
and
Rn=e−r(na0)(c0+c1r+…+cn−1rn−1)
The explicit, and normalised, forms of a few of these states are
R1(r)=1√4π2a−3/20e−r/a0,R2(r)=1√4π2(2a0)−32[1−r2a0◻e−r(2a0).
Remember these are normalised to
∫∞0Rn(r)∗Rm(r)dr=δnm
Notice that there are solution that do depend on θ and φ as well, and that we have not looked at such solutions here!