# Classical vs. quantum energy of the hydrogen atom

If I have an electron and a proton and calculate the classical energy which I get by bringing the electron from infinity to the distance of a Bohr radius to the proton, I get 27.2 eV, but the electron energy of the ground state of hydrogen is 13.6 eV, so it's exactly half of the classical energy. Right? Is there any intuitive explanation why that is the case? And what would be the energy I need to spend to bring a proton to a distance of 1 Bohr radius to another proton?

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Concerning the factor $\frac{1}{2}$: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy

$$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$

Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.$^1$

However we should also take the kinetic energy $\langle T\rangle>0$ into account! We know from the virial theorem that the kinetic energy

$$\tag{2}\langle T\rangle~=~-\frac{1}{2}\langle U\rangle~>~0$$

is minus half the potential energy for the $1/r^2$ Coulomb force.

Hence the total energy becomes half the Coulomb potential energy:

$$\tag{3} E ~=~ \langle T\rangle+ \langle U\rangle ~=~-\langle T\rangle ~=~\frac{1}{2}\langle U\rangle~<~0,$$

which is (up to sign conventions) the Rydberg energy.

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$^1$ Here OP's estimate is helped by the fact that the expectation value

$$\tag{4}\langle\frac{1}{r}\rangle ~=~\frac{1}{a_0}$$

in the ground state happens to be the inverse Bohr radius without any non-trivial dimensionless number appearing in eq. (4)! [Note for comparison, that e.g. $\langle r\rangle =\frac{3}{2}a_0$.]

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Can you provide an intuitive understanding of the virial theorem? –  garyp Jul 10 '14 at 23:42
@garyp: You may enjoy reading e.g. this and this Phys.SE posts. –  Qmechanic Jul 10 '14 at 23:56