## Three Dimensional Virial Theorem for the Hydrogen Atom

**Catherine Therese J. Quiñones**

The virial theorem is a general theorem relating the potential energy (V) and the kinetic energy (T) in a bound system. A simple physical example is a small object orbiting around another object bound by a force as in the case of a hydrogen atom. The average kinetic energy and potential energies of a system of particle that interact by Coulomb forces are related by

[eq] \langle T \rangle = -\frac{1}{2} \langle V \rangle [/eq] (1)

Since, the Hamiltonian H of the given system is

[eq] \langle H \rangle = \langle T \rangle + \langle V \rangle = E_n [/eq] (2)

Thus substituting Eqn (1) to Eqn (2) yields,

[eq] -\frac{1}{2}\langle V \rangle + \langle V \rangle = E_n [/eq] (3)

[eq] \frac{\langle V \rangle}{2} = E_n [/eq] (4)

Now, we will derive the expectation value of [eq]\frac{1}{r}[/eq] in the unperturbed state of the of a hydrogen atom. We can use the virial theorem to easily solve the expectation value since the system can be considered a bound system with the electron orbiting around the proton which is bound by the Coulombic force.

For a hydrogen atom, the potential energy is expressed as

[eq] V = -\frac{e^2}{4\pi\epsilon_0} \frac{1}{r} [/eq] (5)

where [eq]e[/eq] is the charge of the electron and the proton, [eq]r[/eq] represents the separation distance between the two charges and [eq]\epsilon_0[/eq] is the permittivity of free space . The negative sign indicates that the force is attractive.

The allowed energies [eq]E_n[/eq] is given by

[eq]E_n = – \Bigg[\frac{m}{2\hbar^2}\Bigg(\frac{e^2}{4\pi\epsilon_0}\Bigg)^2\Bigg]\Bigg{\frac{1}{n^2}\Bigg} [/eq] (6)

where [eq]m[/eq] is the mass of the particle, [eq]\hbar[/eq] is Planck’s constant over [eq]2\pi[/eq] and [eq]n= 0,1,2,3,..[/eq] which indicates the quantization of the energy level. The solution is very straight forward. All we need is to plug in eqn (5) and (6) to eqn (4). Hence,

[eq]-\frac{e^2}{4\pi\epsilon_0}\Big\langle\frac{1}{r}\Big\rangle = -2\Bigg[ \frac{m}{2\hbar^2}\Bigg(\frac{e^2}{4\pi\epsilon_0}\Bigg)^2\Bigg] \Bigg{\frac{1}{n^2}\Bigg} [/eq] (7)

[eq]\Big\langle\frac{1}{r}\Big\rangle=\Bigg(\frac{me^2}{4\pi\epsilon_0\hbar^2}\Bigg)\frac{1}{n^2}[/eq] (8)

Note that the term inside the parenthesis is just [eq]\frac{1}{a_0}[/eq], where [eq]a_0[/eq] is the Bohr radius . Hence we can write the expectation value of [eq]\frac{1}{r}[/eq] as,

[eq]\Big\langle\frac{1}{r}\Big\rangle = \frac{1}{a_0n^2}[/eq] (9)

Thus we have derived the expectation value, [eq]\Big\langle\frac{1}{r}\Big\rangle[/eq], of the hydrogen atom in the unperturbed state using the virial theorem.

June 2nd, 2010 at 6:44 pm

In Eq (3), isn’t this the *total* energy (kinetic + potential) of an electron in the n-th energy level? It is negative, which doesn’t to make sense for a kinetic energy…

June 14th, 2010 at 9:55 pm

Gekko, definitely you’re right. Kinetic energy is definitely positive. Equation (3) actually refers to the total energy and this post has been updated accordingly.

Thanks for pointing this out!

June 21st, 2011 at 2:28 am

Sorry, bad characters..

Shouldn’t that be 2T=-V, V is negative, and T is positive..

June 22nd, 2011 at 3:07 am

Hi everyone! Thank you for your comments! I already updated this post.

January 24th, 2012 at 9:14 pm

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November 16th, 2013 at 8:15 pm

Should n go from 1, 2, 3 , … instead of 0, 1, 2 , 3? I thought that the energy of a hydrogen atom is -13.6 GeV at the ground state (n=0), not -infinity.