A.15 The virial theorem

The virial theorem says that the expectation value of the kinetic energy of stationary states is given by

$$\fbox{$\displaystyle \langle T\rangle = {\textstyle\frac{1}{2}} \langle {\skew0\vec r}\cdot \nabla V\rangle $}$$ (A.74)

Note that according to the calculus rule for directional derivatives, ${\skew0\vec r}\cdot\nabla{V}=r\partial{V}/\partial{r}$.

For the $V=\frac12 c_x x^2 + \frac12 c_y y^2+ \frac12 c_z z^2$ potential of a harmonic oscillator, $x\partial{V}/\partial{x}+y\partial{V}/\partial{y}+z\partial{V}/\partial{z}$ produces $2V$. So for energy eigenstates of the harmonic oscillator, the expectation value of kinetic energy equals the one of the potential energy. And since their sum is the total energy $E_{n_xn_yn_z}$, each must be $\frac12 E_{n_xn_yn_z}$.

For the $V={\rm constant}/r$ potential of the hydrogen atom, $r\partial{V}/\partial{r}$ produces $-V$, So the expectation value of kinetic energy equals minus one half the one of the potential energy. And since their sum is the total energy $E_n$, $\langle{T}\rangle=-E_n$ and $\langle{V}\rangle=2E_n$. Note that $E_n$ is negative, so that the kinetic energy is positive as it should be.

To prove the virial theorem, work out the commutator in

$$\frac{{\rm d}\langle{\skew0\vec r}\cdot\vec p\rangle}{{\rm... ...ac{{\rm i}}{\hbar}\langle[H,{\skew0\vec r}\cdot\vec p]\rangle$$

using the formulae in chapter 4.5.4,

$$\frac{{\rm d}\langle{\skew0\vec r}\cdot\vec p\rangle}{{\rm... ...\langle T\rangle - \langle{\skew0\vec r}\cdot\nabla V\rangle,$$

and then note that the left hand side above is zero for stationary states, (in other words, states with a precise total energy).