A.17 The virial theorem

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

$$\fbox{$\displaystyle \big\langle T\big\rangle = {\textstyle\frac{1}{2}} \langle {\skew0\vec r}\cdot \nabla V\rangle $}$$ (A.75)

Now for the $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12c_xx^2+\frac12c_yy^2+\frac12c_zz^2$ 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 $\frac12E_{n_xn_yn_z}$.

For the $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ constant$\raisebox{.5pt}{$/$}$$r$ potential of the hydrogen atom, note that according to the calculus rule for direc­tional derivatives, ${\skew0\vec r}\cdot\nabla{V}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r\partial{V}$$\raisebox{.5pt}{$/$}$$\partial{r}$. Therefore $r\partial{V}$$\raisebox{.5pt}{$/$}$$\partial{r}$ produces $\vphantom0\raisebox{1.5pt}{$-$}$$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 $E_n$, $\langle{T}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-E_n$ and $\langle{V}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2E_n$. 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

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

using the formulae in chapter 4.5.4,

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

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