Past exam of the advanced quantum physics course of the University of Cambridge / 2016 / Paper 1 / 1 / c

Using a spherical Bohr-radius volume for the atom, the calculations below give a cutoff electromagnetic vacuum energy density of about $6.2\times 10^{15}\mathrm{J}{\mathrm{m}}^{-3}$, compared with a hydrogen binding-energy density of about $3.5\times 10^{12}\mathrm{J}{\mathrm{m}}^{-3}$. The cutoff vacuum estimate is about $1.8\times 10^3$ times larger. Alternative reasonable definitions of atomic volume change this ratio by a factor of a few, not the main order-of-magnitude conclusion.

The cutoff result is a bare zero-point mode sum. It is not a classical radiation bath incident on the atom, nor a calculation of an observable absolute vacuum energy without a subtraction or renormalization prescription.

In a large periodic quantization volume $\mathcal{V}$, one wavevector state occupies $(2\pi {)}^3/\mathcal{V}$ in wavevector space. There are two transverse photon polarizations, and each oscillator contributes $\hslash ck/2$. A lower wavelength cutoff ${\lambda }_{\min }$ corresponds to $k_{\max }=2\pi /{\lambda }_{\min }$, so the energy density isFor ${\lambda }_{\min }=10^{-10}\mathrm{m}$,The factor two is polarization, not an extra independent count of positive and negative frequencies. The continuum wavevector integral already counts independent travelling-wave modes in the usual field quantization convention. The quartic cutoff dependence also shows why an assumed shortest wavelength is crucial.

Take the hydrogen ground-state binding-energy magnitude as $E_B\simeq 13.6\mathrm{eV}=2.18\times 10^{-18}\mathrm{J}$, and use $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$ as its characteristic radius. A definite volume convention isIt givesA quantum atom has no sharp outer surface. If instead its characteristic volume is taken as $a_0^3$, the estimate is $1.47\times 10^{13}\mathrm{J}{\mathrm{m}}^{-3}$ and the ratio is about $4.2\times 10^2$. Both conventions place the two densities roughly three orders of magnitude apart at order-of-magnitude accuracy. The comparison uses the magnitude of the binding energy, not its negative sign relative to ionization.