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

The open shell contains one $f$ electron, so $L=3$ and $S=1/2$. For a less-than-half-filled shell, Hund's rules select the smaller total angular momentum, $J=L-S=5/2$, giving the free-ion ground level ${}^2{F}_{5/2}$. The Landé g-factor isIn a magnetic field weak compared with the spin-orbit splitting, the electronic Zeeman shifts are $\delta E_{m_J}=g_J{\mu }_{B}B{m}_J$, for $m_J=-5/2,-3/2,\dots ,5/2$. Thus six equally spaced magnetic substates appear. At $B=1\mathrm{T}$, the adjacent-level spacing isThe complete lowest-to-highest multiplet spread is five spacings,Giving both quantities removes the ambiguity in the word splitting. This is the free-ion, weak-field result; no crystal-field doublet or hyperfine structure is specified.

Hermiticity supplies the reverse matrix elements, also equal to the real number $t$. In an orthonormal basis for the degenerate eigenspace, the perturbation iswhere ${\mathbf{J}}_3$ is the all-ones matrix. Its symmetric vector $(1,1,1)/\sqrt{3}$ has eigenvalue $2t$. Every vector whose components sum to zero has eigenvalue $-t$; that subspace has dimension two. For example, $(1,-1,0)/\sqrt{2}$ and $(1,1,-2)/\sqrt{6}$ form an orthonormal basis for it.

The first-order energies are thereforeThe trace of the shifts is $2t-t-t=0$, agreeing with the zero diagonal. The ordering depends on the sign of $t$; the formula does not assume that the symmetric state is the lowest. Diagonalizing the perturbation inside the degenerate space is essential: the original diagonal entries alone would incorrectly predict no splitting.

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.