Past exam of the condensed matter physics course of the University of Oxford / 2014 / 2

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

Using $c=2700\mathrm{m}{\mathrm{s}}^{-1}$ gives the electron-acoustic recoil lengthThe temperature ratio therefore fixesIt follows that $n=4/a^3=9.48\times 10^{28}{\mathrm{m}}^{-3}$, $k_F=1.4105\times 10^{10}{\mathrm{m}}^{-1}$ and $k_D=1.7771\times 10^{10}{\mathrm{m}}^{-1}$. Substitution givesThe resulting Fermi energy is about $7.58\mathrm{eV}$ and the temperature ratio is indeed $240$. These values follow from the supplied effective sound speed and free-electron/Debye approximations; they are not independent precision measurements of copper's parameters.

An electron losing energy to an acoustic phonon has, in a normal process, ${\mathbf{k}}^{\prime }={\mathbf{k}}_F-{\mathbf{k}}_{\mathrm{ph}}$, with possible addition of a reciprocal-lattice vector for an Umklapp process. The energy transfer is at most of order $\hslash {\omega }_D=k_B{\theta }_D$, much smaller than $E_F=k_B{T}_F$. Hence the magnitudes change very little even though the momentum direction can change substantially. At $T\sim {\theta }_D\ll T_F$, thermal vacancies near the Fermi surface allow emission into some lower-energy states; exactly at zero temperature emission into an occupied Fermi sea would be blocked by the Pauli exclusion principle.

Using $k^{\prime }=(1-\delta )k_F$ and the free-electron dispersion, energy conservation givesSince $\delta \ll 1$, the bracket is $2\delta +O({\delta }^2)$. Rearranging yieldsEquivalently $\delta \simeq \hslash c{k}_{\mathrm{ph}}/(2E_F)$. For a Debye-cutoff phonon in the preceding model, $\delta \simeq {\theta }_D/(2T_F)=1/480=0.00208$, confirming the approximation. Near-elastic scattering constrains the radial distance in wave-vector space, not the scattering angle alone.

The electron-acoustic recoil length compares electronic recoil with the acoustic quantum energy associated with the same momentum. For transferred wave number $q$,The dispersions cross at $q=1/\lambda$; there the reduced electron de Broglie wavelength is $\lambda$ and its free-particle group velocity is $2c$. Thus $\lambda$ is the length scale at which an electron's localization/recoil energy and a corresponding acoustic excitation energy become comparable. At $q\lambda \gg 1$ the electron recoil energy is large compared with the phonon energy; this explains the separation of electronic and acoustic energy scales when lattice wave numbers exceed $1/\lambda$.

Formally it is half the Reduced Compton wavelength with the speed of light replaced by the speed of sound. That is an analogy of scales, not a relativistic electron mass or a pair-production threshold: sound does not replace the limiting speed in electron dynamics. Also it is a reduced-wavelength convention; an ordinary de Broglie wavelength $2\pi /q$ at the crossing is $2\pi \lambda$. No scattering lifetime appears, so $\lambda$ is not an electron mean free path.