Lamb-Mössbauer factor ($f$)

The Lamb-Mössbauer factor $f$ is the fraction in the coherent elastic line. In the harmonic-lattice approximation it is a Debye–Waller factor. Write the displacement of the emitting or absorbing atomic nucleus relative to its equilibrium site as $\mathbf{u}$, put $u_{\parallel }=\hat{\mathbf{k}}\cdot \mathbf{u}$, and define the angular wavenumber $k=E_{\gamma }/(\hslash c)$. Choose the equilibrium site so that $⟨u_{\parallel }⟩=0$. Its directional displacement variance is ${\sigma }_{\parallel }^2=⟨u_{\parallel }^2⟩$. A harmonic solid in thermal equilibrium has a Gaussian distribution for this displacement, whose characteristic function gives the elastic amplitudeTaking the squared magnitude gives the intensity fractionThe factor $1/2$ belongs to the amplitude exponent; it disappears when the amplitude is squared. For an isotropic solid, $⟨u_{\parallel }^2⟩=⟨|\mathbf{u}{|}^2⟩/3$, so an equivalent expression is $f=\exp [-k^2⟨|\mathbf{u}{|}^2⟩/3]$. Directional and three-dimensional mean squared displacements must not be substituted for one another without this factor.

An Einstein solid approximation replaces the vibrational spectrum by one positive angular frequency $\omega$. For the directional quantum harmonic oscillator, with the same effective mass $M$ as in $E_R$,Here $k_B$ is the Boltzmann constant and $T$ is temperature. This follows by taking the expected value of $u_{\parallel }^2$ in each oscillator energy eigenstate, $\hslash (n+1/2)/(M\omega )$, and averaging with the Boltzmann distribution. Substitution givesThe zero-temperature limit retains zero-point energy and a nonzero displacement variance. Cooling therefore increases the recoil-free fraction towards a limit below one for finite $E_R/(\hslash \omega )>0$. At high temperature, the exponent tends to $-2E_R{k}_{B}T/(\hslash \omega {)}^2$. This is a single-frequency approximation to a harmonic lattice's coherent elastic weight, not a complete description of all zero-net-energy processes of an isolated oscillator. Real vibrational spectra, directional binding and anharmonic motion can change its predictions.

Extend the stipulated teaching system by choosing $\hslash \omega =0.020\mathrm{eV}$, so $\eta =E_R/(\hslash \omega )=0.1$. Then $f(0)=0.90484$, while $f=0.80542$ at $k_{B}T=\hslash \omega$. If source and absorber are independent, the fraction of events eligible for recoil-free emission followed by recoil-free absorption contains the product $f_s{f}_a$; this gives $0.81873$ for two zero-temperature copies. It is not the total probability of absorption, which also depends on the number of resonant atomic nuclei, geometry and spectral line overlap.