Recoil-free nuclear transition

A recoil-free nuclear transition emits or absorbs a gamma ray while leaving the internal vibrational state of the surrounding solid unchanged. It belongs to the elastic part of the process: no phonon is created or destroyed. The photon still carries momentum $p=E_{\gamma }/c$, where $c$ is the speed of light. The emitting solid takes momentum $-\mathbf{p}$; the absorbing solid takes $+\mathbf{p}$. Conservation of momentum holds for each event.

Why does an isolated atomic nucleus pose a problem? Start with an excited state at rest and let $E_0$ be its transition energy, with nuclear mass $M$ and $E_0\ll M{c}^2$. Conservation of momentum gives a recoil momentum of magnitude $p$; its nonrelativistic kinetic energy is thereforeBy conservation of energy, emission spends part of $E_0$ on nuclear recoil, whereas absorption must supply both $E_0$ and the final kinetic energy. To leading order,Thus the relevant comparison is $2E_R$ against the overlapping spectral line widths, rather than $E_R/E_0$ alone. If $2E_R$ greatly exceeds the widths, the isolated-nucleus lines have negligible overlap. Finite Lorentzian line shape tails mean that this is suppression, rather than an absolute prohibition on absorption.

For a bound atomic nucleus, the recoil-free channel transfers momentum to the center of mass of the emitting or absorbing solid, with negligible kinetic energy $p^2/(2M_{\mathrm{solid}})$. Other channels exchange phonons and carry substantial energy into or out of the internal vibrations. A large $M_{\mathrm{solid}}$ makes the elastic channel's energy cost small; the vibrational quantum state determines its probability. Merely replacing $M$ by a large mass in the free-particle formula does not calculate that probability.

Here is a deliberately stipulated teaching system, independent of historical isotope data: $E_0=20\mathrm{keV}$, $M{c}^2=10^{11}\mathrm{eV}$, and a single-line natural linewidth ${\Gamma }_E=10^{-7}\mathrm{eV}$ measured as an energy full width at half maximum. These give $E_R=0.002\mathrm{eV}$ and $2E_R/{\Gamma }_E=40,000$. A solid of $10^{12}$ such nuclear masses has elastic recoil energy $2\times 10^{-15}\mathrm{eV}$, far below ${\Gamma }_E$. In this article ${\Gamma }_E$ is an energy width; the corresponding lifetime is $\tau =\hslash /{\Gamma }_E$, with $\hslash$ the Planck constant divided by $2\pi$.

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.