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RelaxationVsScattering

< Master Equations: quadrupolar case | Index | Simple metals and superconductors >

Neutron magnetic scattering

Let us compare neutron inelastic scattering (e.g. from spin waves, critical) with nuclear relaxation in the corresponding cases. Only conceptual passages are reported here.

Magnetic neutron scattering (from Chatterji, Ed., Elsevier 2006) of neutrons changing wave vector and spin {$({\mathbf k}_0,\sigma_0) \rightarrow ({\mathbf k}_1,\sigma_1)$} is described by the following differential cross-section

{$$\frac {d\sigma}{d\Omega} = \frac 1 {N\Phi \Delta\Omega} W$$}

where {$N$} is the number of scatterers, {$\Phi$} the flux of incident neutrons, {$\Delta\Omega$} the solid angle of the detector and {$W$} the transition rate from state {$({\mathbf k}_0,\sigma_0)$} to state {$({\mathbf k}_1,\sigma_1)$}. Using the Fermi golden rule

{$$\frac {d\sigma}{d\Omega} = \frac 1 {N} \frac{k_1}{k_0} \left(\frac m{2\pi\hbar^2}\right)^2 | \langle {\mathbf k}_1\sigma_1\mid V\mid {\mathbf k}_0\sigma_0 \rangle|^2$$}

Energy is conserved, and a sum over all initial and final states of the sample, with their probabilities, is implied. Hence differentiating with respect to the energy difference {$E$} between incoming and outgoing neutrons

{$$\frac {d^2\sigma}{d\Omega dE} = \frac 1 {N_m} \frac{k_1}{k_0} \frac{g_n^2 r_e^2}{2\pi\hbar}\sum_{\alpha\beta}(\delta_{\alpha\beta}-\hat{Q}_\alpha\hat{Q}_\beta)\int\langle{\mathbf D}_\alpha(-{\mathbf Q},0){\mathbf D}_\beta({\mathbf Q},t) e^{-iEt/\hbar}dt, \qquad (\alpha,\beta)=x,y,z $$}

where {${\mathbf Q}={\mathbf k}_1-{\mathbf k}_0$} is the scattering wave-vector, {$g_n=1.9132$} is the neutron g-factor, and {$r_e=e^2/m_ec^2=2.82e^{-15}$} m is the classical radius of the electron. The magnetic matrix element {${\mathbf D}={\mathbf M}({\mathbf Q})/2\mu_B$} is the normalized space Fourier transform of the total magnetization operator {${\mathbf M}({\mathbf r})$}

{$$ {\mathbf M}({\mathbf Q}) = \int {\mathbf M}({\mathbf r}) e^{i{\mathbf Q}\cdot{\mathbf r}} d{\mathbf r}$$}

The magnetic differential cross-section is then proportional to the time and space Fourier transform of the self correlation function of the i-th electron magnetic moment {$\mathbf{\mu}_i$} in the unit cell. It is convoluted with the corresponding nuclear correlation function and this gives rise to an elastic term, an inelastic spin term, an inelastic phonon term and a coupled inelastic term (neglected). The magnetic cross-section can be rewritten as

{$$\frac {d^2\sigma}{d\Omega dE} = \frac{k_1}{k_0} \frac{g_n^2 r_e^2}{4\pi\mu_B^2}\sum_{\alpha\beta}\frac{\delta_{\alpha\beta}-\hat{Q}_\alpha\hat{Q}_\beta}{1-e^{-E/k_BT}} \sum_{i,j} e^{-[W_i({\mathbf Q})+W_j({\mathbf Q})]}\, \chi^{\prime\prime}_{\alpha\beta i j}({\mathbf Q},E) $$}

in terms of the imaginary part of the magnetic generalized susceptibility

{$$ \chi^{\prime\prime}_{\alpha\beta i j}({\mathbf Q},E) = \frac{2\mu_B^2} \hbar (1-e^{-E/k_BT})\,f^*_i({\mathbf Q})f_j({\mathbf Q})e^{i{\mathbf Q}\cdot({\mathbf r}_j-{\mathbf r}_i)} \,\sum_l e^{i{\mathbf Q}\cdot{\mathbf R}_l} \int_{-\infty}^\infty\, \langle \mu_{0i\alpha}(0)\mu_{lj\beta}(t)\rangle e^{-iEt/\hbar}dt$$}

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Comparison with nuclear relaxation

We can simplify derivation of T_{_2_} by directly introducing the Fermi golden rule in the master equations. Assuming a simple contact scalar coupling, i.e. an instantaneous j^th -electron magnetic field {${\mathbf B}_{ej}=A_{hf}{\mathbf S}_j$} at the on-site nucleus j we have that the nuclear frequency in this field is {$\omega_e=\gamma B_{ej}=\gamma A_{hf} S_j$}. Therefore the result that we found can be rewritten pedantically, with the useless index j, as

{$$ \frac 1 {T_2} = \left(\frac {\gamma A_{hf}}{g\mu_B}\right)^2\int_{-\infty}^\infty\, \frac 1 2 \left[\langle \mu_{jx}(0)\mu_{jx}(\tau)\rangle + \langle \mu_{jy}(0)\mu_{jy}(\tau)\rangle \cos\gamma B\tau+ \langle \mu_{jz}(0)\mu_{jz}(\tau)\rangle \right] d\tau$$}

where we supposed time inversion symmetry ({$\int_{-\infty}^\infty\ f(t) dt=2\int_0^\infty f(t) dt$}). We can recognize self components of the same susceptibility, the local one, integrated over all {$\mathbf Q$}

{$$ \chi^{\prime\prime}_{\alpha\beta j j}(\gamma B) = \frac{2\mu_B^2} \hbar (1-e^{-E/k_BT})\, \int_{-\infty}^\infty\, \langle \mu_{j\alpha}(0)\mu_{j\beta}(\tau)\rangle e^{-i\gamma B\tau}dt $$}

Inserting the nuclear Larmor frequency {$\omega=\gamma B$}, the detailed balance for the nuclei {$1-\exp(-\hbar\omega/k_BT)=k_BT/\omega$} and dropping the index j, we may rewrite again

{$$ \frac 1 {T_2} = (\gamma A_{hf})^2 k_B T \frac {\chi^{\prime\prime}_{xx}(\omega) +\chi^{\prime\prime}_{yy}(-\omega) + \chi^{\prime\prime}_{zz}(0)} \omega $$}

Coupling to more electron moments becomes significant when dealing with critical scattering. With a tensor coupling the electron field is {$B_{e\alpha}=\sum_{k\beta} A_{\alpha\beta}({\mathbf r}_k)S_\beta({\mathbf r}_k)$} and the first term in the relaxation rate above is written as

{$$ \frac 1 {T_2} = \left(\frac \gamma {g\mu_B}\right)^2 \sum_{lk\alpha\beta} A_{x\alpha}({\mathbf r}_k)A_{x\beta}({\mathbf r}_l)\int_{-\infty}^\infty\, \langle \mu_{k\alpha}(0)\mu_{l\beta}(\tau)\rangle e^{-i\zeta_{\alpha\beta}\gamma B\tau}d\tau + \cdots$$},

where {$\zeta_{\beta\beta^\prime}$} keeps track of the sign. Substituting the space Fourier transform of these quantities we obtain the complete expression

{$$ \frac 1 {T_2} = \left(\frac \gamma {g\mu_B}\right)^2 \frac 1 \omega \sum_{\mathbf Q} \left[ A^2_{xx}({\mathbf Q})\chi^{\prime\prime}_{xx}({\mathbf Q},\omega) + A^2_{yy}({\mathbf Q})\chi^{\prime\prime}_{yy}({\mathbf Q},-\omega)+A^2_{zz}({\mathbf Q})\chi^{\prime\prime}_{zz}({\mathbf Q},0)\right]$$}

At the magnetic critical point, for {$T/T_c-1\rightarrow 0$}, the susceptibility {$\chi^{\prime\prime}({\mathbf Q},\omega)$} diverges and so in general does the relaxation rate (the so-called critical divergence).

However {$A_{\alpha\beta}({\mathbf Q})=\sum_k A_{\alpha\beta}({\mathbf r}_k) e^{i{\mathbf Q}\cdot{\mathbf r}_k}$} plays the role of the structure factors {$f_j({\mathbf Q})$} in the scattering cross-section, and it may mask the divergence. For instance it vanishes if the nucleus is coupled only to two equal spins belonging to opposite sublattices, hence the onset of antiferromagnetism that leads to a dominant diverging component {$\chi^{\prime\prime}({\mathbf Q}_{AF},\omega)$} is not seen by those nuclei.

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