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Knight

< Simple metals and superconductors | Index | Korringa relaxation >

Frequency shift in a simple metal

The field at the nuclear probe is larger than the applied one, {$\mu_0 H$}, because of the polarization of conduction electrons, with density {$n$}, whose average spin is zero in the absence and {$\chi_P H/n$} in the presence of the field.

The NMR frequency will be shifted to higher values, by an amount which is experimentally quantified as {$ \omega=\gamma \mu_0 H (1+K)$} A typical situation is shown in the Figure on the left. The non-vanishing average electron spin produces an additional field {$\Delta B=\mu_0\Delta H$} at the nucleus, because of the contact Fermi hyperfine interaction. The positive sign of this interaction determines the overall sign of the shift. In a crude, but adequate approximation the Pauli susceptibility {$\chi = \mu_B^2 g(\epsilon_F) = \mu_B^2 \frac {3 n} {2 \epsilon_F} $} determines a fraction of the hyperfine field {$B_{hf}= \mu_0 \,\frac 2 3 \langle |\Psi(0)|^2\rangle_F \,\mu_B $}

(averaged over the Fermi surface) to appear at the nucleus. The fraction is obtained by replacing {$\mu_B$} by {$\langle\mu\rangle=\chi H v_c$}, {$v_c=1/n$} being the volume occupied by one conduction electron. This yields a Knight shift to the magnetic field

{$ \Delta H \equiv K H = \mu_0 \, \langle |\Psi(0)|^2\rangle_F\ \frac {\mu_B^2}{\epsilon_F} H = B_{hf}\, \frac 3 2\frac {\mu_B} {\epsilon_F} H. $}

It amounts to saying that, apart from a numerical factor of order 1, one has

{$ K \epsilon_F = \mu_B B_{hf} $}

The Knight shift parameter may be than calculated as

{$ K = \frac {\Delta H} {H} \approx \frac {\mu_B B_{hf}}{\epsilon_F} $}

< Simple metals and superconductors | Index | Korringa relaxation >