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KnightShift

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Sketch of Knight shift calculation for a localized moment system. Imagine to have a single crystal of a magnetic material in an external field above the ordering transition. Assume a given muon site as the origin of the coordinates. Consider the lattice of the magnetic ions at {$\mathbf{r}_i, i=1,\cdots,N$}, with magnetic moments {$g\mu_B\mathbf J_i$}, for simplicity one per unit cell in volume {$v_c$}, and neglect for the time being the hyperfine coupling to the nearest neighbors. The dipolar field is

{$$ \tag{1} B_{\mbox{dip}}^\alpha = - \frac {g\mu_B\mu_0}{4\pi v_c} \sum_{i=1}^N \sum_{\beta=x,y,z} D_i^{\alpha \beta} J_i^\beta $$}

where {$\alpha,\beta=x,y,z$} and the dimensionless dipolar tensor is defined as

{$$ D_i^{\alpha \beta} = \frac {v_c}{r_i^3} \frac {3 r_i^\alpha r_i^\beta - r_i^2 \delta_{\alpha\beta}}{r_i^5} $$}

Each tensor {$D_i$} is traceless and symmetric, hence it is defined by five constants

{$$ D_i = \left(\begin{matrix} D_i^{xx} & D_i^{xy} & D_i^{xz}\\ D_i^{xy} & D_i^{yy} & D_i^{yz} \\ D_i^{xz} & D_i^{yz} & -(D_i^{xx}+D_i^{yy})\end{matrix}\right) $$}

In the paramagnetic phase the magnetic moment ({$g\mu_B{\mathbf J} = {\mathbf M} v_c = \chi v_c {\mathbf H}$} in Eq. (1)) is uniform, hence

{$$ \tag{2} B_{\mbox{dip}}^\alpha = - \frac {\chi\mu_0}{4\pi} \sum_{i=1}^N \sum_{\beta=x,y,z} D_i^{\alpha \beta} H^\beta $$} In this simple case the term in front of the external field {$\mathbf H$} is the Knight shift tensor {$\mathbf K$}. Then we can write {$ {\mathbf B}_{\mbox{dip}} = {\mathbf K}\cdot {\mathbf H}$}, with

{$$ K^{\alpha \beta} = - \frac {\chi\mu_0}{4\pi} \sum_{i=1}^N \sum_{\beta=x,y,z} D_i^{\alpha \beta}$$}

The sum of symmetric traceless tensors is a symmetric traceless tensor, therefore the dipolar Knight shift is determined by five constants. They may be chosen as the two independent principal values of {$K$}, that suffice to write the tensor in its principal axis reference frame, plus the three Euler angles that rotate the latter to a generic orientation in the lab frame.

If {$\boldsymbol\chi$} is itself a tensor, then

{$$ \tag{3} K^{\alpha \beta} = - \frac {\mu_0}{4\pi} \sum_{i=1}^N \sum_{\beta=x,y,z} D_i^{\alpha \eta} \chi^{\eta \beta} $$}

Furthermore if there are {$n$} magnetic ions per cell, each displaying single ion anisotropy with different local orientations, then one must consider a local atomic susceptibility tensor and introduce a further sum over the unit cell

{$$ \tag{4} K^{\alpha \beta} = - \frac {\mu_0}{4\pi} \sum_{j=1}^{N_c} \sum_{i=1}^{n} \sum_{\beta=x,y,z} D_{{\mathbf r}_i+{\mathbf R}_j}^{\alpha \eta} \chi_i^{\eta \beta} $$}

In this case the experimental macroscopic susceptibility tensor corresponds to {${\boldsymbol\chi}=\sum_{i=1}^{n}{\boldsymbol\chi}_i$}

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