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NewFMuF

< FMu in longitudinal field | Index | Magnetically ordered materials >

LiF is not compatible with the FMuF idea.

Rocksalt a=0.402 nm, group 225, F=(0 0 0), Li=(0.5 0 0), Let's consider the first octant of the cell, made by a tetrahedron of F (1,2,3,4) and a tetrahedron of Li (1,2,3,4), rotatde by 90 degrees with respect to each other. DFT with muonium in a neutral cell seems to find a configuration of four equivalent muon sites in each octant, in a tetrahedron (red dots) that mimics that of the F, with F-µ distance of 1.19 A (close to that identified by Brewer with the straight molecule model). The potential between the four minima is very shallow. Can there be tunneling between the four sites? Distances between the muon site m1=0.1709(1 1 1) in cell units, the closest to the origin (F1), and the nuclei are as follows: d(m1-F1)=0.119 nm d(m1-F2)=d(m1-F3)=d(m1-F4)=0.199 nm d(m1-Li1)=d(m1-Li2)=d(m1-Li3)=0.1641 nm d(m1-Li4)=0.229 nm d(m1-F5)=0.347 nm Hence the dipolar field from F2, F3, F4 is 0.21 times that from F1 and the dipolar field from Li1, Li2, Li3 is roughly 0.19 that of F1 (taking into account that the 7Li moment is roughly twice smaller).

Let us neglect all other nuclear spins but the nearest F to begin with. Calling {$I_k, k=1,..,4$} the F spins and {$S$} the muon spin We have the following spin Hamiltonian in each of the four muon sites:

{$ H_{k} = B_d \left[ 3 {\mathbf I}_k \cdot {\hat r}_k {\mathbf S} \cdot {\hat r}_k - {\mathbf I}_k \cdot {\mathbf S} \right] $}

Here {$B_d=\mu_0\hbar^2\gamma_\mu\, ^{19}\gamma/d^3$} and taking x, y, z along the cubic axes, the the unit vectors {${\hat r}_k$} along which the F-µ bonds lie are {$1/sqrt3 \hat{u}_k$} with u_k = [1 1 1]; [-1 -1 1]; [-1 1 -1]; and [1 -1 -1]. Plugging these numbers in we get

{$ H_{k} = B_d \left[ u_{kx}u_{ky}(I_{kx}S_{ky}+I_{ky}S_{kx}) + u_{ky}u_{kz}(I_{ky}S_{kz}+I_{kz}S_{ky}) + u_{kz}u_{kx}(I_{kz}S_{kx}+I_{kx}S_{kz})\right] $}

If the DFT solution were right and tunneling produced a narrowing condition the hamiltonian would be the average (sum/4) of the four hamiltonians in each site. Since {$\sum_k u_{k\alpha}u_{k\beta}=0$} it is easily seen that this sum vanishes.

< FMu in longitudinal field | Index | Magnetically ordered materials >