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Ba2CuGe2O7Summary from [2]: non centrosymmetric tetragonal AF with incommensurate nearly AF cycloid. Space group P(-4)21m. Cu in the ab plane describe a perfect square in the diagonal direction, the structure replicates along the c axis. In-plane AF exchange along diagonal of ab plane, J∥ ~ 1 meV per bond. Along c J⊥ ~ 26 µeV. Antisymmetric exchange path yields Dzyaloshinsky-Moriya Cone phase [1], a = 0.8466 nm , c = 0.5445 nm
Two-k structure, for Cu-O-Ge-O-Cu bonds along (1,1,0) the DM vector component Dy points along the (1,-1,0) and the component Dz is along (0,0,1) with alternating sign. This results in a cycloid with propagation vector q along (1,1,90) and spins confined in the (1,-1,0) plane, with deviation β=9,7° from one Cu spin to the next along the edge of the cube and nearly antiparallel in the center site
The muon is bound to Oxygen. Preliminary DFT indicate a site bound to Cu Oxygen (C site) and a second site bound to Ba Oxygen (B site). Both are along the direction joining two such O bound to the same cation at the same z cell coordinate. The incommensurate cycloid magnetic structure generates infinite inequivalent sites from each of these. B and C sites give rise to two families each (for a total of four families). Consider the (1 1 0) cycloid domain and the C site. Two Cu-C vectors lie almost within the Cu (1 1 0) plane of the cycloid and two lie almost at 90 degrees from it (see the animated figure). Let's call them C∥ and C⊥, respectively. The dipolar fields at these two sites, mostly due to their nn Cu, is very different. Now consider the B sites. Each is approximately symmetric between two nn Cu belonging to different sublattices (nearly antiferromagnetic, i.e. collinear and opposite on the local scale). One site family is nearly within the cycloid Cu (110) plane, B0 and another family lies outside the Cu (110) plane, B1. Let us first estimate local fields at the C sites. We consider in first approximation only the contribution of the nn Cu, both dipolar and isotropic hypefine terms. Let's treat this in a general form for the cycloid problem. The Cu spin lies in a given plane π = (1 1 0) in our case. Let's chose the reference frame such that r, joining C and the nn Cu, is along z, and it contains y (that amounts to fixing the perpendicular to z in π).. To start with the normal to the π plane is {$ \hat n = \sin\theta\hat x + \cos\theta \hat z $} Then we choose two orthogonal axes in the π plane, ξ as the direction that is perpendicular to both n and r, which is already fixed as y and ξ as the direction perpendicular to both n and ξ {$ \hat \xi = -\cos\theta\hat x + sin\theta\hat z $} The spin unit vector s can be written with any orientation in the π plane as {$ \hat s = \cos\phi\hat \xi+ \sin\phi\hat y$} and finally, defining the dipolar field B0 = μ0 m/r3 , the hyperfine field Bhf , and their ratio y = Bhf/B0 , we can write the two contributions to the muon field {$\mathbf{B}= \frac {B_0} 2 \left [(3 \hat r\cdot\hat s \hat r - \hat s) + 2y \hat s\right]$} One can compute the modulus of B or directly its ratio to B0 {$ \begin{eqnarray*}x = \frac B {B_0} &=& \frac 1 2 \left( (2y-1)^2+\sin^2\theta\sin^2\phi (12y+3) \right)^{1/2} \\&=& \sqrt{a+b\sin^2\phi}\end{eqnarray*}$} with {$a=\frac 1 4 (2y-1)^2$} and {$b=\frac 1 4 [(12y+3)\sin^2\theta)]$} When φ varies the spectrum of this cycloid is limited beween {$x_m=\sqrt a$}, for φ=0, and {$x_M=\sqrt{a+b}$}, for φ=π/2, and it is obtained as {$p(x)=\left|{\frac {d\phi}{dx}\right|$}, i.e. {$p(x) = \frac 1 {\left|\frac{dx}{d\phi}\right|} = \frac x {\sqrt{(x^2-a)(a+b-x^2)}} $} Since the reference frame is now fixed suppose the initial muon direction along {${\hat s}_\mu=[\sin\Theta\cos\Phi \sin\Theta\sin\Phi \cos\Theta]$}. Then the precession cone around the field direction {$\hat b$} has aperture {$\psi$} given by {$\cos\psi={\hat b}\cdot {\hat s}_\mu$} The transverse fraction is {$f_t=1-\cos^2\psi$} and the longitudinal fraction is {$f_l=\cos^2\psi$}. Thus we may obtain the transverse and longitudinal fraction of the spectrum (red and blue, respectively in the figures below). By FFT we recover the muon asymmetry (right panels, with 0.25 maximum asymmetry)  Left: FFT, Right: Asymmetry. Dipolar and hyperfine fields, orientations in the legend. Warning Bhf in T, not mT and the experimental result may have some aliasing of the missing high frequency part of the spectrum. Height of peaks is arbitrary.  Left: FFT, Right: Asymmetry. Dipolar and hyperfine fields, orientations in the legend. Note that the top row has y=Bhf/Bd=0.5, hence is equivalent to a SDW, with P(t)=J0(γBdt) References [1] S. Mühlbauer et al. Phys. Rev. B 84, 180406(R) (2011) [2] S. Mühlbauer et al. Phys. Rev. B 86 024417 (2012) |
