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Rotations in experiments
Second experiment:
Ref frame [x,y,z] is that of the cif file of the Dy-cene crystal;
ref frame [X;Y;Z] is that of the lab, see Figure 1
Euler angles are {$\alpha,\beta,\gamma$} such that {${\cal R}(\alpha,\beta,\gamma) X_i = x_i, \quad X_i=X,Y,Z$}. Figure 2 shows that
{$$ {\cal R}(\alpha,\beta,\gamma) = R_z(\gamma)R_N(\beta)R_Z(\alpha)$$}
where {$R_N(\beta)=R_Z(\alpha)R_X(\beta)R_Z(-\alpha)$} and {$R_z(\gamma)=R_N(\beta)R_Z(\gamma)R_N(-\beta)$}. Substituting
{$$ \begin{align*} {\cal R}(\alpha,\beta,\gamma) &= R_N(\beta)R_Z(\gamma)R_N(-\beta)\,\,\, R_N(\beta)\,\,\, R_Z(\alpha)\\ &= R_Z(\alpha)R_X(\beta)R_Z(-\alpha)\,\,\,R_Z(\gamma)\,\,\,R_Z(\alpha)\\ &= R_Z(\alpha)R_X(\beta)R_Z(\gamma)\end{align*} $$}
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Figure 1: {$\mathbf{B}_1=B_1\hat X, \mathbf{B}_0=B_0 \hat Z$}
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Inserting the definition of {$R_X(\beta) = \begin{bmatrix}1& 0 & 0\\ 0 & \cos\beta & -\sin\beta\\0 &\sin\beta&\cos\beta \end{bmatrix}$}
and {$R_Z(\phi) = \begin{bmatrix} \cos\beta & -\sin\beta & 0 \\\sin\beta&\cos\beta&0\\ 0&0&1 \end{bmatrix}$} one gets
{$$ {\cal R}(\alpha,\beta,\gamma) = \begin{bmatrix}-\sin\alpha\sin\gamma\cos\beta+\cos\alpha\cos\gamma& -(\sin\alpha\cos\beta+\sin\gamma\cos\alpha)&\sin\alpha\sin\beta\\ \sin\alpha\cos\gamma+\sin\gamma\cos\alpha\cos\beta & -\sin\alpha\sin\gamma+\cos\alpha\cos\beta\cos\gamma& -\sin\beta\cos\alpha\\\sin\beta\sin\gamma& \sin\beta\cos\gamma&\cos\beta\end{bmatrix}$$}
The fit to the spectrum yields that the polar angles of {$\mathbf B_0$} in {$x,y,z$} are {$\theta_0,\phi_0$} implying {$\cos\theta_0=\hat Z\cdot\hat z = \hat Z\cdot{\cal R}\hat Z$} and {$\sin\theta_0\cos\phi_0=\hat Z\cdot\hat x = \hat Z\cdot{\cal R}\hat X$}, i.e.
{$$\begin{align*} \cos\theta_0 &= \begin{bmatrix} 0&0& 1\end{bmatrix}{\cal R}(\alpha,\beta,\gamma)\begin{bmatrix} 0\\0\\ 1\end{bmatrix}\\ \sin\theta_0\cos\phi_0 &=\begin{bmatrix} 1&0& 0\end{bmatrix} {\cal R}(\alpha,\beta,\gamma)\begin{bmatrix} 0\\0\\ 1\end{bmatrix}\end{align*}$$}
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Fig. 2 Euler angles and the rotations [from wikipedia, under Creative Commons]'
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i.e.
{$$\begin{align*} \cos\theta_0 &= {\cal R}_{33} = \cos\beta\\ \sin\theta_0\cos\phi_0 &={\cal R}_{13}=\sin\alpha\sin\beta \end{align*}$$}
This simply means that {$\theta_0=\beta, \phi_0=\pi/2-\alpha$} or {$\theta_0=\beta, \phi_0=\alpha-\pi/2$}.
First experiment
the sample rotated was {$\pi/2$} around X by {$R_X(\pi/2)$}. Let's work it out, calling {$\hat x^\prime,\hat y^\prime,\hat z^\prime$} the orientation of the cell in this case: {$\cos\theta_1=\hat Z \cdot\hat z^\prime, \sin\theta_1\cos\phi_1=\hat Z \cdot \hat x^\prime$}. Since {$\hat x_i^\prime = R_X(\pi/2)\hat x_i = R_X(\pi/2){\cal R}(\alpha,\beta,\gamma)\hat X_i$} let's work ou first the product of the two matrices:
{$$ \begin{align*}
{\cal R}^\prime(\alpha,\beta,\gamma) &=
R_X(\pi/2){\cal R}(\alpha,\beta,\gamma)\\
& = \begin{bmatrix}1&0&0\\0&0&1\\0&-1&0\end{bmatrix} \,
\begin{bmatrix} {\cal R}_{11}&{\cal R}_{12}&{\cal R}_{13}\\{\cal R}_{21}&{\cal R}_{22}&{\cal R}_{23}\\{\cal R}_{31}&{\cal R}_{32}&{\cal R}_{33}\end{bmatrix} \\
&=
\begin{bmatrix} {\cal R}_{11}&{\cal R}_{12}&{\cal R}_{13}\\{\cal R}_{31}&{\cal R}_{32}&{\cal R}_{33}\\-{\cal R}_{21}&-{\cal R}_{22}&-{\cal R}_{23} \end{bmatrix}
\end{align*} $$}
Therefore
{$$\begin{align*} \cos\theta_1 &= \begin{bmatrix} 0&0& 1\end{bmatrix}{\cal R}^\prime(\alpha,\beta,\gamma)\begin{bmatrix} 0\\0\\ 1\end{bmatrix}\\ \sin\theta_1\cos\phi_1 &=\begin{bmatrix} 0&0& 1\end{bmatrix} {\cal R}^\prime(\alpha,\beta,\gamma)\begin{bmatrix} 1\\0\\ 0\end{bmatrix}\end{align*}$$}
or
{$$\begin{align*} \cos\theta_1 &= {\cal R}^\prime_{33} = -{\cal R}_{23} = \sin\beta\cos\alpha\\ \sin\theta_1\cos\phi_1 &={\cal R}^\prime_{31}=-{\cal R}_{21}=-(\sin\alpha\cos\gamma+\cos\alpha\cos\beta\sin\gamma) \end{align*}$$}
To replicate the spectrum it should suffice to calculate it for
{$$\begin{align*}\theta_1&=\cos^{-1}\left( \sin\theta_0\sin\phi_0\right) \\ \phi_1 &= \cos^{-1}\left( -\frac{\pm\cos\phi_0\cos\gamma+\sin\phi_0\cos\theta_0\sin\gamma}{\sqrt{1-\sin^2\theta_0\sin^2\phi_0}}\right)\end{align*}$$}
i.e in the ideal condition that the experimental rotation were exactly by {$\pi/2$} around X there is one angle to vary, {$\gamma$}, starting from the best fit of {$\theta_0,\phi_0$}, to obtain the best fit of the first experiment. The two determination ({$\pm\cos\phi_0\cdots$}) reproduce each other with a phase {$\ne \pi/2$}
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{$\cos\phi_1$} vs. the unknown angle {$\gamma$} in the first experiment
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Generation of the spectrum
1H spectrum due to one Lorentzian per proton, centered at the Knight shift appropriate for that proton. Each proton is between 3 and 5 Angstrom from the nearest Dy ion, and nearest Dy atoms are 11 Angstrom from each other, hence protons are strongly coupled to a single Dy ion. The Knight shift is calculated as
{$$ \nu_i = \gamma B_0 (1+K) = \gamma (B_0 + B_{iz})$$}
where {$B_{iz}= \mathbf{B}_i\cdot \mathbf{B}_0/B_0$} is the secular component of the Dy dipolar field at the {$i-th$} proton position. Hence
{$$K_i=B_{iz}/B_0=\mathbf{B}_d\cdot \mathbf{B}_0/B_0^2$$}
In the jupyter notebook Run-01-18/Dysprosium-cene-2-perpendicular.ipynb
this is done by generating the dipolar field of the {$N$} protons in the crystal (cif) reference frame once, and then, for each magnetic field direction {$\theta,\phi$}, generate the corresponding magnetic field
{$$\mathbf{B}_0 = B_0 (\sin\theta\cos\phi \hat{x} + \sin\theta\sin\phi\hat{y} + \cos\theta\hat{z})$$}
Assuming probability {$W$} for Dy spin up (frequency shifts {$=+\gamma B_0 K_i$}) and {$1-W$} for Dy spin down (frequency shifts {$=-\gamma B_0 K_i$}) the 1H unnormalized spectrum is
{$$f(\nu) = W \sum_{i=1}^{N} \frac {\delta \nu}{(\nu -\nu_0(1+K_i))^2+\delta \nu^2} + (1-W) \sum_{i=1}^{N} \frac {\delta \nu}{(\nu -\nu_0(1-K_i))^2+\delta \nu^2} $$}
where {$\nu_0=\gamma B_0$}.
The fluorine spectrum is the same but each fluorine is coupled to many Dy ions
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