• View
• Edit
• History
• Print

Dy-ceneSpectrum

Indice

Dipolar field calculation

Dy-cene is composed of molecular units made of C,H and Dy, separated by C,F units and smaller proportion of crystallization solvent (ignored in the following)

We know the 40 F position vectors {$\mathbf{R}_{F,j}$} of an F unit, the 10 coordinates of the nn Dy {$\mathbf{R}_{Dy,k}$} surrounding each F unit.

H are coupled to one central Dy and their nn Dy are ignored since they are farther away. We know the vector components {$\mathbf{R}_{H,j}$} from each of the 67 H to their central Dy.

At a given temperature {$T$} and external field value {$B_0$} the probability of Dy being spin up is {$W$} (and it may further depend on thermal history, below the blocking temperature)

Therefore the H spectrum is calculated as follows. The reference frame is that of the crystal, in which the external field drection will be defined by two angles, {$\theta,\phi$} in a first experiment, and by three {$\theta, \phi,\gamma$} when the sample is rotated by {$\pi/2$} around the coil direction. Defining {$\mathbf m$} the moment of a spin-up Dy in the crystal frame, the dipolar field on the j-th H is

{$$ \mathbf{B}_{dH,j} = \frac {\mu_0}{4\pi |\mathbf{R}_{H,j}|^3} \left( 3\frac {\mathbf{R}_{H,j}\,\mathbf{R}_{H,j}\cdot \mathbf m}{|\mathbf{R}_{H,j}|^2} -\mathbf{m}\right)$$}

Assuming that this dipolar field is much smaller than the external field, the frequency for each proton is corrected by the secular part of the dipolar field

{$$\nu_j = ^1\!\!\gamma\,\left |\mathbf{B}_0+\frac{\mathbf{B}_0\cdot\mathbf{B}_{dH,j}} {B_0}\right| = ^1\!\!\nu (1+K_j) $$}

in a spin-up Dy unit, where {$^1\!\!\nu=^1\!\!\gamma B_0$} and the orientation dependent Knight shift constant of each H is {$ K_j(\theta,\phi)={\mathbf{B}_0(\theta,\phi)\cdot\mathbf{B}_{dH,j}}/{B_0^2}$}. For protons in a spin-down Dy unit, since {$\mathbf m \rightarrow - \mathbf m$} one has {$\mathbf{B}_{dH,j} \rightarrow - \mathbf{B}_{dH,j},\, \forall j$}, hence the H frequency is {$\nu_j = ^1\!\!\nu (1-K_j)$}. It is convenent to define an inverse field vector of modulus {$1/B_0$}, the configuration factors {$u_l = \delta_{l,+}$} for spin-up and {$d_l = u_l-1 = -\delta_{l,-}$} for spin-down, with l = +,-, to redefine the Knight shift constant as

{$$ \mathbf{b}(\theta,\phi) = \frac {\mathbf{B}_0}{B_0^2} \qquad K_{j,l}(\theta,\phi) = (2u_l -1)\mathbf{b}(\theta,\phi)\cdot\mathbf{B}_{dH,j} = \pm \mathbf{b}(\theta,\phi)\cdot\mathbf{B}_{dH,j}$$}

The H spectrum {$S_H(\nu)$} is simulated as a sum of Lorentzians (unnormalized, since experiment and theory have a scale fit factor, but with same width {$\lambda_H$} for simplicity) that reads

{$$\begin{equation}S_H(\nu) = \sum_{l=+,-} \sum_{j=1}^{67}\frac {\lambda_H P_l} {[2\pi(\nu-\nu_H(1+ K_{j,l}))]^2+\lambda_H^2} \end{equation}$$}

with {$P_+ = W, P_- = 1-W$}. Eq. (1) is straightforward to calculate numerically, requiring only 67 separate dot products and 134 Lorentzians to generate the spectrum for a new orientation, just twice as many those that compose the fully polarized spectrum ({$W=1$}).

Fluorines see 10 Dy molecules each (actually they see a sum over many more, but we assume that the 10 nn provides a reasonable picture, say 80% of the effect). Each configuration of Dy moments will produce a different dipolar field. Let's follow again the proton scheme ad explicit the {$\theta,\phi$} dependence of the external field, Defining

{$$\mathbf{R}_{F,jk} = \mathbf{R}_{F,j} - \mathbf{R}_{Dy,k} $$}

the dipolar field on the j-th F from the k-th Dy moment is

{$$ \mathbf{B}_{dF,j,k} = \frac {\mu_0}{4\pi |\mathbf{R}_{F,jk}|^3} \left( 3\frac {\mathbf{R}_{F,jk}\,\mathbf{R}_{F,jk}\cdot \mathbf m}{|\mathbf{R}_{F,jk}|^2} -\mathbf{m}\right)$$}

which reverses sign if the k-th Dy moment is reversed. These are 40x10x3 = 1200 scalar values (3 components, 10 Dy and 40 F)that can be calculated in advance. We must now consider the l-th configuration of the 10 Dy ions, conveniently represented by an up-spin factor {$u_l= \lbrace u_{l,k}\rbrace$} with {$u_{l,k}=1,0$} for k = up,down and a down-spin factor {$d_l= \lbrace d_{l,k}\rbrace$} with {$d_{l,k}=u_{l,k}-1 = 0,-1$}. Thus the total dipolar field on the j-th F in configuration l is

{$$ \mathbf{B}^{(l)}_{dF,j} = \sum_{k=1}^{10} (2u_{l,k} -1) \mathbf{B}_{dF,j,k}, \qquad n_l = \sum_{k=1}^{10} {u_{l,k}}, \qquad P_l = W^{n_l}\,(1-W)^{10-n_l} $$}

with probability {$P_l$} and number of spin-up Dy moments {$n_l$}. The first two can be calculated in advance (before specifying the field orientation) and they entail storing an array of size 1024x40x3 = 121920, {$B^{(l,\alpha)}_{dF,j},\quad \alpha=x,y,z$}, and an array of size 1024, {$n_l$}. Hence the F spectrum {$S_F(\nu)$} will be

{$$\begin{equation}S_F(\nu) = \sum_{l=1}^{2^{10}} \sum_{j=1}^{40} \frac {\lambda_F P_l} {[2\pi(\nu-^{19}\!\!\nu(1+K_{j,l}))]^2+\lambda_F^2} \end{equation}$$}

Now the 40960 Knight shift constants, one for each j (F) and l (configuration), are obtained by a dot product with the inverse field vector

{$$\begin{equation}K_{j,l}(\theta,\phi) = \sum_{k=1}^{10} (2u_{l,k} -1) \mathbf{b}(\theta,\phi)\cdot\mathbf{B}_{dF,j,k} = \sum_{k=1}^{10} \mathbf{b}^{(l)}_k(\theta,\phi)\cdot\mathbf{B}_{dF,j,k} \end{equation}$$}

where {$\mathbf{b}^{(l)}_k(\theta,\phi) = (2u_{l,k} -1) \mathbf{b}(\theta,\phi)$}. The brute force F spectrum {$S_F(\nu)$} of Eq. (2) requires recalculating 40960 dot products (611 times more than {$S_H$}) for each new orientation.It amounts to calculating 121920 products and sums to then add 40960 Lorentzians, i.e. 40960x140 data function calculations, taking 2.4 s on the XPS.

top

Fluorine histogram trick

The calculation can be extended to rotation angle {$\gamma$}. It requires

The raw version would require the full cost at each parameter extraction, {$W,B_0,\mbox{scale},\lambda_H/\lambda_F,\theta,\phi,\gamma$}, in particular the 122849 frequencies requiring each a 40 sums of products. For each of those, ~130 lorentzian values should be computed, each involving a square, a sum, a division.

It is mandatory to reduce this as much as possible for each individual parameter. E.g. {$\mbox{scale}$} should just multiply a stored value, without redoing previous steps, {$B_0$} should just renormalize {$\omega_F$} and redo the last step.

Additionaly, the last step can be speeded up by histogramming the frequencies over the ~130 frequency bins {$\Delta\omega$} (assuming that {$\lambda_F\gg\Delta\omega$}) with weights . top

Indice