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QuadrupoleZeeman

< The quadrupolar interaction | Index | The master equations: how to calculate the relaxation rate >

Assume that we have a quadrupolar interaction with {$\hbar \nu_Q=3e^2qQ/[2I(2I-1)]$}. Consider {$\hat x, \hat y, \hat z$} as the reference frame of the principal axes of the electric field gradient {$\mathbf V$}, with {$eq=V_{zz}$} and {$\eta=(V_{xx}-V_{yy})/V_{zz}$}. Now add a local field at the nucleus {$\mathbf{B}_n$} (in the following we imagine a scalar hyperfine field) in a generic direction as show in the figure. The Hamiltonian for this problem is given by Eq. 2 in the previous page

We want now to calculate the spectrum for this case. Let's start assuming that the initial state is just after a nutation of the {$I$} nuclear spin in the rf coil direction, e.g. {$|x m\rangle$}, such that {$I_x|x m\rangle = m |x m\rangle$}. The FID signal will be

{$$\langle x m|I_x(t)|x m\rangle$$}

with {$|x m \rangle= \sum_{r=1}^{2I+1} \langle r|x m\rangle |r\rangle$}, where {$c_{rm}=\langle r|x m\rangle$} are the coefficients of {$|x m\rangle$} in the {$\cal H $} basis, with {$\cal H|r\rangle=\hbar\omega_r|r\rangle$} . Furthermore

{$$I_x(t)= (e^{i\cal H t / \hbar}I_x e^{-i\cal H t / \hbar})$$}

This can be rewritten as

{$$I_x(t)= (I_x)_{rs} e^{i\omega_{rs}t}$$}

in terms of the differences {$\omega_{rs}=\omega_r-\omega_s$} between the eigenfrequencies of the Hamiltonian. Summarizing

{$$\langle x m|I_x(t)|x m\rangle = \sum_{r\ne s=1}^{2I+1} {c_{rm}^*c_{sm} } (I_x)_{rs} e^{i\omega_{rs}t}$$}

If we want now to simulate this spectrum numerically we need just to compute the eigenfrequency differences and store for each of them the weight

{$${c_{rm}^*c_{sm} }(I_x)_{rs} $$}

Grouping together {$\omega$} and {$-\omega$} terms in {$I_x(t)$}, we get {$\Re(c_{rm}^*c_{sm})(I_x)_{rs}\cos\omega_rs t+\Im(c_{rm}^*c_{sm})(I_x)_{rs}\sin\omega_rs t)$}. This implies that the quadrature spectrum (the modulus of the amplitude) is

{$$ |c_{rm}^*c_{sm}|(I_x)_{rs}$$}

When dealing with powders each crystallite is differently oriented and {$c^x_{rm}=\langle r|x m\rangle$}, {$c^y_{rm}=\langle r|y m\rangle$}, {$c^z_{rm}=\langle r|z m\rangle$} sample the response in three orthogonal directions. The total FID will be

{$$ \sum_{\alpha=x,y,z}\left[\Re({c^\alpha_{rm}}^*c^\alpha_{sm}) \cos \omega_{rs} t+\Im({c^\alpha_{rm}}^*c^\alpha_{sm}) \sin \omega_{rs} t \right] (I_\alpha)_{rs} $$}

which requires rephasing to get a real amplitude.

See matlab/nmrnqr/readme

< The quadrupolar interaction | Index | The master equations: how to calculate the relaxation rate >