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HMu

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In the case of a two spin state under dipolar interactions one has

{$ \frac {\cal H} \hbar = -\omega_d(2I_zS_z-I_xS_x-I_yS_y) = \omega_d \begin{bmatrix} - \frac 1 2 \quad & \quad 0\quad & \quad 0\quad & \quad 0\quad \\ 0 & \frac 1 2 & \frac 1 2 & 0 \\ 0 & \frac 1 2 & \frac 1 2 & 0 \\ 0 & 0 & 0 & -\frac 1 2\end{bmatrix} $}

The eigenvalues are {$\omega_1=\omega_4=-\frac{\omega_d} 2 $} and {$ \omega_2=0, \omega_3=\omega_d$}

The unitary transformation that rotates the eigenstates from the basis of {$I_z, S_z$} onto that of {$\cal H$} is

{$ U=\begin{bmatrix} \quad 1 \quad & \quad 0\quad & \quad 0\quad & \quad 0\quad \\ 0 & \frac 1 {\sqrt 2} & \frac 1 {\sqrt 2} & 0 \\ 0 & \frac 1 {\sqrt 2} & - \frac 1 {\sqrt 2} & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} $}

so that the time evolution operator {$e^{-i{\cal H}t/\hbar}$} in the basis of {$I_z,S_z$} will be

{$ E(t)=U^{-1}\begin{bmatrix} \quad \alpha \quad & \quad 0\quad & \quad 0\quad & \quad 0\quad \\ 0 & \alpha^{-2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \alpha \end{bmatrix} U $}

with {$\alpha=e^{-i \omega_d t/2}$}

Let us consider longitudinal detection, i.e. the initial muon spin and the detector axis are parallel to {$\hat z$}. Let us choose here the muon Pauli sigmas as the inner 2x2 components in the tensor product and the proton Pauli matrices as the 4x4 outer components. For an arbitrary angle {$\theta$} between the dipole axis and {$\hat z$} one has that both the density matrix at time {$t=0$} and the observed quantity are given by

{$1_e\otimes I_\xi = \begin{bmatrix} \cos\theta & \sin\theta & \quad 0\quad & \quad 0\quad \\ \sin\theta & -\cos\theta & 0 & 0\\ 0 & 0 & \cos\theta & \sin\theta \\ 0 & 0 & \sin\theta & -\cos\theta\end{bmatrix} $}

with {$I_\xi= \cos\theta I_z + \sin\theta I_x$}. Then the observed muon polarization is given by

{$P(t)= Tr E(t) I_\xi E(-t) I_\xi $}

It is straightforward to calculate

{$ P(t)= \frac 1 2 \left(\cos^2\theta \,(1+\cos\omega_d t) \,+ \,\sin^2\theta\,(\cos\frac {\omega_d} 2 t + \cos\frac 3 2 \omega_d t) \right) $}

and, averaging over {$\theta$} for a polycrystalline sample one gets

{$ P_z(t)=\frac 1 6 \left(1+\cos\omega_d t \,+ 2(\cos\frac{\omega_d}2 t + \cos\frac 3 2 \omega_d t) \right) $}

Remember that the {$\gamma$} ratios are not those that we use to compute precession frequencies ({$\nu_\mu=\frac{\gamma_\mu}{2\pi}B_\mu$}), hence if you fitted a local field instead of a dipolar paramenter {$\omega_d$}, you will have

{$ r = \left( \frac{\mu_0 \gamma_F \hbar}{4\pi B_\mu} \right)^{1/3}$}

jupyter notebook FMu and FMuF with sympy, run under python3, debugging not complete

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