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Hamiltonian

Consider a muonium atom with a pseudo-dipolar interaction, i.e. the usual magnetic dipole field of the electron moment, with electron and muon at finite distance r, averaged over the electron wave function. The latter is zero for the s wave function, but it is non vanishing for the excited p states of Mu in vacuum, as well as for bond-center (BC) Mu in Si. The first case has a vanishing contact term, but in a crystalline environment the LCAO is an intermediate case with partial s and p symmetries. In zero external field this leads to

{$$ \frac {\cal H} h = \nu_0 {\mathbf I}\cdot{\mathbf S} + \Delta\nu{\mathbf I}\cdot\left(\matrix{-\frac 1 2 & 0 & 0\cr 0 & -\frac 1 2 & 0 \cr 0 & 0 & 1}\right)\cdot{\mathbf S} $$}

The zero field case is particularly simple: following the isotropic Mu calculation the spin matrix is simply

{$$ \frac {\cal H} h = \frac 1 4 \left(\matrix{\nu_0+\Delta\nu & 0 & 0 & 0 \cr 0 & -(\nu_0+\Delta\nu) & 2\nu_0-\Delta\nu & 0 \cr 0 & 2\nu_0-\Delta\nu & -(\nu_0+\Delta\nu) & 0 \cr 0 & 0 & 0 & \nu_0+\Delta\nu}\right) $$}

Eigenstates are classified with the total angular momentum J, as triplet J=1, M=±1 (the first and the last on the diagonal of the matrix) and two further states that mix J=1, M=0 with J=0, with eigenvalues

{$$\begin{eqnarray} \nu_{1,4} &=&\frac 1 4 \nu_0+\Delta\nu\\ \nu_2 &=& \frac 1 4 \nu_0\\ \nu_3&=& -\frac 3 4 \nu_0+\frac 1 2 \Delta\nu \end{eqnarray} $$}

respectively, in units of h. Defining the first state as that of the electron as before and reminding that now the quantization is along the anisotropy axis

{$$|+\, +\rangle = \left(\begin{array} 1\\0\\0\\0\end{array}\right); \qquad |+\, -\rangle = \left(\begin{array} 0\\1\\0\\0\end{array}\right); \qquad |-\, +\rangle = \left(\begin{array} 0\\0\\1\\0\end{array}\right); \qquad |-\, -\rangle = \left(\begin{array} 0\\0\\0\\1\end{array}\right)$$}

the eigenvectors can be readily calculated as

{$$ \begin{eqnarray}| 1\rangle &=& | +\, +\rangle; \\ | 2\rangle &=& \frac 1 {\sqrt{2}} [| +\, -\rangle + | -\, +\rangle]; \\ | 3\rangle &=& \frac 1 {\sqrt{2}} [| +\, -\rangle -| -\, +\rangle]; \\ | 4\rangle &=& | -\, -\rangle \end{eqnarray}$$}

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Density matrix

Following always the isotropic Mu calculation, and the equations numbered therein, we now define as θ the angle between the anisotropy z and the initial muon spin η and choose

{$$\sigma_{\mu\eta}(\theta)= \cos\theta\,\sigma_{\mu z} +\sin\theta\,\sigma_{\mu x}$$}

Dropping the muon unity matrix as in eq. (2) we can then write (c=cosθ, s=sinθ)

{$$ \rho_0(\theta)= \frac 1 4 \left( \matrix{ c &s&0&0 \\ s & -c & 0 & 0\\ 0&0&c&s\\ 0&0&s&-c}\right)$$}

With this initial density matrix we wish to calculate the polycrystalline average of the longitudinal spin polarization {$P_{\eta \eta}(\theta,t) $} over θ. Using eq. (6) we have

{$$ \begin{eqnarray} P_{\eta\eta}(\theta,t) = \frac 1 4 \sum_{n,m=1}^4 \ e^{-i2\pi (\nu_n-\nu_k) t} |\langle k |\sigma_{\mu\eta}\otimes1_e| n\rangle |^2 = \frac 1 2 \left[\cos^2\theta + \sin^2\theta \left(\cos\frac \pi 2 \Delta\nu t + \cos 2\pi(\nu_0 - \frac {\Delta\nu} 4 ) t\right)\right]\end{eqnarray}$$}

where we have used

{$$ \begin{eqnarray}|\langle 1 |\sigma_{\mu z}| 1\rangle|^2 &=& |\langle 4 |\sigma_{\mu z}| 4\rangle|^2 &=& 1 \\ |\langle 2 |\sigma_{\mu z}| 2\rangle|^2 &=&|\langle 3 |\sigma_{\mu z}| 3\rangle|^2&=&|\langle 1 |\sigma_{\mu z}| 2\rangle|^2&=&|\langle 1 |\sigma_{\mu z}| 3\rangle|^2&=&|\langle 4 |\sigma_{\mu z}| 2\rangle|^2&=&|\langle 4 |\sigma_{\mu z}| 3\rangle|^2&=&0; \\ |\langle 2 |\sigma_{\mu z}| 3\rangle|^2 &=& 1\end{eqnarray}$$}

and (keeping Pauli matrix algebra in mind)

{$$ \begin{eqnarray}|\langle 1 |\sigma_{\mu x}| 1\rangle|^2 &=& |\langle 2 |\sigma_{\mu x}| 2\rangle|^2&=& |\langle 3 |\sigma_{\mu x}| 3\rangle|^2&=& |\langle 4 |\sigma_{\mu x}| 4\rangle|^2& =&0 \\ |\langle 1 |\sigma_{\mu x}| 2\rangle|^2 &=&|\langle 1 |\sigma_{\mu x}| 3\rangle|^2&=&|\langle 3 |\sigma_{\mu x}| 4\rangle|^2&=&|\langle 3 |\sigma_{\mu x}| 4\rangle|^2&=& 1 \\ |\langle 1 |\sigma_{\mu x}| 4\rangle|^2 &=& 0\end{eqnarray}$$}

Since {$\langle \cos^2\theta\rangle=\frac 1 3,\quad \langle \sin^2\theta\rangle=\frac 2 3$}

{$$ \begin{eqnarray} P_{zz}(t)=\langle P_{\eta\eta}(\theta,t) \rangle = \frac 1 3 \left[1 + 2 \left(\cos\frac \pi 2 \Delta\nu t + \cos 2 \pi (\nu_0 - \frac {\Delta\nu}4) t\right)\right]\end{eqnarray}$$}

where we have now reserved the label η=z for the initial muon spin direction, since the anisotropy axis is in any direction in the polycrystal.

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AnisotropicMuonium

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