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We will treat this example a bit pedantically, as a useful exercise. Let us treat formally the general case, specializing for the time being to the geometry of a typical instrument, the General Purpose Spectrometer (GPS) of PSI. In the absence of an applied field the Hamiltonian is simply

{$ (1) \qquad\qquad \frac {\cal H} h = \frac {\gamma_\mu} {2\pi} B_\mu\, I_\zeta$}

where {$\hat\zeta$} is the direction of the local field {$\mathbf{B}_\mu$}. We consider a single muon site.

We need to specify the geometry in detail. Only four out of five GPS detectors are shown for clarity. The direction {$\hat\zeta$} is defined by the polar angles {$\theta,\phi$} in the GPS reference frame {$\hat x,\hat y,\hat z$}, shown in the figure. Likewise the initial muon spin {$\mathbf{ I_R}$} lies along {$\hat Z$}, in the {$\hat y,\hat z$} plane, forming an angle {$\Theta$} with the {$-\hat z$} direction. The spin rotator can select {$\Theta\approx 0 \,\mbox{or}\,50$} degrees. We need two additional axes that for an orthogonal frame with {$\hat\zeta$}, say, {$\hat \xi, \hat \eta$}. We chose them such that: {$\hat\xi$} lies in the {$x,y$} plane, and is obtained as {$\hat \xi={\cal R}_z(\phi)\hat y$} {$\hat\eta$} in the plane defined by {$\hat z$} and {$ \mathbf{B}_\mu$}, and is obtained as {$\hat\eta={\cal R}_\xi(\frac \pi 2 -\theta)\hat z$}

With this notation we can write the muon polarization observed in the forward-backward detectors along {$z$} as:

{$ P_z(t)= Tr [\rho \sigma_z(t)], $}

where {$\rho=\frac 1 2 (1+\sigma_Z)$}, when written in the basis of {$\sigma_Z$}. It is however more convenient to express all matrices in the basis of {$\sigma_\zeta$} (and of the Hamiltonian, Eq. 1), {$ |m\rangle = |\pm\rangle$}. The required transformations are:

{$ \begin{eqnarray} \sigma_Z&=&\sigma_\xi \sin\Theta \cos\phi-\sigma_\eta (\cos\Theta\sin\theta + \sin\Theta\cos\theta\sin\phi) + \sigma_\zeta(\sin\Theta\sin\theta\sin\phi-\cos\Theta\cos\theta)\\ \sigma_z&=&\sigma_\eta\sin\theta+\sigma_\zeta\cos\theta\\ \sigma_y&=&\sigma_\xi\cos\phi -\sigma_\eta\cos\theta\sin\phi+\sigma_\zeta\sin\theta\sin\phi\end{eqnarray} $}

and, since {$Tr(1+\sigma_Z)\sigma_z(t)= Tr\sigma_Z\sigma_z(t)$} (Pauli matrices are traceless), the forward-backward polarization turns out to be:

{$ P_z(t) = \frac 1 2 \left[ \langle + | \sigma_Z | +\rangle\langle + | \sigma_z | + \rangle + \langle - | \sigma_Z | -\rangle\langle - | \sigma_z | - \rangle + \langle + | \sigma_Z | -\rangle\langle - | \sigma_z | + \rangle e^{-i\gamma_\mu B_\mu t}+ \langle - | \sigma_Z | +\rangle\langle + | \sigma_z | - \rangle e^{i\gamma_\mu B_\mu t}\right]. $}

The first two terms represent the constant, longitudinal polarization, due to the field component parallel to the initial muon spin direction, while the last two terms correspond to the transverse polarization, which precesses around the local field.

An alternative, perhaps more intuitive interpretation of this expression is to consider the first matrix element in each term of the sum as the projection of the initial muon state {$|\Psi\rangle$} over the eigenstates of the Hamiltonian, which is the weight of the corresponding expectation value of the observed quantity, {$\sigma_z$}. We could then simply write {$P_z(t)=\langle \Psi(t)|\sigma_z|\Psi(t)\rangle$}, with {$|\Psi(t)\rangle=\sum_m c_m e^{i\omega_m t}|m\rangle, \qquad m=\pm$}. The density matrix is recovered as {$\rho_{mn}= c_m^*c_n$}, leading to the same expression obtained above.

After a little algebra, substituting the expressions for the relevant {$\sigma$} operators and recalling that {$\sigma_\zeta$} is diagonal, while {$\langle +| \sigma_\xi| -\rangle = 1$}, {$\langle +| \sigma_\eta| -\rangle = -i$}, we get

{$ \begin{eqnarray} (2) \qquad\qquad P_z(t) &=& \cos\theta(\sin\Theta\sin\theta\sin\phi - \cos\Theta\cos\theta) +\\ &&\sin\theta\left[\sin\Theta\cos\phi \sin\gamma_\mu B_\mu t - (\cos\Theta\sin\theta + \sin\Theta\cos\theta\sin\phi )\cos\gamma_\mu B_\mu t \right]\end{eqnarray}. $}

The same calculation yields

{$(3)\qquad\qquad \begin{eqnarray} P_y(t) &=& Tr [\rho \sigma_y(t)]\\ &=& \sin\phi\sin\theta(\sin\Theta\sin\theta\sin\phi - \cos\Theta\cos\theta) + \\ && \left[\sin\Theta(\cos^2\phi+\cos^2\theta\sin^2\phi) +\cos\Theta\sin\theta\cos\theta\sin\phi \right]\cos\gamma_\mu B_\mu t + \cos\Theta\sin\theta\cos\phi\sin\gamma_\mu B_\mu t .\end{eqnarray}$}

We end by summarizing in a table the most used configurations, distinguishing for each observation direction {$\alpha=y,z$} the constant, longitudinal component, from the transverse, precessing component.

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