TN-49 "Muon polarisation in electromagnetic fields": Pierre Depommier
(Wed, 3 Jan 2001)

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\title{TRIUMF Experiment E614 \\
       Technical Note No. 49 \\
       {~} \\
       Muon polarisation in electromagnetic fields}

\author{ {~} \\
       Pierre Depommier \\
       {~} \\
       {\sl Department of Physics, University of Montreal}, \\
       {\sl P.O. Box 6128, Station ``Downtown'', Montreal, Quebec, 
       Canada H3C 3J7} \\
       {\sl pom@lps.umontreal.ca}
       }
\date {January 3, 2001}
\maketitle

\section{Introduction}
 
This note was originally written for the KEK-E246 experiment (Search
for a transverse polarization of the muon in the $K^+ \to \pi^0 \mu^+ \nu$
decay), following extensive discussions with Peter Gumplinger who wrote the
KEK-E246 and TRIUMF-E614 Monte Carlo codes, including the rotation of the 
muon polarization in a magnetic field by using the Bargmann-Michel-Telegdi 
(BMT) equation\cite{BAR59}. 

This note will also be of interest for the TRIUMF-E614 experiment. 
Its purpose is mainly pedagogical. 

It starts with the description of the polarization in the rest 
system of the particle. It is based on the density matrix formalism and the 
Stokes vector (a three-vector).

Then the relativistic treatment of the polarization is presented.
If the particle, initially at rest, is submitted to a Lorentz boost, the 
polarization vector becomes a four-vector. It reduces to a three-vector only 
in the rest system of the particle. 
 
The Bargmann-Michel-Telegdi (BMT) equation, which describes the motion of the 
polarization vector (a four-vector) in an electromagnetic field, is 
introduced. An expression is obtained for the motion of the Stokes 
vector\cite{STO52} (defined in the rest system of the particle, a 
three-vector).

As an application, the special case of a uniform magnetic field is considered
and an analytical solution is given. A computer program based on this
analytical solution has been written. This program has been used to test a
subroutine of the Monte Carlo simulation, which applies to the more 
general case of a non-uniform (slowly varying) magnetic field.  
 
In the experiments we are interested in, polarized muons are produced, 
then they go through
several materials and through electromagnetic fields. Finally, they are slowed 
down to rest before they decay. One is therefore faced with several problems:
\begin{enumerate}
\item{Knowledge of the initial polarization.}
\item{Rotation of the polarization vector in electromagnetic fields.}
\item{Depolarization, or change of magnitude of the polarization vector.}
\end{enumerate}
This note deals with the second problem, considering the case of a pure and
uniform magnetic field. As said before its purpose is mainly pedagogical. It 
has also been used to check the Monte Carlo codes written by Peter Gumplinger 
for the KEK-E246 and TRIUMF-E614  experiments.

\section{General formalism.}
Let us consider an object with two polarization degrees of freedom. It
can be a spin-1/2 particle (electron, muon, proton, etc...) but also a
photon (even if in that case the situation is completely different). 
The general formalism for the description of the polarization of such 
an object has been discussed by Tolhoek \cite{TOL56}.

The state of polarization can be written as:
\begin{equation} 
\vert \phi \rangle = a_1 \vert 1 \rangle +~ a_2 \vert 2 \rangle 
\hspace{1.0cm}\hbox{or}\hspace{1.0cm}
\phi = \left( \begin{array}{c} a_1 \\ a_2 \end{array} \right) 
\end{equation}
where $a_1$ and $a_2$ are complex numbers subject to the normalization
condition:
\begin{equation} 
\vert a_1 \vert^2 + \vert a_2 \vert^2 = 1 
\end{equation}
$\vert 1 \rangle$ and $\vert 2 \rangle$ are two basic polarization states. 
For a spin-1/2 particle in its rest frame they could be spin-up and spin-down. 
For a photon they could be two circularly-polarized states of different 
helicities, or two linear polarization states orthogonal to each other.

One defines the expectation value $\vec \xi = \phi^\dagger \vec \sigma \phi$  
(the components of $\vec \sigma$ are the Pauli matrices):
\begin{equation} 
\xi_x = a_1 a_2^* + a_2 a_1^* \hspace{1.0cm}
\xi_y = i ( a_1 a_2^* - a_2 a_1^* ) \hspace{1.0cm}
\xi_z = a_1 a_1^* - a_2 a_2^* 
\end{equation}
These three real numbers are the components of a vector. It is well known 
that the spin-1/2 can be described either with a two-component spinor 
($SU_2$) or a three-component vector ($R_3$). This vector has unit length:
\begin{equation} 
\xi = \sqrt{\xi_x^2 + \xi_y^2 + \xi_z^2} = 1 
\end{equation}
Hence two parameters are needed to describe the polarization of a spin-1/2
particle or a photon. Note that one uses the Pauli matrices for the photon,
although it is not a spin-1/2 particle. The use of the Pauli ($2 \times 2$)
matrices is related to the fact that there are two polarisation states.

For a single particle this vector $\vec \xi$ carries information in terms of
expectation values. Things become closer to physical reality if one considers
ensembles of particles, like polarized beams or polarized targets.

For a beam of particles, or a target, one must take averages over an ensemble
of particles. One introduces the density matrix:
\begin{equation} 
\rho = \left( \begin{array}{cc} 
\overline{a_1 a_1^*} & \overline{a_1 a_2^*} \\ & \\
\overline{a_2 a_1^*} & \overline{a_2 a_2^*}
\end{array} \right) = \frac{1}{2} \bigg( 1 + \vec \sigma \cdot \vec \xi 
\bigg) 
\end{equation}
where $\vec \xi$ is a three-component (real) vector:
\begin{equation} 
\xi_x = \overline{a_1 a_2^*} + \overline{a_2 a_1^*} \hspace{1.0cm}
   \xi_y = i ( \overline{a_1 a_2^*} - \overline{a_2 a_1^*} ) \hspace{1.0cm}
   \xi_z = \overline{a_1 a_1^*} - \overline{a_2 a_2^*} 
\end{equation}  
The vector $\vec \xi$ now carries statistical information. It can be 
identified with the Stokes vector which was
introduced many years ago (well before quantum mechanics and the discovery of
the electron) to describe the polarization of light\cite{STO52}.
The three components of this vector are measurable quantities.

The Stokes vector does not have a unit length. Rather:
\begin{equation} 
\xi^2 = 1 - 4 ( \overline{a_1 a_1^*} ~ \overline{a_2 a_2^*}
                 - \overline{a_1 a_2^*} ~ \overline{a_2 a_1^*} ) 
\end{equation}
According to the well-known inequality:
\begin{equation} 
\vert \vec a_1 \cdot \vec a_2 \vert \le \vert \vec a_1 \vert ~ 
\vert \vec a_2 \vert 
\end{equation}
one finds that:
\begin{equation} 
\xi \le 1 
\end{equation}
For an unpolarized state, where the phases and the moduli in the ensemble
are taken at random (subject to the normalization condition):
\begin{equation} 
\overline{a_1 a_2^*} = \overline{a_2 a_1^*} = 0 \hspace{1.0cm}
   \overline{a_1 a_1^*} = \overline{a_2 a_2^*} = \frac{1}{2} 
\end{equation}
And $\vec \xi$ is a null vector.

One can also write:
\begin{equation} 
\vec \xi = \vec n P 
\end{equation}
where $\vec n$ is a unit vector which defines the state of polarization
and $P$ is the degree
of polarization. Now, three parameters are needed to describe the polarization
of an ensemble of spin-1/2 particles or photons.

For a pure state (all particles are in the same state of polarization) we can 
remove the horizontal bars:
\begin{equation} 
P = 1 \hspace{2.0cm} \rho = 
\frac{1}{2} \bigg( 1 + \vec \sigma \cdot \vec n \bigg) 
\end{equation}
For an unpolarized state:
\begin{equation} 
P = 0 \hspace{2.0cm} \rho = \frac{1}{2} 
\end{equation}
For $0 < P < 1$ one has a state of partial polarization.
Such a state can be obtained in several ways, for example:
\begin{enumerate}
\item{by mixing incoherently a completely polarized ensemble and an unpolarized
ensemble (with weights $P$ and $1 - P$):
\begin{equation} 
\rho = \frac{1}{2} \bigg( 1 + P \vec \sigma \cdot \vec n \bigg) = 
P \frac{1}{2} \bigg( 1 + \vec \sigma \cdot n \bigg) + 
\bigg( 1 - P \bigg) \frac{1}{2} 
\end{equation}}
\item{by mixing incoherently two completely polarized ensembles of opposite
polarizations (with weights $(1 + P)/2$ and $(1 - P)/2$):
\begin{equation} 
\rho = \frac{1}{2} \bigg( 1 + P \vec \sigma \cdot \vec n \bigg) = 
\frac{1 + P}{2} \frac{1}{2}\bigg( 1 + \vec \sigma \cdot n \bigg)
+ \frac{1 - P}{2} \frac{1}{2} \bigg( 1 - \vec \sigma \cdot n \bigg) 
\end{equation}}
\end{enumerate}
\section{Motion of polarization in a magnetic field}
The solution is well-known in the case of non-relativistic quantum mechanics.
It results directly from the evolution equation:
\begin{equation} 
i \frac{d}{dt} (\phi^\dagger \vec \sigma \phi) =
\phi^\dagger [ \vec \sigma, H ] \phi 
\end{equation}
with:
\begin{equation} 
H = \frac{eg}{2m} \vec \sigma \cdot \vec {\cal B} 
\end{equation}
where $e$ is the elementary electric charge ($\pm \vert e \vert$) for 
positron and 
electron, respectively), $g$ the gyromagnetic ratio, $m$ the mass and
$\vec {\cal B}$ the magnetic field. Hence:
\begin{equation} 
\frac{d \vec \xi}{dt} = \frac{eg}{2m} ( \vec {\cal B} 
\times \vec \xi ) 
\label {EQ1}
\end{equation}
where $\vec \xi$ is the Stokes vector. This is, in fact, a classical equation.

\section{Relativistic case}
Let us now consider the relativistic case. 
The precession of the polarization of particles moving in a homogeneous
electromagnetic field has been studied by Bargmann, Michel and Telegdi
\cite{BAR59}. The purpose of the paper was to give a fully relativistic 
treatment of the problem. 

In the relativistic treatment one introduces two four-vectors (velocity and
polarization):
\begin{equation} 
u^\mu = \left( \begin{array}{c} u_0 \\ \vec u \end{array} \right)
\hspace{2.0cm} 
   u_\mu = \left( \begin{array}{r} u_0 \\ -\vec u \end{array} \right) 
\hspace{2.0cm}\hbox{(polar vector)} 
\end{equation}
with $u^0 = \gamma$ and $\vec u = \gamma \vec v$. And $\gamma$ is the usual
relativistic factor $\gamma = E/m$.
\begin{equation} 
s^\mu = \left( \begin{array}{c} s_0 \\ \vec s \end{array} \right)
\hspace{2.0cm} 
   s_\mu = \left( \begin{array}{r} s_0 \\ -\vec s \end{array} \right) 
\hspace{2.0cm}\hbox{(axial vector)} 
\end{equation}
These four-vectors have unit length:
\begin{equation} 
u^\mu u_\mu = 1 \hspace{2.0cm} s^\mu s_\mu = -1 
\end{equation}
and the following relation holds:
\begin{equation} 
s^\mu u_\mu = 0 
\end{equation}
or:
\begin{equation} 
s^0 = \vec s \cdot \vec v 
\end{equation}
In the rest frame of the particle $\vec v = 0$ and $s^0 = 0$.

One has the following relations between the four-vector $s^\mu$ and the
three-vector $\vec \xi$:
\begin{equation} 
s^0 = \frac{\vec \xi \cdot \vec p}{m} \hspace{1.0cm}
\vec s = \vec \xi + \frac{(\vec \xi \cdot \vec p) ~\vec p}{m~ (E + m)} 
\end{equation}
where $\vec p$ is the momentum of the particle and $E$ its total energy.
The latter equation can also be written as:
\begin{equation} 
s^0 = \beta \gamma \vert \vec \xi_L \vert \hspace{1.0cm} 
\vec s = \gamma \vec \xi_L + \vec \xi_T 
\end{equation}
where $\vec \xi_L$ and $\vec \xi_T$ are the longitudinal and transverse
components of $\vec \xi$, along the momentum $\vec p$ and orthogonal to it,
respectively. One recognizes the transformation property of a vector
(pseudovector) under a Lorentz transformation.

One now has to obtain the relativistic equations. The relativistic equation
of motion is:
\begin{equation} 
\frac{du^\mu}{d\tau} = \frac{1}{m} f^\mu 
\end{equation}
where $f$ is the four-vector force and $\tau$ is the proper time:
\begin{equation} 
\tau = t/\gamma 
\end{equation}
For a particle in an electromagnetic field one has:
\begin{equation} 
\frac{du^\mu}{d\tau} = \frac{e}{m} (F^{\mu \nu} u_\nu) 
\end{equation}
where $F^{\mu \nu}$ is the electromagnetic tensor.
One could naively write for $s^\mu$, by analogy with Equation \ref{EQ1}:
\begin{equation} 
\frac{ds^\mu}{d\tau} = \frac{eg}{2m} (F^{\mu \nu} s_\nu) 
\end{equation}
But it is easy to show that the equation:
\begin{equation} 
\frac{ds^\mu}{d\tau} u_\mu + s^\mu \frac{du_\mu}{d\tau} = 0 
\end{equation}
would not be satisfied.
More details can be found in Jackson's book \cite{JACK} (see also refs. 
\cite{KONO,ITZY}). A recent paper \cite{COST} deals with the problem and gives 
interesting historical references.

One finally gets the BMT (Bargmann-Michel-Telegdi) equation\cite{BAR59}:
\begin{equation} 
\frac{ds^\mu}{d\tau} = \frac{e}{m} \bigg[ \frac{g}{2} 
F^{\mu \nu} s_\nu + \bigg( \frac{g}{2} - 1 \bigg) 
(s_\alpha F^{\alpha \beta} u_\beta ) u^\mu \bigg] 
\end{equation}
The problem of spin rotation in an electromagnetic field has also been 
discussed by Lyuboshits\cite{LYU80}.
He obtains a formula for the motion of the Stokes vector $\vec \xi$:
\begin{equation} 
\frac{d \xi}{dt} = \vec \Omega \times \vec \xi 
\end{equation}
where:
\begin{equation} 
\vec \Omega (t) = - \frac{e}{2 m_0 c} g \bigg[ \vec {\cal B} - 
\frac{\gamma - 1}{\gamma} \bigg( \frac{\vec v}{v} \bigg)
\bigg( \vec {\cal B} \cdot \frac{\vec v}{v} \bigg) + 
\bigg( \vec {\cal E} \times \frac{\vec v}{c} \bigg) \bigg] -
(\gamma - 1) \bigg[ \bigg( \frac{\vec v}{v} \bigg) \times \frac{d}{dt} 
\bigg( \frac{\vec v}{v} \bigg) \bigg] 
\end{equation}  
Or equivalently:
\begin{equation} 
\vec \Omega (t) = - \frac{eg}{2 m_0 c \gamma} \bigg[ \vec {\cal B}_L + 
\gamma \vec {\cal B}_T + \gamma \bigg( \vec {\cal E} \times \frac{\vec v}{c} 
\bigg) \bigg]
- (\gamma - 1) \bigg[ \bigg( \frac{\vec v}{v} \bigg) \times \frac{d}{dt} 
\bigg( \frac{\vec v}{v}  \bigg) \bigg]
\nonumber 
\end{equation}  
\begin{equation} 
= - \frac{eg}{2 m_0 c \gamma} \vec {\cal B}^*
- (\gamma - 1) \bigg[ \bigg( \frac{\vec v}{v} \bigg) \times \frac{d}{dt} 
\bigg( \frac{\vec v}{v}  \bigg) \bigg] 
\end{equation}  
Where $\vec {\cal B}_L$ and $\vec {\cal B}_T$ are the longitudinal and 
transverse components of the magnetic field (with respect to the particle 
velocity) and $\vec {\cal B}^*$ the magnetic field in the rest system of the 
particle. Therefore the interpretation of the first term is simple. The 
second term represents the Thomas precession\cite{MOLL}.

It is not difficult to show that this formula is equivalent to the one given 
by Buon\cite{BUO85}:
\begin{equation} 
\vec \Omega (t)= - \frac{e}{m_0 c \gamma} \bigg[ (1 + a \gamma) 
\vec {\cal B}_T
+ (1 + a) \vec {\cal B}_L + \bigg( a + \frac{1}{\gamma + 1} \bigg)
\bigg( \vec {\cal E} \times \frac{\vec v}{c} \bigg) \gamma \bigg] 
\label{buon}
\end{equation}  
where $a$ is the anomaly:\cite{PDG00}
\begin{equation} 
a = \frac{g}{2} - 1 = 0.0011659160 \pm 0.0000000006 
\end{equation}

\section{Special case: pure magnetic field}
\subsection{Rotation of the velocity}
One uses $c = 1$.
\begin{equation} 
\frac{d \vec p}{dt} = e \bigg( \vec v \times \vec {\cal B} + 
\vec {\cal E} \bigg)
\hspace{2.0cm}\frac{dE}{dt} = e \vec v \cdot  \vec {\cal E} 
\end{equation}
Here $e$ is the elementary charge. It can be positive or negative.

If there is no electric field:
\begin{equation} 
\frac{d \vec p}{dt} = e \bigg( \vec v \times \vec {\cal B} \bigg)
\hspace{1.0cm}(\vec p = m_0 \gamma \vec v)
\hspace{2.0cm}\frac{dE}{dt} = 0 
\end{equation}
\begin{equation} 
\vec p \cdot \frac{d \vec p}{dt} = 
e m_0 \gamma \bigg[ \vec v \cdot (\vec v \times \vec {\cal B}) \bigg] = 0 
\end{equation}
Therefore $\vert \vec p \vert$, $E$, $\vert \vec v \vert$ and $\gamma$
are constant in time. And:
\begin{equation} 
\frac{d \vec v}{dt} = \frac{e}{m_0 \gamma} 
( \vec v \times \vec {\cal B} ) 
\end{equation}
In a homogeneous (and independent of time) magnetic field 
$\vec {\cal B} = \vec N {\cal B}$,
where $\vec N$ is a unit vector:
\begin{equation} 
\frac{d \vec v}{dt} 
= \frac{e{\cal B}}{m_0 \gamma} (\vec v \times \vec N) 
= i \omega (\vec \Sigma \cdot \vec N) \vec v \hspace{1.0cm}
(\omega = \frac{e{\cal B}}{m_0 \gamma}) 
\end{equation}
Note that $\omega_0$ and $\omega$ can be positive or negative.

The vector $\vec \Sigma$ is defined as:
\begin{equation} 
\Sigma_x = \left( \begin{array}{rrr} 
 0 & 0 & 0 \\ 0 & 0 &-i \\ 0 &+i & 0 \end{array} \right) \hspace{1.0cm}
\Sigma_y = \left( \begin{array}{rrr} 
 0 & 0 &+i \\ 0 & 0 & 0 \\-i & 0 & 0 \end{array} \right) \hspace{1.0cm}
\Sigma_z = \left( \begin{array}{rrr} 
 0 &-i & 0 \\+i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) 
\end{equation}
The solution is:
\begin{equation} 
\vec v (t) = \exp \bigg[ i \omega (\vec \Sigma \cdot \vec N) t \bigg] 
\vec v_0
= \bigg[ 1 + i (\vec \Sigma \cdot \vec N) \sin \omega t 
- (\vec \Sigma \cdot \vec N)^2 (1 - \cos \omega t) \bigg] \vec v_0 
\end{equation}
Let us introduce the unit vectors:
\begin{equation} 
\vec k (t) = \frac{\vec v}{v_0}\hspace{2.0cm}
   \vec k_0 = \frac{\vec v_0}{v_0} 
\end{equation}
Then:
\begin{equation} 
\vec k (t) = \bigg[ 1 + i (\vec \Sigma \cdot \vec N) \sin \omega t 
- (\vec \Sigma \cdot \vec N)^2 (1 - \cos \omega t) \bigg] \vec k_0 
\end{equation}
Let us define the angle $\chi$ ($0 \le \chi \le \pi$):
\begin{equation} 
\vec N \cdot \vec k = \vec N \cdot \vec k_0 = \cos \chi 
\end{equation}
Then:
\begin{equation} 
{\cal B}_L = {\cal B} \cos \chi\hspace{2.0cm}
   {\cal B}_T = {\cal B} \sin \chi 
\end{equation}
One gets:
\begin{equation} 
\vec k (t) = \vec k_0 \cos \omega t + \vec N \cos \chi (1 - \cos \omega t) 
-  (\vec N \times \vec k_0) \sin \omega t 
\end{equation}
Let us define two unit vectors:
\begin{equation} 
\frac{\vec k_0 -\vec N \cos \chi}{\sin \chi} = \vec P 
\end{equation}
\begin{equation} 
\frac{\vec N \times \vec k_0}{\sin \chi} = \vec Q 
\end{equation}
It is easy to verify that $\vec N$, $\vec P$ and $\vec Q$ form an
orthogonal system of reference. Then:
\begin{equation}
\vec k (t) = \vec N \cos \chi + (\vec P \cos \omega t - \vec Q \sin \omega t)
\sin \chi 
\end{equation}
The integration gives:
\begin{equation} 
\vec r (t) = \bigg( \frac{v_0}{\omega} \bigg) \bigg\{ 
\vec N \omega t \cos \chi + \bigg[ \vec P \sin \omega t - \vec Q
(1 - \cos \omega t) \sin \chi \bigg] \bigg\} 
\end{equation}
given that for $t = 0$ one has $\vec r = 0$.

The angle $\chi$ is constant and also the magnitude of $\vec k$:
\begin{equation} 
\vert \vec k \vert = k = k_0 
\end{equation}
The vector $\vec k$ precesses around the vector $\vec N$ at the frequency
$\omega/2 \pi$.

\subsection{Rotation of the polarization vector}
\begin{equation} 
\frac{d \vec \xi}{dt} = \vec \xi \times \vec \Omega 
\end{equation}
with\cite{BUO85}:
\begin{equation} 
\vec \Omega (t) = \frac{e}{m_0 \gamma} \bigg[ (1 + a) \vec {\cal B}_L 
+ (1 + \gamma a) \vec {\cal B}_T \bigg]
= \frac{e}{m_0 \gamma} \bigg[ (1 + \gamma a) \vec {\cal B} + 
\frac{a (1 - \gamma)}{v^2} (\vec {\cal B} \cdot \vec v) \vec v \bigg] 
\end{equation}
\nonumber
\begin{equation} 
= \omega \bigg[ (1 + \gamma a) \vec N - a (\gamma - 1) 
\vec k \cos \chi \bigg] 
\end{equation}
Here again the electric charge $e$ and the frequency $\omega$ can be positive
or negative.

We recall the expression of $\vec k$:
\begin{equation} 
\vec k (t) = \vec N \cos \chi + (\vec P \cos \omega t - \vec Q \sin \omega t)
\sin \chi 
\end{equation}
The modulus $\Omega = \vert \vec \Omega \vert$ is constant in time:
\begin{equation} 
\Omega^2 = 
\omega ^2 \bigg[ (1 + \gamma a)^2 -
a (\gamma - 1)(2 + a + \gamma a) \cos^2 \chi \bigg] 
\nonumber
\end{equation}
\begin{equation} 
= \omega^2 \bigg[ (1 + a)^2 + a (\gamma - 1) 
(2 + a + \gamma a) \sin^2 \chi \bigg] 
\end{equation} 
The expression for $\vec \Omega$ can be written:
\begin{equation} 
\vec \Omega (t) = \omega \bigg\{ \vec N +
a \bigg[ (\cos^2 \chi + \gamma \sin^2 \chi) \vec N  
- (\gamma - 1) \sin \chi \cos \chi 
(\vec P \cos \omega t - \vec Q \sin \omega t)\bigg] \bigg\} 
\end{equation}
We define:
\begin{equation} 
\lambda = \omega a \sqrt{\cos^2 \chi + \gamma^2 \sin^2 \chi} 
\end{equation}
\begin{equation} 
\cos \eta = \frac{\cos^2 \chi + \gamma \sin^2 \chi} 
{\sqrt{\cos^2 \chi + \gamma^2 \sin^2 \chi}} \hspace{2.0cm}
\sin \eta = \frac{(\gamma - 1) \sin \chi \cos \chi} 
{\sqrt{\cos^2 \chi + \gamma^2 \sin^2 \chi}} 
\end{equation}
This gives:
\begin{equation} 
\vec \Omega = \omega \vec N + \lambda \bigg[ \vec N \cos \eta -
(\vec P \cos \omega t - \vec Q \sin \omega t) \sin \eta \bigg] 
\end{equation}
The length of the vector $\vec \Omega$ is given by:
\begin{equation} 
\Omega^2 = \omega^2 + 2 \lambda \omega \cos \eta + \lambda^2 
\end{equation}
It is independent of time.

One can write:
\begin{equation} 
\frac{d \vec \Omega}{dt} = \lambda \omega ( \vec P \sin \omega t + 
\vec Q \cos \omega t ) \sin \eta 
\end{equation}
and:
\begin{equation} 
\vec N \times \vec \Omega = - \lambda ( \vec P \sin \omega t +
\vec Q \sin \omega t ) \sin \eta 
\end{equation}
Therefore:
\begin{equation} 
\frac{d \vec \Omega}{dt} = \omega (\vec \Omega \times \vec N) 
\end{equation}
The vector $\vec \Omega$ precesses around the vector $\vec N$. The angle
$\kappa$ between $\vec \Omega$ and $\vec N$ is given by:
\begin{equation} 
\cos \kappa = \frac{\omega + \lambda \cos \eta} 
{\sqrt{\omega^2 + 2 \lambda \omega \cos \eta + \lambda^2}} 
\end{equation}
and in the reference system formed by the vectors $\vec N$, $\vec P$ and 
$\vec Q$, the components of $\vec \Omega - \omega \vec N$ are: 
\begin{equation} 
\vec \Omega - \omega \vec N = \lambda \vec M 
\end{equation}
where $\vec M$ is a unit vector:
\begin{equation} 
\vec M = \left( \begin{array}{c} 
  \cos \eta \\
- \sin \eta \cos \omega t \\ 
  \sin \eta \sin \omega t \end{array} \right) 
\end{equation}
It can be shown that the vector $\vec \xi$ makes a constant angle with the
vector $\vec M$:
\begin{equation} 
\frac{d}{dt} \bigg[ (\vec \Omega - \omega \vec N) \cdot \vec \xi \bigg]
= \frac{d \vec \Omega}{dt} \cdot \vec \xi + (\vec \Omega - \omega \vec N)
\cdot \frac{d \vec \xi}{dt} 
\end{equation}
\begin{equation} 
= \omega (\vec \Omega \times \vec N) \cdot \vec \xi + (\vec \Omega -
\vec N) \cdot (\vec \xi \times \Omega) = \omega \bigg[ (\vec \Omega \times
\vec N) \cdot \vec \xi - \vec N \cdot (\vec \xi \times \vec \Omega) \bigg]
= 0 
\end{equation}

Therefore the motion of the polarization vector can be depicted in the
following way:
\begin{itemize}
\item{The vector $\vec \Omega$ precesses around the magnetic field at the
frequency $\omega/2 \pi$.}
\item{The polarization vector $\vec \xi$ precesses around the vector 
$\vec M$ at the frequency $\lambda/2 \pi$.}
\end{itemize}

\subsection{An analytical solution}
The motion of the polarization vector is given by:
\begin{equation} 
\frac{d \vec \xi}{dt} = \vec \xi \times \vec \Omega
= i (\vec \Sigma \cdot \vec \Omega) \vec \xi 
\end{equation}
(the $\vec \Sigma$ matrices have been defined before).

With:
\begin{equation} 
\vec \Omega = \omega \vec N + \lambda \vec M 
\end{equation}
the equation of motion contains two terms:
\begin{equation} 
\frac{d \vec \xi}{dt} = \vec \xi \times \vec \Omega
= i (\vec \Sigma \cdot \vec \Omega) \vec \xi
= i \omega (\vec \Sigma \cdot \vec N) + i \lambda (\vec \Sigma \cdot \vec M) 
\end{equation}

In the reference system formed by the vectors $\vec N$, $\vec P$ and $\vec Q$,
the components of $\vec \Omega$ are: 
\begin{equation} 
\vec \Omega (t) = \left( \begin{array}{c} 
  \omega + \lambda \cos \eta \\
- \lambda \sin \eta \cos \omega t \\ 
  \lambda \sin \eta \sin \omega t \end{array} \right) 
\end{equation}

Let us define a new vector $\vec \xi^\prime$ such that:
\begin{equation} 
\vec \xi^\prime = A \vec \xi \hspace{2.0cm} \vec \xi = A^{-1} \vec \xi^\prime 
\end{equation}
The equation of motion for $\vec \xi^\prime$ writes:
\begin{equation} 
\frac{d \vec \xi^\prime}{dt}
= \bigg[ i A (\vec \Sigma \cdot \vec \Omega) A^{-1} 
- A \frac{d A^{-1}}{dt} \bigg] \vec \xi^\prime 
\end{equation}
Let us take:
\begin{equation} 
A (t) = \left( \begin{array}{ccc}
1   &      0        &        0        \\
0   & \cos \omega t & - \sin \omega t \\
0   & \sin \omega t &   \cos \omega t
\end{array} \right) 
\end{equation}
After some calculations:
\begin{equation} 
\frac{d \vec \xi^\prime}{dt} = \lambda 
\left( \begin{array}{ccc}
   0        &         0   & \sin \eta \\
   0        &         0   & \cos \eta \\
- \sin \eta & - \cos \eta &       0   
\end{array} \right) \vec \xi^\prime
= i \lambda (\vec \Sigma \cdot \vec R)
\vec \xi^\prime 
\end{equation}
with:
\begin{equation} 
\vec R = \left( \begin{array}{c} \cos \eta \\ - \sin \eta \\ 0
\end{array} \right) 
\end{equation}
The solution is:
\begin{equation} 
\vec \xi^\prime (t) = \exp \bigg[ i \lambda (\vec \Sigma \cdot \vec R) \bigg]
\vec \xi^\prime (0) 
= \bigg[ 1 + i (\vec \Sigma \cdot \vec R) \sin \lambda t
+ (\vec \Sigma \cdot \vec R)^2 (\cos \lambda t - 1) \bigg] \vec \xi^\prime (0) 
\end{equation}
Since $A(0) = 1$, $\vec \xi^\prime (0) = \vec \xi (0)$.
Therefore:
\begin{equation} 
\vec \xi (t) = 
\bigg[ A^{-1} + i A^{-1}(\vec \Sigma \cdot \vec R) \sin \lambda t +
A^{-1} (\vec \Sigma \cdot \vec R)^2 (\cos \lambda t - 1) 
\bigg] \vec \xi (0) 
\end{equation}
and:
\begin{equation} 
\vec \xi (t) = \Bigg[
\left( \begin{array}{ccc}
1   &      0          &        0        \\
0   &   \cos \omega t &   \sin \omega t \\
0   & - \sin \omega t &   \cos \omega t 
\end{array} \right) 
\nonumber
\end{equation}
\begin{equation} 
+ \sin \lambda t
\left( \begin{array}{ccc}
    0                     &  0   &  \sin \eta \\
- \sin \eta \sin \omega t & - \cos \eta \sin \omega t 
&   \cos \eta \cos \omega t \\ 
- \sin \eta \cos \omega t & - \cos \eta \cos \omega t
& - \cos \eta \sin \omega t 
\end{array} \right) 
\nonumber
\end{equation}
\begin{equation} 
+ (\cos \lambda t - 1)
\left( \begin{array}{ccc}
\sin^2 \eta                         & \sin \eta \cos \eta         &   0 \\
  \sin \eta \cos \eta \cos \omega t &   \cos^2 \eta \cos \omega t & 
\sin \omega t \\
- \sin \eta \cos \eta \sin \omega t & - \cos^2 \eta \sin \omega t &
\cos \omega t
\end{array} \right) \Bigg] \vec \xi (0) 
\end{equation}
Finally, we get for the components of $\vec \xi (t)$:
\begin{eqnarray} 
\xi_n(t) & = &
\xi_n(0) \bigg[ 1 + \sin^2 \eta (\cos \lambda t - 1) \bigg] 
\nonumber \\ 
         &   & + \xi_p(0) \bigg[ \sin \eta \cos \eta (\cos \lambda t - 1) 
\bigg] 
\nonumber \\
         &   & + \xi_q(0) \bigg[ \sin \eta \sin \lambda t \bigg] 
\end{eqnarray}
\begin{eqnarray} 
\xi_p(t) & = &
\xi_n(0) \bigg[ - \sin \eta \sin \omega t \sin \lambda t + \sin \eta \cos \eta
\cos \omega t (\cos \lambda t - 1) \bigg] 
\nonumber \\
         &   & + \xi_p(0) \bigg[ \cos \omega t 
- \cos \eta \sin \omega t \sin \lambda t
+ \cos ^2 \eta \cos \omega t (\cos \lambda t - 1) \bigg] 
\nonumber \\
         &   & + \xi_q(0) \bigg[ \sin \omega t 
+ \cos \eta \cos \omega t \sin \lambda t
+ \sin \omega t (\cos \lambda t - 1) \bigg] 
\end{eqnarray}
\begin{eqnarray} 
\xi_q(t) & = &
\xi_n(0) \bigg[ - \sin \eta \cos \omega t \sin \lambda t - \sin \eta \cos \eta
\sin \omega t (\cos \lambda t - 1) \bigg] 
\nonumber \\
         &   & + \xi_p(0) \bigg[ - \sin \omega t 
- \cos \eta \cos \omega t \sin \lambda t
- \cos^2 \eta \sin \omega t (\cos \lambda t - 1) \bigg] 
\nonumber \\
         &   & + \xi_q(0) \bigg[ \cos \omega t 
- \cos \eta \sin \omega t \sin \lambda t
+ \cos \omega t (\cos \lambda t - 1) \bigg] 
\end{eqnarray}
To get the polarization components in the $(x, y, z)$ frame one has to
use the transformation:
\begin{equation} 
\left( \begin{array}{c} 
\xi_x (t) \\ \xi_y (t) \\ \xi_z (t) \end{array} \right) =
\left( \begin{array}{ccc} N_x & P_x & Q_x \\
                          N_y & P_y & Q_y \\
                          N_z & P_z & Q_z \end{array} \right) 
\left( \begin{array}{c} 
\xi_n (t) \\ \xi_p (t) \\ \xi_q (t) \end{array} \right) 
\end{equation}
The initial polarization can be transformed from the $(x, y, z)$ frame to the
$(n, p, q)$ frame by using the inverse matrix.

We now calculate the longitudinal polarization. The vector $\vec k$ can
be written:
\begin{equation} 
\vec k = k_n \vec N + k_p \vec P + k_q \vec Q 
\end{equation}
with:
\begin{eqnarray} 
k_n & = & \cos \chi 
\nonumber \\ 
k_p & = & + \sin \chi \cos \omega t 
\nonumber \\
k_q & = & - \sin \chi \sin \omega t 
\end{eqnarray}
The longitudinal polarisation is:
\begin{equation} 
P_l (t) = \vec k \cdot \vec \xi = 
k_n \xi_n (t) + k_p \xi_p (t) + k_q \xi_q (t) 
\end{equation}
or:
\begin{eqnarray} 
P_l(t) & = & \xi_n(0)\bigg[ \cos \chi \sin \eta \sin (\eta + \chi) 
\cos \lambda t \bigg] 
\nonumber \\
       &   & + \xi_p(0) \bigg[ \sin \chi + \cos \eta \sin (\eta + \chi) 
(\cos \lambda t - 1) \bigg] 
\nonumber \\
       &   & + \xi_q(0) \bigg[ \sin (\eta + \chi) \sin \lambda t \bigg] 
\end{eqnarray}
For $t = 0$:
\begin{equation} 
P_l(0) = \xi_n(0) \cos \chi + \xi_p(0) \sin \chi 
\end{equation}
In fact:
\begin{equation} 
\vec \xi (0) = \xi_p(0) \vec P + \xi_n(0) \vec N 
\end{equation}
\begin{equation} 
\vec \xi (0) \cdot \vec k_0 = \xi_n(0) \cos \chi + \xi_p(0) \sin \chi 
\end{equation}
and finally:
\begin{equation} 
P_l (t) = P_l(0) + \sin (\eta + \chi)\bigg \{ \bigg[ \xi_n(0) \sin \eta
+ \xi_p(0) \cos \eta \bigg] (\cos \lambda t - 1) + \xi_q(0) \sin \lambda t 
\bigg \} 
\end{equation}
\subsection{Special cases}
\subsubsection{No anomaly}
We have $a = 0$ and therefore $\lambda = 0$. The formulae simplify to:
\begin{eqnarray} 
\xi_n (t) & = & \xi_n(0) 
\nonumber \\ 
\xi_p (t) & = & \xi_p(0) \cos \omega t + \xi_q(0) \sin \omega t 
\nonumber \\
\xi_q (t) & = & - \xi_p(0) \sin \omega t + \xi_q(0) \cos \omega t 
\nonumber \\
P_l (t)   & = & P_l (0) 
\end{eqnarray}
\subsubsection{Initial velocity orthogonal to magnetic field}
We have $\chi = \pi/2$, $\lambda = \omega \gamma a$, $\eta = 0$.
\begin{eqnarray} 
\xi_n (t) & = & \xi_n(0) 
\nonumber \\ 
\xi_p (t) & = & \xi_p(0) \cos (1 + \gamma a)\omega t + 
\xi_q(0) \sin (1 + \gamma a)\omega t 
\nonumber \\
\xi_q (t) & = & - \xi_p(0) \sin (1 + \gamma a)\omega t + 
\xi_q(0) \cos (1 + \gamma a)\omega t 
\nonumber \\ 
P_l (t)   & = & P_l (0) + \bigg[ \xi_p(0) (\cos \lambda t - 1)
+ \xi_q(0) \sin \lambda t \bigg] 
\end{eqnarray}
\subsubsection{Initial velocity along magnetic field}
We have $\chi = 0$, $\lambda = \omega a$, $\eta = 0$. 
\begin{eqnarray} 
\xi_n (t) & = & \xi_n(0) 
\nonumber \\ 
\xi_p (t) & = & \xi_p(0) \cos (1 + a)\omega t + 
\xi_q (t) \sin (1 + a)\omega t 
\nonumber \\ 
\xi_q (t) & = & - \xi_p(0) \sin (1 + a)\omega t + 
\xi_q(0) \cos (1 + a)\omega t 
\nonumber \\ 
P_l (t)   & = & P_l (0) 
\end{eqnarray}
\subsubsection{Particle at rest}
We have $\gamma = 1$ and therefore $\lambda = \omega a$. The polarization
vector precesses around the magnetic field at the frequency $\omega (1 + a)/
2 \pi$. 

\subsection{The program POLAR}
This program calculate the rotation of the polarization vector in a constant
(and time-independent) magnetic field.

The input data are:
\begin{itemize}
\item{the number of values to be printed.}
\item{the value of $\gamma$ (relativistic factor), the value of the anomaly
$a$ and the value of the Larmor frequency ($\omega_0 = e {\cal B}/m_0$).}
\item{the polar angles ($\alpha$, $\beta$) of the initial velocity.}
\item{the polar angles ($\theta$, $\phi$) of the magnetic field.}
\item{the polar angles ($\psi$, $\zeta$) of the initial polarization vector.}
\end{itemize}
The conventions for ($\alpha$, $\theta$, $\psi$) and ($\beta$, $\phi$, $\zeta$)
are as follows:

\begin{table}[h]
\begin{center}
\begin{tabular}{||c|c|c||} \hline \hline
Axis & $\alpha$, $\theta$, $\psi$ & $\beta$, $\phi$, $\zeta$ \\ \hline \hline
$x$  &  $90^o$                    &  $~0^o$                  \\
$y$  &  $90^o$                    &  $90^o$                  \\
$z$  &  $~0^o$                    &  any value               \\ \hline \hline
\end{tabular}    
\end{center}
\end{table}

The output data are functions of the angle $\omega t$:

\begin{itemize}
\item{the components of the velocity in the laboratory frame.}
\item{the coordinates of the particle in the laboratory frame.}
\item{the components of the polarization vector $\vec \xi$
in the rest frame of the particle.}
\end{itemize}

\subsection{Data}
From ``Review of Particle Physics'':\cite{PDG00}
electron cyclotron frequency/field.
\begin{equation} 
\omega^e_{cycl}/{\cal B} = e/m_e = 1.758 820 174(71) \times 10^{11}  
\hspace{0.1cm}\hbox{rad s$^{-1}$ T$^{-1}$} 
\end{equation}
Using the electron and muon masses:
\begin{equation} 
m_e = 0.510 998 902\hspace{0.1cm}\hbox{MeV} 
\end{equation}
\begin{equation} 
m_\mu = 105.658 357\hspace{0.1cm}\hbox{MeV} 
\end{equation}
One gets:
\begin{equation} 
\omega^\mu_{cycl}/{\cal B} = e/m_\mu = 0.850 624 60 \times 10^9
\hspace{0.1cm}\hbox{rad s$^{-1}$ T$^{-1}$}  
\end{equation}  

\section{Appendix: the ``magic'' momentum}
In $(g - 2)$ experiments\cite{ROB00} a special value of the muon momentum is 
chosen in order to eliminate the effect of the electric field which is used in 
the accelerator for vertical focussing.

\subsection{Rotation of the particle velocity}
In the presence of transverse electric and magnetic fields (orthogonal to the 
particle velocity):
\begin{equation}
\vec v \cdot \vec {\cal B} = \vec v \cdot \vec {\cal E} = 0
\end{equation}
the equations describing the particle motion:
\begin{equation} 
\frac{d \vec p}{dt} = e \bigg( \vec v \times \vec {\cal B} + 
\vec {\cal E} \bigg)
\hspace{2.0cm}\frac{dE}{dt} = e (\vec v \cdot  \vec {\cal E}) 
\end{equation}
can be simplified (introducing $\vec \beta = \vec v$ and 
$E = m_0 \gamma \vec \beta$, $c = 1$):
\begin{equation}
m_0\frac{d\gamma}{dt} = 0 \hspace{1.0cm} \gamma = \hbox{ constant}
\hspace{1.0cm}
m_0 \gamma \frac{d \vec \beta}{dt} = 
e \bigg( \vec \beta \times \vec {\cal B} + \vec {\cal E}_T \bigg)
\end{equation}
Using the identity:
\begin{equation}
\vec \beta \times (\vec \beta \times \vec {\cal E}) = 
\vec \beta~(\vec \beta \cdot \vec {\cal E})
- \beta^2~\vec {\cal E}
\end{equation}
The longitudinal and transverse components of the electric field are:
\begin{eqnarray}
\vec {\cal E}_L & = & \frac{1}{\beta^2}~
\vec \beta~(\vec \beta \cdot \vec {\cal E}) \nonumber \\
\vec {\cal E}_T & = & \vec {\cal E} - \frac{1}{\beta^2}~
\vec \beta~(\vec \beta \cdot \vec {\cal E}) =
- \frac{1}{\beta^2}~ \vec \beta \times (\vec \beta \times \vec {\cal E}) 
\end{eqnarray}
\begin{equation}
m_0 \gamma \frac{d \vec \beta}{dt} = e \bigg( \vec \beta \times \vec {\cal B} -
\frac{1}{\beta^2}~\vec \beta \times (\vec \beta \times \vec {\cal E}) \bigg) =
e~\vec \beta \times \bigg( \vec {\cal B} -
\frac{1}{\beta^2}~ (\vec \beta \times \vec {\cal E}) \bigg) 
\end{equation}
\begin{equation}
\frac{d \vec \beta}{dt} = \frac{e}{m_0 \gamma}~
\vec \beta \times \bigg( \vec {\cal B} -
\frac{1}{\beta^2}~ (\vec \beta \times \vec {\cal E}) \bigg) 
\end{equation}
or:
\begin{equation}
\frac{d \vec \beta}{dt} = \vec \beta \times \vec \omega_c
\end{equation}
with:
\begin{equation}
\vec \omega_c = \frac{e}{m_0 \gamma} \bigg( \vec {\cal B} -
\frac{1}{\beta^2}~ (\vec \beta \times \vec {\cal E}) \bigg) 
\end{equation}
and, since:
\begin{equation}
\frac{1}{\beta^2} = \frac{\gamma^2}{\gamma^2 - 1}
\end{equation}
\begin{equation}
\vec \omega_c = \frac{e}{m_0 \gamma} \bigg( \vec {\cal B} -
\frac{\gamma^2}{\gamma^2 - 1}~ (\vec \beta \times \vec {\cal E}) \bigg) 
\end{equation}
To be compared with Farley and Picasso\cite{FAR79}, page 266:
\begin{equation}
\vec \omega_c = \frac{e}{m_0} \bigg( \frac{\vec {\cal B}}{\gamma} -
\frac{\gamma}{\gamma^2 - 1}~ (\vec \beta \times \vec {\cal E}) \bigg) 
\end{equation}

\subsection{Rotation of the polarization vector}
The formula for the rotation of the polarization vector is\cite{BUO85}, for
a positive muon:
\begin{equation}
\vec \omega_s = \frac{e}{m_0 \gamma} \bigg[ (1 + a \gamma)
\vec {\cal B}_T
+ (1 + a) \vec {\cal B}_L + \bigg( a + \frac{1}{\gamma + 1} \bigg)
\bigg( \vec {\cal E} \times \frac{\vec v}{c} \bigg) \gamma \bigg]
\end{equation}
and with $\vec {\cal B}_L = 0$, $\vec {\cal B}_T = \vec {\cal B}$, 
and changing notations:
\begin{equation}
\vec \omega_s = \frac{e}{m_0 \gamma} \bigg[ (1 + a \gamma)
\vec {\cal B}
- \bigg( a + \frac{1}{\gamma + 1} \bigg)
\bigg( \vec \beta \times \vec {\cal E} \bigg) \gamma \bigg]
\end{equation}

where $a$ is the anomaly:\cite{PDG00}
\begin{equation}
a = \frac{g}{2} - 1 = 0.0011659160 \pm 0.0000000006
\end{equation}
to be compared with Farley and Picasso\cite{FAR79}, page 266:
\begin{equation}
\vec \omega_s = \frac{e}{m_0} \bigg[ 
\frac{\vec {\cal B}}{\gamma} - \frac{1}{\gamma + 1} 
(\vec \beta \times \vec {\cal E}) + a \bigg( \vec {\cal B} - 
(\vec \beta \times \vec {\cal E}) \bigg) \bigg] 
\end{equation}

\subsection{The ``magic momentum''}

The difference between the two frequencies is:
\begin{equation}
\Delta \omega = \omega_s - \omega_c = 
\frac{e}{m_0 c \gamma} \bigg[ (1 + a \gamma) \vec {\cal B} 
- \bigg( a + \frac{1}{\gamma + 1} \bigg) 
\bigg( \vec \beta \times \vec {\cal E} \bigg) \gamma 
- \vec {\cal B}
+ \frac{\gamma^2}{\gamma^2 - 1} (\vec \beta \times \vec {\cal E}) \bigg] 
\end{equation}
\begin{equation}
\Delta \omega = 
\frac{e}{m_0} \bigg[ a \vec {\cal B} 
- \bigg( a + \frac{1}{\gamma + 1} - \frac{\gamma}{\gamma^2 - 1} \bigg) 
\bigg( \vec \beta \times \vec {\cal E} \bigg) \bigg] 
\end{equation}
\begin{equation}
\Delta \omega = 
\frac{e}{m_0} \bigg[ a \vec {\cal B} 
- \bigg( a  - \frac{1}{\gamma^2 - 1} \bigg) 
\bigg( \vec \beta \times \vec {\cal E} \bigg) \bigg] 
\end{equation}
identical to the formula of ref. \cite{ROB00}.

This frequency will not depend on the electric field if:
\begin{equation}
a  - \frac{1}{\gamma^2 - 1} = 0
\end{equation}
or:
\begin{equation}
\gamma = \sqrt{1 + \frac{1}{a}} = 29.3 
\end{equation}
The corresponding momentum (the ``magic momentum'') is 3.09 GeV/c.

\section{Acknowledgments}
The author wishes to thank Dr. Jan Govaerts (Louvain-La-Neuve) who has read
the manuscript and made useful comments. And also Peter Gumplinger for many
interesting discussions. I also want to mention that Louis Michel has read 
this technote a few weeks before he passed away.
 
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