## Euler angle

with the help of the post changing frame, we are now good to use the Euler angle.

recall $V_R = R_n ( - \theta ) V_S$

for the rotating frame axis is rotating positive with the static frame.

the Euler angle is performed on 3 steps

1. rotate on $Z_S$, the z-axis with $\alpha$, which is $R_{zS} ( - \alpha )$. the x-axis and the y-axis is now different, we notate this frame with a 1 .
2. rotate on $Y_1$, the y-axis in the 1- frame  by angle $\beta$, which is $R_{y1} ( - \beta )$. the new axis is notated by 2.
3. rotate on $Z_2$, the z-axis in the 2-frame by angle $\gamma$, which is $R_{z2} ( - \gamma )$. the new axis is notated by R.

The rotating frame is related with the static frame by: $V_R = R_{z2} ( - \gamma ) R_{y1} ( - \beta ) R_{zS} ( - \alpha ) V_S$

or $R_R ( \alpha, \beta, \gamma ) =$ $R_{z2} ( - \gamma )$ $R_{y1} ( - \beta )$ $R_{zS} ( - \alpha )$ for each rotation is on a new frame, the computation will be ugly, since, after each rotation, we have to use the rotation matrix in new coordinate.

There is another representation, notice that: $R_{y1} ( -\beta ) =$ $R_{zS} ( - \alpha )$ $R_{yS} ( - \beta )$ $R_{zS} ( \alpha)$

which mean, the rotating on y1 -axis by $\beta$ is equal to rotate it back to $Y_S$  on zS -axis and rotated it by $\beta$ on yS – axis, then rotate back the $Y_S$ to $Y_1$ on zS – axis. i use a and b for the axis between the transform.

and we have it for the z2-axis. $R_{z2} ( -\gamma ) = R_{y1} ( - \beta ) R_{z1} ( - \gamma ) R_{y1} ( \beta )$

by using these 2 equation and notice that the z1-axis is equal to zS-axis. $R_R ( \alpha , \beta, \gamma ) = R_{zS} ( - \alpha ) R_{yS} (- \beta ) R_{zS} ( - \gamma )$

which act only on the the same frame. ## Rotation operator on x, y in Matrix form

in the J.J. Sakurai’s book, the formalism of finding the matrix representation of rotation operator is general, but quite long and detail. A general treatment is necessary for understanding the topic, but i think, who will use arbitrary rotation? so,  here i give a simple and direct calculation on $J_x$ and $J_y$, for use-ready.

the method is diagonalization. because we already knew the matrix form of the angular momentum operator. which is not given in J.J.Sakurai’s book.

recall that the formalism: $f(M) = P \cdot f(D) \cdot P^{-1}$

since $D$ is diagonal matrix, thus $f(D)_{ij} = f(\lambda_i) \delta_{ij}$

so, we have to find out the $P$ for $J_x$ and $J_y$.

i am still trying to obtain the equation, but…..

anyway, using program can solve it without headache. ( but typing Latex is )here are some result. $J_x(\frac{1}{2}) = \begin {pmatrix} \cos \left( \frac {\theta}{2} \right) & - i \sin \left( \frac{\theta}{2} \right) \\ -i \sin ( \frac{\theta}{2} ) & \cos (\frac {\theta}{2}) \end {pmatrix}$ $J_y(\frac{1}{2}) = \begin {pmatrix} \cos \left( \frac {\theta}{2} \right) & - \sin ( \frac{\theta}{2} ) \\ \sin ( \frac{\theta}{2} ) & \cos (\frac {\theta}{2}) \end {pmatrix}$

## on Diagonalization (reminder)

since i don’t have algebra book on my hand, so, it is just a reminder, very basic thing.

for any matrix $M$ , it can be diagonalized by it eigenvalue $\lambda_i$  and eigen vector $v_i$, given that it eigenvectors span all the space. thus, the transform represented by the matrix not contractive, which is to say, the dimension of the transform space is equal to the dimension of the origin space.

Let denote, D before Diagonal matrix, with it elements are eigenvalues. $D_{ij} = \lambda_i \delta_{ij}$

P be the matrix that collect the eigenvectors: $P_{i j} = \left( v_i \right)_j = \begin {pmatrix} v_1 & v_2 & ... & v_i \end {pmatrix}$

Thus, the matrix $M$ is : $M = P \cdot D \cdot P^{-1}$

there are some special case. since any matrix can be rewritten by symmetric matrix $S$ and anti-symmetric matrix $A$. so we turn our focus on these 2 matrices.

For symmetric matrix $S$, the transpose of $P$ also work $S =P \cdot D \cdot P^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

which indicated that $P^T = P^{-1}$. it is because, for a symmetric matrix, $M = M^T$ ,  the eigenvalues are all different, then all eigenvector are all orthogonal, thus $P^T \cdot P = 1$.

For anti-symmetric matrix $A$ $A = P \cdot D \cdot P^{-1}$

since the interchange of row or column with corresponding exchange of eigenvalues in D still keep the formula working. Thus, the case $P = P^T$ never consider.

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For example, the Lorentz Transform $L = \gamma \begin {pmatrix} \beta & 1 \\ 1 & \beta \end {pmatrix}$

which has eigenvalues: $D = \gamma \begin {pmatrix} \beta-1 & 0 \\ 0 & \beta+1 \end {pmatrix}$ $P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end {pmatrix}$

the eigenvector are the light cone. because only light is preserved in the Lorentz Transform.

and it is interesting that $L = P \cdot D \cdot P^{-1} = P^{-1} \cdot D \cdot P = P^T \cdot D \cdot (P^T)^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

another example is the Rotation Matrix $R = \begin {pmatrix} cos(\theta) & - sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}$ $D = \begin {pmatrix} Exp( - i \theta) & 0 \\ 0 & Exp(i \theta) \end {pmatrix}$ $P = \begin {pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$

the last example to give is the $J_x$ of the spin-½ angular momentum $J_x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ $D = \begin {pmatrix} -1 & 0 \\ 0 & 1 \end {pmatrix}$ $P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}$