Review on rotation

The rotation of a vector in a vector space can be done by either rotating the basis vector or the coordinate of the vector. Here, we always use fixed basis for rotation.

For a rigid body, its rotation can be accomplished using Euler rotation, or rotation around an axis.

Whenever a transform preserves the norm of the vector, it is a unitary transform. Rotation preserves the norm and it is a unitary transform, can it can be represented by a unitary matrix. As a unitary matrix, the eigen states are an convenient basis for the vector space.

We will start from 2-D space. Within the 2-D space, we discuss about rotation started by vector and then function. The vector function does not explicitly discussed, but it was touched when discussing on functions. In the course, the eigen state is a key concept, as it is a convenient basis. We skipped the discussion for 3-D space, the connection between 2-D and 3-D space was already discussed in previous post. At the end, we take about direct product space.

In 2-D space. A 2-D vector is rotated by a transform R, and the representation matrix of R has eigen value

$\exp(\pm i \omega)$

and eigenvector

$\displaystyle \hat{e}_\pm = \mp \frac{ \hat{e}_x \pm i \hat{e}_y}{\sqrt{2}}$

If all vector expand as a linear combination of the eigen vector, then the rotation can be done by simply multiplying the eigen value.

Now, for a 2-D function, the rotation is done by changing of coordinate. However, The functional space is also a vector space, such that

1. $a* f_1 + b* f_2$ still in the space,
2. exist of  unit and inverse of addition,
3. the norm can be defined on a suitable domain by $\int |f(x,y)|^2 dxdy$

For example, the two functions $\phi_1(x,y) = x, \phi_2(x,y) = y$, the rotation can be done by a rotational matrix,

$\displaystyle R = \begin{pmatrix} \cos(\omega) & -\sin(\omega) \\ \sin(\omega) & \cos(\omega) \end{pmatrix}$

And, the product $x^2, y^2, xy$ also from a basis. And the rotation on this new basis was induced from the original rotation.

$\displaystyle R_2 = \begin{pmatrix} c^2 & s^2 & -2cs \\ s^2 & c^2 & 2cs \\ cs & -cs & c^2 - s^2 \end{pmatrix}$

where $c = \cos(\omega), s = \sin(\omega)$. The space becomes “3-dimensional” because $xy = yx$, otherwise, it will becomes “4-dimensional”.

The 2-D function can also be expressed in polar coordinate, $f(r, \theta)$, and further decomposed into $g(r) h(\theta)$.

How can we find the eigen function for the angular part?

One way is using an operator that commutes with rotation, so that the eigen function of the operator is also the eigen function of the rotation. an example is the Laplacian.

The eigen function for the 2-D Lapacian is the Fourier series.

Therefore, if we can express the function into a polynomial of $r^n (\exp(i n \theta) , \exp(-i n \theta))$, the rotation of the function is simply multiplied by the rotation matrix.

The eigen function is

$\displaystyle \phi_{nm}(\theta) = e^{i m \theta}, m = \pm$

The D-matrix of rotation (D for Darstellung, representation in German)  $\omega$ is

$D^n_{mm'}(\omega) = \delta_{mm'} e^{i m \omega}$

The delta function of $m, m'$ indicates that a rotation does not mix the spaces. The transformation of the eigen function is

$\displaystyle \phi_{nm}(\theta') = \sum_{nm} \phi_{nm'}(\theta) D^n_{m'm}(\omega)$

for example,

$f(x,y) = x^2 + k y^2$

write in polar coordinate

$\displaystyle f(r, \theta) = r^2 (\cos^2(\theta) + k \sin^2(\theta)) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta)$

where $a_0 = 2 + 2k, a_{2+} = a_{2-} = 1-a, a_{other} = 0$.

The rotation is

$\displaystyle f(r, \theta' = \theta + \omega ) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta) D^n_{mm}(\omega) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta + \omega)$

If we write the rotated function in Cartesian form,

$f(x',y') = x'^2 + k y'^2 = (c^2 + k s^2)x^2 + (s^2 + k c^2)y^2 + 2(k-1) c s x y$

where $c = \cos(\omega), s = \sin(\omega)$.

In 3-D space, the same logic still applicable.

The spherical harmonics $Y_{lm}$ serves as the basis for eigenvalue of $l(l+1)$, eigen spaces for difference $l$ are orthogonal. This is an extension of the 2-D eigen function $\exp(\pm n i \theta)$.

A 3-D function can be expressed in spherical harmonics, and the rotation is simple multiplied with the Wigner D-matrix.

On above, we show an example of higher order rotation induced by product space. I called it the induced space (I am not sure it is the correct name or not), because the space is the same, but the order is higher.

For two particles system, the direct product space is formed by the product of the basis from two distinct space (could be identical space).

Some common direct product spaces are

• combining two spins
• combining two orbital angular momentum
• two particles system

No matter induced space or direct product space, there structure are very similar. In 3-D rotation, the two spaces and the direct product space is related by the Clebsch-Gordon coefficient. While in 2-D rotation, we can see from the above discussion, the coefficient is simply 1.

Lets use 2-D space to show the “induced product” space. For order $n=1$, which is the primary base that contains only $x, y$.

For $n=2$, the space has $x^2, y^2, xy$, but the linear combination $x^2 + y^2$ is unchanged after rotation. Thus, the size of the space reduced $3-1 = 2$.

For $n = 3$, the space has $x^3, y^3, x^2y, xy^3$, this time, the linear combinations $x^3 + xy^2 = x(x^2+y^2)$ behave like $x$ and $y^3 + x^2y$ behave like $y$, thus the size of the space reduce to $4 - 2 = 2$.

For higher order, the total combination of $x^ay^b, a+b = n$ is $C^{n+1}_1 = n+1$, and we can find $n-1$ repeated combinations, thus the size of the irreducible space of order $n$ is always 2.

For 3-D space, the size of combination of $x^ay^bz^c, a + b+ c = n$ is $C^{n+2}_2 = (n+1)(n+2)/2$. We can find $n(n-1)/2$ repeated combination, thus, the size of the irreducible  space of order $n$ is always $2n+1$.

Changing of frame II

Few things have to say in advance.

1. A Vector is NOT its coordinate
2. A vector can only be coordinated when there is a frame.
3. A frame is a set of “reference” vectors, which span the whole space. Those reference vectors are called basis of a frame.
4. a transformation is on a vector or its coordinate. And it can be represented by a matrix.
5. A Matrix should act on a coordinate or basis, but not a vector.

where

$\hat{\alpha} = \begin {pmatrix} \hat{\alpha_1} \\ . \\ \hat{\alpha_n} \end{pmatrix}$ is the column vector of  basis reference vector.

$\vec{u_{\alpha}}$ is the coordinate column vector in $\alpha$ basis.

$\vec{U}$ is the vector in space

$\vec{V}$ is the transformed vector in space.

$G$ and $H$ are the matrix of transform.

$G \cdot H \cdot G^{-1}$ has the same meaning of $H$, only the matrix representation of the transform is different due to different basis.

the Euler’s rotation can be illustrated by series of the diagram. each rotation of frame can be made by each $G$ . but when doing real calculation, after we apply the matrix G  on the coordinate, the basis changed. when we using the fact that  a matrix can be regard as a frame transform or vector transform. we have follow:

This diagram can extend to any series of frame rotation. and the $V_s \rightarrow X_s \rightarrow V_2 \rightarrow V_s$ triangle just demonstrate how 2 steps frame transform can be reduced to the vector transform in same frame.

i finally feel that i understand Euler angle and changing of frame fully. :D

HERE is a note on vector transform and frame transform.

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.