Review on rotation

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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.

Implementation of Rotation

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A vector \vec{r} in space has to be expressed using basis \hat{e} as a reference and coordinate \vec{v}, such that,

\displaystyle \vec{r} = \hat{e} \cdot \vec{v} = \sum_{i} \hat{e}_i v_i

A transform of a vector R(\vec{r}) is done by

\vec{r'} = R(\vec{r}) = \hat{e} \cdot R \cdot \vec{v}

This transform can be view as

(\hat{e}\cdot R) \cdot \vec{v} = \hat{e} \cdot ( R \cdot \vec{v})

The first one is change the basis and keep the coordinate, and the later is change the coordinate and keep the basis.

The Euler angle, defined as,

\vec{r'} = \hat{e} \cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot R_{z}(\gamma) \cdot \vec{v}

Since the rotational matrix is fixed and not based on any basis, how to understand this transform can be viewed differently.

The implementation of the rotation usually like this:

\vec{v'} = R_{z}(\alpha) \cdot R_{y}(\beta) \cdot R_{z}(\gamma) \cdot \vec{v}

And the rotation order is \alpha \to \beta \to \gamma. In this order, the rotation is using body (or current, intrinsic) axis.

If the rotation order is \gamma \to \beta \to \alpha, the rotation is using fixed (or global, external, Laboratory) axis.

To understand this, we have to add back the basis \hat{e}. When using the body axis, the left most matrix actually act on the basis (operationally)

\hat{e}\cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot \vec{v} \to \hat{e'} \cdot R_y(\beta) \cdot \vec{v}

When using the fixed axis,

\hat{e}\cdot R_{z}(\alpha) \cdot R_{y}(\beta) \cdot \vec{v} \to \hat{e} \cdot R_z(\alpha) \cdot \vec{v'} \to \hat{e} \cdot \vec{v''}

Thus, if we using Euler angle \alpha \to \beta \to \gamma in fixed frame, the correct matrix is

R_{z}(\gamma) R_y(\beta) R_z(\alpha)

In the following, we use fixed frame, unless specified. 

Since using Euler angle \alpha \to \beta \to \gamma in changing frame, the rotation \gamma is fixed on the new axis \hat{n}(\theta = \beta , \phi = \alpha) . However, the matrix

R(\hat{n}, \omega) = R_z(\phi) \cdot R_y(\theta)\cdot R_z(\omega)

is NOT a rotation on the axis with the angle. As we can see that, for a zero rotation \omega = 0 , any vector will still be transformed by R_z(\phi) \cdot R_y(\theta) .

Now, suppose we have a rotation axis \hat{n} and an rotation angle \omega , the rotation matrix can be constructed as follow.

Notice that the rotational matrix R is a unitary transform, its eigen-vectors are P,

\displaystyle R\cdot P = P \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & a + i b & 0 \\ 0 & 0 & a-i b \end{pmatrix}

a^2 + b^2 = 1

|\textrm{Arg}(a + i b)| = \omega

Using \omega, solve a = \cos(\omega) , b = \sin(\omega).

Since the eigen vector of R_z(\alpha) are

\displaystyle p_0 = \hat{e_z}, p_{\pm} = \frac{1}{2}(\hat{e_x}\pm i \hat{e_y})

Since p_0 \to \hat{n}(\theta, \phi), thus,

\displaystyle p_{\pm} \to \frac{1}{2}( \hat{e_x'} \pm i \hat{e_y'} )

such that

\hat{e_i'} = R_z(\phi)\cdot R_y(\theta) \cdot \hat{e_i}

Notice that the corresponding eigen vector for the eigen value a \pm i b is p_{\mp}.

Then the rotational matrix is

R = P\cdot D\cdot P^{-1}

Another to look at this problem is that, an operator, can be “decomposed” into its eigen vectors and eigen-values, using Dirac notation,

\displaystyle R(\omega) = \sum_{i} |v_i \rangle \omega_i \langle v_i |

The rotation round the vector can be transformed use R_z(\omega) from (0,0) \to (\theta, \phi) , Thus, the vector  has to be transform, so the operator

\displaystyle R(\hat{n}, \omega) = \sum_{i} R_z(\phi) R_y(\theta) |v_i \rangle \omega_i \langle v_i | R_y(-\theta) R_z(-\phi)

R(\hat{n}, \omega) = R_z(\phi) \cdot R_y(\theta) \cdot  R_z(\omega) \cdot R_y(-\theta) \cdot R_z(-\phi)

This is a relation between fixed frame rotation to a body rotation! \hat{n} can be the body z-axis, y-axis, or x-axis in particular.

For example, Let \hat{n} be the rotated y-axis, such that

\displaystyle \hat{n} = R_z(\alpha) \cdot \hat{y} , (\theta, \phi) = (\frac{\pi}{2}, \alpha+\frac{\pi}{2})

\displaystyle R(\hat{y'}, \omega) = R_z(\alpha+\frac{\pi}{2}) \cdot R_y(\frac{\pi}{2}) \cdot R_z(\omega) \cdot R_y(-\frac{\pi}{2}) \cdot R_z( -\alpha - \frac{\pi}{2})

Notice that

R_y(\frac{\pi}{2}) \cdot R_z(\omega) \cdot R_y(-\frac{\pi}{2}) = R_x(\omega)

R_z(\frac{\pi}{2}) \cdot R_x(\omega) \cdot R_z(-\frac{\pi}{2}) = R_y(\omega)

The above formula are not difficult to imagine, decompose the middle operator into the eigen-system and use Dirac notation. For example, the rotate of the z-axis around y-axis with 90 degree is the x-axis. Thus,

\displaystyle R(\hat{y'}, \omega) =R_{yB}(\omega) = R_z(\alpha) \cdot R_y(\omega) \cdot R_z(-\alpha)

We recovered the relation of the Lab frame to body frame for y-axis!

Also, given a vector \vec{r} rotated by \omega to \vec{r'} around unit vector \hat{n} is

\vec{r'} = \vec{r} \cos(\omega) + \hat{n} (\hat{n}\cdot \vec{r}) (1-\cos(\omega)) + \hat{n} \times \vec{r} \sin(\omega)

Changing of frame II

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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.


\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

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


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


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

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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}

detail treatment on Larmor Precession and Rabi Resonance

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a treatment on Larmor Precession and Rabi resonance

the pdf is a work on this topic. it goes through Larmor Precession and give example on spin-½ and spin-1 system.

then it introduce Density matrix and gives some example.

The Rabi resonance was treated by rotating frame method and using density matrix on discussion.

the last topic is on the relaxation.

the purpose of study it extensively, is the understanding on NMR.

the NMR signal is the transverse component of the magnetization.

on angular momentum adding & rotation operator

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the angular momentum has 2 kinds – orbital angular momentum L , which is caused by a charged particle executing orbital motion, since there are 3 dimension space. and spin S , which is an internal degree of freedom to let particle “orbiting” at there.

thus, a general quantum state for a particle should not just for the spatial part and the time part. but also the spin, since a complete state should contains all degree of freedom.

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

when we “add” the orbital angular momentum and the spin together, actually, we are doing:

J = L \bigotimes 1 + 1 \bigotimes S

where the 1 with L is the identity of the spin-space and the 1 with S is the identity of the 3-D space.

the above was discussed on J.J. Sakurai’s book.

the mathematics of L and S are completely the same at rotation operator.

R_J (\theta) = Exp( - \frac {i}{\hbar} \theta J)

where J can be either L or S.

the L can only have effect on spatial state while S can only have effect on the spin-state. i.e:

R_L(\theta) \left| s \right> = \left| s\right>

R_S(\theta) \left| x \right> = \left| x\right>

the L_z can only have integral value but S_z can be both half-integral and integral. the half-integral value of Sz makes the spin-state have to rotate 2 cycles in order to be the same again.

thus, if the different of L and S is just man-made. The degree of freedom in the spin-space is actually by some real geometry on higher dimension. and actually, the orbital angular momentum can change the spin state:

L \left| s \right> = \left | s' \right > = c \left| s \right>

but the effect is so small and

R_L (\theta) \left| s\right > = Exp( - \frac {i}{\hbar} \theta c )\left| s \right>

but the c is very small, but if we can rotate the state for a very large angle, the effect of it can be seen by compare to the rotation by spin.

\left < R_L(\omega t) + R_S(\omega t) \right> = 2 ( 1+ cos ( \omega ( c -1 ) t)

the experiment can be done as follow. we apply a rotating magnetic field at the same frequency as the Larmor frequency. at a very low temperature, the spin was isolated and T_1 and T_2 is equal to \infty . the different in the c will come up at very long time measurement and it exhibit a interference pattern.

if c is a complex number, it will cause a decay, and it will be reflected in the interference pattern.

if we find out this c, then we can reveal the other spacial dimension!


the problem is. How can we act the orbital angular momentum on the spin with out the effect of spin angular momentum? since L and S always coupled.

one possibility is make the S zero. in the system of electron and positron. the total spin is zero.

another possibility is act the S on the spatial part. and this will change the energy level.


an more fundamental problem is, why L and S commute? the possible of writing this

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

is due to the operators are commute to each other. by why?

if we break down the L in to position operator x and momentum operator p, the question becomes, why x and S commute or p and S commute?

[x,S]=0 ?

[p,S]=0 ?

[p_x, S_y] \ne 0 ?

i will prove it later.


another problem is, how to evaluate the Poisson bracket? since L and S is not same dimension. may be we can write the eigenket in vector form:

\begin {pmatrix} \left|x, t \right> \\ \left|s\right> \end {pmatrix}

i am not sure.



For any vector operator, it must satisfy following equation, due to rotation symmetry.

[V_i, J_j] = i \hbar V_k   run in cyclic


where J is rotation operator. but i am not sure is it restricted to real space rotation. any way, spin is a vector operator, thus

$latex [S_x, L_y] = i \hbar S_z = – [S_y, L_x] $

so, L, S is not commute.

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