## 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 $2n+1$.

## Spherical Harmonics and Fourier Series

Recently, I read a very interesting article on the origin of spherical harmonics. I like to summarize in here and add some personal comments.

Starting from Laplace equation

$\nabla^2 \phi(\vec{r}) = 0$

The Laplacian can be separated into radial and spherical part.

$\nabla^2 = \nabla_r^2 + \nabla_\Omega^2$

The solution is called harmonics, and it can be separated into radial part and angular part too,

$\phi(\vec{r}) = R(r) \Theta(\Omega)$

Since the Laplacian is coordinate-free, therefore, the solution is also coordinate free and is rotational invariant. We will come back to this point later.

A homogeneous function of degree n has property,

$f(t\vec{r}) = t^n f(\vec{r})$

In the case of homogeneous harmonics of degree n,

$\phi_n(\vec{r}) = r^n \Theta_n(\Omega)$

Here, the radial part is $R_n(r) = r^n$

Substitute this homogeneous harmonics into the Laplace equation, the $\nabla_r^2$ will produce a coefficient related to the order, and the radial part can be extracted.

$0 = f(r) ( \nabla_\Omega^2 - g(n) ) \Theta(\Omega)$

we have an eigenvalue problem for the angular part

$\nabla_\Omega^2 \Theta = g(n) \Theta$

The eigen function for 2-D Laplacian is the Fourier Series, and that for 3-D is the Spherical Harmonics. In this sense, Fourier Series is a “polar harmonics”.

In 3-D, the angular part of the Lapacian is proportional to the angular momentum operator, $-\hbar^2 \nabla_\Omega^2 = L^2$, where $\hbar$ is the reduced Planck constant, which has the dimension of angular momentum.

$L^2 Y_{lm}(\theta, \phi) = l(l+1) \hbar^2 Y_{lm}(\theta, \phi)$

Here, from the previous discussion, before we solve the equation, we know that the harmonic has maximum order of $l$. The $m$ is the degeneracy for same eigenvalue $l(l+1)$

As we mentioned before, the harmonics should be rotational invariant, such that any direction should be equal. However, when we look at the Spherical Harmonics, the poles are clearly two special points and the rotation around the “z-axis” has limited rotational symmetry with degree $l$. How come?

According to the article, the solution is not necessarily to be separated into $\theta, \phi$, such that

$\displaystyle Y_{lm}(\theta,\phi) = \sqrt{\frac{2l+1}{4\pi} \frac{(l-m)!}{(l+m)!}}P_{lm}(\cos\theta) e^{im\phi}$

I quote the original,

“It is not immediately obvious that we can separate variables and assume exponential functions in the φ direction. We are able to do this essentially because the lines of fixed θ are circles. We could also simply assume this form and show the construction succeeds. This organization is not forced, but separating the variables is so useful that there are no competitive options. A disadvantage of this organization is that it makes the poles into special points.”

The limited rotational symmetry with degree of $l$ is due to the limited “band-width” that restricted by the order of the homogeneous function. The relation between the band width and the order of the harmonics can be understood that the number of “sector” or “node” on the circle/sphere is proportional to the order, thus, the “resolution” is also limited by the order and thus the “band-width”.

Since the Platonic solid is coordinate-free that they are the most symmetry. In the next post, I will show the relation between Spherical Harmonics and Platonic solid. This is related to the section 3.2 in the article,

“One would like to have an uniform discretization for the sphere, with all portions equally represented. From such an uniform discretization we could construct a platonic solid. It is known, however, that there are only a few platonic solids, and the largest number of faces is 20 (icosohedron) and largest number of vertices is 20 (dodecahedron). If we want to discretize the sphere with many points, we cannot do it uniformly. Instead we set the goal of using the fewest points to resolve the Spherical Harmonics up to some degree. Since the Spherical Harmonics themselves are “fair” and “uniform”, this gives a good representation for functions on the sphere. “

As the Fourier Series and Spherical Harmonic are closely related, they should share many properties. For instant, they are orthonormal and form a basis. This leads to the Discrete Fourier Transform and also the “Spherical Transform”,

$\displaystyle f(\Omega) = \sum_{\alpha} a_\alpha \Theta_\alpha(\Omega)$

where $\alpha$ is the id of the basis. One can use the Parseval theorem,

$\displaystyle \int |f(\Omega)|^2 d\Omega = \sum_{\alpha} a_\alpha^2$

Also, the convolution using discrete Fourier transform can also be applied on the spherical harmonics.

Notice that, the Discrete Fourier Transform can “translate” to Continuous Fourier Transform. However, the order of the spherical harmonics is always discrete.

The time-independent Schrödinger equation is

$(-\frac{\hbar^2}{2m}\nabla^2 + V ) \Psi = E \Psi$

Using the Laplacian in spherical coordinate. and Set $\Psi = R Y$

$\nabla^2 R Y - \frac{2m}{\hbar^2}(V-E) R Y = 0$

$\nabla^2 = \frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d}{dr}) - \frac{1}{r^2} L^2$

The angular part,

$L^2 Y = l(l+1) Y$

$\frac{d}{dr}(r^2\frac{dR}{dr}) - l(l+1)R - \frac{2mr^2}{\hbar^2}(V-E) R = 0$

To simplify the first term,

$R = \frac{u}{r}$

$\frac{d}{dr}(r^2 \frac{dR}{dr})= r \frac{d^2u}{dr^2}$

A more easy form of the radial function is,

$\frac{d^2u}{dr^2} + \frac{l(l+1)}{r^2} u - \frac{2m}{\hbar^2} (V-E) u = 0$

The effective potential $U$

$U = V + \frac{\hbar^2}{m} \frac{l(l+1)}{r^2}$

$\frac{d^2u}{dr^2} + \frac{2m}{\hbar^2} (E - U) u = 0$

We can use Rungu-Kutta method to numerically solve the equation.

The initial condition of $u$ has to be 0. (home work)

I used excel to calculate a scattered state of L = 0 of energy 30 MeV. The potential is a Wood-Saxon of depth 50 MeV, radius 3.5 fm, diffusiveness 0.8 fm.

Another example if bound state of L = 0. I have to search for the energy, so that the wavefunction is flat at large distance. The outermost eigen energy is -7.27 MeV. From the radial function, we know it is a 2s orbit.

## Poisson Equation

a Poisson Equation with boundary has a general solution by Green’s theorem. the Green’s theorem stated that a volume integral of Laplacian can be converted to a surface integral. combined this with Green’s function. the general solution of a Poisson equation:

$\nabla^2 \Phi = \rho$

is:

$\Phi(\vec{r}) = \frac{1}{4\pi} \int{\rho(\vec{r'}) G(\vec{r},\vec{r'})d^3x'} + \frac{1}{4\pi}\oint{G(\vec{r},\vec{r'}) \frac{\partial \Phi(\vec{r'})}{\partial n'}-\Phi(\vec{r}) \frac{\partial G(\vec{r},\vec{r'})}{\partial n'}d^2x'}$

usually, the boundary condition is given and this make the solution unique. There are 2 boundary conditions that will give unique solution.

1. Dirichlet – given the potential on the boundary, this requires the Green’s  function be Zero on the boundary.
2. Neumenn – given the field, or the change of the potential on the boundary, this requires the change of the Green’s function be $- 4\pi / S$ on the surface. where S is the total surface area.

Thus, the problem of solving the Poisson equation is equivalent with finding the Green’s function with certain requirement for boundary condition. the Green function should obey this equation:

$\nabla^2 G(\vec{r},\vec{r'})=-4\pi \delta(\vec{r}-\vec{r'})$

because a complete solution of a differential equation must include the homogeneous solution. in this case, the Laplace equation:

$\nabla^2 \Phi=0$

thus, the Green’s function must contain a part from it. for example.

$G(\vec{r},\vec{r'}) = \frac{1}{|\vec{r}-\vec{r'}|} + F(\vec{r},vec{r'})$

with

$\nabla^2 F(\vec{r}, \vec{r'}) = 0$

the method of finding the Green’s function is from the Laplace equation. when we solving Laplace equation in certain coordinate, we get the eigen function for the Laplacian.  The eigen function span the space of function. Thus, we can construct the Green’s function from eigen function.

## 3-D spherical infinite well

the potential is

$V(r,\theta,\phi) = \begin{pmatrix} 0 & |r|

The Laplacian in spherical coordinate is:

$\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \frac{L^2}{r^2}$

since the $L$ is the reduced angular momentum operator, if we set the solution be:

$\psi(r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi)$

Then the angular part was solved and the radial part becomes:

$L^2 Y_{lm} = l(l+1) Y_{ml}$

$\left(r^2 \frac{d^2}{dr^2} + 2 r \frac{d}{dr}+(k^2 r^2 - l(l+1))\right)R(r) = 0$

$k^2 = 2 m E/ \hbar^2$

The radial equation is the spherical Bessel function.

The solution was common written as:

$R(r) = j_l( k r) = \left( - \frac{r}{k} \right)^l \left(\frac{1}{ r} \frac{d}{dr}\right)^l \frac{sin(k r)}{kr}$

The Boundary condition fixed the k and then the energy,

$j_l ( k_{nl} a ) = 0$

the all possible root are notated as n. thus the quantum numbers for this system are:

• n , the order of root
• l , the angular momentum

We can see in here, the different between Coulomb potential and spherical infinite well:

• there is no restriction on n and l, therefore, there will be 1s, 1p, 1d, 1f orbit.
• the energy level also depend on angular momentum, since it determined the order of spherical Bessel function.

we can realized the energy level by the graph of Bessel function. we set some constants be 1, the root are :

$k_{nl} a = \frac{1}{\hbar} \sqrt{ 2 m a^2} \sqrt{E_{nl}} = \pi \sqrt{E_{nl}}$

Thus, we plot

$j_l( \pi \sqrt{E_{nl}})$

Since $k_{nl}$ is a scaling factor to “force” the function to be zero at the boundary. Interestingly, the spherical Bessel function is not normalizable or orthogonal with $r^2$, i.e.

$\int_{0}^{\infty} j_l(r) j_{l'}(r) r^2 dr$

is diverged. Of course, the spherical Bessel function is a “spherical wave” that propagating in space, same as plane wave, which is also not normalizable or orthogonal. However, in the infinite spherical well, with the scaling factor $k$ for same $l$, they are orthogonal! And for difference $l$, the spherical part are already orthogonal. Thus, the wave function for difference energy are orthogonal. i.e. the eigen states are orthogonal! Power of math! In the following plots, the left is $l = 0$ and the right is $l = 1$.

In fact, even for Woods-Saxon potential, the numerical solution of the eigen wave-function are also orthogonal for same angular momentum.

## Lorentz Force and Stress tensor

$F_{ij} =\begin {pmatrix} 0 & D_1 & D_2 & D_ 3\\ -D_1 & 0& H_3 & -H_2 \\ -D_2 & - H_3 & 0 & H_1 \\ -D_3 & H_2 & -H_1 & 0 \end {pmatrix}$

$G_{ij} =\begin {pmatrix} 0 & B_1 & B_2 & B_ 3\\ -B_1 & 0& -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end {pmatrix}$

The field equation are:

$\partial /\partial x_i F_{i j } = - J_i$

$\partial /\partial x_i G_{i j } =0$

That is the result from last time.

the conservation of charge is:

$\partial/\partial x_i J_i = 0$

thus the 4-Laplacian of the F-field is :

$\partial^2/\partial x_i^2 F_{ij} = 0$

The physical meaning of the simplification of the field equation by the field tensor is, a gradient in the tensor field is equation to the minus of 4-current, or zero. recall that the gradient in 3-D vector space, the conservation of charge density is :

$\nabla \cdot \vec{J} = - \partial \rho/\partial t$

we have the same form in the 4-D tensor space. the creation of field is conservation of the 4-charge displacement, if we integrate the 4-current. i dun know what physical meaning of the G-field. personally, i believe that the F-field and G-field can be related by some transform.

we have another interesting things. we can write the Lorentz force into the field tensor:

$\vec{f}_j = \frac{d\vec{P}_j}{d\tau}=q \frac{d\vec{X}_j}{d\tau}F_{ij}$

the reason why we can write this, i don’t know. any physical meaning? i don’t know. may be we can think in this way, the force depends on the motion of the 4-vector and the field and the charge. thus, it is natural to multiple them together to get the force. But why not the G field? never the less, the field tensor reduce the number of Field qualities into 2.

the Lorents Force can be more simple

$\vec{f}_j = J_i F_{ij}$

that the force is created by the field and the current.

The Electromagnetic stress tensor can also be related with the 4-force by :

$\vec{f}_j = - \partial/\partial x_i T_{ij}$

Thus, combined with the Field tensor:

$\partial/\partial x_iT_{ij}+ J_iF_{ij} = 0$

## Laplacian in spherical coordinate

the Momentum operator in spherical coordinate

$\nabla^2 = \frac {1}{r^2}\frac {\partial } { \partial r} \left ( r^2 \frac {\partial} {\partial r} \right ) - \frac {1}{r^2} L^2$

where L is the Reduced angular momentum operator. the minus sign is very important for giving a correct sign. the original angular momentum operator J is related by:

$J=\hbar^2 L$

by compare the Laplacian in spherical coordinate, the L is

$L^2 = - \frac {1}{sin(\theta)} \frac {\partial}{\partial \theta} \left( sin(\theta) \frac {\partial}{\partial \theta} \right ) - \frac {1}{sin(\theta)} \frac{\partial^2} {\partial \phi ^2}$

But this complicated form is rather useless, expect you are mathematic madman.

we can start from classical mechanic

$\vec{L} = \vec {r} \times \vec{p}$

$L_x = y \frac {\partial} {\partial z} - z \frac {\partial}{\partial y }$

$L_y = z \frac {\partial} {\partial x} - x \frac {\partial}{\partial z }$

$L_z = x \frac {\partial} {\partial y} - y \frac {\partial}{\partial x }$

with the change of coordinate

$\begin {pmatrix} x \\ y \\ z \end{pmatrix} = \begin {pmatrix} r sin(\theta) cos(\phi) \\ r sin(\theta) sin(\phi) \\ r cos(\theta) \end{pmatrix}$

and the Jacobian Matrix $M_J$, which is used for related the derivatives.

since

$\frac {\partial}{\partial x} = \frac {\partial r}{\partial x} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial x} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial x} \frac {\partial} {\partial \phi}$

$\frac {\partial}{\partial y} = \frac {\partial r}{\partial y} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial y} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial y} \frac {\partial} {\partial \phi}$

$\frac {\partial}{\partial z} = \frac {\partial r}{\partial z} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial z} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial z} \frac {\partial} {\partial \phi}$

which can be simplify

$\nabla_{(x,y,z)} = M_J^T \nabla_{(r, \theta, \phi )}$

$M_J = \frac {\partial ( r, \theta, \phi) }{\partial (x,y,z)}$

$M_J^{\mu\nu} = \frac {\partial \mu}{\partial \nu}$

then, we have

$L_x = i sin(\phi) \frac {\partial }{\partial \theta} +i cot(\theta) cos(\phi) \frac { \partial }{\partial \phi}$

$L_y =-i cos(\phi) \frac {\partial }{\partial \theta} + i cot(\theta) sin(\phi) \frac { \partial }{\partial \phi}$

$L_z = - i \frac {\partial }{\partial \phi}$

However, even we have the functional form, it is still not good.  we need the ladder operator

$L_+ = L_x + i L_y = Exp(i \phi) \left( \frac {\partial }{\partial \theta} + i cot(\theta) \frac { \partial }{\partial \phi} \right)$

$L_- = L_x - i L_y = Exp(-i \phi) \left( \frac {\partial }{\partial \theta} - i cot(\theta) \frac { \partial }{\partial \phi} \right)$

notice that

$L_+^\dagger = L_-$

so, just replacing $i \rightarrow -i$.

when we looking for the Maximum state of the spherical Harmonic $Y_{max}(\theta, \phi)$

$L_+ Y_{max}(\theta,\phi) = 0 *)$

use the separable variable assumption.

$Y_{max}(\theta, \phi) = \Theta \Phi$

$L_+ \Theta \Phi = 0 = - Exp(i \phi) \left( \frac {d\Theta}{d \theta} \Phi + i cot(\theta) \frac { d\Phi}{d\phi} \right) \Theta$

$\frac {tan(\theta)}{\Theta} \frac { d \Theta} {d \Theta } = - \frac {i}{\Phi} \frac {d \Phi} {d \phi} = m$

the solution is

$Y_{max}(\theta,\phi) = sin^m(\theta) Exp(i m \phi )$

$L^2 Y_{max}(\theta, \phi) = m(m+1) Y_{max}(\theta,\phi)$

an application on Hydrogen wave function is here.