## Rotational Band

For deformed nuclei, it can be rotated in various angular momentum in Laboratory frame. Assume rigid body rotation, the energy is

$\displaystyle E_J = \frac{1}{2}I\omega^2 = \frac{1}{2I}J^2$

In QM, that becomes

$\displaystyle H = \sum_{i=1}^{3} \frac{\hbar^2}{2I_i} J_i^2$

For axial symmetry, $I_1 = I_2 = I$

$\displaystyle H = \frac{\hbar^2}{2I} (J^2 - J_3^2) + \frac{\hbar^2}{2I_3}J_3^2$

Remember, in deformed nuclei, the projection of $J$ along the symmetry axis in the body-frame is $K$ . The expected value of the Hamiltonian with state $|Nn_z m_l K \rangle$ in the body-frame is proportional to $J(J+1)$ for $J^2$ and $K$ for $J_3$. i.e.

$\displaystyle E_J = \frac{\hbar^2}{2I} J(J+1) + E_K$

From body-frame to Lab-frame, we should apply the Wigner D-Matrix to the intrinsic wave function. ( I am not sure the following equation is correct, but the idea is rotating the body-frame wavefunction with Wigner D-Matrix to get the Lab-frame wave function. In Lab frame the total angular momentum must be a good Quantum number as rotational symmetry restored, so as $J_z = M$. The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

$\displaystyle |JMK\rangle = \sum_{M} D_{MK}^{J} |Nn_zm_lK\rangle$

However, the Wigner D-Matrix does not conserve parity transform:

$\displaystyle D_{MK}^J \rightarrow (-1)^{J+K} D_{M-K}^{J}$

In order to restored the parity, we need to include $\pm K$ in the Lab-frame wave function.

$\displaystyle |JMK\rangle = \sum_{M} \left( D_{MK}^J \pm (-1)^{J+K} D_{M-K}^J \right) |Nn_zm_lK\rangle$

where + for positive parity, – for negative parity.

From the above equation, for $K^\pi = 0^+$ ($0^-$), $J$ must be even (odd). For $K > 0$, $J = K, K+1, K+2, ...$.

We can see for $K = 1/2$, the $J = 5/2, 9/2, 11/2$ are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect $\Delta K = 1$.

$\displaystyle \langle JMK | H'(\Delta K = 1) |JMK \rangle$

$\displaystyle \rightarrow \langle D_{MK}^J | H' | D_{MK}^J \rangle+ \langle D_{M-K}^J |H'| D_{MK}^J \rangle + \langle D_{MK}^J | H' | D_{M-K}^J \rangle+ \langle D_{M-K}^J |H'| D_{M-K}^J \rangle$

The term with $\Delta K = 0$ vanished. And since $\latex K = 1$, the only non-zero case is $K = 1/2$.

A possible form of the $H' (\Delta K = 1) = \frac{1}{2} \omega (J_+ + J_-)$. These are the ladder operator to rise or lower the m-component by 1. In 19F case, we can think it is a single proton on top of 18O core.  A rotation core affect the proton with an additional force, similar to Coriolis force on earth.

## Reason for n>l in atomic orbit

If we study atomic structure, we will find the principle quantum number must larger than azimuthal quantum number (the number of angular momentum), i.e.

$n > l$

But most text book give this result using mathematics that there is no solution for the radial equation when $n \leq l$. Some text book solves the radial equation using power series, and argue that the series has to be terminated at power of $n-1$ because the number of node is $n-1$. The exactly solution is related to Laguerre polynomial, and the polynomial is only defined for $n \geq l+1$.

Also, in nuclear physics, there is no restriction and $n$ and $l$ can take any integers. Why these two cases are different? of course, the key point is the potential difference. Coulomb potential is used in atomic orbit. Wood-Saxon potential (or finite square well ) is used in nuclear orbit.

The Schrodinger equation for potential $V(r)$ is

$\displaystyle \left( \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right) - \frac{1}{r^2} L^2 - \frac{2m}{\hbar^2} V(r) \right) \psi(r, \Omega) = -\frac{2mE}{\hbar^2} \psi(r,\Omega)$

Separate the radial and spherical part $\psi(r, \Omega) = R(r) Y(\Omega)$

$\displaystyle L^2 Y(\Omega) = l(l+1) Y(\Omega)$

$\displaystyle \left( \frac{1}{r^2} \frac{d}{d r}\left( r^2 \frac{d}{d r}\right) - \frac{l(l+1)}{r^2} - \frac{2m}{\hbar^2} V(r) \right) R(r) = -\frac{2mE}{\hbar^2} R(r)$

The radial equation further reduce by using $u(r) = r R(r)$

$\displaystyle \frac{d^2 u}{dr^2} - \frac{2m}{\hbar^2} U(r) u(r) = \frac{2mE}{\hbar^2} u(r)$

we can see, the effective potential is

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

For Coulomb potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

Introduce dimensionless quantities, or Express everything using the fine structure constant. $a_0$ is Bohr radius,

$\displaystyle \rho = \frac{2r}{a_0}, \lambda = \frac{e^2}{4\pi \epsilon_0} \sqrt{ - \frac{m}{2\hbar^2 E}} = n$

The equation becomes

$\displaystyle \frac{d^2u}{d\rho^2} + \left(\frac{\lambda}{\rho} - \frac{l(l+1)}{\rho^2}\right) u = \frac{1}{4} u$

The simplified effective potential,

$\displaystyle U(\rho) = \frac{n}{\rho} - \frac{l(l+1)}{\rho^2}$

Now, we related the principle quantum number and azimuthal quantum number in one simply formula.

Since the repulsive part due to the rotation is getting larger and larger, the minimum point of the effective potential is

$\displaystyle \rho_{min} = \frac{2l(l+1)}{n}$

When $l = n$, $\rho_{min} = 2(n+1) , r_{min} = (n+1)a_0$.

I plot the case for $n=2$

The bottom line is $l = 0$. We can see, when $l = 2$, the minimize point of the potential is at $r = 3 a_0$. The potential is vary shallow, and the bound state is very diffuses. Since the 2nd derivative of the wave function $u$ around the minimum point is

$\displaystyle \frac{d^2u}{d\rho^2} = \left( \frac{l(l+1)}{\rho^2} - \frac{n}{\rho} + \frac{1}{4} \right) u \approx \frac{1}{4} \frac{l(l+1) - n^2}{l(l+1)} u = k u$

For $l = n-1$ , $k$ is negative, that give oscillating wave function.

For $l = n$, $k$ is positive, that gives explosive wave function that cannot be normalized.

Therefore, $l < n$.

In fact, if we plot the effective potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

we can find that when $l \geq 1$, the potential is also very shallow. For such a picture, we can see the wave function must be oscillating that the number of node must be larger then 1, so the principle quantum number should be larger than 1.

In nuclear orbit, the effective potential using Wood-Saxon potential is

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

where $R \approx 1.25 A^{1/3} [fm]$ and $a = 0.6 [fm]$. For simplicity, we use

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{l(l+1)}{r^2}$

The position of minimum potential is no simple formula. I plot R=10,  a=1

We can see, for any $l < l_{u}$ there is a “well” that can bound a nucleon.

I haven’t solve the finite spherical well, but the infinite spherical well is solved. The solution is the spherical Bessel function of first kind. that function only has 1 parameter, which is the angular momentum. To determine the energy is simply find the node position, and the number of node is related to the principle quantum number. Therefore, the principle quantum number and angular momentum is unrelated. A similar result should be applied well in the case of finite spherical well like Wood-Saxon type, that, for a fixed angular momentum, there can be many solution with different number of nodes.

Wood-Saxon potential, the effective potential is always sloping, especially when outside of the well, that forces that wave function to be decay faster. However, the Coulomb potential is a long range force, the effective potential can be slowly changing for long distance, that allow the wave function to be oscillating for long range.

## A little summery of what I am doing

My current position is developing a computational program that can measure the system of polarized target automatically and repeatedly. The program needs to connect with the microwave generator, voltage supply, power meter, and oscilloscope. That is purely technical and nothing special.

Later, after the program started to collecting data, another program is needed to analysis the data. Although there are many program that can do analysis, but those program are not so easy to use, in both control and display. So, I make an analysis program. The program is simply read the 2-D time-domain signal, and then determine some parameters of a specific function that fit the signal. However, the signal could be very noisy, so FFTW and wavelet analysis were implemented. That’s why the wavelet analysis appeared in this blog.

After that, the sample of polarized target is made from various pentacene derivatives, that no known energy level exist. One way to find out the energy levels of the singlet excited states is to measure the absorption spectrum. However, the energy of the triplet state is difficult to measure. And the energy level of the triplet state is critical for Dynamic Nuclear Polarization to be happened. Thus, one way to find out is doing computation chemistry.

My last chemistry class was like 15 years ago, when I was junior high school. But the basic of computation chemistry is solving the Schrodinger equation. That is what theoretical nuclear physicists do! The variation method, the Hartree-fock method, I heard it and somehow know it, but never do it with my own hand and computer. That is why I revisit the Hartree-fock method, and found out my previous understanding is so naive.

In the course of studying Hartree-Fock method, and one problem is evaluating the overlap integral

$\displaystyle \int \psi_a(r_1) \psi_b(r_2) \frac{1}{r_{12}} \psi_c(r_1) \psi_d(r_2) dr_1dr_2$

For a 3-D system, the integral involves product of multiple spherical harmonics. That is really troublesome. Therefore, I move to study the spherical harmonics, and the related rotational invariant, Wigner D-matrix, Clebcsh-Gordon series and Fourier series.  The spherical harmonics arises from solving the Laplace equation in spherical coordinate. A general theory of the solution of Lapalace equation involves Legendre polynomial, which is a special case for Hypergeometric function.  And a very interesting connection is that the elliptical function of the 1st and 2nd kind are also two special cases of hypergeometric function, that, the solution of Laplace equation for elliptical boundary condition is elliptical functions. That connects spherical harmonics and elliptical function! Wow!

I am now a bit off-track, that I am very interesting on the function of all functions. We know that there are many elementary functions, such as sin, cos, and Log. And even more kind of special functions, such as

• Hermite — solving 1-D harmonic oscillator
• Laguerre — the radial function of hydrogen
• Legendre — the solution of the “$\theta$” of  Laplace equation
• Gamma — a continuation of factorial
• Bessel — solution of 3-D infinite square well
• Elliptic — magnetic field of a solenoid
• Dirac delta
• Gaussian or Error function

For discrete argument

• Clebsch-Gordon
• Factoral
• Binomial

As far as I know, the Hypergemetric function is like “mother of functions”, although not all special functions can be expressed as it. I am driven by curiosity, so, I am not sure where I will go. For instance, the transformation of Hypergeometric function is very interesting.

So, for now, as an ending, I found one article is very interesting. The History and Future of Special Functions, by Stephen Wolfram.

## 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 Platonic Solids

The Platonic solids has the most symmetric for discrete rotation in 3-D space. They are symmetry in all the faces, lines, and vertexes. The Spherical harmonics also has discrete rotational symmetry on the z-axis for all order. However, for small order, the spherical harmonics also has other symmetric axis.

For example,

$\displaystyle Y_{32}(\theta, \phi) = \sqrt{\frac{105}{32\pi}} \cos\theta \sin^2\theta \exp(2i\phi)$

The spherical plot and spherical contour plot for $1 + Y_{32}$ are

The plot clearly shows the tetrahedron’s symmetry. In fact, the tetrahedron, hexahedron (cubic), and octahedron share the same symmetry.

The tetrahedron inscribe itself.

The hexahedron and octahedron inscribe each other.

The dodecaherdo and icosahedron also inscribe each other.

We can systematically construct the tetrahedron, hexahedron (octahedron), and dodecahedron.

The tetrahedron has 120 degree rotation symmetry around a vertex. And we can guess, $Y_{33}$ should be a suitable candidate. Thus, we can form the tetrahedron from

$f(\theta, \phi) = Y_{30}(\theta,\phi) + Y_{33}(\theta, \phi)$

However, at angle $\tan(\theta/2) = \sqrt{2}, \phi = \pi/3$, the value of $f$ is less then $\theta = \phi = 0$.  To restore the symmetry,

$\displaystyle f_t(\theta, \phi) = Y_{30}(\theta,\phi) + \sqrt{\frac{8}{5}} Y_{33}(\theta, \phi)$

The plots for $f_t$ are show below.

Similarly, we can from the octahedron using $Y_{4m}$ with a coefficient.

$\displaystyle f_h(\theta, \phi) = Y_{40}(\theta,\phi) + \sqrt{\frac{10}{7}} Y_{44}(\theta, \phi)$

For dodecahedron, the use of $Y_{5,5}$ seem to be the logical choice. However, it turns out it cannot has the correct symmetry because $Y_{50}$ is not “even”.

$\displaystyle f_d(\theta, \phi) = Y_{60}(\theta,\phi) + \sqrt{\frac{28}{11}} Y_{65}(\theta, \phi)$

The Mathematica code for generating the graph is

GraphicsGrid[{{
SphericalPlot3D[1 + f(θ,φ), {θ,  0, π}, {φ, 0, 2 π}],
ParametricPlot3D[{Cos[φ] Sin[θ], Sin[φ] Sin[θ], Cos[θ]}, {θ,  0, π}, {φ, 0, 2 π},
ColorFunction -> Function[{x, y, z, u, v},  Hue[Abs[1 + f(u,v)]]],
ColorFunctionScaling -> False, Mesh -> None, PlotPoints -> 200]
}}, ImageSize -> 600]

The code for calculating the coefficient is

Solve[
SphericalHarmonicY[n, 0, k, 0] + a SphericalHarmonicY[n, n, k, 0]
== SphericalHarmonicY[n, 0, 0, 0]
, a]

Where $n$ is the order of the spherical harmonic, $k$ is the angle for the next face or vertex for fixed $\theta = 0$.

Tetraherdon $k = 2 \tan^{-1}(\sqrt{2})$.

Octaherdon $k = \pi/2$.

Dodecahedron $k = 2 \tan^{-1}( (1+\sqrt{5})/2)$.

## 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.

## Product of Spherical Harmonics

One mistake I made is that

$\displaystyle Y_{LM} = \sum_{m_1 m_2} C_{j_1m_1j_2 m_2}^{LM} Y_{j_1m_1} Y_{j_2m_2}$

because

$\displaystyle |j_1j_2JM\rangle = \sum_{m_1m_2} C_{j_1m_1j_2 m_2}^{LM} |j_1m_1\rangle |j_2m_2\rangle$

but this application is wrong.

The main reason is that, the $|j_1j_2JM\rangle$ is “living” in a tensor product space, while $|jm \rangle$ is living in ordinary space.

We can also see that, the norm of left side is 1, but the norm of the right side is not.

Using the Clebsch-Gordon series, we can deduce the product of spherical harmonics.

First, we need to know the relationship between the Wigner D-matrix and spherical harmonics. Using the equation

$\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} Y_{lm'}(\hat{r}) D_{m'm}^{l}(R)$

We can set $\hat{r} = \hat{z}$ and $R(\hat{x}) = \hat{r}$

$Y_{lm}(\hat{z}) = Y_{lm}(0, 0) = \sqrt{\frac{2l+1}{4\pi}} \delta_{m0}$

Thus,

$\displaystyle Y_{lm}(\hat{r}) = \sqrt{\frac{2l+1}{4\pi}} D_{0m}^{l}(R)$

$\Rightarrow D_{0m}^{l} = \sqrt{\frac{4\pi}{2l+1}} Y_{lm}(\hat{r})$

Now, recall the Clebsch-Gordon series,

$\displaystyle D_{m_1N_1}^{j_1} D_{m_2 N_2}^{j_2} = \sum_{jm} \sum_{M} C_{j_1m_1j_2m_2}^{jM} C_{j_1N_1j_2N_2}^{jm} D_{Mm}^{j}$

set $m_1 = m_2 = M= 0$

$\displaystyle D_{0N_1}^{j_1} D_{0 N_2}^{j_2} = \sum_{jm} C_{j_10j_20}^{j0} C_{j_1N_1j_2N_2}^{jm} D_{0m}^{j}$

rename some labels

$\displaystyle Y_{l_1m_1} Y_{l_2m_2} = \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm}$

We can multiply both side by $C_{l_1m_1l_2m_2}^{LM}$ and sum over $m_1, m_2$,  using

$\displaystyle \sum_{m_1m_2} C_{l_1m_1l_2m_2}^{lm}C_{l_1m_1l_2m_2}^{LM} = \delta_{mM} \delta_{lL}$

$\displaystyle \sum_{m_1m_2} C_{l_1m_1l_2m_2}^{LM} Y_{l_1m_1} Y_{l_2m_2} = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2L+1)}} C_{l_10l_20}^{l0} Y_{LM}$

From the Clebsch-Gordan coefficient, all m-values are zero, that mean the projections on the z-axis are zero for all angular momentum, thus, $l_1 + l_2 = L + even$ or $|l_1 \pm l_2 |=L$ or $l_1 + l_2 + L = even$.

In Mathematica,

FunctionExpand[Sum[ClebschGordan[{l1, m1}, {l2, m2}, {L, M} ] SphericalHarmonicY[l1, m1, \[Theta], \[Phi]] SphericalHarmonicY[l2, m2, \[Theta], \[Phi]], {m1, -l1, l1, 1 }, {m2, -l2, l2, 1}]]
FunctionExpand[ Sqrt[((2 l1 + 1) (2 l2 + 1))/(4 \[Pi] (2 L + 1))] ClebschGordan[{l1, 0}, {l2, 0}, {L, 0}] SphericalHarmonicY[ L, M, \[Theta], \[Phi]]]