## Nilsson Orbital using diagonalization method

Long time ago, I tried to tackle the Nilsson orbital by solving the Hamiltonian analytically. However, the Hamiltonian is without LS coupling. This times, I redo the calculation according to the reference B. E. Chi, Nuclear Phyiscs 83 (1966) 97-144.

The Hamiltonian is

$\displaystyle H = \frac{P^2}{2m} + \frac{1}{2}m\left( \omega_\rho^2 (x^2+y^2) + \omega_z^2 z^2 \right) + C L\cdot S + D L\cdot L$

using

$\displaystyle \omega_\rho = \omega_0 \left(1+\frac{2}{3}\delta\right)^{\frac{1}{2}} = \omega \left(\frac{3+2\delta}{3-4\delta}\right)^{1/6}$

$\displaystyle \omega_z = \omega_0 \left(1-\frac{4}{3}\delta\right)^{\frac{1}{2}} = \omega \left(\frac{3-4\delta}{3+2\delta}\right)^{1/3}$

$\displaystyle \beta = \frac{4}{3}\sqrt{\frac{\pi}{5}}\delta$

$\displaystyle r^2 Y_{20}(\theta, \phi) = \frac{1}{4}\sqrt{\frac{5}{\pi}} (3z^2-r^2)$

The Hamiltonian becomes

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 +\frac{1}{2} m \omega_0^2 r^2 - \frac{1}{2} m\omega_0^2 r^2 \beta Y_{20} + C L\cdot S + D L\cdot L$

Set $x_i^2 \rightarrow x_i^2 \frac{\hbar}{m \omega_0}$, and $r^2 \rightarrow \rho^2 \frac{\hbar}{ m \omega_0}$

$\displaystyle \frac{H}{\hbar\omega_0} = \frac{1}{2}(-\nabla^2 + \rho^2) - \rho^2 \beta Y_{20} - 2 \kappa L\cdot S - \mu \kappa L\cdot L$

Set

$\displaystyle \frac{H_0}{\hbar\omega_0} = \frac{1}{2}(-\nabla^2 + \rho^2) - 2 \kappa L\cdot S - \mu \kappa L\cdot L$

and the perturbation is

$\displaystyle \frac{H_p}{\hbar\omega_0} = - \rho^2 Y_{20}$

The wavefunction for the spherical harmonic is

$\displaystyle |Nljk\rangle = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2) \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}$

$\displaystyle A = \sqrt{\frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!}}$

The diagonal elements are

$\displaystyle \frac{1}{\hbar \omega_0 }\langle Nljk|H_0|Nljk\rangle = N + \frac{3}{2} - \kappa \langle L\cdot S \rangle - \mu \kappa l(l+1)$

where $\langle L \cdot S \rangle = \frac{1}{2} ( j(j+1) - l(l+1) - \frac{3}{4} )$

The off diagonal elements are

$\displaystyle \frac{1}{\hbar \omega_0 }\langle Nljk|H_p|Nljk\rangle = - \langle Nljk| r^2 Y_{20}|Nljk\rangle$

( I will evaluate this integral in future )

The rest is diagonalization the Hamiltonian

$\displaystyle H = H_0 + \beta H_p$

Here is the calculation for the 2nd harmonic for $\kappa = 0.05, \mu = 0$

The component of each orbital can be directly taken from the eigenvalue. Here is the [521]1/2 state. $\kappa = 0.05, \mu(N=3) = 0.35, \mu(N=4) = 0.625, \mu(N=5) = 0.63$

## Spectroscopic factor (revisit)

For transfer reaction, the wave function of nucleus B = A + 1 can be written as

$\displaystyle \Psi_{J_B m_B}(A+1) = \sum_{A'nlj} \beta_{nlj}(B,A) A[\Psi_{J_A'}(A) \phi_{nlj}(r)]_{J_B m_B} \\ = \sum_{A' nlj} \beta_{nlj}(B,A') \frac{1}{N_A} \sum_{m_A' m}C_{J_A M_A j m}^{J_B M_B} \Psi_{J_A' m_A'}(A) \phi_{nljm}(r)$

where the $N_A = _{A+1}C_1$ due to anti-symetrization between the single nucleon and the core nucleus A, the Clebsh-Gordon coefficient is come from the angular coupling, and $\beta_{nlj}(B,A)$ is the square root of the spectroscopic factor for orbital $nlj$ between nuclei B and A.

We can see that, to find out the spectroscopic factor, we can integrate out the core nucleus A

$\displaystyle \beta_{nlj}(B,A) = \langle \Psi_{J_B mB} | A [\Psi_{J_A} \phi_{nlj}]_{J_B m_B} \rangle \\ = \sum_{m_A m} C_{J_A m_A j m}^{J_B m_B} \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle \\ = \sum_{m_A m} (-1)^{J_A - J + m_B} \sqrt{2J_B+1} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix} \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle$

In the last step, we write the Clebsh-Gordon coefficient in Winger 3j-symbol. Using the Wigner-Eckart theorem

$\displaystyle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle \\ = \frac{(-1)^{J_B-m_B}}{\sqrt{2J_B+1}} \begin{pmatrix} J_B & j & J_A \\ -m_B & m & m_A \end{pmatrix} \langle J_B || a_{nljm}^\dagger ||J_A \rangle \\ = \frac{(-1)^{-m_B - J_A - j }}{\sqrt{2J_B+1}} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix} \langle J_B || a_{nljm}^\dagger ||J_A \rangle$

Thus,

$\displaystyle \beta_{nlj}(B,A) = \sum_{m_A m} (-1)^{-2J} \sqrt{2J_B+1} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix}^2 \langle J_B|| |a_{nljm}^\dagger ||J_A\rangle$

using the identity

$\displaystyle \sum_{m_A m} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix}^2 = \frac{1}{2J_B+1}$

we have

$\displaystyle \beta_{nlj}(B,A) = \frac{(-1)^{2J}}{\sqrt{2J_B+1}} \langle J_B|| |a_{nljm}^\dagger ||J_A\rangle$

replacing the reduced matrix element in the Wigner-Eckart theorem with previous result,

$\displaystyle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle = \frac{(-1)^{J_B-m_B}}{\sqrt{2J_B+1}} \begin{pmatrix} J_B & j & J_A \\ -m_B & m & m_A \end{pmatrix} \frac{\sqrt{2J_B+1}}{(-1)^{2j}} \beta_{nlj}(B,A)$

$\displaystyle (2J_B+1)\begin{pmatrix} J_B & j & J_A \\ -m_B & m & m_A \end{pmatrix}^2 \beta_{nlj}^2(B,A) \\=\langle J_A m_A |a_{nljm} |J_B m_B \rangle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle$

For A(d,p) B reaction, the nucleus A is in the ground state, we can sum all the $m_B$ and $m$ states, then sum all the $J_B$, notice that

$\displaystyle \sum_{J_B m_B} |J_B m_B \rangle \langle J_B m_B | = 1$

we have

$\displaystyle \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) =\langle J_A m_A |\sum_{m} a_{nljm}a_{nljm}^\dagger |J_A m_A \rangle$

recall

$a_{nljm}a_{nljm}^\dagger = 1 - a_{nljm}^\dagger a_{nljm} \\ \langle J_A m_A | \sum_{m} a_{nljm}^\dagger a_{nljm}|J_A m_A \rangle = n_{nlj}(A)$

The last equality mean the number of nucleon in orbital $nlj$ in nucleus A. Thus, we obtain

$\displaystyle \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) = 2j+1 - n_{nlj}(A)$

For B(p,d)A reaction, the nucleus B is in the ground state and we can sum all $J_A, m_A, m$,

$\displaystyle \sum_{J_A} \beta_{nlj}^2(B,A) = n_{nlj}(B)$

Thus, the adding A(d,p)B and removing A(p,d)C reaction from nucleus A, the sum of spectroscopic factors

$\displaystyle \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) + \sum_{J_C} \beta_{nlj}^2(A,C) = 2j+1$

## Sum rules of Clebsch-Gordon Coefficient

Since the CG coefficient is already normalized.Thus

$\displaystyle \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 1$

Since the number of $M$ is $2J+1$, as $M = -J, -J+1, ... J$. Thus,

$\displaystyle \sum_M \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 2J+1$

At last, the number of dimension of the coupled space or (tensor product space) is equation to $(2j_1 +1) (2j_2+1)$, i.e.

$\displaystyle \sum_J (2J+1) = (2j_1+1)(2j_2+1)$

Thus,

$\displaystyle \sum_{JM, m_1 m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = (2j_1+1)(2j_2+1)$

## Winger 6-j and 9-j symbol

The meaning of 3-j symbol is same as Clebsch-Gordan coefficient. So, we skip in here.

I am not going to construct the 6-j symbol from 3-j symbol. In here, I just state the meaning and usage in Mathematica.

The 6-j symbol is the coupling between 3 angular momenta, $j_1, j_2, j_3$.

There are 2 ways to couple these 3 angular momenta. First,

$j_1 + j_2 + j_3 \rightarrow j_{12} + j_3 \rightarrow J$

the other way is

$j_ 1 + j_2 + j_3 \rightarrow j_1 + j_{23} \rightarrow J$

The 6-j symbol is

$\begin{pmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{pmatrix}$

We can see that there are 4 vector-sum must satisfy.

$\Delta(j_1, j_2, j_{12})$

$\Delta(j_2, j_3, j_{23})$

$\Delta(j_1, j_{23}, J)$

$\Delta(j_{12}, j_3, J)$

If we draw a line to connect these 4 vector-sum, we have:

In Mathematica, there is a build in function

$\textrm{SixJSymbol}[ \left\{j_1, j_2, j_{23} \right\}, \left\{j_3, J , j_{23} \right\}]$

The 9-j symbol is the coupling between 4 angular momenta, $j_1, j_2, j_3, j_4$.

The 9-j symbol can be used in coupling 2 nucleons, $l_1, s_1, l_2, s_2$.

The 9-j symbol is

$\begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

We can see, each row and column must satisfy the vector-sum.

Unfortunately, there is no build in function in Mathematica. The formula for 9-j symbol is

$\displaystyle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix} \\ = \sum_{g} (-1)^{2g} (2g+1) \begin{pmatrix} l_1 & s_1 & j_1 \\ j_2 & J & g \end{pmatrix} \begin{pmatrix} j_2 & s_2 & j_2 \\ s_1 & g & S \end{pmatrix} \begin{pmatrix} L & S & J \\ g & l_1 & l_2 \end{pmatrix}$

Where $g$ sum all possible value, which can be calculate using the 6 couplings inside the 3 6-j symbols.To check your result, the coupling between $d_{5/2}$ and $f_{7/2}$ to from a $L = 5, S = 0, J = 0$ state is

$\begin{pmatrix} 2 & 1/2 & 5/2 \\ 3 & 1/2 & 7/2 \\ 5 & 0 & 5 \end{pmatrix} = \frac{1}{2\sqrt{770}}$

## Hartree method for Helium ground state

After long preparation, I am ready to do this problem.

The two electron in the helium ground state occupy same spacial orbital but difference spin. Thus, the total wavefunction is

$\displaystyle \Psi(x,y) = \frac{1}{\sqrt{2}}(\uparrow \downarrow - \downarrow \uparrow) \psi(x) \psi(y)$

Since the Coulomb potential is spin-independent, the Hartree-Fock method reduce to Hartree method. The Hartree operator is

$F(x) = H(x) + \langle \psi(y)|G(x,y) |\psi(y) \rangle$

where the single-particle Hamiltonian and mutual interaction are

$\displaystyle H(x) = -\frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 x} = -\frac{1}{2}\nabla^2 - \frac{Z}{x}$

$\displaystyle G(x,y) = \frac{e^2}{4\pi\epsilon_0|x-y|} = \frac{1}{|x-y|}$

In the last step, we use atomic unit, such that $\hbar = 1, m=1, e^2 = 4\pi\epsilon_0$. And the energy is in unit of Hartree, $1 \textrm{H} = 27.2114 \textrm{eV}$.

We are going to use Hydrogen-like orbital as a basis set.

$\displaystyle b_i(r) = R_{nl}(r)Y_{lm}(\Omega) \\= \sqrt{\frac{(n-l-1)!Z}{n^2(n+l)!}}e^{-\frac{Z}{n}r} \left( \frac{2Z}{n}r \right)^{l+1} L_{n-l-1}^{2l+1}\left( \frac{2Z}{n} r \right) \frac{1}{r} Y_{lm}(\Omega)$

I like the left the $1/r$, because in the integration $\int b^2 r^2 dr$, the $r^2$ can be cancelled. Also, the $i = nlm$ is a compact index of the orbital.

Using basis set expansion, we need to calculate the matrix elements of

$\displaystyle H_{ij}=\langle b_i(x) |H(x)|b_j(x)\rangle = -\delta \frac{Z^2}{2n^2}$

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle$

Now, we will concentrate on evaluate the mutual interaction integral.

Using the well-known expansion,

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{4\pi}{2l+1} \frac{r_<^l}{r_>^{l+1}} Y_{lm}^{*}(\Omega_1)Y_{lm}(\Omega_2)$

The angular integral

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle \\ = \big( \int Y_i^{*}(x) Y_{lm}^{*}(x) Y_j(x) dx \big) \big( \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy \big)$

where the integral $\int dx = \int_{0}^{\pi} \int_{0}^{2\pi} \sin(\theta_x) d\theta_x d\phi_x$.

From this post, the triplet integral of spherical harmonic is easy to compute.

$\displaystyle \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy = \sqrt{\frac{(2l+1)(2l_k+1)}{4\pi (2l_h+1)}} C_{l0l_k0}^{l_h0} C_{lm l_km_k}^{l_hm_h}$

The Clebsch-Gordon coefficient imposed a restriction on $l,m$.

$\displaystyle \langle R_i(x) R_h(y)| \frac{r_<^l}{r_>^{l+1}} | R_j(x) R_k(y) \rangle \\ = \int_0^{\infty} \int_{0}^{\infty} R_i(x) R_h(y) \frac{r_<^l}{r_>^{l+1}} R_j(x) R_k(y) y^2 x^2 dy dx \\ = \int_0^{\infty} R_i(x) R_j(x) \\ \left( \int_{0}^{x} R_h(y) R_k(y) \frac{y^l}{x^{l+1}} y^2dy + \int_{x}^{\infty} R_h(x)R_k(x) \frac{x^l}{y^{l+1}} y^2 dy \right) x^2 dx$

The algebraic calculation of the integral is complicated, but after the restriction of $l$ from the Clebsch-Gordon coefficient, only few terms need to be calculated.

The general consideration is done. now, we use the first 2 even states as a basis set.

$\displaystyle b_{1s}(r) = R_{10}(r)Y_{00}(\Omega) = 2Z^{3/2}e^{-Zr}Y_{00}(\Omega)$

$\displaystyle b_{2s}(r) = R_{20}(r)Y_{00}(\Omega) = \frac{1}{\sqrt{8}}Z^{3/2}(2-Zr)e^{-Zr/2}Y_{00}(\Omega)$

These are both s-state orbital. Thus, the Clebsch-Gordon coefficient

$\displaystyle C_{lm l_k m_k}^{l_h m_h} = C_{lm00}^{00}$

The radial sum only has 1 term. And the mutual interaction becomes

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = 4\pi \frac{1}{r_>} Y_{00}^{*}(\Omega_1)Y_{00}(\Omega_2)$

The angular part

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle = \frac{1}{4\pi}$

Thus, the mutual interaction energy is

$G_{ij}^{hk} = \displaystyle \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle = \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle$

$G_{ij}^{hk} = \displaystyle \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle \\ \begin{pmatrix} G_{11}^{hk} & G_{12}^{hk} \\ G_{21}^{hk} & G_{22}^{hk} \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} G_{11}^{11} & G_{11}^{12} \\ G_{11}^{21} & G_{11}^{22} \end{pmatrix} & \begin{pmatrix} G_{12}^{11} & G_{12}^{12} \\ G_{12}^{21} & G_{12}^{22} \end{pmatrix} \\ \begin{pmatrix} G_{21}^{11} & G_{21}^{12} \\ G_{21}^{21} & G_{21}^{22} \end{pmatrix} & \begin{pmatrix} G_{22}^{11} & G_{22}^{12} \\ G_{22}^{21} & G_{22}^{22} \end{pmatrix} \end{pmatrix} \\= \begin{pmatrix} \begin{pmatrix} 1.25 & 0.17871 \\ 0.17871 & 0.419753 \end{pmatrix} & \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0439857 & 0.0171633 \end{pmatrix} \\ \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0438957 & 0.0171633 \end{pmatrix} & \begin{pmatrix} 0.419753 & 0.0171633 \\ 0.0171633 & 0.300781 \end{pmatrix} \end{pmatrix}$

We can easy to see that $G_{ij}^{hk} = G_{ji}^{hk} = G_{ij}^{kh} = G_{hk}^{ij} = G_{ji}^{kh}$. Thus, if we flatten the matrix of matrix, it is Hermitian, or symmetric.

Now, we can start doing the Hartree method.

The general solution of the wave function is

$\psi(x) = a_1 b_{1s}(x) + a_2 b_{2s}(x)$

The Hartree matrix is

$F_{ij} = H_{ij} + \sum_{h,k} a_h a_k G_{ij}^{hk}$

The first trial wave function are the Hydrogen-like orbital,

$\psi^{(0)}(x) = b_{1s}(r)$

$F_{ij}^{(0)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.25 & 0.17871 \\ 0.17817 & 0.419753 \end{pmatrix}$

Solve for eigen system, we have the energy after 1st trial,

$\epsilon^{(1)} = -0.794702 , (a_1^{(1)}, a_2^{(1)}) = (-0.970112, 0.242659)$

After 13th trial,

$\epsilon^{(13)} = -0.880049 , (a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$

$F_{ij}^{(13)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.15078 & 0.155932 \\ 0.155932 & 0.408748 \end{pmatrix}$

Thus, the mixing of the 2s state is only 3.7%.

Since the eigen energy contains the 1-body energy and 2-body energy. So, the total energy for 2 electrons is

$E_2 = 2 * \epsilon^{(13)} - G = -2.82364 \textrm{H} = -76.835 \textrm{eV}$

In which ,

$G = \langle \psi(x) \psi(y) |G(x,y) |\psi(x) \psi(y) \rangle = 1.06354 \textrm{H} = 28.9403 \textrm{eV}$

So the energies for

From He to He++.  $E_2 = -2.82364 \textrm{H} = -76.835 \textrm{eV}$
From He+ to He++, $E_1^+ = -Z^2/2 = 2 \textrm{H} = -54.422 \textrm{eV}$.
From He to He+, is $E_1 = E_2 - E_1^+ = -0.823635 \textrm{H} = -22.4123 \textrm{eV}$

The experimental 1 electron ionization energy for Helium atom is

$E_1(exp) = -0.90357 \textrm{H} = -24.587 \textrm{eV}$
$E_1^+(exp) = -1.99982 \textrm{H} = -54.418 \textrm{eV}$
$E_2(exp) = -2.90339 \textrm{H} = -79.005 \textrm{eV}$

The difference with experimental value is 2.175 eV. The following plot shows the Coulomb potential, the screening due to the existence of the other electron, the resultant mean field, the energy, and $r \psi(x)$

Usually, the Hartree method will under estimate the energy, because it neglected the correlation, for example, pairing and spin dependence. In our calculation, the $E_2$ energy is under estimated.

From the $(a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$, we can see, the mutual interaction between 1s and 2s state is attractive. While the interaction between 1s-1s and 2s-2s states are repulsive. The repulsive can be easily understood. But I am not sure how to explain the attractive between 1s-2s state.

Since the mass correction and the fine structure correction is in order of $10^{-3} \textrm{eV}$, so the missing 0.2 eV should be due to something else, for example, the incomplete basis set.

If the basis set only contain the 1s orbit, the mutual interaction is 1.25 Hartree = 34.014 eV. Thus, the mixing reduce the interaction by 5.07 eV, just for 3.7% mixing

I included the 3s state,

$\epsilon^{(13)} = -0.888475 , (a_1^{(13)}, a_2^{(13)}, a_3^{(13)}) = (0.981096, -0.181995, -0.06579)$

The mutual energy is further reduced to 1.05415 Hartree = 28.6848 eV. The $E_2 = -77.038 \textrm{eV}$. If 4s orbital included, the $E_2 = -77.1058 \textrm{eV}$. We can expect, if more orbital in included, the $E_2$ will approach to $E_2(exp)$.

## 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 $2n+1$.

## 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]]]