## Kinematics of Transfer Reaction

Another surprise that no post on the kinematics of transfer reaction!

Suppose the reaction is $A(B, 1)2$, where $A$ is the incident particle, $B$ is the target nucleus that is at rest, $1, 2$ are the outgoing particle.

The Q-value for the reaction is the difference between initial mass and final mass

$Q = m_A+ m_B - m_1-m_2$

When the incident energy is too low, lower then the Q-value (we will see the relationship later) , the reaction is impossible, because the nuclei-1 and 2 cannot be created.

The equation for the reaction is

$P_A + P_B = P_1 + P_2$

where $P_i$ are 4-momenta. The number of freedom is 2. The total unknown is 8 from $P_1, P_2$, the masses of particle-1, and -2 give 2 constrains, the equations give 4 constrains.

$P_A = ( E_A , 0, 0, k_A) = ( m_A + T_A, 0, 0, \sqrt{2 m_A T_A + T_A^2})\\ P_B = (m_B, 0, 0, 0)$

where $T_A$ is the incident energy. We first transform the system into center of momentum (CM) frame. Since the total momentum is zero in the CM frame, we construct a CM 4-momentum

$P_{cm} = (P_A+P_B)/2 = ( m_A+ m_B + T_A, 0, 0, k_A)$

The Lorentz beta is $\beta = k_A/(m_A+m_B+T_A)$.

The invariance mass of the CM 4-momentum is unchanged, which is equal to the total energy of the particles in the CM frame. We can check

$P_A' = ( \gamma E_A - \gamma \beta k_A, 0, 0, -\gamma \beta E_A + \gamma k_A) \\ P_B' = ( \gamma m_B , 0, 0, -\gamma \beta m_B )$

The total energy in CM frame is

$E_{tot} = \gamma( m_A+m_B + T_A) - \gamma \beta k_A = \sqrt{(m_A+m_B+T_A)^2 - k_A^2}$

After the reaction, the momenta of the particle-1 and -2 are the same an opposite direction in the CM frame.Balancing the energy and momentum, the momentum is

$\displaystyle k = \frac{1}{2 E_{tot}} \sqrt{(E_{tot}^2 - (m_1+m_2)^2)(E_{tot}^2 - (m_1-m_2)^2)}$

In the equation, if either of the bracket is negative, the momentum cannot be formed. The smaller term is

$E_{tot}^2 - (m_1+m_2)^2 = (m_A+m_B+T_A)^2 - k_A^2 - (m_1+m_2)^2 \\ =(m_A+m_B+T_A)^2 - k_A^2 - (m_A+m_B-Q)^2 \\ = 2m_BT_A + 2(m_A + m_B)Q - Q^2$

If we set the line for this quantity be zero,

$latex 2m_BT_A + 2(m_A + m_B)Q – Q^2 = 0 \\ T_A = \frac{Q^2 – 2(m_A+m_B)}{2m_B} \\ = \frac{1}{2m_B} ( (Q-m_A-m_B)^2 – (m_A+m_B)^2) \\ = \frac{1}{2m_B} ( (m_1+m_2)^2 – (m_A+m_B)^2)$

Thus, if the $T_A$ is smaller then that value, the reaction is impossible.

The 4-momenta of the outgoing particles can be constructed.

$P_1' = ( \sqrt{m_1^2 - k^2} , k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , k \cos(\theta)) \\ P_2' = ( \sqrt{m_2^2 - k^2} , -k \cos(\phi) \sin(\theta) ,- k \sin(\phi) \sin(\theta) , -k \cos(\theta))$

Back to Lab frame,

$P_1 = ( \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta), k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta))$

For simplicity, set $\phi = 0$,

$P_1 = ( \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta), k \cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta))$

We can see, the locus in the momentum space is

$( k cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta))$

This is an ellipse with length $k (1, \gamma)$, and centered at $(0, \gamma \beta \sqrt{m_1^2 + k^2})$. If the relativistic effect is small, it is a circle.

Here is an example,

## Finite Spherical Square Well with spin-orbital potential

Last time, we studied the finite spherical square well and calculated the energy levels for difference angular momentum. In that case, the good quantum numbers (quantities that is commute with the Hamiltonian, i.e. conserved, invariance with time) are the principle quantum number $n$, which dictated the number of node and the angular momentum $l$, and the spin state $s$. A eigen state can be denoted as $|nl m_l s m_s\rangle$. The degeneracy for each state is $2(2l+1)$.

With the spin-orbital potential

$\displaystyle V_{so} = a L \cdot S$

The angular momentum $m_l$ and spin $m_s$ are not good quantum numbers. The sum, total angular momentum $j = l + s$ and it z-component $m_j$ are. The new eigen state is $|n j m_j l s\rangle$.

The Hamiltonian inside the well is

$\displaystyle H = H_0 + V_{so} = -\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{d}{dr}r^2\frac{d}{dr} - |V_0| + \frac{\hbar^2}{2m} \frac{1}{r^2}L^2 + a L\cdot S$

The energy for $H_0$ dose not change, as the $L^2$ is still a good quantum number. The spin-orbital coupling is evaluated like

$\displaystyle L\cdot S = \frac{1}{2}(J^2 - L^2 - S^2)$

then,

$\displaystyle \langle n j m_j l s| L\cdot S |n j m_j l s\rangle = \frac{1}{2} ( j(j+1) - l(l+1) - s(s-1) )$

Since the spin has only 2 spin states, thus,

$\displaystyle \langle L\cdot S \rangle \\ = \frac{1}{2} l , j = l+\frac{1}{2} \\ = -\frac{1}{2}(l+1), j = l - \frac{1}{2}$

Note that, the mean energy of the L-states are unchanged. The degeneracy for $j_+ = l+1/2$ is $2(l+1)$, for $j_- = l-1/2$ is $2l$. Also, the s-states are not affected, because $l = 0$.

Thus, the wave vector inside the well $k$ becomes

$\displaystyle k = \sqrt{\frac{2m}{\hbar^2 }(|V_0|- \beta \langle L\cdot S\rangle) - |E|)}$

the wave vector outside the well $k'$

$\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2 }(\beta \langle L\cdot S\rangle) + |E|)}$

And the rest is matching the boundary condition.

We can also add one more term, an additional $\gamma L^2$ term so that the centroid of the L-state is shifted. The energy for this potential is $\gamma l(l+1)$.

Here is the energy levels for $\beta = -1$ and $\gamma = -0.5$.

We can see, we reproduced the shell ordering 0s1/2, 0p3/2, 0p1/2, 0d5/2, 1s1/2, 0d3/2, 0f7/2, 1p3/2, 1p1/2, 0f5/2….

The magic number 2, 8, 20, and 40 are reproduced. May be, if I adjust the strength of the spin-orbit coupling, my get the magic number 28 and 50. Since the energy level of the s-states are fixed, I need to adjust the shift of the centroid of the others, and the splitting.

Here is the result of $\beta = -1.2, \gamma = -0.2$

It seems that the magic number of 28, which is from the isolation of 0f7/2, is recreated. And the isolation of 0g9/2, created the magic number of 50.

## 1-D Harmonic Oscillator

It is quite surprised that This is not here!

The Hamiltonian is

$\displaystyle H=\frac{P^2}{2m} + \frac{m \omega^2}{2} X^2$

Lets define creation operator $A^\dagger$ and destruction operator $A$, or the ladder operators, such that,

$\displaystyle A^\dagger =\sqrt{\frac{1}{2m}} P + i \sqrt{\frac{m \omega^2}{2}} X$

$\displaystyle A =\sqrt{\frac{1}{2m}} P - i \sqrt{\frac{m \omega^2}{2}} X$

Then we can see

$\displaystyle H = A^\dagger A + \frac{\hbar \omega}{2} = A A^\dagger - \frac{\hbar \omega}{2}$

and the commutative relations

$\displaystyle A^\dagger A - A A^\dagger = -\hbar \omega$

$\displaystyle HA^\dagger - A^\dagger H = \hbar \omega A^\dagger$

$\displaystyle HA - A H = -\hbar \omega A$

From the last two equations, we can see

$\displaystyle H |n\rangle = E_n |n\rangle \\ A H|n\rangle = (HA+\hbar \omega A) |n\rangle = E_n A|n\rangle \\ HA|n\rangle = (E_n-\hbar \omega) A|n\rangle$

This shows that, if $|n\rangle$ is a state with energy $E_n$, then $A|n\rangle$ is also a state with energy $E_n - \hbar \omega$. Similar for $A^\dagger |n\rangle$, the energy will be $E_n + \hbar \omega$.

Suppose the ground state is $|0\rangle$, which is is lowest energy state. If we apply the destruction operator, it will be gone, i.e.

$\displaystyle A|0 \rangle = 0 \\ A^\dagger A|0\rangle = 0 \\ \left(H- \frac{\hbar\omega}{2}\right)|0\rangle \\ H|0\rangle = \frac{\hbar\omega}{2}|0\rangle = E_0 |0\rangle$

Thus, the ground state energy is $\hbar \omega /2$!

We can donate

$\displaystyle A^\dagger |n\rangle = U_n |n+1\rangle \\ A |n\rangle = D_n |n-1\rangle$

Using $\displaystyle A^\dagger A - A A^\dagger = -\hbar \omega$, we have

$\displaystyle (A^\dagger A - A A^\dagger ) |n \rangle \\ = A^\dagger D_n |n-1 \rangle - A U_n|n+1 \rangle \\ = (U_{n-1}D_n- U_n D_{n+1} )|n\rangle = - \hbar \omega |n\rangle$

or

$\displaystyle U_n D_{n+1} = U_{n-1}D_n + \hbar \omega$

Consider above equation on $|0\rangle$, such that $A|0\rangle = 0$, thus,

$\displaystyle (A^\dagger A - A A^\dagger ) |0 \rangle = - U_0 D_{1} )|0\rangle = - \hbar \omega |0\rangle$

Thus, $U_0 D_1 = \hbar \omega$, then, in general, we have

$\displaystyle U_n D_{n+1} = (n+1) \hbar \omega$

Then

$\displaystyle A^\dagger A |n\rangle = U_{n-1} D_n |n\rangle \\ A A^\dagger |n\rangle = U_{n} D_{n+1} |n\rangle$

The simplest solution is

$U_n = \sqrt{\hbar \omega}\sqrt{n+1} \\ D_n =\sqrt{\hbar \omega}\sqrt{n}$

And, if we treat the state $|n\rangle$ contains $n$ oscillators, we have a number operator $N = A^\dagger A$, such that

$A^\dagger A |n \rangle = n \hbar \omega |n \rangle$

some people will normalized the ladder operators with $\sqrt{\hbar \omega}$. For me, I don’t want to make the starting point be so “artificial”. We can summarized the 1-D Harmonic oscillator as

Now, we are going to solve the wave function. The equation is

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \phi = E \phi = \frac{2n+1}{2}\hbar \omega \phi$

$\displaystyle \frac{d^2\phi}{dx^2} - \frac{m^2 \omega^2}{\hbar^2} x^2 \phi = (2n+1)\frac{m \omega}{\hbar} \phi$

Define $\alpha = \sqrt{\frac{\hbar}{m \omega}}$, the dimension of $\alpha$ is length. Set $u = x/ \alpha$, then

$\displaystyle \frac{d^2\phi}{du^2} + ((2n+1) - u^2) \phi = 0$

Since we know that the function must decay outside the well, set $\phi = f(y) \exp(-y^2/2)$, then

$\displaystyle \frac{d^2f}{du^2} -2 u \frac{df}{du} +2 nf = 0$

The solution is Hermite polynomial $H_n(u)$ of order $n$.

Assume

$\displaystyle H_n(x) = \sum_{i = 0}^{\infty} a_i x^i$

sub into the Hermite equation,

$\displaystyle a_{i+2} = \frac{2(i-n)}{(i+1)(i+2)} a_i$

First,  the even and odd series are separated. (I don’t know why, ) the Hermite polynomial has either even terms or odd terms, but not mixed. When $i = n$, the series terminated.

When $n = 0$, $a_2 = 0$ and all later even terms are zero. The equation is $f''(x) - 2 x f'(x) = 0$, which only support even function. Thus, $H_0(x) = a_0$ is a constant. Thus, we can see, when $n$ is even (odd), Hermite polynomial is even (odd) with only even (odd) $n$.

Hermite polynomial are orthogonal with $\exp(-x^2)$, thus, the wave function $\phi_n(x) = H_n(x) \exp(-x^2/2)$ are orthogonal.

## Finite Spherical Square Well

Hello, everyone, in order to calculate deuteron by Hartree-Fock method, I need a basis. The basis of infinite spherical square well is too “rigid”, that it has to “extension” to non-classical region. Beside of the basis of Wood-Saxon potential. The finite spherical square well is a good alternative. The radial equation is basically the same as finite spherical square well.

The potential is

$\displaystyle V(r) = -|V_0|, r\leq a$

$\displaystyle V(r) = 0, r > a$

Within the well, the wave vector is

$\displaystyle k = \sqrt{\frac{2m}{\hbar^2} (|V_0 |-|E|) }$

, outside the well, the wave vector is

$\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2} |E| }$

The solution is spherical Bessel function. Since the Bessel function of the first and second kind are oscillating like sin or cosine function. To form a decay function when $r > a$, we need the Hankel function with complex argument.

$\displaystyle h_n( i \kappa r) = h_n(x) = - (i x)^n \left( \frac{1}{x} \frac{d}{dx}\right)^n \left(\frac{\exp{(-x)}}{x} \right)$

To make it real, we need a factor $i^n$.

The boundary conditions are continuity and differential continuity.

$\displaystyle j_l(ka) = A i^l h_l(i\kappa a)$

$\displaystyle \left(\frac{d}{dr}j_l(kr)\right)_{r=a} = A i^l \left(\frac{d}{dr}h_l(i\kappa a) \right)_{r=a}$

These two conditions solved for two parameters $A$ and $E$. However, I cannot find an analytical solution to the energy $E$.

If the potential depth is 60 MeV for proton. Radius is 1 fm for a light nuclei. Set $\hbar = 1, m = 1$.

The result is follow,

The 1st column is 1s, 1p, 1d, 1f, and 1g. The 2nd column is 2s, 2p, 2d, 2f, and 2g. The 3rd column is 3s, 3p, and 3d.

By compare with infinite square well,

The energies are lower in finite well, because the wave functions can spread-out to non-classical region, so that the wave length is longer and energy is lower.

In this example, the 3d and 2g orbits are bounded (of course, all orbits in infinite well are bound). This is not because of the depth of the well, but the boundary of the well. In other word, to bring down an orbit, the wave function has to spread out, that is connected with the neutron-halo.

## Wave function in momentum space

The wave function often calculated in spatial coordinate. However, in experimental point of view, the momentum distribution can be extracted directly from the experimental data.

The conversion between momentum space and position space is the Fourier transform

$\displaystyle \phi(\vec{k}) = \frac{1}{\sqrt{2\pi}^3} \int Exp\left(-i \vec{k}\cdot \vec{r} \right) \phi(\vec{r}) d\vec{r}$

Using the plane wave expansion

$\displaystyle Exp(i k\cdot r) = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l(\hat{k}\cdot\hat{r})$

or

$\displaystyle Exp(i k\cdot r) = 4\pi \sum_{l=0}^\infty \sum_{m=-l}^{l} i^l j_l(kr) Y_{lm}(\Omega_k) Y_{lm}^{*}(\Omega_r)$

Thus,

$\displaystyle \phi(\vec{k}) = \frac{4\pi}{\sqrt{2\pi}^3} \sum_{l=0}^\infty (-i)^l \sum_{m=-l}^{l} \int j_l(k r) Y_{lm}(\Omega_k) Y_{lm}^*(\Omega) \phi(\vec{r}) r^2 dr d\Omega$

where $j_l (x)$ is spherical Bessel function. Usually, due to conservation of angular momentum, the angular part can be separated from the spatial part. Let assume the wave function in position space is

$\phi(\vec{r}) = \psi(r) Y_{l_r m_r}(\Omega)$

$\phi(\vec{k}) = \psi(p) Y_{l_k m_k}(\Omega_k)$

Then we have

$\displaystyle \psi(k) = \frac{4\pi}{\sqrt{(2\pi)^3}} \int j_l(k r) \psi(r) r^2 dr$

$\displaystyle \phi(\vec{k}) = \psi(k) (-i)^l Y_{lm}(\Omega_k)$

where $l = l_r = l_k, m=m_r = m_k$, due to the orthogonality of spherical harmonics.

For s, p, d, f-state, the spherical Bessel function is

$\displaystyle j_0(x) = \frac{\sin(x)}{x}$

$\displaystyle j_1(x) = \frac{\sin(x)}{x^2} - \frac{\cos(x)}{x}$

$\displaystyle j_2(x) = \sin(x)\frac{2-x^2}{x^3} - \cos(x)\frac{3}{x^2}$

$\displaystyle j_3(x) = \sin(x)\frac{15-6x^2}{x^4} - \cos(x)\frac{15-x^2}{x^3}$

For Hydrogen-like wave function, the non-normalized momentum distribution is

$\displaystyle \psi_{10}(k) = \frac{4 Z^{5/2}}{(k^2 + Z^2)^2}$

$\displaystyle \psi_{20}(k) = \frac{16 \sqrt{2}Z^{5/2}(4k^2-Z^2)}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{21}(k) =\sqrt{\frac{2}{3}}\frac{64 k Z^{7/2}}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{30}(k) = \frac{36 \sqrt{3} Z^{5/2} (81k^3-30k^2Z^2+Z^4)}{(9k^2 + Z^2)^4}$

$\displaystyle \psi_{31}(k) = \frac{144 \sqrt{6} Z^{7/2} (9k^3-kZ^2)}{(9k^2 + Z^2)^4}$

Here is the plot for momentum distribution $(\psi_{nl}(k))^2 k^2$.

It is interesting that, the number of node decrease with higher angular momentum. But be-aware that it is only in atomic case, not a universal.

The higher the principle quantum number, the smaller of the spread of momentum. This is because, the spread of position wave function getting larger, and the uncertainty in momentum space will be smaller. This is a universal principle.

We also plot the Hydrogen radial function in here $\psi(r)^2 x^2$, for reference,

## Hartree-Fock for Helium excited state II

This times, we will show the energy level diagram for helium atom. We already calculated the ground state, the 1s2s singlet and triplet excited states. We will calculate higher excited states, for example, 1s3s, 1s2p or 1s3p, etc, and included in the diagram.

The most difficult part is the evaluation of the mutual interaction matrix element

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

The angular integral was evaluated in the last post. And the radial integral is done using Mathematica.

I will use hydrogen 1s, 2s, 2p, 3s, 3p, 4s, and 4p states for basis. Thus, the diagram will contain some of the excited states from some possible combination. The 1s4s and 1s4p state will not be calculated, because there is no room for mixing with higher excited states but only for lower excited states. This will make the eigen state be unbound.

During the calculation, one of the tricky point is the identify of the n-th s-state. It is because the mixing is always there, and the mixing can be very large. In order to determine the principle quantum number, we have to check the wave function can see how many peaks in there.

Here is the energy level diagram obtained by Hartree-Fock method using limited hydrogen wave functions as basis set.

## Evaluation of mutual interaction

We are going to evaluation the integral

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

Recalling the multi-pole expansion,

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

and the basis

$b_{nlm}(\vec{r}) = R_{nl}(r) Y_{lm}(\Omega)$

Set an averaged basis

$\displaystyle b_{nl}(\vec{r}) = R_{nl}(r) \frac{1}{\sqrt{2l+1}} \sum_{m=-1}^{l}Y_{lm}(\Omega)$

$\displaystyle \Gamma_{ij}^{hk}(l) = \frac{4 \pi}{2l+1} \sum_{m=-l}^{l} \frac{1}{\sqrt{(2l_i+1)(2l_j+1)(2l_h+1)(2l_k+1)}} \\ \sum_{m_i,m_j}\int Y_{l_i m_i}^{*}(\Omega_1) Y_{l m}^{*}(\Omega_1) Y_{l_j m_j}(\Omega_1) d \Omega_1 \\ \sum_{m_h,m_k} \int Y_{l_h m_k}^{*}(\Omega_2) Y_{l m}(\Omega_2) Y_{l_k m_k}(\Omega_2) d \Omega_2$

In the angular integrals, using Wigner 3-j symbol and the integral

$\displaystyle\int Y_{l_1 m_1}(\Omega) Y_{l m}(\Omega) Y_{l_2 m_2}(\Omega) d \Omega \\= \frac{\sqrt{(2l_1+1)(2l+1)(2l_2+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_1 & l & l_2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix}$

$Y_{lm}^{*}(\Omega) = (-1)^m Y_{l(-m)}(\Omega)$

$\displaystyle \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix} = (-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l & l_2 \\ -m_1 & -m & -m_2 \end{pmatrix} \\ =(-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l_2 & l \\ m_1 & m_2 & m \end{pmatrix}$

we have

$\displaystyle\int Y_{l_i m_i}^{*}(\Omega) Y_{l m}^{*}(\Omega) Y_{l_j m_j}(\Omega) d \Omega \\= (-1)^{m_j} \frac{\sqrt{(2l_i+1)(2l+1)(2l_j+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_i & l & l_j \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l & l_j \\ m_i & m & -m_j \end{pmatrix}$

$\displaystyle\int Y_{l_h m_h}^{*}(\Omega) Y_{l m}(\Omega) Y_{l_k m_k}(\Omega) d \Omega \\= (-1)^{m_h} \frac{\sqrt{(2l_h+1)(2l+1)(2l_k+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_h & l & l_k \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l & l_k \\ -m_h & m & m_k \end{pmatrix}$

Thus,

$\displaystyle \Gamma_{ij}^{hk}(l) = \sum_{m=-l}^{l} \sum_{m_i,m_j } (-1)^{m_j} \begin{pmatrix} l_i & l_j & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l_j & l \\ m_i & -m_j & m \end{pmatrix} \\ \sum_{m_h,m_k} (-1)^{m_h} \begin{pmatrix} l_h & l_k & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l_k & l \\ -m_h & m_k & m \end{pmatrix}$

The 3-j symbol restricted that

$l_i+l_j+l = even$

$l_h+l_k+l = even$

$m_i-m_j + m = 0, |m_i| \leq l_i , |m_j| \leq l_j$

$-m_h + m_k +m = 0, |m_h| \leq l_h , |m_k| \leq l_k$

I guess it is the most simplified formula for the angular part.

The total integral is

$\displaystyle G_{ij}^{hk} = \sum_{l=0}^\infty \Gamma_{ij}^{hk}(l) \langle R_i(x) R_h(y) | \frac{r_<^l}{r_>^{l+1}} |R_j(x) R_k(y) \rangle$

The angular integral imposes condition for $l$.

I am not sure this is a correct way to treat the problem.

First, the averaged basis is still an energy eigen state. It is not the eigen state for the angular part. So, this averaging could introduces an error and we should reminded that this is an approximation. But in the perturbation view point, this averaged basis is still valid.

Second thing is, the sum

$\displaystyle P_{(l_i m_i) (l_j m_j)}^{(l_h m_h)(l_k m_k)}(l) = \sum_{m=-l}^{l} \langle Y_{l_i m_i}|Y_{lm}^*|Y_{l_j m_j}\rangle \langle Y_{l_h m_h}|Y_{lm}|Y_{l_k m_k}\rangle$

is not symmetry for exchange of $i, j, h,k$ in general. For example,

$\displaystyle \frac{-1}{3}=P_{(1-1) (00)}^{(11)(00)}(1) {\neq} P_{(00)(1-1)}^{(11)(00)}(1) = 0$

This is a very uprising result that the mutual interaction dependent on the magnetic quantum number. Thus, in detail, we should use $n l m m_s$ as a basis.

Third, the sum $P_{ij}^{hk}(l)$ is depend on $l$. The mutual interaction require us to sum all possible $l$.

Fourth, the coupling between 1s2p triplet state, the total spin is $S = 1$, total L is $L = 1$, and the total angular momentum can be  $J = 0, 1, 2$. In our treatment, we did not coupled the angular momentum in the calculation explicitly. In fact, in the integral of the spherical harmonic, the coordinate are integrated separately, and the coupling seem to be calculated implicitly. I am not sure how to couple two spherical harmonics with two coordinates.