## Reduction of Matrix

In previous post, the number of combination between the core $\Psi_{J_A}$ and the single particle state $phi_{nlj}$ becomes massive as the number of m-states are huge. To simplify the coupling, assuming the interaction between the state $\Psi_{J_AM_A}\phi_{nljm}$ and $\Psi_{J_A'M_A'}\phi_{n'l'j'm'}$ are the same for all $M_A, m, M'_A, m'$. The Hamiltonian matrix can be reduced.

To demonstrate the idea, suppose the core only has a ground state. And the single-particle state has degenerated state with different m-state, say, $j = 0,1$

Lets the coupled states be

$\displaystyle \begin{pmatrix}\Psi_{00}\phi_{00} \\ \Psi_{00}\phi_{1,-1} \\ \Psi_{00}\phi_{1,0} \\ \Psi_{00}\phi_{1,1} \end{pmatrix}$

The Hamiltonian matrix is

$\displaystyle H = \begin{pmatrix} E_1 & V & V & V \\ V & E_2 & 0 & 0 \\ V & 0 & E_2 & 0 \\ V & 0 & 0 & E2 \end{pmatrix} \Rightarrow \begin{pmatrix} E_1 & \sqrt{3}V \\ \sqrt{3}V & E_2 \end{pmatrix}$

To give a concrete example, say $E_1 = 0, E_2 = 1, V= 1$

$\displaystyle H = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}$

The eigenstates and eigenvalues are

$\displaystyle e_1 = 2.3028, v_1 = ( 0.6011, 0.4614, 0.4614, 0.4614)$

$\displaystyle e_2 = -1.3028, v_2 = ( -0.7992, 0.3470, 0.3470, 0.3470)$

$\displaystyle e_3 = 1.000, v_3 = ( 0.0000, 0.4082, 0.4082, 0.8165)$

$\displaystyle e_4 = 1.000, v_4 = ( 0.0000, 0.7071, 0.7071, 0.0000)$

We can see, there are 2 states the energies do not change and the coefficient just re-configured.

Let’s look at the eigen system of this matrix:

$\displaystyle H = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$

$\displaystyle e_1 = 2.3028, v_1 = ( 0.6011, 0.7992)$

$\displaystyle e_2 = -1.3028, v_2 = ( -0.7992, 0.6011)$

I wish to prove it more general as a theorem, but I can just worked on few cases.

## 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

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

## Expressions of Total Nuclear Hamiltonian

In the past, I always write the total Hamiltonian of N+1 system as

$\displaystyle H = H_N + H_1 + V_{N1}$

The 3 terms on the right are the Hamiltonian of the N-nucleon core, the Hamiltonian of the single-nucleon, and the interaction between the core and the single particle. And the wave function can be written as

$\displaystyle \chi_k = \sum \beta_{ij} \Psi_i \phi_j$

with

$\displaystyle H_N \Psi_i = \epsilon^{N}_i \Psi_i, H_1 \phi_j = \epsilon^{1}_j \phi_j$

so far so good.

Breaking down the above Hamiltonian,

$\displaystyle H = \sum_{i}^{N+1} K_i + \sum_{i\neq j}^{N+1} V_{ij}$

$\displaystyle H_N = \sum_{i}^{N} K_i + \sum_{i\neq j}^{N} V_{ij}$

then

$\displaystyle H = H_N + K_1 + \sum_{i}^{N} V_{i1}$

Thus,

$\displaystyle H_1 = K_1 + \sum_{i}^{N} V_{i1} - V_{N1}$

Thus, we can see, the term $V_{N1}$ is NOT really the sum of all interaction between the single-nucleon and the nucleons in the core. What is it? or What is $H_1$?

Ideally, in the single particle picture (or independent particle model) , we imagine the single-nucleon is bounded by an mean field created by the core, i.e.

$\displaystyle H_{N1} = H_N + K + \bar{V} = H_N + H_1$

$\displaystyle H_{N1} \Psi_i \phi_j = \epsilon^{N1}_{ij} \Psi_i \phi_j$

Adding a 2-body residual interaction, we restore the total Hamiltonian,

$H = H_N + H_1 + R, R = V_{1N}$.

Thus, the $V_{1N}$ is the residual interaction between the single-nucleon and the core. This residual interaction, is created after the mean field!

$\displaystyle H = H_N + K_1 + \bar{V} + \sum_{i}^{N} V_{i1} - \bar{V} = H_N + H_1 + V_{1N}$

$\displaystyle H_1 = K_1 + \bar{V}$

$\displaystyle V_{1N} = \sum_{i}^{N} V_{i1} - \bar{V}$