Reduction of Matrix

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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)

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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)

multiple the adjoint

\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

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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}