From Mean field calculation to Independent particle model

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The independent particle model (IPM) plays a fundamental role in nuclear model. The total Hamiltonian of a nucleus is:

H = \sum{\frac{P_i^2}{2m_i}} + \sum_{i,j}{V_{ij}}

The mean field is constructed by Hartree-Fork method, so that, the total Hamiltonian,

H = \sum{h_i} + \sum_{i,j}{V_{ij} - U_i}, h_i = \frac{P_i^2}{2m_i} + U_i

The second term is called residual interaction. The residual should be as small as possible. Under the mean filed, each nucleon can be treated as independent particle model that experience by h_i, so that,

h_i \left|\phi_i\right> = \left|\phi_i\right> \epsilon_i,

where \epsilon_i is the single particle energy, and \left|\phi_i\right> is the single particle wave function.


Hartree-Fock method

Start with a trial single particle function \phi_i, construct a first trial total wavefuction \Psi_0

\Psi_0=\frac{1}{\sqrt{A!}} \left| \begin{array}{ccc} \phi_1(r_1) & \phi_1(r_2) & ... \\ \phi_2(r_1) & \phi_2(r_2) & ... \\ ... & ... & \phi_A(r_A) \end{array} \right|

where A is the number of particle. Using variation method,

\delta \left<\Psi_0|H|\Psi_o\right> = 0 \Rightarrow \left<\delta \Psi_0|H|\Psi_0\right>

H = \sum\limits_{i} h_i + \sum\limits_{i\neq j} V_{ij}

we don’t need to variate the ket and bar, since they are related.

The variation can be made on

  1. \phi_i or
  2. the particle-hole excitation.

For the particle-hole excitation,

\left|\delta \Psi_0\right> = \sum \eta_{kt} \left|\Psi_{kt}\right>

\left|\Psi_{kt}\right>=\left|1,2,...,t-1, t+1,...,A,k\right>

Then, the variation of the energy becomes,

\sum \eta_{kt} \left< \Psi_{kt}|H| \Psi_0 \right> =0

\left<\Psi_{kt}|H|\Psi_0\right> = \left<k|\sum h |t\right>+ \sum\limits_{r} \left< kr | V_{ij}| rt \right>=0

we used the one-body expression for the two-body interaction,

\sum \limits_{ij} V_{ij} = \sum \limits_{\alpha \beta \gamma \delta} \left|\alpha \beta\right> V_{\alpha \beta \gamma \delta} \left<\gamma \delta\right|

V_{\alpha \beta \gamma \delta} = \left<\alpha \beta |V_ij |\gamma \delta\right>

Thus, we take out \left<k\right| and \left|t\right>, we get the Hartree-Fock single particle Hamiltonian,

h_{HF} = h + \sum \limits_{r} \left< r | V_{ij} | r \right>

using this new single particle Hamiltonian, we have a better single particle wavefunction

h_{HF} \phi_i^1 = \epsilon_i^1 \phi_1^1

U = \sum \limits_{i} h_{HF}

with this new wavefunction, the process start again until convergence. After that, we will get a consistence mean field (in the sense that the wavefunction and the potential are consistence), the single particle energy and the total binding energy.

I am not sure how and why this process can minimized the mean field U

 

 

 

 

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Very short introduction to Partial-wave expansion of scattering wave function

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In a scattering problem, the main objective is solving the Schrödinger equation

H\psi=(K+V)\psi=E\psi

where H is the total Hamiltonian of the scattering system in the center of momentum, K is the kinetic energy and V is the potential energy. We seek for a solution \psi,

\displaystyle \psi_{k}^{+}(r)=e^{i\vec{k}\cdot \vec{r}}+f(\theta)\frac{e^{ikr}}{kr}

The solution can be decomposed

\displaystyle \psi_{k}^{+}(r)=R_{l}(k,r)Y_{lm}(\theta,\phi)=\frac{u_{l}(k,r)}{kr}Y_{lm}(\theta,\phi)

The solution of u_{l}(k,r) can be solve by Runge-Kutta method on the pdf

\displaystyle \left(\frac{d^2}{d\rho^2} + 1 - \frac{l(l+1)}{\rho^2} \right)u_{l}(k,\rho)=U(\rho)u_{l}(k,\rho)

where \rho=kr, k=\sqrt{2\mu E}/\hbar, \mu=(m_1+m_2)/(m_1 m_2) and U=V/E.


For U = 0, the solution of u_l is

\displaystyle u_{l}(k,r)=\hat{j}_l(\rho) \xrightarrow{r\rightarrow \infty} \sin(r') = \frac{e^{ir'}-e^{-ir'}}{2i}

where r' = kr-l\pi/2 and \hat{j}_l is the Riccati-Bessel function. The free wave function is

\displaystyle \phi_k(r)=e^{i\vec{k}\cdot\vec{r}}=\sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ikr}i^l (e^{ir'}-e^{-ir'})

where P_l(x) is the Legendre polynomial.

Note that, if we have Coulomb potential, we need to use the Coulomb wave instead of free wave, because the range of coulomb force is infinity.


For U\neq 0, the solution of u_l(r<R) can be found by Runge-Kutta method, where R is a sufficiency large that the potential V is effectively equal to 0.  The solution of u_l(r>R) is shifted

\displaystyle u_{l}(k,r>R)=\hat{j}_l(\rho)+\beta_l \hat{n}_l(\rho) \xrightarrow{r\rightarrow \infty} \frac{1}{2i}(S_l e^{ir'}-e^{-ir'})

where S_l is the scattering matrix element, it is obtained by solving the boundary condition at r = R. The scattered wave function is

\displaystyle \psi_k(r)=\sum\limits_{l=0} P_l(\cos(\theta)) (2l+1) i^l \frac{u_l(r)}{kr}

put the scattered wave function and the free wave function back to the seeking solution, we have the f(\theta)

 \displaystyle f(\theta) = \sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ik} (S_l - 1)

and the differential cross section

\displaystyle \frac{d\sigma}{d\Omega}=|f(\theta)|^2.


In this very brief introduction, we can see

  • How the scattering matrix S_l is obtained
  • How the scattering amplitude f(\theta) relates to the scattering matrix

But what is scattering matrix? Although the page did not explained very well, especially how to use it.

Meson Theory on Strong Nuclear Force

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The theory is very difficult, I only state the result.

There are 4 main mesons for strong nuclear force.

type mass [MeV] range symmetry force type
 \pi  135  long  pseudo scalar  -\sigma_{1}\cdot\sigma_{2},-S_{12}
 \sigma  400 ~ 2000  medium  scalar -1, -L\cdot S
 \rho  775  short  vector  -2\sigma_1\cdot\sigma_2, +S_{12}
 \omega  783  short vector  +1, -3L\cdot S

Mean free path of a nucleon inside a nucleus

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The mean free path is the average distance between 2 collisions. We simply copy the things in Bohr and Mottelson, Bertulani and Banielewicz, and John Lilley in here.

Since a nucleus has finite size, the mean free path can be view as transparency of nucleon scattering. Since the total cross section is smaller for higher energy, the mean free path is proportional to energy.

The wave number K under a complex optical potential V+iW is

K=\sqrt{\frac{2m}{\hbar^2}(E-V-iW)}=k_r+ik_i.

There are two solutions, one has imaginary k_r,

k_r=\frac{\sqrt{m}}{\hbar}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}

k_i=\frac{m}{\hbar^2}\frac{W}{k_r}=\frac{1}{2\lambda}

where m is the nucleon mass, E is the incident energy, v is the velocity inside the nucleus, and \lambda is the mean free path,

\lambda=\frac{\hbar}{2W\sqrt{m}}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}\approx\frac{\hbar}{W\sqrt{2m}}\sqrt{(E-V)}

Since the imaginary potential W\approx-15-0.07E, V\approx-39+0.11E from 150 MeV to 400 MeV. The calculated mean free path for proton is shown in here. (Take \hbar = 197.33 [MeV\cdot fm/c], proton mass m=938.272 [MeV/c], the \lambda is in fm.)

Screenshot from 2016-01-18 18:57:43.png

The blue line on the plot is exact calculation, the purple line is approximation W << E-V.

Screenshot from 2016-01-18 18:37:48

This plot is taken from S.S.M. Wong, showing the radial shapes of the volume term (simialr to central term) of proton-nucleus optical potential.

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in J. Killey, the k_r is taken as,

k_r = \frac{\sqrt{2m(E-V)}}{\hbar},

which is the wave number without W. The result only for weak W << E-V.