Drawing energy level with Latex

February 29, 2016

GoLuckyRyan computer , 4 Comments

First, make sure you have the tikz, so that you can

\usepackage{tikz}

without any error.

in linux,

sudo apt-get install texlive-pictures pgf

with the tikz package, we can use \draw to draw line or arrow, \node to draw a text. I will demonstrate to draw a simple energy levels scheme. The advantage of using latex is that the energy levels can be drawn accurately and once the template is set, it is very easy.

I define new command

\newcommand{\levS}[4]{
\draw [level](#3,#1) -- (#3+\len,#1) ;
\draw (#3+\len +0.1,#1) -- (#3+\len +0.3, #1 + #4) -- (#3+\len +0.6, #1 + #4);
\node[right] at (#3+\len +0.6,#1+#4) {#1, #2};
}

with usage,

\levS{enery}{spin-parity}{vertical_shift}{offset_of_number}

with this, we can draw something like this.

22O_levels.png

I attached the template in here:

https://drive.google.com/file/d/1uP5k2cTSubQhLG4S4wJuLdRHBeb5dRu0/view?usp=sharing

Fermi and Gamow-Teller Transition

February 29, 2016

GoLuckyRyan Basic , , , , , , , , , , Leave a comment

The beta decay is caused by the weak interaction. The weak interaction is very short range, because the mediate particles, the W^{\pm} and Z^0 bosons are 80 GeV and 91 GeV respectively. The effective range is like 10^{-3} fm. So, the interaction can assumed to be a delta function and only the coupling constant matter. The Fermi coupling constant is

1.17 \times 10^{-11} (\hbar c)^2~ \mathrm{MeV^2}

The fundamental process of beta decay is the decay of quark.

\displaystyle u \xrightarrow{W^+} d + e^+ + \nu_e

Since a pion is made from up and down quark, the decay of pion into position and electron neutrino is also due to weak interaction.

The Hamilton of the beta decay is

\displaystyle H_w(\beta^{\pm})=G_V \tau_{\mp} + G_A \sigma \tau_{\mp}

where G_V is the vector coupling constant, the term is called Fermi transition. The \tau_{\pm} is the isospin ladder operator. The beta+ decay changes the isospin from +1/2 (neutron) to -1/2 (proton). The G_A is the axial coupling constant, the term is called Gamow-Teller transition. \sigma is spin operator. Because of this operator, the Gamow-Teller transition did not preserve parity.

The G_A is different from G_V, which is caused by the effect of strong interaction. The Goldberger-Trieman relation

\displaystyle g_A = \frac{G_A}{G_V} = \frac{f_\pi g_{\pi N}}{M_N c^2} = -1.3

where f_\pi \sim 93~\textrm{MeV} is the pion decay constant. g_{\pi N} \sim 14 \times 4\pi is the coupling constant between pion and nucleon.  This, we can see the effect of the strong interaction, in which pion is the meson for strong nuclear force.

 

The transition probability can be estimated by Fermi-Golden rule

\displaystyle W(p_e)=\frac{2\pi}{\hbar}|\left< \psi_f|H|\psi_0\right> |^2 \rho(E_f)

the final state wavefunction

\displaystyle \left|\psi_f\right> = \frac{1}{\sqrt{V}} e^{ik_e r} \frac{1}{\sqrt{V}} e^{ik_{\nu}r} \left|j_f m_f\right>

\displaystyle e^{ikr} = \sum \limits_{L}\sqrt{4\pi (2L+1)} i^L j_L(kr) Y_{L0}(\theta)

using long wavelength approximation, the spherical Bessel function can be approximated by the first term.

\displaystyle j_L(kr) \sim \frac{(kr)^L}{(2L+1)!!}

\displaystyle \left| \psi_f\right>=\frac{1}{V}(1 + i \sqrt{\frac{4\pi}{3}} Y_{10} + ...) \left|j_f m_f\right>

The first term 1, or L=0 is called allowed decay, so that the orbital angular momentum of the decayed nucleus unchanged. The higher order term, in which the weak interaction have longer range has very small probability and called L-th forbidden decay.

The density of state is

\displaystyle \rho(E_f) = \frac{V}{2\pi^2 \hbar^7 c^3} F(Z,E_e)p_e^2 (E_0-E_e) ( (E_0-E_e)^2-(m_{\nu} c^2)^2)^2

where the F(Z, E_e) is the Fermi function.

The total transition probability is the integration with respect to the electron momentum.

\displaystyle W = \int W(p_e) dp_e = \frac{m_e^5 c^4}{2 \pi^3 \hbar^7} f(Z,E_0) |M|^2

where f(Z,E_0) is the Fermi integral. The half-life

\displaystyle T_{1/2} = \frac{\ln{2}}{W}

To focus on the beta decay from the interference of the density of state, the ft-value is

\displaystyle ft = f(Z,E_0) T_{1/2} =\frac{2\pi^3\hbar^7}{m_e^5 c^4} \frac{\ln{2}}{|M|^2}

The ft-value could be difference by several order.

There is a super-allowed decay from 0^{+} \rightarrow 0^{0} with same isospin, which the GT does not involve. an example is

\displaystyle ^{14}\mathrm{O} \rightarrow ^{14}\mathrm{N} + e^+ + \nu_e

The ft-value is 3037.7s, the smallest of known.

Fermi Gamow-Teller
\Delta S=0 \Delta S=1
J_f=J_i + L J_f=J_i + L+1
T_f=T_i + 1

 

transition L \log_{10} ft_{1/2} \Delta J \Delta T \Delta \pi
Fermi GT
Super allowed 3.1 ~ 3.6 0^+ \rightarrow 0^+ not exist 0 no
allowed 0 2.9 ~ 10 0 (0), 1 0, 1 ; T_i=0\rightarrow T_f=0 forbidden no
1st forbidden 1 5 ~ 19 (0),1 0, 1, 2 0,1 yes
2nd forbidden 2 10 ~18 (1), 2 2, 3 no
3rd forbidden 3 17 ~ 22 (2), 3 3, 4 yes
4th forbidden 4 22 ~ 24 (3), 4 4, 5 no

The () means not possible if either initial or final state is zero. i.e 1^{-} \rightarrow 0^+ is not possible for 1st forbidden.

 

Gamma Transition

February 28, 2016

GoLuckyRyan Basic, theory , , , , , , , Leave a comment

 

The gamma decay brings a excited nucleus to a lower energy state by emitting a photon. Photon carry angular momentum with intrinsic spin of 1. Its parity is positive. Thus, we have three constrain from conversation laws immediately.

E_f = E_i + \hbar \omega

J_f^{\pi_f} = J_i^{\pi_i} + L

where E is the energy of the nucleus, \hbar \omega and L are the photon energy and angular momentum respectively. We also have to consider the parities of electric and magnetic transition are different.

 

To calculate the transition rate, we can start from a classical equation of power emission from an antenna, since the photon energy is quantized, the transition rate [number of photon emitted per time] is the power divided by a photon energy.

T(qL) = \frac{2(2L+1)}{\epsilon_0 L [(2L+1)!!]^2 \hbar} (\frac{\omega}{c})^{2L+1} B(qL)

where qL is the electromagnetic multipole with angular momentum L, and B(qL) the the reduced transition probability, it is equal to the square of the magnitude of the transition matrix element M_{fi}(qL).

In the electric transition, the multipole is

qL = e r^L

we assume the transition is conduced by a single nucleon and the rest of the nucleus is unaffected. The transition matrix element than can be written as

M_{fi}(qL) = \left<j_f m_f|e r^L Y_{LM}(\Omega)| j_i m_i\right>

The single particle wave function can be written as

\left|j m\right>= R_{nl}(r) [Y_l \times \chi_{1/2}]_{jm}

The matrix elements becomes,

M_{fi}(qL) = e \int_{0}^{\infty} R_{n_f l_f}^* r^L R_{n_i l_i} r^2 dr \times \left<Y_{LM}\right>

To evaluate the radial integral, we make another assumption that the nucleus is a sphere of uniform density with radius R=r_0 A^{1/3},

R_{nl}(r) = \frac{\sqrt{3}}{R^{3/2}}, for r<R, so that \int_{0}^{R} |R_{nl}(r)|^2 r^2 dr = 1

Then the radial integral is

\left<r^L \right>=\frac{3}{R}\int_{0}^{R} r^{L+2} dr = \frac{3}{L+3} r_0^L A^{L/3}

The reduced transitional probability

B_{sp}(qL)=\sum \limits_{M m_f} |\left<j_f m_f|e r^L Y_{LM}|j_i m_i\right>|^2

= e^{2} \left< r^{L} \right> ^{2} \sum \limits_{M m_f} \left<Y_{LM}\right>

the angular part could be assumed as 1/4\pi as the total solid angle is 4\pi. Thus, with these three assumptions, we have the Weisskopf single particle estimation for the L-pole reduced electric transition probability

B_W(EL) = \frac{1}{4\pi}(\frac{3}{L+3})^2 r_0^{2L} A^{2L/3} [e^2 fm^{2L}]

For the magnetic transition, we have to take into account of the spin and orbit angular momentum. The single particle reduced transition probability

B_{sp}(ML) = \sum \limits_{M m_f} |\left<j_f mf| qL |j_i m_i\right>|^2

the result,

B_{sp}(ML)=L(2L+1) \left< r^{(L-1)} \right> ^2

\sum \limits_{M m_f} ((g_s - \frac{2g_l}{L+1}) \left< [ Y_{L-1} \times \vec{s} ]_{LM} \right> \left< [ Y_{L-1} \times \vec{j} ]_{LM} \right> )^2

The term

L(2L+1)(g_s \frac{2g_l}{L+1})^2 \sim 10

and the rest of the angular part assumed to be 1/4\pi again, then

B_W(ML) = \frac{10}{\pi}(\frac{3}{L+3})^2 r_0^{(2L-2)} A^{(2L-2)/3} \mu_N^2 fm^{2L-2}

and notice that \mu_N = e\hbar / (2m_p).

 

Some results can be deduced form the calculation

B_{sp}(ML)/B_{sp}(EL) \sim 0.3 A^{-2/3}

B_{sp}(EL)/B_{sp}(E(L-1)) \sim \frac{1}{7} \times 10^7 A^{-2/3} E_\gamma^{-2}

T(E1) = 1.0 \times 10^{14} A^{2/3} E_\gamma^3

T(E2) = 7.3 \times 10^{7} A^{4/3} E_\gamma^5

T(E3) = 34 A^{2} E_\gamma^7

T(M1) = 3.1 \times 10^{13} A^{0} E_\gamma^3

T(M2) = 2.2 \times 10^{7} A^{2/3} E_\gamma^5

T(M3) = 10 A^{4/3} E_\gamma^7

The deviation from the single particle limit indicates a strong collective state.

 

0 \rightarrow 0, forbidden

1^+ \rightarrow 0^+, M1

2^+ \rightarrow 0^+,  E2

 

 

 

 

Shell model calculation and the USD, USDA, and USDB interaction

February 28, 2016

GoLuckyRyan Basic, theory , , , , , , , , , , , , , , , , , , , , , Leave a comment

Form the mean field calculation, the single particle energies are obtained. However, the residual interaction is still there that the actual state could be affected. Because the residual interaction produces the off-diagonal terms in the total Hamiltonian, and that mixed the single particle state.

The Shell Model calculation can calculate the nuclear structure from another approach. It started from a assumed nuclear Hamiltonian, with a basis of wavefunctions. The Hamiltonian is diagonalized with the basis, then the eigenstates are the solution of the wavefunctions and the nuclear structure, both ground state and excited states. The basis is usually the spherical harmonic with some radial function. Or it could be, in principle, can take from the result of mean field calculation. Thus, the Shell Model calculation attacks the problem directly with only assumption of the nuclear interaction.

However, the dimension of the basis of the shell model calculation could be very huge. In principle, it should be infinitely because of the completeness of vector space. Fro practical purpose, the dimension or the number of the basis has to be reduced, usually take a major shell. for example the p-shell, s-d shell, p-f shell. However, even thought the model space is limited, the number of basis is also huge. “for ^{28}Si the 12-particle state with M=0 for the sum of the j_z quantum numbers and T_z=0 for the sum of the %Latex t_z$ quantum numbers has dimension 93,710 in the m-scheme” [B. A. Brown and B. H. Wildenthal, Ann. Rev. Nucl. Part. Sci. 38 (1998) 29-66]. Beside the huge dimensions and the difficult for diagonalizing the Hamiltonian, the truncation of the model space also affect the interaction.

We can imagine that the effective interaction is different from the actual nuclear interaction, because some energy levels cannot be reached, for example, the short range hard core could produce very high energy excitation. Therefore, the results of the calculation in the truncated model space must be “re-normalized”.

There are 4 problems in the shell model calculation:

  • the model space
  • the effective interaction
  • the diagonalization
  • the renormalization of the result

The shell model can also calculate the excited state with 1\hbar \omega (1 major shell). This requires combination of the interactions between 2 major shell.

For usage, say in the code OXBASH, user major concern is the choice of the interaction and model space. The shell model are able to calculate

  • The binding energy
  • The excitation energies
    • The nucleons separation energies
  • The configuration of each state
  • The magnetic dipole matrix elements
  • The Gamow-Teller (GT) transition
  • The spectroscopic factor
  • …… and more.

 

The W interaction (or the USD) for the s-d shell was introduced by B.H. Wildenthal around 1990s. It is an parametric effective interaction deduced from fitting experimental energy levels for some s-d shell nuclei. Before it, there are some theoretical interactions that require “no parameter”, for example the G-matrix interaction is the in-medium nucleon-nucleon interaction.

The problem for the USD interaction is the interpretation, because it is a black-box that it can reproduce most of the experimental result better than theoretical interactions, but no one know why and how. One possible way is translate the two-body matrix elements (TBME) back to the central, spin-orbit, tensor force. It found that the central and spin-orbit force are similar with the theoretical interactions, but the tensor force could be different. Also, there could be three-body force that implicitly included in the USD interaction.

In 2006, B.A. Brown and W.A. Richter improved the USD interaction with the new data from the past 20 years [B.A. Brown, PRC 74, 034315(2006)]. The new USD interaction is called USDA and USDB. The difference between USDA and USDB is the fitting (something like that, I am not so sure), but basically, USDA and USDB only different by very little. Since the USDB has better fitting, we will focus on the USDB interaction.

The single particle energy for the USDB is

  • 1d_{3/2} = 2.117
  • 2s_{1/2} = -3.2079
  • 1d_{5/2} = -3.9257

in comparison, the single particle energies of the neutron of 17O of 2s_{1/2} = -3.27 and 1d_{5/2} = -4.14.

Can to USD interaction predicts the new magic number N=16?

Yes, in a report by O. Sorlin and M.-G. Porquet (Nuclear magic numbers: new features far from stability) They shows the effective single particle energy of oxygen and carbon using the monopole matrix elements of the USDB interaction. The new magic number N=16 can be observed.

ESPS_new_magic.png

Maximum Likelihood

February 28, 2016

GoLuckyRyan computer, Math , Leave a comment

In data analysis, especially the number of data is small, in order to found out the parameter of the distribution, which fit the data the best, maximum likelihood method is a mathematical tool to do so.

The ideal can be found in Wikipedia. For illustration, I generate 100 data points from a Gaussian distribution with mean = 1, and sigma = 2.

In Mathematica,

Data = RandomVariate[NormalDistribution[1, 2], 100]
MaxLikeliHood = Flatten[Table[{
 mean,
 sigma,
 Log[Product[PDF[NormalDistribution[mean, sigma], Data[[i]]] // N, {i, 1,100}]],
 },
 {mean, -3, 3, 1}, {sigma, 0.5, 3.5, 0.5}], 1]

This calculate the a table of mean form -3 to 3, step 1, sigma from 0.5 to 3.5 step 0.5. To find the maximum of the LogProduct in the table:

maxL=MaxLikeliHood[[1 ;; -1, 3]];
Max[%]
maxN = Position[maxL[[1 ;; -1, 3]], %]
maxL[[Flatten[maxN]]]

The result is

{{1,2.,-217.444}}

which is the correct mean and sigma.

 

From Mean field calculation to Independent particle model

February 28, 2016

GoLuckyRyan Basic, Math, nuclear, theory , , Leave a comment

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

 

 

 

 

Single electron – single proton continuous solid effect

February 27, 2016

GoLuckyRyan Math, theory , , , , , Leave a comment

I made this long time ago, did not posted or published anywhere.

I planned to made a more beautiful journal paper in near future, may be this year.

CW Solid effect

somethings need to be clean up and more detail is needed.

The calculation is based on Hartman-Hahn and Tim Wenckebach, the difference is, they use Fermi-Golden Rule to approximate the polarization transfer rate, but I solved it exactly with computer and also deduced the analytical solution for high-frequency truncated Hamiltonian. This reveal the validity on the frequency truncation. And I will add a comparison with experimental data in the planned-to-do paper.

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

February 26, 2016

GoLuckyRyan Basic, nuclear, Scattering, theory , , , , , , , , , , Leave a comment

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.

Scalar, vector, axial vector, pseudoscalar

February 10, 2016

GoLuckyRyan Basic , , , Leave a comment

type \Delta \pi example
scalar + charge, mass
pseudoscalar – helicity h=\vec{s} \cdot \hat{p}, magnetic charge
vector – \vec{r}, \vec{p}
axial vector + \vec{L}=\vec{r}\times\vec{p}, \vec{s}