Coulomb Energy | Nuclear Physics 101

Coulomb Displacement energy for 11Be and 11B

January 3, 2021

GoLuckyRyan Basic 11B, 11Be, beta decay, Coulomb Energy, IAS, isobaric analog state, isospin Leave a comment

In this post, I calculated the Coulomb energy when a 1s1/2 neutron change to a 1s1/2 proton, based on Woods-Saxon potential with parameters of $R = 1.25A^{1/3}$ fm and $a = 0.67$ fm. The result is 1.15 MeV. far from the experimental value of 1.83 MeV. I speculated the reason is the incorrect wavefunction. But the true reason could be the isospin.

The 11Be ground state is T=3/2 and Tz = 3/2. The 1st T=3/2 state (or the IAS state) of 11B has Tz = 1/2. The IAS is not simply changing the neutron to a proton. As the isospin ladder operator is

$T_- = T_-(1) + T_-(2) + T_-(3)$

where the index 1,2,3 for the 3 neutrons in the 1s1/2 and 0p3/2. The effect of the ladder operator can be illustrated as

We can see, the energy of the 1s1/2 neutron to 1s1/2 proton is only 1/3 of the IAS energy. The other 2/3 is comes from 0p3/2 neutron to 0p3/2 proton. Since the 1s1/2 neutron should not contribute any Coulomb energy, the Coulomb energy of the p3/2 shell should be the Coulomb energy of 10Be.

The mass of 10Be is 9325.504 MeV. The mass of 10B is 9324.436 MeV. Thus, the Coulomb displacement energy of 10Be is 1.965 MeV.

The Coulomb energy of the T=3/2 IAS of 11B should be

$\displaystyle \Delta = \frac{1}{3} 1.15 + \frac{2}{3} 1.97 = 1.7$ MeV

This estimation is much closer to the 1.83 MeV experimental value.

We can also work backward, the Coulomb energy of changing 1s1/2 neutron to 1s1/2 proton should be 1.55 MeV.

Coulomb Energy for a s-orbital under Woods-Saxon potential

April 7, 2020

GoLuckyRyan Basic Coulomb Energy, polylogarithm function, Woods-Saxon Leave a comment

Pick up from the Coulomb Energy calculation, For beta-decay, a neutron change to a proton and emit an electron. I assume the proton-orbital is as same as that of the neutron. In particular, I am working on 11Be beta-decay.

The outer most neutron is in the 1s1/2 orbital with separation energy of 0.502 MeV.

First, I have to find the 1s1/2 neutron wave function using Woods-Saxon potential. Since it is the s-orbital, the spin-orbital part can be ignored. And only 3 parameters left: the potential depth, the half-depth radius, and the diffuseness parameter. The diffuseness parameter is fixed at 0.67 fm. The half-depth radius is $R = 1.25 A^{1/3} = 2.69304 ~\textrm{fm}$ for $A = 10$ . ( May be I should use $A = 11$ ? ). I adjusted the depth to give the -0.502 MeV neutron separation energy. The depth is -51.56 MeV.

The proton/charge distribution is also same as the Woods-Saxon potential, but with a normalization that

$\displaystyle 4\pi \int_0^\infty \rho_0(x) x^2 dx = 1$

In general, a normalized spherical Woods-Saxon potential is

$\displaystyle \rho_0(x) = \frac{1}{1+\exp{\frac{x-R}{a}}} \frac{-1}{4\pi a^3 Li_3(-e^{R/a}) }$

,where $Li_n(z)$ is the polylogarithm function.

The charge distribution and the squared s-wave function is

Annotation 2020-04-07 122922.png

Now, we have to integrate the wave function and the charge distribution

$\displaystyle V(y) = 4\pi \int_0^y \rho_0(x) \frac{x^2}{y} dx + 4\pi \int_y^\infty \rho_0(x) x dx$

The potential is follow

Annotation 2020-04-07 123454.png

The Coulomb potential of the Woods-Saxon shape is “smaller” than that of the Uniform sphere is expected.

The Coulomb energy with Z = 4 is then

$\displaystyle W = Z \frac{e^2}{4\pi \epsilon_0} 4\pi \int_0^\infty V(y) \phi(y)^2 y^2 dy = 1.15~\textrm{MeV}$

The Coulomb energy using the potential of a uniform sphere with Z = 4 is

$\displaystyle W_{U} = 1.24~\textrm{MeV}$

The experimental Coulomb displacement energy for the 11Be (T = 3/2 ) state can be extracted using the mass and the excited state of 11B(T=3/2). The excitation energy of 11B(T=3/2) = 12.554 MeV. The experimental Coulomb energy is

$\displaystyle W_e = M(11B) + Ex(11B) - M(11Be) + m_n - m_p = 1.827~\textrm{fm}$

The calculation is ~ 0.7 MeV smaller. The difference should be due to the shape of the wave function, and also co-related to the shape of the potential, as the Woods-Saxon shape does not accurately descript light nuclei. (or, take the electron mass into account ?? )

I changed the WS parameter $a = 0.3$ fm. So that the wave function is more concentrated toward the center that the 1st node is around 1.9 fm (the potential depth becomes -59.2 MeV). The Coulomb energy with uniform sphere is 1.47 MeV.

a [fm]	Coulomb Energy with Uniformly charge sphere [MeV] (R=1.25(11)^3 = 2.78 fm)
0.67	1.24
0.3	1.47
0.1	1.62
0.02	1.65

R [fm]	Coulomb Energy with Uniformly charge sphere [MeV] ( a= 0.67 fm)
2.78	1.24
2.5	1.28
2.1	1.34

After a few trials, I found that when R=2.2 fm, a = 0.1 fm, The Coulomb energy of the wavefunction of the Woods-Saxon potential and Uniformly charge sphere is 1.85 MeV. The 1st node of the 1s1/2 wavefunction is around 1.5 fm. But the depth of the potential is -98 MeV, which is too deep.

Coulomb Energy for two protons system

March 5, 2020

GoLuckyRyan Basic Coulomb, Coulomb Energy, spherical Bessel, spherical harmonic Leave a comment

From this post and this post, we already have the framework to calculate the energy. The 1-body intersection is zero, as we do not interested on the kinematics energy. And also, suppose we already have the solution for the wave function for a fixed mean-field. Although it is unrealistic, it is a good starting point.

In fact, we have to solve the Schrödinger equation (using Hartree-Fock or numerical solution) with a realistic NN-interaction with and without Coulomb potential, and subtract the difference to get the Coulomb energy.

We have to evaluate the integral

$\displaystyle W = \left<12||12\right> - \left<12||21\right>$

When the 2 protons are in same orbital, the total wavefunction is

$\displaystyle \Psi(1,2) = \frac{1}{\sqrt{2}}\left(\uparrow \downarrow - \downarrow \uparrow \right) \phi(r_1) \phi(r_2)$

The exchange term is zero as the Coulomb operator does not act on the spin part.

$\displaystyle \left<12||12\right> = \sum_{L}^{\infty} \sum_{M=-L}^{L} \frac{4\pi}{2L+1}\int_{0}^{\infty} \int_{0}^{\infty} \phi^2(x) \phi^2(y) \frac{r_<^L}{r_>^{L+1}} x^2 y^2 dx dy \\ \int Y_{lm}^*(1) Y_{LM}^*(1) Y_{lm}(1) d\Omega_1 \int Y_{lm}^*(2) Y_{LM}(2) Y_{lm}(2) d\Omega_2$

For the angular part, from this post or this post, we have

$\displaystyle \int Y_{l_1m_1}^* Y_{LM}^* Y_{l_2m_2} d\Omega = \sqrt{\frac{(2L+1)(2l_1+1)}{4\pi(2l_2+1)}} C_{L0l_10}^{l_20} C_{LMl_1m_1}^{l_2m_2}$

For $l_1 = l_2$ , the Clebsh-Gordon coefficient dictated that $2l \geq L$ , $L = even$ , $M = 0$ .

We use the potential is 3D spherical infinite well with radius $a = 2.2$ fm. the wave function is spherical Bessel function ,

$\phi_n(r ) = \begin{matrix} A j_n(k r) & r < a \\ 0 & r > a \end{matrix}, j_n(ka) = 0$ ,

where $A$ is the normalization factor.

Screenshot 2020-03-05 at 17.48.19.png

The first non-zero root of spherical Bessel Function take the form,

$r_0(n) = 2.96519 + 0.701921\sqrt{n} + 0.993301 n, n < 500$

The energy for the $n$ -th s-state is

$\displaystyle E_n = \frac{\hbar k^2}{2 m_p^2}, k a = r_0(n)$

In Mathematica, the radial calculation can be done as

a=2.2;
R[r_]:=Piecewise[{{SphericalBesselJ[n, r0[n] r], 0 < r < a}, {0, r > 0}}];
A=(NIntegrate[(R[r])^2 r^2, {r, 0, ∞}]])^(-½);
w = NIntegrate[(R[x]R[y])^2 y^L/x^(L+1) x^2 y^2 , {x, 0, ∞}, {y,0, x}] + NIntegrate[(R[x]R[y])^2 x^L/y^(L+1) x^2 y^2 , {x, 0, ∞}, {y,x, ∞}]
W=1.44 A^4 w

The result for first n s-orbital is

Screenshot 2020-03-05 at 17.59.37.png

I also extract the radius for uniformly charged sphere. The trend is reasonable as higher principle number, the wave function is close to the boundary $a = 2.2 \textrm{fm}$ .

Next time, we will work on Woods-Saxon potential.

Coulomb energy for generic charge distribution

March 4, 2020

GoLuckyRyan Basic Coulomb, Coulomb Energy Leave a comment

Classically, for an arbitrary charge distribution $\rho(r,\theta, \phi)$ , the potential energy at position $\vec{r}_0$ is

$\displaystyle V(r_0) = \int\int\int{\frac{\rho(r,\theta,\phi)}{|\vec{r}-\vec{r}_0|} r^2 \sin\theta d\phi d\theta dr}$

Suppose the charge distribution is spherical symmetric, i.e. $\rho(r,\theta,\phi) = \rho(r )$ . And the distance $|\vec{r}-\vec{r}_0|^2 = r^2 + r_0^2 - 2 rr_0 \cos\theta$ .

$\displaystyle V(r_0) = 2\pi\int{\rho(r)}r^2\int{\frac{1}{\sqrt{r^2 + r_0^2 - 2 rr_0 \cos\theta}}\sin\theta d\theta} dr$

The angular part,

$\displaystyle \int_{0}^{\pi}{\frac{1}{\sqrt{r^2 + r_0^2 - 2 rr_0 \cos\theta}}\sin\theta d\theta } \\ = \int_{1}^{-1}\frac{1}{\sqrt{r^2+r_0^2 - 2 r r_0 x}} dx \\ = \frac{-1}{2 rr_0} \int_{1}^{-1} \frac{d(r^2+r_0^2 - 2rr_0 x)}{\sqrt{r^2+r_0^2 - 2 r r_0 x}} \\ = \frac{1}{2 rr_0} \int_{(r-r_0)^2}^{(r+r_0)^2} y^{-\frac{1}{2}} dy = \frac{r + r_0 - |r - r_0|}{rr_0}$

The last step is a bit tricky. So, the angular part is

$\displaystyle \int_{0}^{\pi}{\frac{1}{\sqrt{r^2 + r_0^2 - 2 rr_0 \cos\theta}}\sin\theta d\theta } = \frac{r + r_0 - |r - r_0|}{rr_0} = \frac{2}{r_>}$

Recall that

$\displaystyle \frac{1}{|\vec{r}-\vec{r}_0|} = \sum_{L}^{\infty}\sum_{M=-L}^{L} \frac{4\pi}{2L+1} \frac{r_<^L}{r_>^{L+1}} Y_{LM}^*(\Omega)Y_{LM}(\Omega_0)$

For $L = 0$ , that reduced to $4\pi/r_>$ .

Thus, for spherical arbitrary charge distribution, the Coulomb potential is,

$\displaystyle V(r_0) = 4\pi\int{\rho(r)} \frac{r^2}{r_>} dr$

Annotation 2020-04-07 081009.png

Lets redo the Coulomb energy for the uniform charge sphere here.

The charge distribution is

$\displaystyle \rho_0(x) = Z \frac{3}{4\pi} \frac{1}{R^3} , x < R$

The potential at position $y$ is

$\displaystyle V(y) = 4\pi \int_0^y \rho_0 (x) \frac{x^2}{y}dx + 4\pi \int_y^\infty \rho_0(x) x dx$

For $y < R$

$\displaystyle V(y) = Z \frac{3}{R^3} \left( \int_0^y \frac{x^2}{y}dx + \int_y^R x dx \right) = Z \frac{3}{R^3} \left( \frac{y^2}{3} + \frac{R^2}{2} - \frac{y^2}{2} \right) \\ = \frac{Z}{R} \left( \frac{3}{2} - \frac{y^2}{2 R^2} \right)$

For $y > R$

$\displaystyle V(y) = 4\pi \int_0^R \rho_0 (x) \frac{x^2}{y}dx = \frac{Z}{y}$

The result is the same.

The Classical Coulomb energy for another charge distribution $\rho_0(\vec{r}_0)$ is

$\displaystyle W = \int V(r_0) \rho_0(\vec{r}_0) d\vec{r}_0$

Quantum Mechanically, the Coulomb energy between two wavefunction,

$\displaystyle W = \left<\psi_1(r_1) \psi_2(r_2)|V_{12}| \psi_1(r_1) \psi_2(r_2) \right> \\ = \int e \psi_1^2(r_1) \left(\int \psi_2^2(r_2) \frac{e}{|\vec{r}_1 - \vec{r}_2|} d\vec{r}_2 \right) d\vec{r}_1 \\ = \int \rho_1(r_1) V(r_1) d\vec{r}_1$

We can see the expression is consistent with the Classical case.

For fermion N-body, the total wave function (in the 3rd part of this post) is,

$\displaystyle \Psi = \frac{1}{\sqrt{N!}}\sum_{n}^{N!} (-1)^{p_n} P_n \phi_a(1)\phi_b(2)...\phi_z(N) = (N!)^{-1/2}\sum_{n}^{N!} (-1)^{p_n} P_n \phi_H$

where $P_n$ is the permutation operator with parity $p_n$ .

The Column energy is

$\displaystyle W = \left<\Psi| V_{12} + V_{23} +.... V_{N-1,N} |\Psi \right> = \frac{N(N-1)}{2} \left<\Psi|V_{12}|\Psi \right>$

$\displaystyle \left<\Psi|V_{12}|\Psi \right> = (N!)^{-1} \sum_{i}^{N!} \sum_{j}^{N!} (-1)^{p_i+p_j+2} \left< P_i(\phi_H^*) |V_{12}| P_j(\phi_H) \right>$

Since the Coulomb operator $V_{12}$ only act on particle 1 and 2, thus, the wave function for particle 3, 4,… N has to be the same. Thus, any none-zero product is exchange between particle 1 and 2. therefore, for any given permutation $P_i$ the other permutation must be $P_j = P_i - P_{12} P_i$

$\displaystyle W = (2(N-2)!)^{-1} \sum_{i}^{N!} \left< P_i(\phi_H^*) |V_{12}| P_i(\phi_H) \right> - \left< P_i(\phi_H^*) |V_{12}|P_{12}P_i(\phi_H) \right>$

In the $N!$ permutation, for a fixed orbital of particle 1 and 2, there are $(N-2)!$ way to permute the $N-2$ particle. Thus

$\displaystyle W = \frac{1}{2} \sum_{a}^{N}\sum_{b\neq a}^{N-1} \left< ab|| ab \right> - \left<ab||ba\right>$

$\displaystyle W = \sum_{a}^{N}\sum_{b>a}^{N} \left< ab|| ab \right> - \left<ab||ba\right>$

Here we use

$\displaystyle \left<\phi_a^*(1)\phi_b^*(2)|V_{12}|\phi_h(1)\phi_k(2) \right> = \left<ab||hk\right>$

for $N=2$

$\displaystyle W = \sum_{a=1}^{2}\sum_{b=2}^{2} \left< ab|| ab \right> - \left<ab||ba\right> = \left< 12|| 12 \right> - \left<12||21\right>$

which is the result we expect.

Electric field, Potential and Energy of Uniform Sphere

February 18, 2020

GoLuckyRyan Basic Coulomb Energy Leave a comment

Here we will derive the equation for the field, potential, and energy of a uniformly charged sphere.

Assume the total charge of the sphere is Z, the radius is R. The electric field outside the sphere is,

$\displaystyle \vec{E(r)} = \frac{Ze}{4\pi\epsilon_0 r^2} \vec{r}$

Inside the sphere, at radius $r$ , the charge outside formed a shell that the electric field cancelled out. Assume the charge density $Ze = \rho \frac{4\pi}{3} r^3$ , and $dq = Ze r^3/R^3$ , thus, the electric field at radius r is,

$\displaystyle \vec{E(r)} = \frac{dq}{4\pi \epsilon_0 r^2} \vec{r}= \frac{Ze r}{4 \pi \epsilon_0 R^3} \vec{r}$

For the potential, outside is integrate the electric field from infinite to radius r, using,

$\displaystyle V(x) - V(y) = \int_{y}^{x} \vec{E(r)} \cdot d\vec{r}$

$\displaystyle V(r>R) = V(\infty) + \int_{\infty}^{r > R} \frac{Ze}{4\pi\epsilon_0 r^2} dr = \frac{Ze}{4\pi\epsilon_0 r}$

$\displaystyle V(r<R) = V(R ) + \int_{R}^{r < R} \frac{Ze r}{4 \pi \epsilon_0 R^3} dr \\ = V(R) + \frac{Ze}{4\pi \epsilon_0 R^3}\frac{R^2 - r^2}{2} \\ = \frac{Ze}{4\pi\epsilon_0 R} \left( \frac{3}{2} - \frac{r^2}{2R^2}\right)$

If I set the term $Ze / 4 \pi \epsilon_0 = 1$ , and $R = 1$ .

Screenshot 2020-02-18 at 18.48.47.png

From this calculation, we can imagine a proton moving from the center of nuclei to the outside, it will gain energy of $Z^2 e^2 / 8 \pi \epsilon_0 R$ . Of course, in the meantime, there is a proton moving from outside to the center.

The self energy is the energy to form the uniform charged sphere. We can imagine the sphere started from nothing, an a tiny bit of charge at the center, and a shell of tiny charge formed on outside, so on, and so on. The electric energy is

$\displaystyle W = \int V(r) dq$

Because the extra charge is always from from outside, the potential is

$\displaystyle V(r ) = \frac{Ze r^3/R^3}{4\pi\epsilon_0 r} = \frac{Ze}{4\pi\epsilon_0 R^3} r^2$

$dq = \rho 4\pi r^2 dr$

$\displaystyle W = \int_0^R \frac{Ze}{4\pi\epsilon_0 R^3} r^2 \rho 4\pi r^2 dr = \frac{Ze\rho}{\epsilon_0 R^3}\int_0^R r^4 dr \\ = \frac{Ze \rho}{\epsilon_0 R^3} \frac{R^5}{5} = \frac{Ze 3 \rho}{4\pi\epsilon_0 R} \frac{4\pi}{3}\frac{R^3}{5} = \frac{3}{5} \frac{Z^2 e^2}{\epsilon_0 R}$

In the last step, we used $Z e = \rho \frac{4\pi}{3} R^3$ .

Coulomb Displacement Energy

February 12, 2020

GoLuckyRyan Basic Coulomb, Coulomb Energy Leave a comment

I went back to my home town for visa renewal and read few books on history ( american revolution, early islam, modern chinese). After I got the visa a back to US, a lot of work has to be done and waiting for me….

The Coulomb displacement energy is the energy difference between isobaric analog states with same isospin. For example, the ground state energy different between 15O and 15N. The only difference between 15O and 15N is the proton in 0p1/2 shell replaced by a 0p1/2 neutron. Both ground state is in isospin $T=1/2$ . The Coulomb displacement energy is defined as,

$\displaystyle \Delta = BE(A, Z-1) - BE(A,Z) \\ ~~~~ = M(A,Z) - M(A,Z-1) + (m_n - m_p)$

The binding energies are

$BE(^{15}\textrm{O}) = 111.95535 ~\textrm{MeV}$
$BE(^{15}\textrm{N}) = 115.4919 ~\textrm{MeV}$

The difference of the binding energy is 3.54 MeV.

Let’s calculate the Coulomb energy based on proton charge distributed uniformly in a sphere with radiu $R=1.25 A^{1/3}$ . For $A = 15, R = 3.083 \textrm{fm}$ . The Coulomb energy for uniform sphere is

$\displaystyle E_c = \frac{3}{5}\frac{e^2}{R} Z(Z-1), e^2 = 1.44 ~\textrm{MeV} \cdot \textrm{fm}$

Thus the Coulomb energies of 15O and 15N are

$E_c(^{15}\textrm{O}) = 15.69 \textrm{MeV}$
$E_c(^{15}\textrm{N}) = 9.81 \textrm{MeV}$

The difference is 4.11 MeV, about 0.5 MeV difference than the binding energy difference!

Should the electron mass add into the Coulomb displacement energy ??? or the electron mass be taken into account in further calculation? say, the Q-value for charge exchange reaction.

Coulomb Energy in Nuclear Unit

February 16, 2017

GoLuckyRyan Basic Coulomb, Coulomb Energy Leave a comment

Coulomb energy between 2 charge $q_1, q_2$ in SI unit is

$\displaystyle U_c = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} [J]$

$\displaystyle U_c = \frac{e^2}{4\pi\epsilon_0} \frac{Z_1 Z_2}{r [m]} [J \cdot m] = \frac{Z_1 Z_2}{r [m]} (\frac{e^2}{4\pi \epsilon_0}) [J \cdot m] = \frac{Z_1 Z_2}{r [m]} 2.30708 \times 10^{-28} [J \cdot m]$