Why triton is unstable?

There may be a simple answer….. if someone know, please leave a massage.

3H is unstable, it will undergo beta- decay, emit an electron and electron anti-neutrino and becomes 3He, which is stable. The Q value is very small, only 18.6 keV, assuming that the anti-neutrino has no mass. The half-life is 12.3 years.

Given that the deuteron is the only stable pn pair. 3H can be made by adding a neutron in a deuteron. And 3He can be made by adding a proton in a deuteron. Since the Coulomb force between protons, it is natural to think that 3He would be less bound than 3H. And it is,

$BE(^{3}He) = 7.718043 MeV$
$BE(^{3}H) = 8.481798 MeV$

The difference is 0.76376 MeV.

The masses are

$M(^{3}He) = m_p + m_n + m_p - BE(^{3}He) = 2808.39 MeV$
$M(^{3}H) = m_p + m_n + m_n - BE(^{3}H) = 2808.92 MeV$

The mass different between proton and neutron is 1.2933 MeV.

$m_n - m_p - m_e = 0.782333 MeV$

The binding energy usually connects to the stability. Higher binding energy should be more stable, and lesser mass for the same atomic number. It seems that the triton is a special case. Triton and 3He against this “common” sense.

Spectroscopic factor (again)

In the last posts on SF, my mind have been biased to the mean field model. I started everything from the mean field. Now, lets forget the mean filed, just starting from experimental view point and any complete basis.

Suppose an arbitrary complete single-particle basis, say, a Woods-Saxon numerical basis, 3D spherical harmonic oscillator, or spherical Gaussian basis, and donate it as $\phi_i$. Using this basis,, we can form the Slater determinate of N-nucleons as

$\displaystyle \Phi_i = \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_1(1) & ... & ... \\ ... & \phi_r(s) & ... \\ ... & ... & \phi_N(N) \end{vmatrix}$.

The solution of any N-body Hamiltonian can be calculated by solving the eigen-system with the matrix elements

$\displaystyle H_{ij} = \left< \Phi_i |H| \Phi_j \right>$

Say, the solution is

$\displaystyle H \Psi_i = E_i \Psi_i$

Experimentally, for the A(d,p)B reaction, we are observing the transition probability

$\displaystyle T = \left< \chi_p \Psi_i(B) |V|\chi_d \Psi_0(A) \right>,$

where $\chi_k$ is the distorted plane wave of  particle $k$ and $V$ is the interaction during the reaction. The integration summed all internal degree of freedom, particularly,

$\displaystyle \left<\Psi_i(B) |V_{BA} | \Psi_0(A) \right > \approx \left<\Psi_i(B) | \Psi_0(A) \right > = \sum_i \alpha_i \phi_i = \psi$

where $\psi$ is called the quasi-particle wave function and it is expressed in the complete basis of $\phi_i$.

And then, the quasi-particle wave function will also be integrated.

$\displaystyle T = \left< \chi_p \psi |V'|\chi_d \right>$

In fact, it is the frame work of many calculations, that require 4 pieces of input,

1. the optical potential for the incoming channel to calculate the distorted wave of the deuteron,
2. the optical potential for the outgoing channel to calculate the distorted wave of the proton,
3. the quasi-particle wave function, and
4. the interaction.

If the Woods-Saxon basis is being used, then we can have shell model picture that, how the quasi-particle “populates” each Woods-Saxon states, and gives the spectroscopic factor.

The Woods-Saxon basis is somewhat “artificial”, it is an approximation. the nature does not care what the basis is, the transition probability is simple an overlap between two nuclei (given that the interaction is not strong, i.e. in the case of Direct reaction). Since the spectroscopic factor is always in reference for a basis, i.e., the quasi-particle wave function has to be projected into a basis, so, the spectroscopic factor is model dependence. If a weird basis is used, the shell model picture is lost. In this sense, SF is pure artificial based on shell model.

In experiment, we mainly observe the scattered proton, and from the angular distribution, the orbital angular momentum is determined. And from the conservation of angular momentum, we will know which orbital the neutron is being added to nucleus A.

But it is somehow weird. Who said the quasi-particle must be in an orbital angular momentum but not many momenta? If the spin of nucleus A is zero, because of conservation of angular momentum, the spin of nucleus B is equal to that of added neutron. But if the spin of nucleus A is not zero,  the neutron can be in many difference spins at once. In general, the conservation of angular momentum,

$\displaystyle \left < \Psi_i(B) | \phi \Psi_0(A) \right >$

could be difference than that of

$\displaystyle \left < \chi_d \Psi_i(B) | V | \chi_p \Psi_0(A) \right >$

We discussed that the spectroscopic factor is artificial. However, if we use a self-consistence basis, i.e. the basis that describes both nucleus A and B very well (is such basis exist?) , then the quasi-particle wave function is being described using the basis of nucleus B. To be explicit, lets say, we have a basis can describe nucleus B as

$\Psi_0(A) = 0.8 \Phi_0 + 0.6 \Phi_1$

$\Psi_i(B) = \sum_{jk} \beta_{jk} \left( \phi_j \otimes \Psi_k(A) \right)$

where the single particle basis capture most of the nucleus A. is the spectroscopic factor meaningful in this sense?

So, is such basis exist? i.e, can it describe a nuclear wave function (almost) perfectly? In ab initial calculation, it can expend the nuclear wave function using some basis and give the overlap. Can the quasi-particle wave function be projected into the “single-particle state” ? How to define the “single-particle state” in ab initial calculation?

The experimental spectroscopic factor is calculated by compare the experimental cross section with that from theory. In the theory, the cross section is calculated by assuming the bound state is from a Woods-Saxon potential without correlation. Thus. the spectroscopic factor is with respect to the Woods-Saxon potential, an approximation.

Some thoughts on the quenching of spectroscopic factor

Spectroscopic factor plays the central role in unfolding the nuclear structure. In the simplest manner, the total Hamiltonian of the nucleus is transformed into a 1-body effective potential and the many-body residual interaction, i.e.,

$\displaystyle H = \sum_i^N \frac{P_i^2}{2m_i} + \sum_{i \neq j}^N V_{ij} = \sum_i^N \left( \frac{P_i^2}{2m_i} + U \right) + \sum_{i\neq j}^N \left( V_{ij} - U \right) \\ = \sum_{i}^N h_i + H_R = H_0 + H_R$

The effective single-particle Hamiltonian has solution:

$\displaystyle h_i \phi_{nlj}(i) = \epsilon_{nlj} \phi_{nlj}(i)$

where $\epsilon_{nlj}$ is the single-particle energy. The solution for $H_0$ is

$\displaystyle H_0 \Phi_k(N) = W_k \Phi_k(N)$

$\displaystyle \Phi_k(N)= \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_{p_1(k)}(1) & \phi_{p_1(k)}(2) & ... & \phi_{p_1(k)}(n) \\ \phi_{p_2(k)}(1) & \phi_{p_2(k)}(2) & ... & ... \\ ... & ... & ... & ... \\ \phi_{p_n(k)}(1) & \phi_{p_n(k)}(2) & ... & \phi_{p_N(k)}(N) \end{vmatrix}$

$\displaystyle W_k = \sum_i^N \epsilon_{p_i(k)}$

where $p_i(k)$ is the set of basis for state $k$ from $\phi_{nlj}$, and $W_k$ is the eigenenergy.

The residual interaction is minimized by adjusted the mean-field $U$. Thus, the residual interaction can be treated as a perturbation. This perturbs the nuclear wave function

$H \Psi_k(N) = W_k \Psi_k(N), \Psi_k(N) = \sum_i \theta_{i}(k) \Phi_i(N) .$

The normalization requires $\sum_i \theta_{i}^2(k) = 1$.

In the Slater determinant $\Phi_k$, a single-particle wave function for a particular orbital can be pull out.

$\displaystyle \Phi_k(N) = \phi_{\mu} \otimes \Phi_{k}(N-1)$

where $\otimes$ is anti-symmetric, angular coupling operator. Thus,

$\displaystyle \Psi_k(N) = \sum_{\mu i} \theta_{\mu i}(k) \phi_{\mu} \otimes \Phi_i(N-1)$

The $\theta_{\mu i}^2 (k)$ is the spectroscopic factor. There are another sum-rule for adding and removing a nucleon. so that the sum is equal to the number of particle in a particle orbital.

I always imagine the quenching is because we did not sum-up the SFs from zero energy to infinity energy (really???), thus, we are always only observing a small fraction of the total wave function. For example,  the total wavefunction would look like this:

$\displaystyle \Psi_k(N) = \phi_{0} \otimes \left(\theta_{00}(k) \Phi_0(N-1) + \theta_{01}(k)\Phi_1(N-1) +.... \right) \\ + \phi_{1} \otimes \left( \theta_{10} \Phi_0(N-1) +... \right) +... .$

In experiment, we observe the overlap between ground-state to ground-state transition

$\displaystyle \left< \phi_0 \Phi_0(N-1) | \Phi_0(N-1) \right> = \theta_{00}(0)$

for ground-state to 1st excited state transition for the same orbital is

$\displaystyle \left< \phi_0 \Phi_1(N-1) | \Phi_0(N-1) \right> = \theta_{01}(0)$

And since we can only observed limited number of excited states, bounded by either or boht :

• experimental conditions, say incident energy
• the excited states that are beyond single-particle threshold.
• finite sensitivity of momentum

Thus, we cannot recover the full spectroscopic factor. This is what I believe for the moment.

Experimentally, the spectroscopic factor is quenched by 40% to 50%. The “theory” is that, the short-range interaction quench ~25%, the long-range interaction quench ~20%. The long- and short-range interaction correlate the single-particle orbital and reduce the degree of “single-particle”.

The short-range interaction is mainly from the “hard-core” of nucleon, i.e., the interaction at range smaller than 1 fm. The long-range interaction is coupling with nearby vibration states of the rest of the nucleus.

For example, from the 19F(d,3He) reaction, the spectroscopic factor for 19F 1s1/2 state is ~0.4, and 0d5/2 is ~0.6.

Thus, the wavefunction of 19F is

$\left|^{19}\textrm{F}\right> \approx \sqrt{0.4} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.6} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right>$

It is worth to note that the above SFs is not re-analysised and the “quenching” is not shown. Many old data had been re-analysised using global optical model and the SF is reduced and show that the sum of SFs is ~ 0.55.

If it is the case for 19F, the wavefunction would become,

$\left|^{19}\textrm{F}\right> \approx \sqrt{0.2} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.3} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right> + \sqrt{0.5} \Psi_k$

Here I use $\Psi_k$ for the “correlated wavefunction” that the single-particle orbital cannot simply pull out. Nevertheless, if $x$ and $y$ are correlated,

$f(x,y) \neq g(x) h(y)$

Am I misunderstood correlation?

My problem is, What does a correlated wave function look like?

In my naive understanding, the Slater determinant $\Phi_k$ is a complete basis for N-nucleon system. A particular single-particle orbital can ALWAYS be pull out from it. If it can not, therefore, the Slater determinant is NOT complete. The consequence is that all theoretical calculation is intrinsically missed the entire CORRELATED SPACE, an opposite of Slater determinant space (of course, due to truncation of vector space, it already missed somethings).

If the theory for correlation is correct, the short-range interaction is always there. Thus, the spectroscopic factor for deuteron 0s1/2 orbital is ~0.8, assuming no long-range correlation. However, we already knew that 96% of deuteron wavefunction is from 0s1/2 and  only 4% is from 1d5/2 due to tensor force. Is it not mean the spectroscopic factor of deuteron 0s1/2 state is 0.96?  Is deuteron is a special case that no media-modification of nuclear force? But, if the short-range correlation is due to the hard core of the nucleon, the media-modification is irrelevant. Sadly, there is no good data such as d(e,e’p) experiment. Another example is 4He(d,p)5He experiment. What is the spectroscopic factor for g.s. to g.s. transition, i.e. the 0p3/2 orbital? is it ~0.6 or ~ 1.0?

Since the experimental spectroscopic factor has model dependency (i.e. the optical potential). Could the quenching is due to incomplete treatment of the short- and long-range correlation during the interaction, that the theoretical cross section is always bigger?

In the very early days, people calibrate their optical potential using elastic scattering for both incoming and out-going channel, and using this to produce the inelastic one. At that time, the spectroscopic factors are close to ~1. But since each optical potential is specialized for each experiment. It is almost impossible to compare the SF from different experiments. Thus, people switch to a global optical potential. Is something wrong with the global optical potential? How is the deviation?

Let me summarize in here.

1. The unperturbed wave function should be complete, i.e. all function can be expressed as a linear combination of them.
2. A particular single-particle orbital can be pull out from the Slater determinate $\Phi_k$.
3. The residual interaction perturbs the wave function. The short-/long-range correlation should be in the residual interaction by definition or by construction of the mean field.
4. The normalization of wave function required the sum of all SF to be 1.
5. Another sum rule of SFs equals to the number of particle.
6. By mean of the correlation, is that many excited states have to be included due to the residual interaction. No CORRELATED space, as the Slater determinant is complete. (pt. 1)
7. Above points (1) to (7) are solid mathematical statements, which are very hard to deny.
8. The logical result for the quenching of the observed SF is mainly due to not possible to sum up all SFs from all energy states for all momentum space.
9. The 2-body residual interaction can create virtual states. Are they the so called collective states?
10. But still, collective states must be able to express as the Slater determinant (pt. 1), in which a particular single-particle orbital can be pull out (pt. 2).
11. May be, even the particular single-particle orbital can be pull out, the rest cannot experimentally observed ? i.e. $\Phi_k(N-1)$ is not experimentally reachable. That go back to previous argument for limitation of experiments (pt. 8).
12. For some simple systems, say doubly magic +1, deuteron, halo-nucleon, very weakly bounded exited state, resonance state, the sum of SF could be close to 1. Isn’t it?
13. The theoretical cross section calculation that, the bound state wave function is obtained by pure single-particle orbital. I think it is a right thing to do.
14. The use of global optical potential may be, could be not a good thing to do. It may be the METHOD to deduce the OP has to be consistence, instead of the OP itself has to be universal. Need more reading from the past.

some thoughts on mass

Lets forget about the Higg mechanism.

Lets start from deuteron, it contains a proton and neutron, which bounded by strong nuclear force. The kinematics energy between the proton and neutron is about 35 MeV. Since the binding energy of deuteron is 2.2 MeV, thus the strong nuclear force is about -37 MeV. According to Einstein $E = mc^2$. This 2.2 MeV is converted to 2.2 MeV/c2 of mass.

In this example, when a system contains internal motion, internal interaction, the “mass” is not just the mass of individual components, but also included the interaction energy.

Thus, what is mass? Only elementary particles had “pure” mass. The mass of all other particles must contains the energy of internal interaction and motion.

So, for the Earth, we know the mass from the orbital motion. But this mass, when the Earth exploded into tiny many pieces, some of the mass was gone because the internal interaction and motion no longer there.

What is the mass of the solar system? We can add all the masses of the sun, the planets, comets, ices, debris, etc, but we also need to add the internal motion and interaction. Although the gravitation force is weak, but still, we need to add this.

What is the mass of the Universe? Again, we can added all visible mass, but we also need to add the dark energy, the dark matter ( I suspect the dark matter is kind of illusion that we don’t fully understood the gravity, can the dark matter is due to gravitation energy, which act as mass?).

Simple model for 4He and NN-interaction

Starting from deuteron, the binding energy, or the p-n interaction is 2.2 MeV.

From triton, 3H, the total binding energy is 8.5 MeV, in which, there are only 3 interactions, two p-n and one n-n. Assume the p-n interaction does not change, the n-n interaction is 4.1 MeV.

The total binding energy of 3He is 7.7 MeV. The p-p interaction is 3.3 MeV.

Notices that we neglected the 3-body force in 3H and 3He. And it is strange that the n-n and p-p interaction is stronger then p-n interaction.

In 4He, the total binding energy becomes 28.3 MeV. I try to decompose the energy in term of 2-body, 3-body, and 4-body interaction.

If we only assume 2-body interaction, the interaction strength from n-p, n-n, and p-p are insufficient. One way to look is the 1-particle separation energy.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n).
The proton separation energy is 19.8 MeV = 2(p-n) + (p-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p).

There is no solution for above 3 equations. Thus, only consider 2-body interaction is not enough.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

Assuming the 2-body terms are the same in 2H, 3H, and 3He, the (n-p-p) and (n-n-p) is 4.03 MeV, which is strange again, as the Coulomb repulsion should make the (n-p-p) interaction smaller then the (n-n-p) interaction. The (n-n-p-p) interaction is -4.03 MeV.

Lets also add 3-body force in 3H and 3He.

The neutron separation energy of 3H is 6.3 MeV = (p-n) + (n-n) + (n-n-p)
The toal energy of 3H is 8.5 MeV = 2(p-n) + (n-n) + (n-n-p)
The toal energy of 3He is 7.7 MeV = 2(p-n) + (p-p) + (n-p-p)
The neutron separation energy of 4He is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 4He 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy of 4He is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

We have 6 equations, with 6 unknown [ (p-n) , (n-n) , (p-p) , (n-n-p), (n-p-p), and (n-n-p-p)]. Notice that the equation from proton separation energy of 3He is automatically satisfied. The solution is

(p-n) = 2.2 MeV
(p-p) = – 4.7 – (n-n) MeV
(n-n-p)  = 4.1 – (n-n) MeV
(n-p-p) = 8 + (n-n) MeV
(n-n-p-p) = 0 MeV

It is interesting that there is redundant equation. But still, the (p-n) interaction is 2.2 MeV, and 4-body (n-n-p-p) becomes 0 MeV. Also the (n-p-p) is more bound than (n-n-p) by 3.9 + 2(n-n) MeV. If (n-p-p) should be more unbound, than (n-n) must be negative and smaller than -1.95 MeV.

Since the interaction strength has to be on the s-orbit (mainly), by considering 2H, 3H, 3He, and 4He exhausted all possible equations (I think). We need other way to anchor either (n-n), (p-p), (n-p-p), and (n-n-p) interactions.

Use the Coulomb interaction, the Coulomb interaction should add -1.44 MeV on the NN pair (assuming the separation is a 1 fm). Lets assume the (n-n) – (p-p) = 1.44 MeV

(n-n) = -1.63 MeV
(p-p) = – 3.07 MeV
(n-n-p)  = 5.73 MeV
(n-p-p) = 6.37  MeV

The (p-p) is more unbound than (n-n) as expected, but the (n-p-p) is more bound than (n-n-p) by 0.64 MeV. This is surprising! We can also see that, the 3-body interaction play an important role in nuclear interaction.

According to this analysis, the main contribution of the binding energies of 3H and 3He are the 3-body force.

In 3H:  (n-n) + 2(n-p) + (n-n-p) = -1.6 + 4.4 + 5.7 = 8.5 MeV
In 3He: (p-p) + 2(n-p) + (n-p-p) = -3.1 + 4.4 + 6.4 = 7.7 MeV

Worked on the algebra, when ever the difference  (n-n) – (p-p)  > 0.8 MeV, the (n-p-p) will be more bound that (n-n-p). Thus, the average protons separation should be more than 1.8 fm. I plot the interactions energies with the change of Coulomb energy below.

The (n-n) and (p-p) are isoscalar pair, where tensor force is zero. While the (n-p) quasi-deuteron is isovector pair. Thus, the difference between (n-n) and (n-p) reflect the tensor force in s-orbit, which is 3.8 MeV. In s-orbit, there is no spin-orbital interaction, therefore, we can regard the tensor force is 3.8 MeV for (n-p) isovector pair.

Following this method, may be, we can explore the NN interaction in more complex system, say the p-shell nuclei. need an automatic method. I wonder the above analysis agreed present interaction theory or not. If not, why?

Stopping power and Bethe-Bloch Formula

I am so surprised that this topic is not in this blog.  But I am not always posting what I did. Anyway…

The Stopping power is energy loss per length, in unit of MeV/cm.

$\displaystyle S = -\frac{dE}{dx}$

Since the stopping power is proportional to density, it is often normalized with the density in literature and the unit becomes MeV/(ug/cm^2).

The Bethe-Bloch formula based on classical argument.

In order to calculate the range, we can integrate the stopping power. Consider the particle moved by $\Delta x$ distance, the energy loss is

$E(x+\Delta x) = E(x) - S(E(x)) \Delta x$

rearrange, gives

$\displaystyle \frac{dE(x)}{dx} = - S(E)$

$\displaystyle \int_{E_0}^{E} \frac{-1}{S(E)} dE = x(E ; E_0)$

This is the relation between particle energy and range.

We can plot $S(E(x))$ to get the Bragg peak.

In here, I use the physical calculator in LISE++, SRIM, and Bethe-Bloch formula to calculate the stopping power for proton in CD2 target.

The atomic mass of deuteron is 2.014102 u. The density of CD2 target is 0.913 g/cm3.

For the Bethe curve, I adjusted the density to be 0.9 mg/cm3, and simple use the carbon charge. And the excitation potential to be $10^{-5} z$, where $z = 6$, so that the Bethe curve agree with the others.

It is well known that the Bethe curve fails at low energy.

Assume the proton is 5 MeV. The energy loss against distance is

and the stopping power against distance is

We can see the Bragg peak is very sharp and different models gives different stopping range. It is due to the rapid decrease of energy at small energy. In reality, at small energy, the microscopic effect becomes very important and statistical. thus, the curve will be smoothed and the Bragg peak will be broaden.

In very short range, the stopping power increase almost linearly, Because the incident energy is large, so that the stopping power is almost constant around that energy range.

For $S(E_0) = h$,

$\displaystyle x = \frac{-1}{h}(E - E_0 )$

$\displaystyle E = E_0 - hx, hx << E_0$

$\displaystyle S(E) = S(E_0 - hx) \approx S(E_0) - \frac{dS(E_0)}{dE}(hx)$

Finite Spherical Square Well

Hello, everyone, in order to calculate deuteron by Hartree-Fock method, I need a basis. The basis of infinite spherical square well is too “rigid”, that it has to “extension” to non-classical region. Beside of the basis of Wood-Saxon potential. The finite spherical square well is a good alternative. The radial equation is basically the same as infinite spherical square well.

The potential is

$\displaystyle V(r) = -|V_0|, r\leq a$

$\displaystyle V(r) = 0, r > a$

Within the well, the wave vector is

$\displaystyle k = \sqrt{\frac{2m}{\hbar^2} (|V_0 |-|E|) }$

, outside the well, the wave vector is

$\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2} |E| }$

The solution is spherical Bessel function. Since the Bessel function of the first and second kind are oscillating like sin or cosine function. To form a decay function when $r > a$, we need the Hankel function with complex argument.

$\displaystyle h_n( i \kappa r) = h_n(x) = - (i x)^n \left( \frac{1}{x} \frac{d}{dx}\right)^n \left(\frac{\exp{(-x)}}{x} \right)$

To make it real, we need a factor $i^n$.

The boundary conditions are continuity and differential continuity.

$\displaystyle j_l(ka) = A i^l h_l(i\kappa a)$

$\displaystyle \left(\frac{d}{dr}j_l(kr)\right)_{r=a} = A i^l \left(\frac{d}{dr}h_l(i\kappa a) \right)_{r=a}$

These two conditions solved for two parameters $A$ and $E$. However, I cannot find an analytical solution to the energy $E$.

In mathematica, the spherical Hankel function is a build in function.

SphericalHankelH1[n, r]

If the potential depth is 60 MeV for proton. Radius is 1 unit for a light nuclei. Set $\hbar = 1, c = 1, m = 1$. (This is equivalent to radius of $\sqrt{(\hbar c)^2 / m} = 6.44 \textrm{fm}$

The result is follow,

The 1st column is 1s, 1p, 1d, 1f, and 1g. The 2nd column is 2s, 2p, 2d, 2f, and 2g. The 3rd column is 3s, 3p, and 3d.

By compare with infinite square well,

The energies are lower in finite well, because the wave functions can spread-out to non-classical region, so that the wave length is longer and energy is lower.

In this example, the 3d and 2g orbits are bounded (of course, all orbits in infinite well are bound). This is not because of the depth of the well, but the boundary of the well. In other word, to bring down an orbit, the wave function has to spread out, that is connected with the neutron-halo.

Historical Review on Solid Polarized Target

by Akira Masaike

this review is on “Proceedings of the 11th International workshop on Polarized Sources and Targets”

the introduction stated that using thermal polarization (brute force method) in low temperature, ~ 0.01K and high magnetic field~10T can produce proton polarization as high as 76%. another way is using dynamical method by paramagnetic materials, (was proposed by Overhauser) the coupling between electron spin and proton spin can polarize proton in higher temperature and lower magnetic field. Abragam proved that by using paramagnetic impurity can also work for non-metallic materials.

Abragam and Jeffries used $La_2 Mg_3 (NO_3)_{12} 24 H_2O$ (or called LMN), containing small percent of neodymium to polarize proton spin by dynamic nuclear polarization at 1963. but the LMN crystal is not good for inelastic reaction because of the low dilution factor ( the ratio of polarized proton to total number of nucleon ), and this produce a lot other un-wanted reactions. and also the crystal is damaged under scattering by relativistic particles.

organic materials are tested and a major breakthrough is by using butanol with water doped with porphyrexide in 1969 by Mango t al.. the polarization is 40% at 1K and 2.5T.  and also, diol with $Cr^{5+}$ were polarized to 45% at 1K and 2.5T by Glättli et al. Masaike et al. was polarized diol up to 80% at temperature lower then 0.5K. Burtanol in $^3He$ cryotat also polarized upto 67%. All these were done in 1969.

deuteron in deuterated organic material can also be polarized. the principle is called “the equal spin temperature model”

$\displaystyle P = B_I( \frac{\mu B }{ 2 k_B T_s} )$

$B_I$ is Brilluin function and $T_s$ is the nuclear spin temperature.

Spin Frozen Target is a technique that the polarized proton spin last long enough with ESR radiation. the theory was developed around 1965 by Schmugge and Jeffries and constructed by Rusell at 1971 in Rutherford laboratory.

$NH_3$ or ammonia has high dilution factor. it was polarized to 70% by doped with ethylene glycol $Cr^{5+}$ complex at 0.5K and 2.5T by Scheffler and Borghini at CERN at 1971. but ammonia has slow growth of polarization and may explode by high intensity proton irradiation at 1983 at CERN.  but later, it was overcomed by Meyer et al. at Bonn. Thanks to Meyer, ammonia becomes a popular polarized target.

High polarization of H, D and $^6Li$ in dilution refrigerator were found in Bonn around 2005. The COMPASS experiment in CERN use target of $latxe ^6LiD$ at polarization 50% at 300mK and 2.5T.

Hydrogen deuteride (HD) has the highest dilution factor and in principle, both can be polarized. the relaxation properties at 0.5K depends on ortho-para and para-ortho conversions of $latxe H_2$ and $latxe D_2$. The polarization for proton and deuteron of 60% and 14% by brute force method at 10mK and 13.5T was done by Grenoble-Orsay group at 2004.

Crystal of Naphthalene (77K) and p-tarphenyl (270K) doped with pentacene have been polarized in 0.3T by Iinuma et al. at 2000. the high temperature and low magnetic field is due to the paramagnetic excited state in pentacene and diamagnetic on ground state. the population of Zeeman  sublevels of the lowest triplet state is 12% for m=+1, 76% for m=0 and 12% for m=-1 regardless of temperature and magnetic field strength. by using ESR radiation and the “integrated solid effect”, the proton can be polarized and remain polarized after the electrons go back to ground state. 70% polarization at liquid nitrogen temperature.

[ (p,d) 3NF ] Installation of Polarization meter

i will join an “measurement of d-p elastic scattering at 300 MeV/nucleon” experiment.

this experiment is going to study the 3-nuclear force, which is the force connects 3 body at same time. since the nuclear force is polarization dependence, or spin dependence, we have to measure the polarization of the beam.

the way to do is measuring the asymmetry of scattering, by spin-orbital coupling. for a spin up particle incident on left, the spin-orbital coupling strength is different from the incident on the right. thus, we will install a plastic scintillator on both side, left and right, up and down, to determine the polarization.

the scintillator was installed at particular angle such that the different is biggest. ( i don’t know how to calculate this angle ) the optimal angle for proton and deuteron are different, so, we use 8 detector in total.

after installation, we put a Co-60 gamma ray source in front of it, and connect the photomultipler tube (PMT) with about 1000V to 2000V, depend on the sensitivity of each PMT. The signal was displayed on CRO, and it shoen a typical signal pattern. although the gamma ray energy is fixed, the detection process in the plastic scintillator is via Compton scattering of electron, that make fluctuation of the signal. Thus, we take count average of about 60 to 100 counts. the typical signal depth is 200mV and fall time is 3uS, raise time is 5 to 7 uS.

footnote:

Co-60 has 1 gamma ray decay of 0.058MeV, then it goes 2 channels beta decay of 0.31MeV with 99.88% and 1.48MeV with 0.12% to Ni-60, the excited Ni-60 will emit 2 gamma ray, of 1.17MeV and 1.33MeV.

Angular distribution of Neutrons from the Photo-Disintegration of the Deuteron

this paper was written on 1949. at that time, deuteron just discovered 20 years. this paper presents a method on detecting the diffraction cross section of the neutron from a disintegrated deuteron by gamma ray of energy 2.76MeV. and by this, they found the photo-magnetic to photo-electric cross section ration. the ratio is 0.295 ± 0.036.

the photo-electric dipole transition and photo-magnetic dipole transition can both be induced by the gamma ray. Photo carry 1 angular momentum, the absorption of photon will excited the spherical ground state $^1S$ into $^3P$. the 2 mechanisms of the disintegrations results 2 angular distributions of the neutrons. by examine the angular distribution, they find out the ratio.

the photo-magnetic cross section is isotropic and the photo-electric cross section is follow a of a $sin^2$ distribution. the average intensity of neutron detected on a angle is:

$I(\gamma ) = \int_{\gamma_1}^{\gamma_2} {(a + b sin^2(\gamma)) sin(\gamma) d\gamma } / \int_{\gamma_1}^{\gamma_2} {sin(\gamma) d\gamma }$

where a is the contribution from the photo-magnetic interaction and b is from photo-electric interaction. and $\gamma_1$ and $\gamma_2$ are the angle span by the finite size of the target and detector. the integration is straight forward and result is:

$I(\gamma) = a+b( 1 - 1/3 ( cos^2(\gamma_1) + cos(\gamma_1) cos(\gamma_2) + cos^2 ( \gamma_2) )$

and the author guided us to use the ration of 2 angle to find the ration of a and b. and the ration of a and b is related to the probability of the magnetic to the electric effect by

$a/b = 2/3 \tau$

. and the photo-magnetic to photo-electric cross section ratio is:

$\tau/(\tau+1)$

the detector was described in detail on 4 paragraphs. basically, it is a cylindrical linear detector base on the reaction $B^{10} ( n,\alpha)Li^7$. it was surrounded by paraffin to slow down fast nuetrons.

on the target, which is heavy water, $D_2 O$, they use an extraordinary copper toriod or donut shape container. it is based on 3 principles:

• The internal scattering of neutron
• Departure from point source
• The angular opening of the γ – ray source

they place the γ – ray source along the axis of the toriod, move it along to create different scattering angle.

they tested the internal scattering of the inside the toriod and found that it is nothing, the toriod shape does not have significant internal scattering.

they test the reflection of neutron form surrounding, base on the deviation from the inverse-square law. and finally, they hang up there equipment about 27meters from the ground and 30 meters from buildings walls. (their apparatus’s size is around 2 meters. They measured 45, 60, 75 and 90 degree intensity with 5 degree angular opening for each.