Simple model for 4He and NN-interaction

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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.

nn-pp.PNG


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? 

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Particle nano-Ampere (pnA)

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I am so confused between the unit of nA, pnA, enA, and pps.


pps = particle per second.

pnA  = particle nanoampere.
The electrical current in nanoamperes (10^{-9}A) that would be measured if all beam ions were singly charged. i.e. neglecting the actual charge state. 1 \textrm{pnA} =  6.25 \times 10^{9} \textrm{ions/second}

enA = pnA × Charge state.

In this sense, enA = nA, the actual current carried by the beam.

In General

\textrm{pnA} = \textrm{pps} \times e \times 10^9

\textrm{enA}  = \textrm{nA} = Z \times \textrm{pps} \times e \times 10^9

where e is electron charge.


For example, 12C with charge state of 6+ at 4 \times 10^{10} pps.

= 4 \times 10^{10} \times e \times 10^{9} = 6.41 \textrm{pnA}

or

= 6 \times 4 \times 10^{10} \times e \times 10^{9} = 38.5 \textrm{nA} = 38.5 \textrm{enA}


An ion beam of 5+ charge state at 1 pnA.

= 1 / e  10^{-9} = 6.24 \textrm{pps}

or

= 5 \textrm{enA}  = 5 \textrm{nA}

Sum of sine-square and n-root of unity

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Suddenly, I encounter this problem: find the sum

\displaystyle \sum_{n=1}^{N} \sin^2\left(x + \frac{2\pi}{N} n \right)

Here is my way of thinking,

since

\displaystyle \sin(a_n) = \frac{1}{2i} (e^{ia_n} - e^{-ia_n} )

where

a_n = x + \frac{2\pi}{N} n

then

\displaystyle \sin^2(a) = \frac{1}{4} (2- e^{2ia} - e^{-2ia} )

The sum break down to calculate

\displaystyle \sum_{n=1}^{N} \sin^2\left(x + \frac{2\pi}{N} n \right)  = \frac{N}{2} - \frac{1}{4} \sum_{n=1}^N (e^{2ia} + e^{-2ia})

The sum is related to n-root of unity as

\displaystyle \sum_{n=1}^N e^{2ia} = 0

because

\displaystyle z^N = 1 \rightarrow z^N-1 = 0 \rightarrow \Sigma_{n=1}^N (z - e^{i 2\pi n / N}) = 0

break down the product,

\displaystyle z^N + \sum_{n=1}^N e^{i 2\pi n/N} z^N + .... + \Sigma_{n=1}^N e^{i 2\pi n /N} = 0

thus,

\displaystyle  \sum_{n=1}^N e^{i 2\pi n/N}  = 0

And since \sqrt{1} = 1, thus, z^{N/2} = 1 and

\displaystyle  \sum_{n=1}^N e^{i 4\pi n/N}  = 0

The graphical imagination is that, the solutions of n-root of unity are the vectors,  isotropically  distributed from center of a unit circle to its circumference. Thus, the add up of all these vectors will form a N-size polygon, and the result of the sum is zero.


I found it is very interesting to transform one question into another question that can be solved graphically.

This problem is originated from 3-phase AC power that N = 3. I wonder, is it the total power be a constant for N >= 3, and the answer is yes.

The ultimate reason for that is, when generating AC power, the motor is rotating at constant speed and a constant power is inputting to the system, which can using 1-phase (N = 2), 2-phase (N = 4), 3-phase (N = 3), and so on, to generate the AC power. Thus, as long as the phase distributed equally, the sum of power must be a constant.

Stopping power and Bethe-Bloch Formula

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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.

SvsE.PNG

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 isEvsX.PNG

and the stopping power against distance is

SvsX.PNG

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)

 

 

Short-range interaction on same nucleons pair

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From “Nuclear structure from a simple perspective” by Richard F. Casten, Chapter 4. The first half discusses the coupling between pp or nn T=1 isovector pair under δ-interaction, which is a good approximation to short-range interaction.

I found that the book makes it a bit complicated. The only 3 cases matter

Capture.PNG

The logic is follow. As the spatial part of the total wavefunction must be symmetric for the δ-interaction to be effective. The spin-isospin part must be antisymmetric as the total wavefunction must be antisymmetric due to identical fermions system. Also, the isospin part must be isovector or symmetric. Thus, the spin part must be antisymmetric or S = 0.

  1. In other word, the δ-interaction has no effect on the S =1 state.
  2. Also, the book said that l_1 + l_2 - J = even.
  3. Most parallel states are either J = J_{min} , J_{max}

Using these 3 reasons, and consider these 3 cases, which actually covered all possible combinations and all cases. Point 2 also means that only half of the coupled states are affected. Since the number of state are even, when J_{max} = even then J_{min} = odd or via versa. If J_{max} is formed by pure S = 1 state, thus, the most affected state is J_{min} as J_{min} is the other most parallel states, in which the overlap of the wavafunctions is maximum.

In the book, the example is d5/2 f7/2, J = 1, 2, 3, 4, 5, 6. This is belong to case 1, thus, the lowest state is J = 1. another example is d5/2 g7/2, which is case 2, the lowest state is J = 6.

For equivalent orbit, either case 1 or case 3. The lowest state is J = 0.


There are one interesting case. when coupling the s1/2. Since the s-orbit is isotropic, thus, no matter how the other orbit, the overlap is always maximum. Since pure S = 1 state is unaffected, thus, we can somehow, imagine the nucleon pair is forming S = 0 pair. In fact, for example, when s1/2 couples with d5/2,

|J = 2 \rangle = \frac{1}{10} |20\rangle + \frac{1}{15\sqrt{2}} |21\rangle

|J = 3 \rangle = \frac{1}{6\sqrt{5}} |21\rangle

The J = 3 state is unaffected by δ-interaction

Sum rule of 9-j symbol

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The 9-j symbol is the coupling coefficient when combining 2 nucleons with state |l_1 s_1 j_1 \rangle and |l_2 s_2 j_2 \rangle to from the state |j_1 j_2 J M \rangle and |L S J M \rangle

|L S J M \rangle = \sum_{j_1 j_2 } |j_1  j_2 J M \rangle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}

or in simpler form

|L S J M \rangle = \sum_{j_1 j_2 } |j_1  j_2 J M \rangle \langle j_1 j_2 | L S \rangle

or reverse.

|j_1 j_2 J M \rangle = \sum_{L S } |L  S J M \rangle \langle L S | j_1 j_2 \rangle

or to say, this is the coefficient between LS coupling and jj coupling scheme.

we can also see that

\langle L S | j_1 j_2 \rangle = \begin{pmatrix} l_1 & l_2 & L \\ s_1 & s_2 & S \\ j_1 & j_2 & J \end{pmatrix}


For example, when p_{1/2} and p_{3/2} coupled to J = 1  . We have

\displaystyle|p_{1/2} p_{3/2} (J = 1) \rangle = \frac{1}{9} |01\rangle -  \frac{1}{6\sqrt{6}} |10\rangle + \frac{1}{12} |11\rangle + \frac{1}{36} |21\rangle

Since for each |LS\rangle, it is (2L+1) (2S+1) degenerated. Thus, the sum rule is

\displaystyle \sum_{LS} \left(\langle L S | j_1 j_2 \rangle\right)^2 (2L+1) (2S+1) = \frac{1}{(2j_1+1)(2j_2+1)}

The sum is equal the a fraction, because the left-hand side is (2j_1+1) (2j_2+1) degenerated for the state |p_{1/2} p_{3/2} (J = 1) \rangle .

Sum rules of Clebsch-Gordon Coefficient

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Since the CG coefficient is already normalized.Thus

\displaystyle \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 1

Since the number of M is 2J+1, as M = -J, -J+1, ... J . Thus,

\displaystyle \sum_M \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 2J+1

At last, the number of dimension of the coupled space or (tensor product space) is equation to (2j_1 +1) (2j_2+1) , i.e.

\displaystyle \sum_J (2J+1) = (2j_1+1)(2j_2+1)

Thus,

\displaystyle \sum_{JM, m_1 m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = (2j_1+1)(2j_2+1)

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