## Particle nano-Ampere (pnA)

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

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

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

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 $S(E_0) = h$,

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

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

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

## Short-range interaction on same nucleons pair

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

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

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

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,

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

## Winger 6-j and 9-j symbol

The meaning of 3-j symbol is same as Clebsch-Gordan coefficient. So, we skip in here.

I am not going to construct the 6-j symbol from 3-j symbol. In here, I just state the meaning and usage in Mathematica.

The 6-j symbol is the coupling between 3 angular momenta, $j_1, j_2, j_3$.

There are 2 ways to couple these 3 angular momenta. First,

$j_1 + j_2 + j_3 \rightarrow j_{12} + j_3 \rightarrow J$

the other way is

$j_ 1 + j_2 + j_3 \rightarrow j_1 + j_{23} \rightarrow J$

The 6-j symbol is

$\begin{pmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{pmatrix}$

We can see that there are 4 vector-sum must satisfy.

$\Delta(j_1, j_2, j_{12})$

$\Delta(j_2, j_3, j_{23})$

$\Delta(j_1, j_{23}, J)$

$\Delta(j_{12}, j_3, J)$

If we draw a line to connect these 4 vector-sum, we have:

In Mathematica, there is a build in function

$\textrm{SixJSymbol}[ \left\{j_1, j_2, j_{23} \right\}, \left\{j_3, J , j_{23} \right\}]$

The 9-j symbol is the coupling between 4 angular momenta, $j_1, j_2, j_3, j_4$.

The 9-j symbol can be used in coupling 2 nucleons, $l_1, s_1, l_2, s_2$.

The 9-j symbol is

$\begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

We can see, each row and column must satisfy the vector-sum.

Unfortunately, there is no build in function in Mathematica. The formula for 9-j symbol is

$\displaystyle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix} \\ = \sum_{g} (-1)^{2g} (2g+1) \begin{pmatrix} l_1 & s_1 & j_1 \\ j_2 & J & g \end{pmatrix} \begin{pmatrix} j_2 & s_2 & j_2 \\ s_1 & g & S \end{pmatrix} \begin{pmatrix} L & S & J \\ g & l_1 & l_2 \end{pmatrix}$

Where $g$ sum all possible value, which can be calculate using the 6 couplings inside the 3 6-j symbols.To check your result, the coupling between $d_{5/2}$ and $f_{7/2}$ to from a $L = 5, S = 0, J = 0$ state is

$\begin{pmatrix} 2 & 1/2 & 5/2 \\ 3 & 1/2 & 7/2 \\ 5 & 0 & 5 \end{pmatrix} = \frac{1}{2\sqrt{770}}$