Good Quantum numbers in L-S coupling

I’m having a hard time understanding a few things about L-S and j-j coupling in 2 (or more) electron atoms. What I picked up from our lectures is the following:

The Hamiltonian can be “broken up” into 3 parts, a Central field part, a residual electrostatic part and a Spin-Orbit part:

$$hat{H} = hat{H}_{CF}+Deltahat{H}_{RE} + Deltahat{H}_{SO}$$

where we can treat $Deltahat{H}_{RE}$ and $Deltahat{H}_{SO}$ as small perturbations. Which of these terms has to be tackled first is decided by their relative size, so that $Deltahat{H}_{RE} > Deltahat{H}_{SO}$ leads to what we call the LS-coupling scheme, where we treat $Deltahat{H}_{RE}$ first, and $Deltahat{H}_{RE} < Deltahat{H}_{SO}$ gives us the jj-coupling scheme, where $Deltahat{H}_{SO}$ is treated first.

In the LS-coupling scheme we find that in order to diagonalise $Deltahat{H}_{RE}$ we have to pick a basis of eigenstates labeled by the quantum numbers $|L,M_L,S,M_Srangle$ as these diagonalise the perturbation $Deltahat{H}_{RE}$. This then leads to energy shift, which depend on $L$ and $M$, $Delta E_{RE} = Delta E_{RE}left(L,Sright)$, and the introduction of term symbols.

Now we treat the smaller perturbation $Deltahat{H}_{SO}$. According to our lectures the following is true:
$$Deltahat{H}_{SO} = sum_i beta_i underline{hat{l}}_i cdot underline{hat{s}}_i$$
and also that:
$$ langle underline{hat{l}}_i cdot underline{hat{s}}_i rangle propto langle underline{hat{L}}cdot underline{hat{S}}rangle = frac{1}{2} langle hat{J}^2 – hat{L}^2 – hat{S}^2 rangle $$

This seems to imply to me that, actually, the basis states of $|L,S,J,M_J rangle$ would diagonalise both the Residual Electrostatic perturbation and also the Spin-Orbit interaction at the same time.

This must obviously be wrong as if it was possible to diagonalise both perturbations at the same time using the same set of eigenfunctions, there would be no need to distinguish between the two cases of L-S coupling and j-j coupling.

I’d greatly appreciate it, if someone could explain to me where I went to wrong, or was able to point me to some resources explaining this in more detail.

How to add numbers to Adobe Illustrator anchor points?

You know the “connect the dots” books for children where you connect the dots from 1 to 100 and a picture becomes visible?

I would love to do the same in Adobe Illustrator.

Is there a way (with any kind of scripting maybe) to add numbers to anchor points?

Is there a way (with any kind of scripting maybe) to color the points based on their anchor index (for example points 1-99 are black, 100-199 are blue etc.)?

Representation of real numbers

Take a look at section 2.2.2 of this book (from Page-15 to 16).
>

$$max f (x)= x sin(10πx)+2.0 … … … … … (2.8)$$
$$s.t. −1 ≤ x ≤ 2$$

2.2.2 Representation and Evaluation

We can use a real number, in the range $[−1,2]$, to represent a solution in Eq. $2.8$ directly. Many operators can handle real number representation. But we use the binary code or binary representation here for two reasons. GAs were originally proposed to be binary code to imitate the genetic encoding of natural organisms. On the other hand, binary code is good for pedagogy. A binary chromosome is necessary to represent a solution x in the scale $[−1,2]$. The same holds for the binary representation of real numbers in a computer.

In binary code, we cannot represent a real number completely correctly, so a
trade-off is necessary. A tolerance needs to be defined by the user, which means the errors below the tolerance are extraneous. If we divide the definition domain into $2^1 = 2$ parts evenly and select the smallest number in the parts to represent any number in the division, we can only represent $−1$ and $0.5$ by $0$ and $1$ respectively. $2^2 = 4$ divisions make the $00$, $01$, $10$, and $11$ represent $−1$, $−0.25$, $0.5$, and $1.25$, respectively. The larger division number we select, the less error there is in representing a real number on binary code. Suppose we use $100$ binary codes to represent a real number in the range $[−1,2]$; the maximum error is $frac{3}{2100} ≈ {2.37}^{−30}$, which would be satisfactory for most users. In this way, we can represent a real number with any accuracy requirements.

In this problem, we use $l = 12$ binary codes to represent one real number as follows, which constitutes a chromosome to be evolved.

Actually, I haven’t understood this text. I know that binary numbers are already able to represent fractions.

So, what are they talking about?

Representation of real numbers in Genetic Algorithm

Take a look at section 2.2.2 of this book (from Page-15 to 16).

2.2.2 Representation and Evaluation

We can use a real number, in the range $[−1,2]$, to represent a solution in Eq. $2.8$ directly. Many operators can handle real number representation. But we use the binary code or binary representation here for two reasons. GAs were originally proposed to be binary code to imitate the genetic encoding of natural organisms. On the other hand, binary code is good for pedagogy. A binary chromosome is necessary to represent a solution x in the scale $[−1,2]$. The same holds for the binary representation of real numbers in a computer.

In binary code, we cannot represent a real number completely correctly, so a
trade-off is necessary. A tolerance needs to be defined by the user, which means the errors below the tolerance are extraneous. If we divide the definition domain into $2^1 = 2$ parts evenly and select the smallest number in the parts to represent any number in the division, we can only represent $−1$ and $0.5$ by $0$ and $1$ respectively. $2^2 = 4$ divisions make the $00$, $01$, $10$, and $11$ represent $−1$, $−0.25$, $0.5$, and $1.25$, respectively. The larger division number we select, the less error there is in representing a real number on binary code. Suppose we use $100$ binary codes to represent a real number in the range $[−1,2]$; the maximum error is $frac{3}{2100} ≈ {2.37}^{−30}$, which would be satisfactory for most users. In this way, we can represent a real number with any accuracy requirements. In this problem, we use $l = 12$ binary codes to represent one real number as follows, which constitutes a chromosome to be evolved.

Actually, I haven’t understood this text. I know that binary numbers are already able to represent fractions.

So, what are they talking about?

How do you solve this [1,5,9,4] pair numbers problem?

So given an array of [1,5,9,4], find all pairs of numbers that can sum/subtract/divide/multiply to another number in the array(that number is not part of the pair), and return the total count of pairs satisfying this condition. (If a pair has 2 math operators that fit to the same number or if a pair can combine to 2 different numbers, dont count it twice).

How do I start this? and how do I explain my asymptotic complexity(Im a noob wit this)?

Geoserver polygon labeling Arabic numbers using SLD

I’ve used SLD to label polygon features with their parcel Ids. The problem is for Ids include ‘.’ and ‘-‘ marks, the ordering of character is messed up. For example 39.1 changes to 139. I’ve checked them in Pgsql and ArcMap which show them correctly.



  
    parcels
    
      parcels

      
        
          Single symbol
          2000
          
             
              #7e7e7e
              0.7
              bevel
            
          
           
            
              
                geom
              
            
            
              pelak_fa
            
            
              Iranian Sans
              11
              normal
            
            
              
                
                  0.5
                  0.0
                
                
                  0
                  5
                
              
            
            
              #523735
            
            150
            50
            yes
          
        
      

    
  

Given a finite collection of numbers, the products obtained by multiplying them in any order are all equal.

How do I use induction to prove the following for $ngeq 3$ ($n$ is the number of numbers in the finite collection)?

1) Given a finite collection of numbers, the products obtained by multiplying
them in any order are all equal.

2) Given a finite collection of numbers, the sums obtained by adding
them in any order are all equal.

The base case can be shown using the commutative and associative laws for addition and multiplication.

How can I stop myself from pronouncing abbreviations and numbers in an L2 in my L1?

(The question title is clumsy. Any help to improve it is appreciated.)

When reading a text in English, occasionally I will meet a symbol. I of course know how to speak that symbol in English, but most of the time I pronounce it as it is in my native language. It can be symbols (!@#$%^&*), letters and abbreviations (WTO, HIV) or numbers. The only exception is when the symbol only has one digit or character (1, 2, 3 or A, B, C) or in a well-known structure that I have got used to pronounce it in English (the United Nation is not abbreviated as “UN” in my language, therefore there is a room in my mind for “UN” to be pronounce as “u-en”).

For example, if I see a sentence “WTO has 162 member states”, I will read it in my mind as “W-tê-O has một-sáu-hai member states”, not “W-T-O has one-six-two member states”

I think the problem is understandable: come first, serve first. Usually, this makes no harm to my comprehension when reading an English article, but sometimes, it does.

Stirling numbers Sum

Let
$$
a_n(i,j)=sum_{k=max(i,j)}^{n}s(n,k) S(k,i) S(k,j),
$$

here $s, S$ are the Stirling numbers if the first and second kind.
I want to simplify the expression. So far I have got the following
begin{gather*}
a_n(n,i)=S(n,i),\
a_n(n-1,i)=binom{n}{2}left(S(n,i)-S(n-1,i) right)
end{gather*}

$
a_n(n-2,i)=frac{1}{4}binom{n}{3}left((3n{-}5)S(n,i){-}6(n{-}1)S(n{-}1,i)+(3n{-}1)S(n{-}2,i) right)
$

I there a closed expression for the sum?