Calculating the Matrix Elements for (^39)K

Determining the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
        m' = \(3 | μ\_1\%1 | 
              \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
        m = 2⟩\) matrix element from the measured lifetime

The lifetime of the L = 1 excited states in (^39)K, τ, has been measured to be 25.8  ns.  All but one of the excited states can spontaneously decay to several ground states. Because \(TraditionalForm
      \`\(| F' = 3, \ m' = 3⟩\)\) decays only to \(TraditionalForm
      \`\(\(| F\) = 2, \ m = 2⟩\)\), the spontaneous lifetime is related to \(TraditionalForm
      \`μ = 
        ⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(3 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩\) by  (1/(Subscript[τ, sp])) = ((4 (k^3))/(3 ℏ)) (μ^2).

\(\(ℏ = 1.054×10\^\(-27\)\  (*\ erg·s\ *) ; \ 
    k = \(2  π\)\/\(766.7×10\^\(-7\)\)\  (*\ cm\^\(-1\)\ *) ; \ 
    τ = 25.8×10\^\(-9\)\  (*\ s\ *) ; 

    μ = \@\(\(3  ℏ\)\/\(4 \( k\^3\) τ\)\)\ 
     (*\ esu·cm\ *) \)\)

Generally, μ is put into Debye (D). 1  D = (10^(-18)) esu·cm.

μ = (μ/(10^(-18)))

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

Calculating the reduced matrix element \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
          \(3 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
            \)⟩\)

Using JET 92.1:
⟨n' (l' s) j' m' | (Subscript[μ, q]^1) | 
              n(l s) j m⟩=(((-1))^(j' - m'))(({{j', 1, j}, {-m', q, m}}))⟨n' (l' s) j' || (μ^1) ||  
                n(l s) j⟩
we can calculate the matrix elements for other transitions between these two F-levels. Knowing the left hand side of the equation for one transition allows us to calculate the reduced matrix element in the F = I + J basis. Rewriting 92.1: 
⟨n' (J' I) F' m' | (Subscript[μ, q]^1) | 
              n(J I) F m⟩=(((-1))^(F' - m'))(({{F', 1, F}, {-m', q, m}}))⟨n' (J' I) F' || (μ^1) ||  
                n(J I) F⟩
The following function calculates the entire coefficient on the right side of the equation:

\(threeJcoeff[fp_, mp_, q_, f_, m_] := 
      \((\(-1\))\)\^\(fp - mp\)×
		
        ThreeJSymbol[{fp, \(-mp\)}, {1, q}, {f, m}]\)

threeJcoeff[3, 3, 1, 2, 2]

This gives \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(3 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \(\@7\) μ\). Using this result and 92.1 repetitively, we an determine all of the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements.

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

We will express all matrix elements in terms of μ as calculated above. The function f232 (2 for the (Subscript[D, 2]) line, 3 for F' = 3 and 2 for F = 2) performs the calculation:

f232[mp_, q_, m_] := threeJcoeff[3, mp, q, 2, m] (Sqrt[7])

Calculations

f232[3, 1, 2]

f232[2, 0, 2]

f232[1, -1, 2]

f232[2, 1, 1]

f232[1, 0, 1]

f232[0, -1, 1]

\(\(f232[1, 1, 0]
\)\)

\(\(
f232[0, 0, 0]\)\)

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(3 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = 1\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(2 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = 1\/\@3\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(1 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = 1\/\@15\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(2 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \@\(2\/3\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 2 \@\( 2\/15\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 1\/\@5\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = \@\(2\/5\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 3, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = \@\(3\/5\)\)
The remaining elements can be determined using the following rule:
⟨F',  m' | (Subscript[μ, q]^1) | F, 
               m⟩= (((-1))^(F' - m'))(({{F', 1, F}, {-m', q, m}}))(Sqrt[7])μ=(((-1))^(-m' - F - 1))(({{F', 1, F}, {m', -q, -m}}))(Sqrt[7])μ=(((-1))^(-F' - F - 2  m' - 1)) 
                  ⟨F',  
                    -m' | (Subscript[μ, -q]^1) | F,  
                    -m⟩
 Since for (^39)K the 2×m will be even, we can write:
\(TraditionalForm
      \`⟨F', \ m' | μ\_q\%1 | F, \ m
        ⟩\)=\(TraditionalForm
      \`\(\((\(-1\))\)\^\(\(-F'\) - F - 1\)\) 
        ⟨F', \ 
          \(-m'\) | μ\_\(-q\)\%1 | F, \ 
          \(-m\)⟩\)
For transitions between these two levels,the quantity \(TraditionalForm
      \`ϕ = 
        \(\((\(-1\))\)\^\(\(-F'\) - F - 1\) = 
          \(\((\(-1\))\)\^\(\(-3\) - 2 - 1\) = 1\)\)\).
The matrix elements can be checked by using the sum-rule:
\(TraditionalForm
      \`∑\+\(m', m
            \)\((\(|⟨F', \ 
                  m' | μ\_q\%1 | F, \ m
                  ⟩ | \))\)\^2 = 
        \(\((\(|⟨F' || μ\^1 || F⟩ | 
                  \))\)\^2\/3 = \(7  μ\)\/3\)\) 
For q = ±1:

1 + (1/15) + (2/3) + (1/5) + (2/5) == (7/3)

For q = 0:

2×((1/3) + (8/15)) + (3/5) == (7/3)

Calculating the other (Subscript[D, 2]) matrix elements

The reduced matrix elements in the (Subscript[D, 2]) line

To calculate the matrix elements between different F-levels of the (Subscript[D, 2])line, we can use JET 104.1:
⟨(J' I) F' || (μ^1) || 
              (J I) F⟩ = (((-1))^(J' + I + F + 1))(Sqrt[(2  F' + 1) (2  F + 1)]){({{J', I, F'}, {F, 1, J}})}⟨J' || (μ^1) || J
                ⟩
The function sixJcoeff evaluates the coefficient on the right side of the equation:

sixJcoeff[jp_, i_, fp_, j_, f_] := 
      (((-1))^(jp + i + f + 1)) 
        (Sqrt[(2 fp + 1) (2  f + 1)]) 
        SixJSymbol[{jp, i, fp}, {f, 1, j}]

We know \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(3 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \(\@7\) μ\), so we can calculate \(TraditionalForm
      \`⟨J' = 
            \(3\/2 || μ\^1 || J = 1\/2\)⟩\ by\ 
        \(evaluating : \)\)

sixJcoeff[3/2, 3/2, 3, 1/2, 2]

This gives \(TraditionalForm
      \`⟨J' = 
            \(3\/2 || μ\^1 || J = 1\/2\)⟩\  = 
        2 \( μ.\)\)We can use this result and 104.1 again to calculate other reduced matrix elements in the (Subscript[D, 2]) line.

g2[fp_, f_] := sixJcoeff[3/2, 3/2, fp, 1/2, f]×2

Calculations

g2[3, 2]

g2[2, 2]

g2[1, 2]

g2[2, 1]

g2[1, 1]

g2[0, 1]

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(3 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \@7\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(2 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \(-\@\(5\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(1 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = 1\/\@2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(2 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1
              \)⟩ = \@\(5\/2\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(1 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1
              \)⟩ = \(-\@\(5\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(0 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1
              \)⟩ = 1\)
To check these results, we can use the following sum-rule:
\(TraditionalForm
      \`∑\+\(F'\)\((
              \(|⟨\((J' I)\) F' || μ\^1 || 
                    \((J\ I)\) F⟩ | \))\)\^2 = 
        \(\((\(|⟨J' || μ\^1 || J⟩ | 
                    \))\)\^2\) 2  F + 1\/\(2  J + 1\)\)
For F = 2,  J = (1/2), we have:

((((Sqrt[7])))^2)+(((-(Sqrt[(5/2)])))^2) + ((((1/(Sqrt[2]))))^2)== (2^2) ((2 (2)  + 1)/(2 (1/2) + 1))

For F = 1, J = (1/2), we have:

((((Sqrt[(5/2)])))^2) +(((-(Sqrt[(5/2)])))^2)+(((1))^2)== (2^2) ((2 (1)  + 1)/(2 (1/2) + 1))

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

With \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 
            \(2 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \(-\@\(5\/2\)\)\) and JET 92.1 we can calculate the matrix elements between F' = 2 and F = 2. Defining f222:

f222[mp_, q_, m_] := 
      threeJcoeff[2, mp, q, 2, m] × -(Sqrt[(5/2)])

Calculations

f222[2, 0, 2]

f222[1, -1, 2]

f222[2, 1, 1]

f222[1, 0, 1]

f222[0, -1, 1]

f222[1, 1, 0]

f222[0, 0, 0]

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = \(-\(1\/\@3\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = \(-\(1\/\@6\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 1\/\@6\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \(-\(1\/\(2\ \@3\)\)\)\)
\(TraditionalForm
      \`\(⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \)\)-(1/2)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 1\/2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 0\)

The ϕ for these two levels is \(TraditionalForm
      \`ϕ = \(\((\(-1\))\)\^\(\(-2\) - 2 - 1\) = \(-1.\)\)\)

Checking these results using the sum-rule \(TraditionalForm
      \`∑\+\(m', m
            \)\(( | ⟨F', \ 
                  m' | μ\_q\%1 | F, \ m
                  ⟩ | )\)\^2 = 
        \(( | ⟨F' || μ\^1 || F⟩ | )
              \)\^2\/3\):
For q = ±1:

(((((-1)/(Sqrt[6]))))^2) + ((((1/(Sqrt[6]))))^2) + (((((-1)/2)))^2) + 
        ((((1/2)))^2) == (1/3) (((-(Sqrt[(5/2)])))^2)

For q=0:

2((((((-1)/(Sqrt[3]))))^2)+ (((((-1)/(2 (Sqrt[ 3])))))^2))+0==(1/3) (((-(Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

f212[mp_, q_, m_] := threeJcoeff[1, mp, q, 2, m] ×(1/(Sqrt[2]))

Calculations

f212[1, -1, 2]

f212[1, 0, 1]

f212[0, -1, 1]

f212[1, 1, 0]

f212[0, 0, 0]

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = 1\/\@10\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \(-\(1\/\(2\ \@5\)\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 1\/\(2\ \@5\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 1\/\(2\ \@15\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = \(-\(1\/\@15\)\)\)
\(TraditionalForm
      \`ϕ = \ \(\((\(-1\))\)\^\(\(-1\) - 2 - 1\) = 1\)\)

((((1/(Sqrt[10]))))^2) + ((((1/(2 (Sqrt[ 5])))))^2) + ((((1/(2 (Sqrt[ 15])))))^2) == 
      (1/3) ((((1/(Sqrt[2]))))^2)

2 (((-(1/(2 (Sqrt[5])))))^2) + (((((-1)/(Sqrt[15]))))^2) == 
      (1/3) ((((1/(Sqrt[2]))))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 1, 
        \ m⟩\) matrix elements

f221[mp_, q_, m_] := threeJcoeff[2, mp, q, 1, m] ×(Sqrt[(5/2)])

Calculations

f221[2, 1, 1]

f221[1, 0, 1]

f221[0, -1, 1]

f221[1, 1, 0]

f221[0, 0, 0]

Results in terms of μ

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = 1\/\@2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = 1\/2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = 1\/\(2 \@ 3\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = 1\/2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = 1\/\@3\)
ϕ = (((-1))^(-2 - 1 - 1)) = 1

((((1/(Sqrt[2]))))^2) + ((((1/(2 (Sqrt[ 3])))))^2) + ((((1/2)))^2) == 
      (1/3) ((((Sqrt[(5/2)])))^2)

2 ((((1/2)))^2) + ((((1/(Sqrt[3]))))^2) == (1/3) ((((Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 1, 
        \ m⟩\) matrix elements

f211[mp_, q_, m_] := 
      threeJcoeff[1, mp, q, 1, m] × -(Sqrt[(5/2)])

Calculations

f211[1, 0, 1]

f211[0, -1, 1]

f211[1, 1, 0]

f211[0, 0, 0]

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = \(-\(\@\(5\/3\)\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = \(-\(\@\(5\/3\)\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = \@\(5\/3\)\/2\)
\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = 0\)
\(TraditionalForm
      \`ϕ = \(\((\(-1\))\)\^\(\(-1\) - 1 - 1\) = \(-1\)\)\)

(((((-1)/2) (Sqrt[(5/3)])))^2) + ((((1/2) (Sqrt[(5/3)])))^2) == 
      (1/3) (((-(Sqrt[(5/2)])))^2)

2 (((((-1)/2) (Sqrt[(5/3)])))^2) + 0 == 
      (1/3) (((-(Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 0, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 1, 
        \ m⟩\) matrix elements

f201[mp_, q_, m_] := threeJcoeff[0, mp, q, 1, m] × 1

Calculations

f201[0, -1, 1]

f201[0, 0, 0]

Results in μ units

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 0, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = 1\/\@3\)

\(TraditionalForm
      \`⟨\((J' = 3\/2, I = 3\/2)\) F' = 0, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = \(-\(1\/\@3\)\)\)
\(TraditionalForm
      \`ϕ = \(\((\(-1\))\)\^\(\(-1\) - 1 - 1\) = \(-1\)\)\)

((((1/(Sqrt[3]))))^2) == (1/3) (((1))^2)

(((((-1)/(Sqrt[3]))))^2) == (1/3) (((1))^2)

Calculating the (Subscript[D, 1]) matrix elements

Determining the \(TraditionalForm
      \`\(, 
        ⟨\((l' = 1, s = 1\/2)\) J' = 
            \(1\/2 || μ\^1 || \((l = 0, s = 1\/2)\) J = 
              1\/2\)⟩\)\) matrix element

To get \(TraditionalForm
      \`\(, 
        ⟨\((l' = 1, s = 1\/2)\) J' = 
            \(1\/2 || μ\^1 || \((l = 0, s = 1\/2)\) J = 
              1\/2\)⟩\)\) and the reduced matrix elements in the (Subscript[D, 1]) line, we must reduce the \(TraditionalForm
      \`⟨\((l' = 1, s = 1\/2)\) J' = 
          \(3\/2 || μ\^1 || \((l = 0, s = 1\/2)\) J = 
            1\/2\)⟩\) to the l basis usingJET 104.1 again: ⟨(l' s) J' || (μ^1) || 
              (l s) J⟩ = (((-1))^(l'' + s + J + 1))(Sqrt[(2  J' + 1) (2  J + 1)]){({{l', s, J'}, {J, 1, l}})}⟨l' = 
                1 || (μ^1) || l = 0⟩

sixJcoeff[1, 1/2, 3/2, 0, 1/2]

With \(TraditionalForm
      \`⟨\((l' = 1, s = 1\/2)\) J' = 
            \(3\/2 || μ\^1 || \((l = 0, \ s = 1\/2)\) J = 
              1\/2\)⟩ = 2  μ\):

 \(TraditionalForm
      \`⟨l' = 
            \(1 || μ\^1 || l = 0\)⟩ = 
        \(\@3\) μ\). 
 
 Then, in μ units, \(TraditionalForm
      \`⟨\((l' = 1, s = 1\/2)\) J' = 
          \(1\/2 || μ\^1 || \((l = 0, \ s = 1\/2)\) J = 
            1\/2\)⟩\) is:

sixJcoeff[1, 1/2, 1/2, 0, 1/2]×(Sqrt[3])

\(TraditionalForm
    \`⟨\((l' = 1, s = 1\/2)\) J' = 
          \(1\/2 || μ\^1 || \((l = 0, \ s = 1\/2)\) J = 
            1\/2\)⟩ = \(-\@2\) μ\)

The reduced matrix elements in the (Subscript[D, 1]) line

With \(TraditionalForm
      \`⟨\((l' = 1, s = 1\/2)\) J' = 
          \(1\/2 || μ\^1 || \((l = 0, \ s = 1\/2)\) J = 
            1\/2\)⟩\)=-(Sqrt[2])μ, the (Subscript[D, 1]) reduced matrix elements are given by ⟨(J' I) F' || (μ^1) || 
              (J I) F⟩ = (((-1))^(J' + I + F + 1))(Sqrt[(2  F' + 1) (2  F + 1)]){({{J', I, F'}, {F, 1, J}})}⟨J' || (μ^1) || J
                ⟩ using the function g1:

g1[fp_, f_] :=  
      sixJcoeff[1/2, 3/2, fp, 1/2, f]×(-(Sqrt[2]))

Calculations

g1[2, 2]

g1[1, 2]

g1[2, 1]

g1[1, 1]

Results

\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 
            \(2 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \(-\@\(5\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 
            \(1 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2
              \)⟩ = \@\(5\/2\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 
            \(2 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1
              \)⟩ = \(-\@\(5\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 
            \(1 || μ\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1
              \)⟩ = 1\/\@2\)
Checking with the sum-rule:
\(TraditionalForm
      \`∑\+\(F'\)\((
              \(|⟨\((J' I)\) F' || μ\^1 || 
                    \((J\ I)\) F⟩ | \))\)\^2 = 
        \(\((\(|⟨J' || μ\^1 || J⟩ | 
                  \))\)\^2\) \(2  F + 1\)\/\(2  J + 1\)\)
For F = 2,  J = (1/2), we have:

(((-(Sqrt[(5/2)])))^2) + ((((Sqrt[(5/2)])))^2) == 
      (((-(Sqrt[2])))^2) ((2 (2) + 1)/(2 ((1/2)) + 1))

For F = 1,  J = (1/2), we have:

(((-(Sqrt[(5/2)])))^2) + ((((1/(Sqrt[2]))))^2) == 
      (((-(Sqrt[2])))^2) ((2 (1) + 1)/(2 ((1/2)) + 1))

Calculating the \(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

f122[mp_, q_, m_] := 
      threeJcoeff[2, mp, q, 2, m]×(-(Sqrt[(5/2)]))

Calculations

f122[2, 0, 2]

f122[1, -1, 2]

f122[2, 1, 1]

f122[1, 0, 1]

f122[0, -1, 1]

f122[1, 1, 0]

f122[0, 0, 0]

Results

\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = \(-\(1\/\@3\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = \(-1\)\/\@6\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 1\/\@6\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \(-\(1\/\(2\ \@3\)\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \(-1\)\/2\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 1\/2\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 0\)

The ϕ for these two levels is \(TraditionalForm
      \`ϕ = \(\((\(-1\))\)\^\(\(-2\) - 2 - 1\) = \(-1.\)\)\)

(((((-1)/(Sqrt[6]))))^2) + ((((1/(Sqrt[6]))))^2) + (((((-1)/2)))^2) + 
        ((((1/2)))^2) == (1/3) (((-(Sqrt[(5/2)])))^2)

2 ((((-(1/(Sqrt[3]))))^2) + (((-(1/(2 (Sqrt[ 3])))))^2)) + 
        (((0))^2) == (1/3) (((-(Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 2, 
        \ m⟩\) matrix elements

f112[mp_, q_, m_] := threeJcoeff[1, mp, q, 2, m]×(Sqrt[(5/2)])

Calculations

f112[1, -1, 2]

f112[1, 0, 1]

f112[0, -1, 1]

f112[1, 1, 0]

f112[0, 0, 0]

Results

\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 2⟩ = 1\/\@2\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = \(-\(1\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 1⟩ = 1\/2\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = 1\/\(2\ \@3\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 2\), \ 
          m = 0⟩ = \(-\(1\/\@3\)\)\)
\(TraditionalForm
      \`ϕ = \ \(\((\(-1\))\)\^\(\(-1\) - 2 - 1\) = 1\)\)

((((1/(Sqrt[2]))))^2) + ((((1/2)))^2) + ((((1/(2 (Sqrt[ 3])))))^2) == 
      (1/3) ((((Sqrt[(5/2)])))^2)

2 (((((-1)/2)))^2) + (((((-1)/(Sqrt[3]))))^2) == 
      (1/3) ((((Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 1, 
        \ m⟩\) matrix elements

Calculations

f121[mp_, q_, m_] :=  
      threeJcoeff[2, mp, q, 1, m]×(-(Sqrt[(5/2)]))

f121[2, 1, 1]

f121[1, 0, 1]

f121[0, -1, 1]

f121[1, 1, 0]

f121[0, 0, 0]

Results

\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(2 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = \(-\(1\/\@2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = \(-\(1\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_\(-1\)\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = \(-\(1\/\(2\ \@3\)\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = \(-\(1\/2\)\)\)
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 2, \ 
          m' = \(0 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = \(-\(1\/\@3\)\)\)
ϕ = (((-1))^(-2 - 1 - 1)) = 1

(((((-1)/(Sqrt[2]))))^2) + (((-(1/(2 (Sqrt[3])))))^2) + 
        (((((-1)/2)))^2) == (1/3) (((-(Sqrt[(5/2)])))^2)

2 (((((-1)/2)))^2) + (((((-1)/(Sqrt[3]))))^2) == 
      (1/3) (((-(Sqrt[(5/2)])))^2)

Calculating the \(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
        m' | μ\_q\%1 | \((J = 1\/2, I = 3\/2)\) F = 1, 
        \ m⟩\) matrix elements

f111[mp_, q_, m_] := threeJcoeff[1, mp, q, 1, m]×(1/(Sqrt[2]))

Calculations

f111[1, 0, 1]

f111[0, -1, 1]

f111[1, 1, 0]

f111[0, 0, 0]

Results

\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_0\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 1⟩ = 1\/\(2\ \@3\)\)
⟨(J' = (1/2), I = (3/2)) F' = 1,  
                m' = 0 | (Subscript[μ, -1]^1) | 
                      (J = (1/2), I = (3/2)) F = 1,  
                m = 1⟩ = (1/(2 (Sqrt[3])))
\(TraditionalForm
      \`⟨\((J' = 1\/2, I = 3\/2)\) F' = 1, \ 
          m' = \(1 | μ\_1\%1 | 
                \((J = 1\/2, I = 3\/2)\) F = 1\), \ 
          m = 0⟩ = \(-\(1\/\(2\ \@3\)\)\)\)
⟨(J' = (1/2), I = (3/2)) F' = 1,  
                m' = 0 | (Subscript[μ, 0]^1) | 
                      (J = (1/2), I = (3/2)) F = 1,  
                m = 0⟩ = 0
\(TraditionalForm
      \`ϕ = \(\((\(-1\))\)\^\(\(-1\) - 1 - 1\) = \(-1\)\)\)

((((1/(2 (Sqrt[ 3])))))^2) + (((((-1)/(2 (Sqrt[ 3])))))^2) == 
      (1/3) ((((1/(Sqrt[2]))))^2)

2 ((((1/(2 (Sqrt[ 3])))))^2) + (((0))^2) == (1/3) ((((1/(Sqrt[2]))))^2)