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I am trying to calculate eq (A3)

ClebschCoefficient[Sd_, Sc_, Sd3_, Sc3_, S_, L_, L3_, J_, J3_] := 
  Module[{ClebschGordan1, ClebschGordan2}, 
   ClebschGordan1 = 
    ClebschGordan[{Sd, Sd3}, {Sc, Sc3}, {S, Sd3 + Sc3}];
   ClebschGordan2 = ClebschGordan[{S, Sd3 + Sc3}, {L, L3}, {J, J3}];
   ClebschGordan1*ClebschGordan2];

(*Function to Calculate the Specific States*)

calculateState[Sd_, Sc_, L_, J_, J3_] := 
  Module[{state}, 
   state = Sum[
     ClebschCoefficient[Sd, Sc, Sd3, Sc3, S3, L, L3, J, J3]*
      Ket[Sd3, Sc3, L3], {Sd3, -Sd, Sd}, {Sc3, -Sc, 
      Sc}, {S3, -(Sd3 + Sc3), Sd3 + Sc3}, {L3, -L, L}];
   state];

I would like the output as

enter image description here

(*Define Quantum Numbers for Each State*)
state1 = calculateState[0, 1/2, 1, 1/2, 1/2];
state2 = calculateState[1, 1/2, 1, 1/2, 1/2];
state3 = calculateState[0, 1/2, 1, 3/2, 3/2];
state4 = calculateState[1, 1/2, 1, 3/2, 3/2];
state5 = calculateState[1, 1/2, 1, 5/2, 5/2];

The output i am getting is

state1 = Ket[0, 1/2, 0]/Sqrt[3]

state2 = -(1/3) Ket[0, 1/2, 0] + 1/3 Sqrt[2] Ket[1, -(1/2), 0] + Ket[1, 1/2, -1]/Sqrt[2]

state3 = Ket[0, 1/2, 1]

state4 = -(Ket[0, 1/2, 1]/Sqrt[3]) + Sqrt[2/3] Ket[1, -(1/2), 1] + Sqrt[3/5] Ket[1, 1/2, 0]

state5 = Ket[1, 1/2, 1]

I don't understand if there is mistake in the code or in the selection rule. If there is mistake in selection rule i would like to know the explanation for it if possible. Thank you in advance.

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  • $\begingroup$ I see at least one inconsistency in your code: according to Eq.(A3) the right hand side is denoted as $|{}^{2S+1}L_J;J_3\rangle$, i.e. it depends on $S$, but the list of arguments of your function calculateState[Sd_, Sc_, L_, J_, J3_] does not contain the $S$ parameter. $\endgroup$
    – yarchik
    Commented Jul 8 at 14:30
  • $\begingroup$ Yeah since it works as a diquark-quark system i have considered the total spin $S = Sd + Sc$. $\endgroup$
    – Anshul Bokade
    Commented Jul 9 at 3:52
  • 1
    $\begingroup$ Yeah i figured it out. I was making the mistake in selection rule of $S$. Making $S$ as an explicit parameter and giving proper value gives me all the states. Thank you for your help. $\endgroup$
    – Anshul Bokade
    Commented Jul 9 at 4:17

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