Number of ways for total spin-1

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The discussion focuses on calculating the total number of ways to achieve a total spin of S = 1 using 2N spin-1/2 particles. For N = 1, the only combination is 1 way, while for N = 2, the calculation leads to 6 ways, though there is confusion regarding the correct approach. Participants explore the method of counting arrangements with positive and negative spins, suggesting that for 2N spins, the total number of ways can be expressed as a combination formula. Corrections are made to ensure the general formula aligns with specific cases, ultimately clarifying that the correct expression is 2N choose N-1. The conversation emphasizes the importance of accurate calculations in quantum spin systems.
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Homework Statement
Enumerate the number of independent ways in which 2N spin-1/2's can be com-
bined to form a total spin-1. Here, N is an integer.
Relevant Equations
Addition of spins, s1 + s2 = S
s1 = s2 = 1/2
To the extent I understood this question, we have to enumerate the total no. of ways to get a total of spin S = 1 from 2N number of spin-1/2's.
Now, I think that by spin-1/2's, the question is referring to s1 = s2 = 1/2 (and not something like 3/2, 5/2, ...).
When N = 1, we have 1/2 + 1/2 => No. of ways = 1
When N = 2, we have 1/2 + 1/2 + 1/2 + 1/2 => No. of ways = 3 + 2 + 1 = 6
Likewise,
For N = N, No. of ways should be given by = (2N-1)+(2N-2)+(2N-3)+...+3+2+1
I am sure I'm missing many things here, this question carries 15 marks.
Kindly help !
 
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tanaygupta2000 said:
When N = 2, we have 1/2 + 1/2 + 1/2 + 1/2
That does not add up to 1, and I don't see how it would lead to 6 anyway.
 
Yes sir.
Can I follow this approach:
  • For two spins => 1/2 + 1/2 = 1 way
  • For four spins => (-1/2) + 1/2 + 1/2 + 1/2 = 4 ways since minus sign can be on any of the 4 spins
  • For six spins => (-1/2) + (-1/2) + 1/2 + 1/2 + 1/2 + 1/2 = 15 ways of arranging minus signs
  • Likewise, for 2N spins, total no. of ways = 2N[C](2N-1) (C --> Combination)
and so on?
Is this method okay as required in the question?
 
Last edited:
tanaygupta2000 said:
Yes sir.
Can I follow this approach:
  • For two spins => 1/2 + 1/2 = 1 way
  • For four spins => (-1/2) + 1/2 + 1/2 + 1/2 = 4 ways since minus sign can be on any of the 4 spins
  • For six spins => (-1/2) + (-1/2) + 1/2 + 1/2 + 1/2 + 1/2 = 15 ways of arranging minus signs
  • Likewise, for 2N spins, total no. of ways = 2N[C](2N-1) (C --> Combination)
and so on?
Is this method okay as required in the question?
Your specific cases look good, but don’t match your general formula (##^{2N}C_{2N-1}##).
 
haruspex said:
Your specific cases look good, but don’t match your general formula (##^{2N}C_{2N-1}##).
Yes sir, sorry I meant ##^{2N}C_{N-1}## number of ways are possible for 2N spins.
 
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