Number of ways for total spin-1

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SUMMARY

The discussion focuses on calculating the number of ways to achieve a total spin of S = 1 using 2N spin-1/2 particles. The correct formula for the number of arrangements is given as 2N choose N-1 (2N[C](N-1)). For N = 1, there is 1 way; for N = 2, there are 6 ways; and for N = 3, there are 15 ways. The participants confirm the validity of this approach and clarify the correct application of combinations in the context of spin arrangements.

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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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