Number of ways for total spin-1

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Homework Help Overview

The discussion revolves around determining the number of ways to achieve a total spin of S = 1 using 2N spin-1/2 particles. Participants are exploring the implications of combining spins and the calculations involved in enumerating these combinations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the number of ways to combine spins, with specific cases for N = 1, 2, and 3 being discussed. There is a focus on the arrangement of spins and the impact of negative signs on the total spin. Questions arise regarding the validity of the proposed general formula and its alignment with specific examples.

Discussion Status

There is an ongoing exploration of different methods to calculate the total number of ways to achieve the desired spin. Some participants are questioning the consistency of the general formula with specific cases, while others are seeking confirmation on their reasoning and calculations.

Contextual Notes

Participants are working within the constraints of a homework problem, which may impose specific requirements for the approach and calculations. There is a noted uncertainty regarding the correct application of combinations in the context of spin arrangements.

tanaygupta2000
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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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