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Probability of states

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider an ideal system of 5 non-interacting spin 1/2 particles in the absence of an external magnetic field. What is the probability that n of the five spins have spin up for each of the cases n = 0, 1, 2, 3, 4, 5?


    2. Relevant equations
    I'm guessing [itex]\frac{N!}{n!(N-n)!}[/itex]


    3. The attempt at a solution

    I've done total number of arrangements Ω = [itex]\frac{N!}{n!(N-n)!}[/itex]

    done this for each case, n=0,1,2 etc.

    Then p = [itex]\frac{1}{Ω}[/itex] for the corresponding Ω to get the probability.

    This is a guess and I'm not really sure if I'm going the right way.

    I'm getting probability-like values, such as p=0.2 for cases n=1 and n=4.

    p=0.1 for cases n=2, n=3.

    Would appreciate some help.

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2014 #2

    vela

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    This isn't the total number of arrangements. You have 5 particles, each of which can be in one of two state. How many arrangements can you make? This is just like asking how many outcomes you can get from flipping 5 coins.

    You also need to figure out what ##\Omega## represents if it's not the total number of arrangements.

    You may want to read about the binomial distribution again.
     
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