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Superposition of states, spin 1/2

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Two particles, their spin are 1/2.

    The hamiltonian is ##H=\gamma s_1 \cdot s_2##

    At t=0, the state ##|\alpha(0)>## is such as ##s_{1z}|\alpha(0)>=\hbar/2 |\alpha(0)>## and ##s_{2z}|\alpha(0)>=\hbar/2 |\alpha(0)>##. Find the state ##|\alpha(0)>##.


    2. The attempt at a solution
    I think that ##|\alpha(0)>=|1,1>## in the base of total angular momentum, i.e. |1/2, 1/2> in the other one. But the correct result seems to be ##|\alpha(0)>=1/ \sqrt2 [|1,1; 1/2, 1/2> + 1/ \sqrt 2 (|1,0; 1/2, 1/2>+|0,0; 1/2, 1/2>)]##.. what's wrong?

    I used the following table:
    pSdDYdr.png
     
  2. jcsd
  3. Sep 15, 2015 #2

    blue_leaf77

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    I don't see why an unnormalized state should be the correct answer, besides which bases does it use on the RHS?

    EDIT: No, I didn't pay a closer attention on the parentheses, yes this state is normalized.
     
    Last edited: Sep 15, 2015
  4. Sep 15, 2015 #3
    I haven't undestood the result in the book. Can you explain me how you would solve this point?
     
  5. Sep 15, 2015 #4
    I haven't understood why in the result appear the states |1,0;1/2,1/2> and |0,0;1/2,1/2>. They don't have s1z=1/2 AND s2z=1/2. They are linear combination of ##s_{1z}=\pm 1/2## AND ##s_{2z}=\pm 1/2##!!! :(
     
  6. Sep 15, 2015 #5

    blue_leaf77

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    Did you post the complete question?
     
  7. Sep 15, 2015 #6
    yes..
     
  8. Sep 17, 2015 #7

    Orodruin

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    You are correct, the only state which is an eigenstate of both ##s_{1z}## and ##s_{2z}## with eigenvalues ##\hbar/2## is the state ##|\uparrow\rangle \otimes |\uparrow\rangle##.
     
  9. Sep 18, 2015 #8
    thanks to all of you!
     
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