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Finding coefficients of superposition of states

  1. Aug 13, 2015 #1
    I have some troubles in finding coefficients of superposition of states.

    I have 2 particles, their spins are s1=3/2 and s2=1/2.
    At t=0, the system is described by |a(0)>=|3/2, 1/2, 1/2, 1/2>

    I have to find |a(t)>.

    I have thought to proceed in the following way:

    1) use the basis |s, s_z> where s=s1+s2 and s_z= s_1z+s_2z and find the expressions of these vectors in function of the "old" basis (old basis: |s1, s_1z, s_2, s_2z> )
    2) find the expression of |a(0)> in this new basis and then find its expression in function of t.

    But something went wrong... For example, if i want to find |s=2, s_z=1>, I have:
    |s=2, s_z=1>=a1 |3/2, 3/2, 1/2, -1/2>+a2 |3/2, 1/2, 1/2, 1/2>

    If I apply the operator J_, I obtain
    [tex]0= \sqrt 3 a_1 |3/2, 1/2, 1/2, -1/2>+ 2 a_2 |3/2, -1/2, 1/2, 1/2>+a_2 |3/2, 1/2, 1/2, -1/2>[/tex].. is it wrong?

    And now, how can I find a1 and a2 (normalized)?
     
  2. jcsd
  3. Aug 13, 2015 #2

    blue_leaf77

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    Time evolution requires the knowledge of the system's Hamiltonian.
     
  4. Aug 13, 2015 #3
    you're right. It is [tex]H=\alpha s_1 \cdot s_2[/tex]
    where alpha is constant. Could you help me?
     
  5. Aug 13, 2015 #4

    atyy

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    Find the eigenstates of the Hamiltonian.
     
  6. Aug 13, 2015 #5

    blue_leaf77

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    Step 1) and 2) are correct.
    I don't see anything wrong, you just haven't founnd the coefficients yet.
    They are called Clebsch-Gordan coefficients. There is a way to calculate them so that the states are normalized, but you can also make use of your internet connection to find a related table.
     
  7. Aug 13, 2015 #6
    I have to calculate them ;)
     
  8. Aug 13, 2015 #7

    blue_leaf77

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    There is one convention in calculating Clebsch-Gordan coefficient, namely the coefficient corresponding to ##|s_{max},s_{max}\rangle## is agreed to be unity. In other words
    $$ |s_{max},s_{max}\rangle = |s_1,s_1\rangle |s_2,s_2\rangle $$
    So, to get the coefficients for ##|2,1\rangle##, you should start by applying lowering operator, ##S_- = S_{1-}+S_{2-}##, to the state ##|2,2\rangle##.
     
  9. Aug 14, 2015 #8
    well, I have tried.

    [tex]j-|2,2>=|2,1|=k_1 |3/2, 1/2, 1/2, 1/2>+k_2 |3/2, 3/2, 1/2, -1/2>[/tex] where |[tex]k_1>=\sqrt 3, k_2=1[/tex]
    if I want to normalize them,[tex] |k_1>=\sqrt 3 /2, k_2=1/2[/tex]

    If I apply j_ again,

    [tex]j- |2,1>=|2,0>=k_1 *k_3 |3/2, -1/2, 1/2, 1/2>+k_1*k_4|3/2, 1/2, 1/2, -1/2>+k_2*k_5|3/2, 1/2, 1/2, -1/2> [/tex]

    where k_3=2, k_4=1, k_5=sqrt 3...

    And so, I haven't obtained the correct result... what's wrong?
     
  10. Aug 14, 2015 #9

    blue_leaf77

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    Remember that ##J_-|j,m\rangle = C_-|j,m-1\rangle##, regardless of whether the lowering operator is a total or a single operator

    Calculate the coefficient which is supposed to be in front of ##|2,1 \rangle ## on the left hand side, and you will find the same answer as you did here. Do the same for your second calculation.
     
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