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Rotation Operator

  1. Jan 26, 2013 #1
    How does finding the rotation operator for a spin 1/2 particle differ from finding that of a spin 1 particle?
     
  2. jcsd
  3. Jan 26, 2013 #2
    if you have the generators (angular momentum operators) then its practically the same thing you just exponentiation.
    Im not sure if thats what you're asking..
     
  4. Jan 26, 2013 #3
    Thanks for your reply, but I was pointing to a different road, my question in other words is how to derive, rotation operator for spin 1? How do we get there?
     
  5. Jan 26, 2013 #4

    fzero

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    How did you find the rotation operator for spin 1/2? You probably started with the rules for the generators [itex]J_i[/itex], namely the commutation relation

    [tex] [J_i,J_j] = i \epsilon_{ijk} J_k[/tex]

    and the eigenvalue conditions

    [tex] J^2 | j,m\rangle = j(j+1) | j,m\rangle,~~~J_3 | j,m\rangle = m | j,m\rangle .[/tex]

    You then have to choose a set of basis vectors to correspond to the states [itex]| j,m\rangle[/itex]. You can choose a basis so that [itex]J_3[/itex] is diagonal. From here, you can use trial and error to find a pair of matrices that have the correct commutation relations with [itex]J_3[/itex] and satisfy the [itex]J^2[/itex] equation. Otherwise you can form the raising and lowering operators

    [tex] J_\pm = \frac{1}{\sqrt{2}} (J_1\pm iJ_2),[/tex]

    [tex] J_3 J_\pm | j,m\rangle = (m\pm 1) | j,m\pm 1\rangle[/tex]

    and note that

    [tex] J_+ | 1,1\rangle = J_- | 1,-1\rangle =0.[/tex]

    These last conditions can be solved with less guesswork.

    Once you have the generators, you can exponentiate them to find the rotation matrices.
     
  6. Jan 26, 2013 #5

    morrobay

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    Can entangled spin 1/2 particles have time rate of change spins ? +-+-+-+- Or is spin a fixed value ?
     
  7. Jan 27, 2013 #6
    Thanks for your reply, but actually I didn't use this way.
    We were trying to solve it without using the matrix-method. We know that rotation operator about some axis of unit vector u equals to exp[i/[itex]\hbar[/itex]*uSθ]
    Then I can manipulate that and use Taylor's expansion to expand the exponential and then separate the terms into odd powers and even powers to end up with using again the Taylor's expansion but now to compile the 'odd' 'even' terms.

    Then what procedure should be done to know spin rotation operator for spin equals 1?

    If not what is the correct thing to do?

    Thank you a lot.
     
    Last edited: Jan 27, 2013
  8. Jan 27, 2013 #7
    It is simple,the generators here are like e(imθ) for rotation about z axis,where m=1/2 for spin 1/2 and m=1 for spin 1.i.e. replace θ/2 by θ.
     
  9. Jan 27, 2013 #8
    Thank you, i tried it and it worked! Thank you andrien.
     
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