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Angle between spins

  1. Oct 23, 2012 #1
    If ##|\alpha>## is spin up, and ##|\beta>## is spin down. Then if angle between those spins and some other up and down spin is ##\theta##, then
    [tex]|\alpha'>=\cos \frac{\theta}{2}|\alpha>+\sin \frac{\theta}{2}|\beta>[/tex]
    [tex]|\beta'>=\sin \frac{\theta}{2}|\alpha>-\cos \frac{\theta}{2}|\beta>[/tex]
    Why?
     
  2. jcsd
  3. Oct 23, 2012 #2

    Bill_K

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    For each value j of angular momentum there are 2j+1 linearly independent states. For example these can be taken as the states with spin projection mz = -j,... +j along the z axis. They form a basis in a 2j+1-dimensional space. We can just as well take for a basis the states with projection ma along any other axis a, and the transformation from one basis to another is a unitary transformation,

    |ma> = Σ|mz><mz|D(j)(α,β,γ)|ma>

    where D(j)(α,β,γ) is a unitary operator whose matrix elements <mz|D(j)(α,β,γ)|ma> are called the rotation matrix. An arbitrary rotation in three dimensions requires three Euler angles α,β,γ to describe.

    For spin 1/2 the space is two-dimensional, just spin up and spin down. The simplest rotation from the z axis to some other axis a is through an angle θ directly down a line of longitude, and the rotation matrix is (almost!) what you have written,

    [tex]\left(\begin{array}{cc}cos θ/2&sin θ/2\\-sin θ/2&cos θ/2\end{array}\right)[/tex]
     
  4. Oct 23, 2012 #3
    Nvm, I had misunderstood the question.
     
  5. Oct 23, 2012 #4
    But why you get ##\frac{\theta}{2}## in matrix if you rotate for angle ##\theta##?
     
  6. Oct 23, 2012 #5
    Wait, the state-space is 2-dimensional, but isn't this problem making reference to a rotation in the real space where this spin-1/2 particle is? I mean, there's a z-axis.
     
  7. Oct 23, 2012 #6

    tom.stoer

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    simple question: are we talking about the angle between the two axes in 3-dim. position space or about the angles between two spin states in 2-dim spin. space?
     
  8. Oct 23, 2012 #7
    Yes, that's what I meant. It's not really that clear from the question.
     
  9. Oct 23, 2012 #8

    Bill_K

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    Tom, All I have tried to do is give a simple answer to a simple question. Please let's not throw confusion at it. Especially since you know how a spinor transforms inside out, forwards and backwards. The appearance of the half angle in the rotation matrix is a result of the mapping between SO(3) and SU(2).
     
  10. Oct 24, 2012 #9
    I think it has to do with a direct spin character i.e. it is true for spin 1/2 something like
    exp(imθ) .
     
  11. Oct 27, 2012 #10
    Ok if I don't know that. I have some up spin. How to get up spin which is rotate for angle ##\theta## from that spin. Can I use Pauli matrices and spherical coordinates and get that result?
     
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