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Spin 1 particle, spinor state

  1. Jul 1, 2010 #1
    Hi there,
    I have a question, something that is confusing me.
    If a particle of spin 1 is measured to have m=1 along the x direction, would the spinor state just be a column vector with (1,0,0), which would also be the spinor if x was infact z. OR would the spinor be determined by multiplying the Sx matrix by a column vector of (a,b,c,d) and letting this equal the eigenvalue (hbar in this case) muliplied by column vector (a,b,c,d) and working out their values and normalising.?
  2. jcsd
  3. Jul 1, 2010 #2


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    Well, I would write down the three-dimensional representation of SU(2) and look for the appropriate eigenvectors (which you call spinor states). An m=1 state along x is described by an eigenvector of S_x with eigenvalue 1, right?
  4. Jul 1, 2010 #3
    I am not sure what you mean by SU(2) ?
  5. Jul 1, 2010 #4
    Simp, welcome to PhysicsForums.
    However, your thread is in the wrong forum, as this is the Quantum Mechanics forum and not the Physics Homework forum. If you need assistance with your homework problems, may I direct you to the appropriate forum: https://www.physicsforums.com/forumdisplay.php?f=152
    Thank you.
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