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Rotation operators

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the matrix representation of the rotation operator
    [itex] R(\phi k) [/itex] using the states |+z> and |-z> as a basis. Using your matrix representation verify that [itex] R^{\dagger}R=1 [/itex]
    3. The attempt at a solution
    Do I need to write [itex] R| \psi> [/itex] in terms of a matrix.
    If I have [itex] |\psi>=a|+z>+b|-z> [/itex]
    Then do I just operate R on [itex] \psi [/itex] and then write this in terms of a matrix.
    are these related to the Pauli spin matrices
     
  2. jcsd
  3. Sep 22, 2012 #2

    vela

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    I'm not sure exactly what you had in mind, but it's probably not the most straightforward way to solve this problem.
    Yes. Remember that the spin operators ##\hat{S}_x##, ##\hat{S}_y##, and ##\hat{S}_z## are generators of rotations. (This is definitely covered in your textbook.) Use that fact to calculate R.
     
  4. Sep 22, 2012 #3
    ok thanks for your help. My book gives the matrix for R and it is in the z basis.
    And I took [itex] R^{\dagger}R [/itex] and it equaled one. But If the matrix wasn't in the
    z basis would I use the roatation matrix to get the answer.
    I would take [itex] S^{\dagger}RS [/itex] and this would give the correct R for the problem.
     
  5. Sep 23, 2012 #4

    vela

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    I don't understand what you're asking. Well, I sorta do, but I'd like you to clarify your question. What is S? How did you find S?
     
  6. Sep 23, 2012 #5
    [itex] s= \[left (begin{array}{cc}<+z|+x>& <+z|-x>\\ <-z|+x>& <-z|-x> \end{array}\right)\][/itex]
    where the bras are what basis I am going to and the kets are the basis that I was in. I don't really know how S is derived though
     
    Last edited: Sep 23, 2012
  7. Sep 23, 2012 #6

    vela

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    Yes, that's the regular method you use to change bases. You're doing the same thing you learned in linear algebra. It's just the notation that's different.
     
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