1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotation of operators

  1. Apr 5, 2008 #1
    Suppose we know the matrix elements of an operator with respect a given cartesian reference frame [itex]L[/itex]. If we know the sequence of rotations going from [itex]L[/itex] to some other reference frame [itex]L'[/itex], what is the expression for the operator in the new reference frame.

    Let [itex]R[/itex] be the required rotation and [itex]\mathcal{D}(R)[/itex] the corresponding rotation operator. We know that the state of the systems changes under active rotation by multiplication [itex]| \psi \rangle \mapsto \mathcal{D}(R) |\psi\rangle[/itex]. In our case we're rotating the environment so the basis states which make up the operator should transform according to [itex]|\phi_i \rangle \mapsto U|\phi_i\rangle[/itex].

    Therefore

    [itex]\hat{O} = \sum_{ij} o_{ij} | \phi_i \rangle\langle \phi_j | \mapsto \sum_{ij} o_{ij} U| \phi_i \rangle \langle \phi_j |U^{\dag} = U \hat{O} U^{\dag} [/itex].

    Am I understanding this correctly?
     
    Last edited: Apr 5, 2008
  2. jcsd
  3. Apr 5, 2008 #2
    [tex]|\psi\rangle \longrightarrow D(R)|\psi\rangle[/tex]
    [tex]O|\psi\rangle\longrightarrow D(R)O|\psi\rangle=D(R)OD^{-1}(R)D(R)|\psi\rangle[/tex]
     
  4. Apr 6, 2008 #3
    So I gather what I said is correct, in other words I could obtain the spin-x operator from the spin-z operator in the following fashion:

    [itex]\hat{S}_x = \mathcal{D}_y(\pi/2) \hat{S}_z \mathcal{D}_y(-\pi/2)[/itex] e.g.??
     
  5. Apr 6, 2008 #4
    Yes, and this is also easily checked explicitly for spin [tex]\tfrac{1}{2}[/tex] because of the Pauli matrix identity [tex]e^{i \theta \hat{n} \cdot \vec{\sigma} } = ...[/tex]. But a more group theoretical treatment is probably best.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotation of operators
  1. Rotation Operator (Replies: 7)

Loading...