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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].


    [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.
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