# Rotation of operators

1. Apr 5, 2008

### jdstokes

Suppose we know the matrix elements of an operator with respect a given cartesian reference frame $L$. If we know the sequence of rotations going from $L$ to some other reference frame $L'$, what is the expression for the operator in the new reference frame.

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

Therefore

$\hat{O} = \sum_{ij} o_{ij} | \phi_i \rangle\langle \phi_j | \mapsto \sum_{ij} o_{ij} U| \phi_i \rangle \langle \phi_j |U^{\dag} = U \hat{O} U^{\dag}$.

Am I understanding this correctly?

Last edited: Apr 5, 2008
2. Apr 5, 2008

### PRB147

$$|\psi\rangle \longrightarrow D(R)|\psi\rangle$$
$$O|\psi\rangle\longrightarrow D(R)O|\psi\rangle=D(R)OD^{-1}(R)D(R)|\psi\rangle$$

3. Apr 6, 2008

### jdstokes

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:

$\hat{S}_x = \mathcal{D}_y(\pi/2) \hat{S}_z \mathcal{D}_y(-\pi/2)$ e.g.??

4. Apr 6, 2008

### lbrits

Yes, and this is also easily checked explicitly for spin $$\tfrac{1}{2}$$ because of the Pauli matrix identity $$e^{i \theta \hat{n} \cdot \vec{\sigma} } = ...$$. But a more group theoretical treatment is probably best.