# Hamiltonian in rotating frame

I have a Hamiltonian for spin in a magnetic field $$\hat{H}(t) = \mathbf{B}(t) \cdot \mathbf{\hat{S}}$$ $\mathbf{B}(t) =B_x(t) \hat{\mathbf{x}} + B_y(t)\hat{\mathbf{y}} + B_z(t)\hat{\mathbf{z}}$ is the magnetic field, and $\mathbf{\hat{S}}$ is the vector spin operator. If I choose new time-dependent axes such that $\mathbf{B}(t) = B_{z'}(t) \hat{\mathbf{z}}'$, how do I find the Hamiltonian in the new frame?

• How about you rewrite the Spin in this new frame. Technically you should find that scalar product is rotation invariant. Which is nice because it means your hamiltonian doesn't depend on the frame, something we would want to have for a good description of a hamiltonian. – Ismasou Sep 17 '18 at 19:52
• Its not actually a scalar product though is it, @Ismasou ? It gives a 2x2 matrix not a scalar, so no reason why it should be invariant? – oweydd Sep 21 '18 at 15:11

You're applying a Rotation Operator into your magnetic field, as follow $$B'=R(\theta)B.$$ So we would apply the same for the vector spin operator $$S'=R(\theta)B.$$
The Hamiltonian becomes $$H=(B')^T.S'=BR(\theta)^TR(\theta)S=B.S$$ $$(B.S)$$ is a scalar product in the spacial sense, yes it is a matrix but it's not a vector, as $$S$$ would be a vector, i.e. 3 Matrices in every direction of 3D space, $$(B.S)$$ is just one matrix. Thus this quantity should be conserved.
• The time dependence case should be the same? Unless I'm missing something, the only difference is the angle will depend on time, but you would still have $R^T R=1$. – Ismasou Sep 22 '18 at 9:04