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Let ##\textbf{R}(\theta)## be a rotation in the plane about angle ##\theta##. Then for the active view, we rotate a vector ##\textbf{v}## thusly $$ \textbf{R}(\theta)[\textbf{v}] = v_i \textbf{R}(\theta)[\textbf{e}_i] = v_i R_{ji} \textbf{e}_j, $$ where I have expanded ##\textbf{R}(\theta)[\textbf{e}_i]## in terms of the basis {##{\textbf{e}_j}##}. For the passive view, on the other hand, we instead transform the basis, rotating each basis vector by ##-\theta##: $$ \bar{\textbf{e}}_i = \textbf{R}(-\theta)[\textbf{e}_i] = \textbf{R}^{-1}(\theta)[\textbf{e}_i] = R_{ji}^T\textbf{e}_j = R_{ij}\textbf{e}_j.$$ Using the orthogonality relation, we can solve for the original basis as ## \textbf{e}_i = R_{ji}\bar{\textbf{e}}_j ##. Then we have $$ \textbf{v} = v_i \textbf{e}_i = v_i R_{ji} \bar{\textbf{e}}_j.$$ So in either the active or passive view, the components of ##\textbf{v}## transform the same way (though it seems strange to de-emphasize that the basis is different between the two cases, which is what happens if we stick to matrices, as many texts do).

Now I will try to relate the Heisenberg and Schrodinger pictures to the above discussion. Let ##\textbf{U}(t)## be a unitary time evolution operator. Then if ##|\psi(0)\rangle## is the state vector at time zero, we can evolve the state as $$|\psi(t)\rangle = \textbf{U}(t)|\psi(0)\rangle.$$ Because ##\textbf{U}(t)|\psi(0)\rangle## is analogous to ##\textbf{R}(\theta)[\textbf{v}]##, evolving the state vector in this way is taking the active view. On the other hand, the expectation value of an observable ##\textbf{A}## is given as $$ \langle \psi | \textbf{A} | \psi \rangle = \langle \psi(0)|{\textbf{U}}^{\dagger}(t) \textbf{A} \textbf{U}(t) | \psi(0)\rangle.$$ In the Heisenberg picture, we associate the time evolution operators with ##\textbf{A}## instead of acting on ##|\psi(0)\rangle## or its corresponding bra, i.e., we define the time dependent operator $$\textbf{A}(t) = \textbf{U}^{\dagger}(t) \textbf{A} \textbf{U}(t),$$ and regard the state vector as unchanging. In this picture, we have done nothing to either the components or the basis vectors of ##|\psi(0)\rangle##. Hence, the Heisenberg picture does not seem to correspond to either the active or the passive view of transformations.

Is all this correct?