= Solution
For a <unitary operator> $W$, $WW^\dagger=I$. Thus for every nonnegative integer $j$,
$$
(W^\dagger H W)^j=W^\dagger H^jW.
$$
Insert this into the convergent <matrix exponential> series in the finite-dimensional qubit setting:
$$
W^\dagger e^{iH}W=\sum_{j=0}^\infty\frac{i^j}{j!}W^\dagger H^jW=\sum_{j=0}^\infty\frac{i^j}{j!}(W^\dagger HW)^j.
$$
\b[Consequently unitary conjugation commutes with the exponential:]
$$
\boxed{W^\dagger e^{iH}W=e^{iW^\dagger HW}.}
$$
For an unbounded self-adjoint <Hamiltonian>, the same identity follows from the <spectral theorem for normal operators>, with the domain transported by $W$; no unbounded power-series manipulation is needed.
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