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$$\langle \psi^{(S)} |\hat{\mathcal{O}}|\psi^{(S)}\rangle\equiv\langle \psi^{(H)} |\hat{\mathcal{O}}|\psi^{(H)}\rangle$$$$\langle \psi^{(S)} |\hat{\mathcal{O}}|\psi^{(S)}\rangle\equiv\langle \psi^{(H)} |\hat{\mathcal{O}}|\psi^{(H)}\rangle$$
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