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quantum basics orthogonality

Generated by the Physics Derivation Graph. Eq. \ref{eq:3849595} is an initial equation. $$x = \langle\psi_{\alpha}| \hat{A} |\psi_{\beta}\rangle \label{eq:3849595}$$ Apply operator in Eq. \ref{eq:3849595} to ket; yields Eq. \ref{eq:4940359}. $$x = \langle\psi_{\alpha}| a_{\beta} |\psi_{\beta} \rangle \label{eq:4940359}$$ Apply operator in Eq. \ref{eq:3849595} to bra; yields Eq. \ref{eq:4349300}. $$x = \langle \psi_{\alpha}| a_{\alpha} |\psi_{\beta}\rangle \label{eq:4349300}$$ Simplify Eq. \ref{eq:4940359}; yields Eq. \ref{eq:2409402}. $$x = a_{\beta} \langle \psi_{\alpha} | \psi_{\beta} \rangle \label{eq:2409402}$$ Simplify Eq. \ref{eq:4349300}; yields Eq. \ref{eq:4934893}. $$x = a_{\alpha} \langle \psi_{\alpha}| \psi_{\beta}\rangle \label{eq:4934893}$$ LHS of Eq. \ref{eq:2409402} is equal to LHS of Eq. \ref{eq:4934893}; yields Eq. \ref{eq:3495045}. $$a_{\beta} \langle \psi_{\alpha} | \psi_{\beta} \rangle = a_{\alpha} \langle \psi_{\alpha} | \psi_{\beta} \rangle \label{eq:3495045}$$ Subtract $$a_{\alpha} \langle \psi_{\alpha} | \psi_{\beta} \rangle$$ from both sides of Eq. \ref{eq:3495045}; yields Eq. \ref{eq:4939583}. $$a_{\beta} \langle \psi_{\alpha} | \psi_{\beta} \rangle - a_{\alpha} \langle \psi_{\alpha} | \psi_{\beta} \rangle = 0 \label{eq:4939583}$$ Combine like terms in Eq. \ref{eq:4939583}; yields Eq. \ref{eq:3494855}. $$( a_{\beta} - a_{\alpha} ) \langle \psi_{\alpha} | \psi_{\beta} \rangle = 0 \label{eq:3494855}$$ Eq. \ref{eq:3494855} is one of the final equations.