Conjugate linear symmetry and its application to <inline-formula><tex-math id="M2">\begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}</tex-math><alternatives><graphic xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="3-20191173_M2.jpg"/><graphic xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="3-20191173_M2.png"/></alternatives></inline-formula>-symmetry quantum theory

The Hamiltonians of classical quantum systems are Hermitian (self-adjoint) operators. The self-adjointness of a Hamiltonian not only ensures that the system follows unitary evolution and preserves probability conservation, but also guarantee that the Hamiltonian has real energy eigenvalues. We call such systems Hermitian quantum systems. However, there exist indeed some physical systems whose Hamiltonians are not Hermitian, for instance, \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry quantum systems. We refer to such systems as non-Hermitian quantum systems. To discuss in depth \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry quantum systems, some properties of conjugate linear operators are discussed first in this paper due to the conjugate linearity of the operator \begin{document}$ {\mathcal{P}}{\mathcal{T}}, $\end{document}including their matrix represenations, spectral structures, etc. Second, the conjugate linear symmetry and unbroken conjugate linear symmetry are introduced for linear operators, and some equivalent characterizations of unbroken conjugate linear symmetry are obtained in terms of the matrix representations of the operators. As applications, \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry and unbroken \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry of Hamiltonians are discussed, showing that unbroken \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry is not closed under taking tensor-product operation by some specific examples. Moreover, it is also illustrated that the unbroken \begin{document}$ {\mathcal{P}}{\mathcal{T}} $\end{document}-symmetry is neither a sufficient condition nor a necessary condition for Hamiltonian to be Hermitian under a new positive definite inner product.