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I am currently studying discrete symmetries in quantum mechanics and have trouble proving that the set of discrete symmetry operators is a group. An operator, $\hat S$, is called a symmetry operator if the system Hamiltonian, $\hat H$, is invariant under the action of $\hat S$,

\begin{equation} \hat S^\dagger \hat H \hat S = \hat H. \tag{1} \label{eq:1} \end{equation}

Most books state that the set of symmetry operators is a group without taking much time to prove it. Associativity and existence of a neutral element is easy to prove since the composition of operators is associative and the identity operator, $\hat I$, acts as neutral element. However, I find it hard to prove, using only equation \eqref{eq:1}, that every discrete symmetry has an inverse $\hat S ^{-1}$. Most resources I have checked simply claims this to be the case and then presents some examples.

How can I prove that every discrete symmetry has an inverse?

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  • $\begingroup$ Related: physics.stackexchange.com/q/19751/2451 and links therein. $\endgroup$ – Qmechanic Apr 25 at 14:08
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    $\begingroup$ The operator $\hat S$ must also preserve $|\langle \psi_1|\psi_2\rangle|$ and so has to be unitary or antiunitary, and therefore has an inverse. $\endgroup$ – mike stone Apr 25 at 14:17
  • $\begingroup$ That must be the answer! Thank you! $\endgroup$ – Michael Iversen Apr 25 at 14:21

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