Wigner’s symmetry representation theorem

Bargmann 8 — symmetry and (anti)unitarity

Chevalier, Proposition 2, step 6. If \(T\) is a symmetry on \(H\), a phase-equivalent (anti)isometry \(A\) is (anti)unitary.

\(T:H\rightarrow H\) is surjective. There is a phase-equivalent (anti)isometry \(A:H\rightarrow H\), which is also surjective: \[T(\varphi)=\psi\implies A(\varphi)=\lambda \psi\implies A(\varphi/\overline\lambda)=\psi,\] where the overline gives the appropriate (anti)linear value. As an (anti)isometry, \(A\) is also injective since \[A(\varphi)=0\implies \Vert A(\varphi)\Vert^2=0\implies \Vert \varphi\Vert^2=0\implies\varphi=0.\]

We define the (anti)linear adjoint: \[\langle \varphi | A^\dagger(\psi) \rangle=\overline{\langle A(\varphi) | \psi \rangle}.\] As an isometry: \[\langle A(\varphi) | A(\psi) \rangle=\overline{\langle \varphi | \psi \rangle}=\overline{\langle \varphi | A^\dagger(A(\psi)) \rangle}\implies A^\dagger A=\mathbb 1_H,\] the identity on \(H\).

Surjectivity (injectivity) implies for there exists a (unique) \(\psi'\) such that \(\psi=A(\psi')\). Hence, \(A A^\dagger(\psi)=A A^\dagger A(\psi')=A(\psi')=\psi\implies A A^\dagger=\mathbb 1_H\). \(A^\dagger\) is the inverse of \(A\), and so \(A\) is (anti)unitary. Hence:

Chevalier, Proposition 2, statement. Let \(H\) be a complex inner product space with \(\mathrm{dim}H\ge 2\) and \(T:H\rightarrow H'\) a mapping preserving the modulus of the inner product. There exists an isometry or an anti-isometry \(A\) on \(H\) which differs from \(T\) only by a phase factor. Two isometries \(A\) and \(A'\) or two anti-isometries \(B\) and \(B'\) satisfy the requirement of the preceding sentence if and only if they differ by a constant phase factor. The mapping \(T\) cannot be phase equivalent to an isometry and to an anti-isometry. If \(T\) is a symmetry (i.e. if \(T\) is surjective) and \(H\) a Hilbert space then A is a unitary or an antiunitary operator.