Wigner’s symmetry representation theorem

Bargmann 6 — checking relation of \(A\) and \(T\)

Chevalier, Proposition 2, step 5: \(A\) and \(T\) differ by a phase factor. Further \(A\) is an (anti)isometry.

We already know that \(A\) is phase-equivalent to \(T\) on \(D=(\varphi_0+[\varphi_0]^\perp) \cup [\varphi_0]^\perp\). Thus for general \(\varphi \in [\varphi_0]^\perp\), we have \(T(\alpha \varphi_0 +\varphi)=\alpha' T(\varphi_0+\varphi/\alpha)=\alpha' \gamma A(\varphi_0+\varphi/\alpha)=\alpha' \gamma A(\alpha\varphi_0+\varphi)/\hat\alpha\), where \(\hat\alpha\) is the identity or conjugate of \(\alpha\), according to (anti)linearity of \(A\). Since \(|\gamma|=1\) and \(|\alpha'|=|\hat\alpha|=|\alpha|\), \(|\alpha' \gamma/\hat\alpha|=1\). \(A\) and \(T\) thus differ by a phase factor. Thus \(A\) is norm-preserving, hence bounded, hence continuous.

From additivity: \(\Vert A(\varphi +\psi)\Vert^2=\Vert A(\varphi)\Vert^2+\Vert A(\psi)\Vert^2+\langle A(\varphi)|A(\psi)\rangle+\langle A(\psi)|A(\varphi)\rangle\).

Also: \(\Vert A(\varphi +\psi)\Vert^2=\langle \varphi +\psi|\varphi +\psi\rangle=\Vert \varphi\Vert^2+\Vert \psi\Vert^2+\langle \varphi|\psi\rangle+\langle \psi|\varphi\rangle\).

Since \(\Vert A(\varphi)\Vert^2=\Vert \varphi\Vert^2\), and similarly for \(\psi\): \(\langle A(\varphi)|A(\psi)\rangle+\langle A(\psi)|A(\varphi)\rangle=\langle \varphi|\psi\rangle+\langle \psi|\varphi\rangle\).

Applying the properties of the inner product: \(\langle A(\varphi)|A(\mathrm i\psi)\rangle+\langle A(\mathrm i\psi)|A(\varphi)\rangle=\langle \varphi|\mathrm i\psi\rangle+\langle \mathrm i\psi|\varphi\rangle=\mathrm i(\langle \varphi|\psi\rangle-\langle \psi|\varphi\rangle)\).

Applying (anti)linearity, and inner product properties: \(\langle A(\varphi)|A(\mathrm i\psi)\rangle+\langle A(\mathrm i\psi)|A(\varphi)\rangle=\hat{\mathrm i}(\langle A(\varphi)|A(\psi)\rangle-\langle A(\psi)|A(\varphi)\rangle)\).

Equating: \(\langle A(\varphi)|A(\psi)\rangle-\langle A(\psi)|A(\varphi)\rangle=\pm (\langle A(\varphi)|A(\psi)\rangle-\langle A(\psi)|A(\varphi)\rangle)\). The plus sign corresponds to linearity and the minus to antilinearity. Hence \(\langle A(\varphi)|A(\psi)\rangle=\langle \varphi|\psi\rangle\) in the linear case, and \(\langle A(\varphi)|A(\psi)\rangle=\langle \psi|\varphi\rangle\), antilinear. This can be abbreviated to \(\langle A(\varphi)|A(\psi)\rangle=\widehat{\langle \varphi|\psi\rangle}\).