Wigner’s symmetry representation theorem

Bargmann 7 — lemma 4

1. The nature of \(A\) can be derived directly from \(T\).

We define for rays: \[\Delta([\varphi_1],[\varphi_2],[\varphi_3])=\frac{\langle \varphi_1 | \varphi_2 \rangle \langle \varphi_2 | \varphi_3 \rangle \langle \varphi_3 | \varphi_1 \rangle}{\Vert \varphi_1 \Vert^2 \Vert \varphi_2 \Vert^2 \Vert \varphi_3 \Vert^2}.\] We note that this quantity is independent of the ray representatives chosen. Further if \(T\) is phase equivalent to \(T'\), \(\Delta([T'(\varphi_1)],[T'(\varphi_2)],[T'(\varphi_3)])=\Delta([T(\varphi_1)],[T(\varphi_2)],[T(\varphi_3)])\).

For an isometry \(A\), \(\langle A(\varphi) | A(\psi) \rangle = \langle \varphi | \psi \rangle\), and thus: \(\Delta([A(\varphi_1)],[A(\varphi_2)],[A(\varphi_3)])=\Delta([\varphi_1],[\varphi_2],[\varphi_3])\).

Similarly for an anti-isometry \(B\), \(\langle B(\varphi) | B(\psi) \rangle = \langle \psi | \varphi \rangle\): \(\Delta([B(\varphi_1)],[B(\varphi_2)],[B(\varphi_3)])=\Delta([\varphi_1],[\varphi_2],[\varphi_3])^*\).

The effect of \(T\) on \(\Delta\), thus tells us the behaviour of the linear/antilinear tranformation equivalent to it.

2. If \(\mathrm{dim}H \ge 2\), an anti-isometry cannot be phase equivalent to an isometry.

There exist two orthonormal vectors \(\varphi,\psi\). We evaluate: \[\Delta\left([\varphi],\left[\frac{\varphi-\psi}{\sqrt 2}\right],\left[\frac{\varphi+(1-\mathrm i)\psi}{\sqrt 3}\right]\right)=\frac{1}{\sqrt 2} \frac{\mathrm i}{\sqrt 6} \frac{1}{\sqrt 6}=\frac {\mathrm i} 6.\] The middle term shows the inner product evaluations in order in accordance with the \(\Delta\)-definition above. An isometry will leave this value the same, while an anti-isometry will give the complex conjugate. An isometry cannot therefore be phase equivalent to an anti-isometry.

3. If two isometries or two anti-isometries are phase equivalent then they differ by a constant factor.

Chevalier refers at this point to a “classical exercise in elementary linear algebra”. This may be true, but I couldn't find a quick google, so I made up my own approach. First, phase equivalent mappings take rays to rays, and (anti)linear mappings are additive. If \(A\) and \(B\) are phase equivalent, \(A([\varphi])=B([\varphi])\implies B(\varphi)=\lambda_\varphi A(\varphi)\), where \(\lambda_\varphi\) may depend on the vector. In fact \(B(\varphi+\psi)=\lambda_\varphi A(\varphi)+\lambda_\psi A(\psi)=\lambda_{\varphi+\psi} (A(\varphi)+A(\psi))\). If \(\varphi,\psi\) are linearly independent, \(\lambda_\varphi=\lambda_{\varphi+\psi}=\lambda_\psi\). Also \[B(\kappa \varphi)=\lambda_{\kappa \varphi}A(\kappa \varphi)=\hat\kappa B(\varphi)=\hat\kappa \lambda_\varphi A(\varphi)=\lambda_\varphi A(\kappa \varphi)\implies \lambda_{\kappa \varphi}=\lambda_\varphi.\] The \(\hat \kappa\) indicates the value appropriate to an (anti)linear mapping. All this shows \(\lambda\) to be constant over \(H\). For (anti)isometries the constant has modulus 1: \[\langle A(\varphi)|A(\psi)\rangle=\langle B(\varphi)|B(\psi)\rangle= |\lambda|^2 \langle A(\varphi)|A(\psi)\rangle,\] dropping the irrelevant subscript.