Wigner's Theorem on Quantum Symmetries

Derives Lemma A9.1. Orthonormality is Equation (A9.1): \(\abs{\braket{\psi_j}{\psi_k}} = \abs{\braket{e_j}{e_k}} = \delta_{jk}\). Completeness: let the unit vector \(\chi\) be orthogonal to every \(\psi_k\). Since \(T\) is surjective, \([\chi] = T[\varphi]\) for some unit \(\varphi\), and then \(\abs{\braket{\varphi}{e_k}} = \abs{\braket{\chi}{\psi_k}} = 0\) for every \(k\), contradicting completeness of \(\set{e_k}\) and \(\norm{\varphi} = 1\).

Derives Lemma A9.2. Expand a unit representative of \(T[u_k]\) in the basis \(\set{\psi_j}\): by Equation (A9.1) tested against each \([e_j]\), its coefficients have moduli \(1/\sqrt2\) at \(j \in \set{1,k}\) and \(0\) elsewhere, so the representative is \((\ee^{\ii\alpha}\psi_1 + \ee^{\ii\beta}\psi_k)/\sqrt2\); multiplying by \(\ee^{-\ii\alpha}\) (same ray) makes it \((\psi_1 + \ee^{\ii\gamma_k}\psi_k)/\sqrt2\) with a uniquely determined \(\gamma_k\). Define \(f_k = \ee^{\ii\gamma_k}\psi_k\); the first relation of Equation (A9.3) holds by construction, and \(f_k \in T[e_k]\) still. For the second: a representative of \(T[u_{jk}]\) likewise reads \((f_1 + c_j f_j + c_k f_k)/\sqrt3\) with \(\abs{c_j} = \abs{c_k} = 1\) after normalizing the \(f_1\)-coefficient; testing against \(T[u_j]\) and \(T[u_k]\) with Equation (A9.1), \(\abs{1 + c_j} = \abs{1 + 1} = 2\) forces \(c_j = 1\), and similarly \(c_k = 1\). For the third: a representative of \(T[(e_j+e_k)/\sqrt2]\) is \((\ee^{\ii\mu}f_j + \ee^{\ii\nu}f_k)/\sqrt2\); testing against \(T[u_{jk}]\), \(\abs{\ee^{-\ii\mu} + \ee^{-\ii\nu}} = 2\) forces \(\mu = \nu\), and the common phase is dropped.

Derives Lemma A9.3. As in Lemma A9.2, a representative of the left side has components only along \(f_1, f_k\) with moduli \(1/\sqrt{1+\abs{\alpha}^2}\) and \(\abs{\alpha}/\sqrt{1+\abs{\alpha}^2}\); normalizing the \(f_1\)-component to be real positive determines the representative and hence \(\chi_k(\alpha)\) uniquely, with \(\abs{\chi_k(\alpha)} = \abs{\alpha}\). Testing against \(T[u_k]\) via Equation (A9.1) gives \(\abs{1 + \chi_k(\alpha)} = \abs{1 + \alpha}\); together with equal moduli this forces \(\Re\chi_k(\alpha) = \Re\alpha\), hence \(\chi_k(\alpha) \in \set{\alpha,\bar\alpha}\).

No mixing within one \(k\): suppose \(\chi_k(\alpha) = \alpha\) and \(\chi_k(\beta) = \bar\beta\) with \(\Im\alpha \ne 0 \ne \Im\beta\). Test the rays Equation (A9.4) for \(\alpha\) and \(\beta\) against each other: Equation (A9.1) demands \(\abs{1 + \bar\alpha\beta} = \abs{1 + \bar\alpha\bar\beta}\), i.e.\ \(\Re\left(\bar\alpha\beta\right) = \Re\left(\bar\alpha\bar\beta\right)\), i.e. \(\Im\alpha\,\Im\beta = 0\) — a contradiction. Real \(\alpha\) satisfy both alternatives, so \(\chi_k\) is the identity on all of \(\C\) or conjugation on all of \(\C\).

No mixing across \(k\): let \(j \ne k\) and suppose \(\chi_j = \mathrm{id}\) while \(\chi_k\) is conjugation, and pick \(\alpha = \beta = \ii\). A representative of \(T[(e_1 + \alpha e_j + \beta e_k)/\sqrt3]\), normalized so that its \(f_1\)-coefficient is real positive, reads \((f_1 + c_j f_j + c_k f_k)/\sqrt3\) with \(\abs{c_j} = \abs{\alpha}\) and \(\abs{c_k} = \abs{\beta}\) (tests against each \([e_m]\)). Testing against the family Equation (A9.4) with target \(j\), at the two values \(\gamma = 1\) and \(\gamma = \ii\), gives \(\abs{1 + \overline{\chi_j(\gamma)}\,c_j} = \abs{1 + \bar\gamma\,\alpha}\), which pins the real and the imaginary part of \(c_j\) and forces \(c_j = \chi_j(\alpha)\) — the within-target dichotomy just proved fixes \(\chi_j\) globally, so \(\chi_j(\gamma)\) is known at both test values — and likewise \(c_k = \chi_k(\beta)\). Testing the resulting representative against \(T[(e_j + e_k)/\sqrt2]\) — the third relation of Equation (A9.3) — yields \(\abs{\chi_j(\alpha) + \chi_k(\beta)} = \abs{\alpha + \beta} = 2\), whereas \(\chi_j(\ii) + \chi_k(\ii) = \ii + \overline{\ii\,} = 0\): a contradiction. (When \(\dim\mathcal{H} = 2\) there is a single \(k\) and this step is vacuous.)

Proof that \(U\) implements \(T\). Derives Theorem 99.3. Let \(\psi = \sum_k a_k e_k\) be an arbitrary unit vector and \(\varphi = \sum_k b_k f_k\) a unit representative of \(T[\psi]\). Testing against each \([e_k]\) gives \(\abs{b_k} = \abs{a_k}\). If only one \(a_n \neq 0\), then \([\psi] = [e_n]\) and \(T[\psi] = [f_n] = [U\psi]\) directly. Otherwise fix an index \(n\) with \(a_n \neq 0\), taking \(n = 1\) whenever \(a_1 \neq 0\), and rephase \(\varphi\) so that \(b_n = \chi(a_n)\) (possible since \(\abs{b_n} = \abs{a_n}\)). Components with \(a_k = 0\) are settled at once: \(\abs{b_k} = \abs{a_k} = 0\) gives \(b_k = 0 = \chi(a_k)\); in particular, if \(n \neq 1\) then \(a_1 = 0\) by the choice of \(n\), and only targets \(k \ge 2\), \(k \neq n\), remain to be pinned. For every such target and every \(\gamma \in \C\) we claim

— with the same global \(\chi\). For \(n = 1\) this is Lemma A9.3 itself. For \(n \ge 2\), the first two paragraphs of the proof of Lemma A9.3, run with anchor \(n\) in place of \(1\) — the needed alignment inputs, the pair \((f_n + f_k)/\sqrt2\) and the triples based at \(n\), are supplied by Equation (A9.3) — yield Equation (A9.6) with some uniform alternative \(\chi^{(n)}_{k} \in \set{\mathrm{id},\ \text{conjugation}}\) in place of \(\chi\). To identify the alternative, test the ray \([(e_n + \ii e_k)/\sqrt2]\) against \(T[(e_1 + e_n + \ii e_k)/\sqrt3] = [(f_1 + f_n + \chi(\ii)\,f_k)/\sqrt3]\) — the latter expansion established exactly as in the no-mixing step of Lemma A9.3 (anchor \(1\), targets \(n\) and \(k\)): Equation (A9.1) demands

and since \(\chi^{(n)}_{k}(\ii) = \pm\ii\) and \(\chi(\ii) = \pm\ii\), the left side equals \(2\) when \(\chi^{(n)}_{k}(\ii) = \chi(\ii)\) and \(0\) otherwise: hence \(\chi^{(n)}_{k}(\ii) = \chi(\ii)\), and the uniform dichotomy forces \(\chi^{(n)}_{k} = \chi\), proving the claim. Applying Equation (A9.1) to the ray Equation (A9.6) against \([\psi]\):

Square both sides and cancel the equal terms \(\abs{a_n}^2 + \abs{\gamma}^2\abs{a_k}^2\):

For \(\chi = \mathrm{id}\), running \(\gamma\) over \(1\) and \(\ii\) recovers both real and imaginary parts of \(\bar a_n a_k\) and \(\bar a_n b_k\) and forces \(b_k = a_k = \chi(a_k)\); for \(\chi\) conjugation, the same two tests force \(b_k = \bar a_k = \chi(a_k)\). Hence \(\varphi = \sum_k \chi(a_k) f_k = U\psi\), so \(T[\psi] = [U\psi]\).

Uniqueness up to phase, and exclusivity. Derives Theorem 99.3. Let \(U\) and \(V\) both implement \(T\) and set \(M = U^{-1}V\), an operator preserving every unit ray: \([M\psi] = [\psi]\). On the basis, \(M e_k = \lambda_k e_k\) with \(\abs{\lambda_k} = 1\).

If \(U\) and \(V\) have the same linearity character, \(M\) is linear: \(M(e_1 + e_k) = \lambda_1 e_1 + \lambda_k e_k\) must be proportional to \(e_1 + e_k\), so \(\lambda_k = \lambda_1\) for all \(k\) and \(M = \lambda_1\identity\): \(V = \lambda_1 U\).

If they had opposite character, \(M\) would be antilinear. Then \(M(e_1 + e_k) = \lambda_1 e_1 + \lambda_k e_k \propto e_1 + e_k\) gives \(\lambda_k = \lambda_1\), while \(M(e_1 + \ii e_k) = \lambda_1 e_1 - \ii\lambda_k e_k \propto e_1 + \ii e_k\) gives \(\lambda_k = -\lambda_1\) — impossible. Hence for \(\dim\mathcal{H}\ge2\) a symmetry transformation is implemented either by unitaries or by antiunitaries, never both, and within its class the implementation is unique up to an overall phase.