Talk:Wigner's theorem

This page is unsatisfactory because it does not actually state the theorem in question; it just talks about it. John Baez (talk) 13:10, 20 April 2013 (UTC)

Rewrite

I rewrote it from scratch. I think about including a proof outline. Suggestions (and corrections and additions) are most welcome. YohanN7 (talk) 08:13, 7 February 2015 (UTC)

Proof

I have somewhat advanced plans on including a detailed proof, specifically Bargmann's proof. The rationale is as follows:

• Since the theorem is of fundamental importance, its proof is not automatically out of place, in spite of it being longer than one line.
• The theorem was left unproved for over 30 years in the eyes of some (but perhaps not in the eyes of Wigner and Bargmann).
• Bargmann wrote the proof (he essentially spelled out Wigner's original proof in detail) perhaps mostly due to his frustration over fellow researchers indicating that proof of the theorem requires heavy machinery.
• The theorem was proved in painful detail by very elementary means, meaning that it is accessible to most visitors of this page. Senior high school mathematics suffices!

I think that the four bullets outweigh that Wikipedia doesn't usually include mathematical proofs. The proof is notable in the very sense of Wikipedia's use of the term, and is also described as such in several of the sources I have read. YohanN7 (talk) 13:28, 19 February 2015 (UTC)

Mistake in preliminaries?

In the subsection Preliminaries of Statement there's something that bugs me, maybe I'm mistaken. When we define the notion of compatibility, we say that it means

$T{\underline {\Psi }}=\{e^{i\alpha }U\Psi |\alpha \in \mathbb {R} \},$

or equivalently

$U\Psi \in T{\underline {\Psi }}.$

I don't see how these two are equivalent. Surely we have, by definition, $U\Psi \in {\underline {\Psi }}^{\prime }={\underline {U\Psi }}$ , since ${\underline {\Psi }}^{\prime }\equiv \{e^{i\alpha }U\Psi \}$. But then we could just look at a specific $T$, namely the one with:

$T:{\underline {\Psi }}\rightarrow \{U\Psi \}\in {\underline {\Psi }}^{\prime }.$

By construction, this means $U\Psi \in T{\underline {\Psi }}$, but we also have $T{\underline {\Psi }}=\{U\Psi \}\neq {\underline {\Psi }}^{\prime }$. Hence the statement in the article is not true.

Am I mistaken or is this an error? --- 82.42.249.115 (talk) 09:14, 24 April 2015 (UTC)

I think it should be

When we define the notion of compatibility, we say that it means

$T{\underline {\Psi }}=\{e^{i\alpha }U\Psi |\alpha \in \mathbb {R} \},$

which implies

$U\Psi \in T{\underline {\Psi }},\quad \forall \Psi .$

YohanN7 (talk) 11:02, 28 April 2015 (UTC)