Is this a “good enough” statement of Wigner's theorem from Quantum Mechanics?

Question

I posted this on math StackExchange and got no replies, so I'm trying my luck here!

I'm a fourth year physics and math student who is writing up a report on some quantum mechanical symmetries and their consequences. The "audience" for my paper are either senior physics majors or first year graduate students in physics. Wigner's theorem's mathematical content is unfortunately beyond the scope of my report so I decided to "water it down". My question is this: is this an acceptably simplified version of Wigner's theorem?

Theorem (Wigner) Let $\Psi, \Phi$ be arbitrary state vectors in a Hilbert space $\mathscr{H}$. Suppose $\delta: \mathscr{H} \mapsto \mathscr{H}$ is bijective and Wigner symmetric (that is, under the mapping $\delta$, the transition probability $\langle \Psi | \Phi \rangle$ is unchanged). Then

1. there exists a linear and unitary or antilinear and antiunitary operator $U$ such that $$\delta(\psi) = U \psi U^{-1},$$
2. $U$ is unique up to an arbitrary phase factor.

Is there anything I'm misunderstanding with this formulation of Wigner's theorem?

Answer 1

$\let\d=\delta \def\sH{\mathscr H}$ Some amendments are needed.

1) This is merely notational. Don't write $\d:\sH\mapsto\sH$ but rather $\d:\sH\to\sH$. "$\to$" is used between domain and codomain of a map. "$\mapsto$" between a point in domain and its image in codomain. Example: $$f:\Bbb C \to \Bbb R,\quad z \mapsto f(z) = |z|^2.$$

2) Transition probability is $|\langle\Psi|\Phi\rangle|^2$. This is the right definition ot transition probability in QM. The difference is also fundamental for the meaning of theorem, as conservation of $\langle\Psi|\Phi\rangle$ is much stronger and is essentially the theorem's thesis.

3) Unitary and antiunitary imply linear and antilinear respectively. So those clauses could and should be suppressed.

4) Not $\d(\psi) = U \psi U^{-1}$, but $\d(\Psi) = U \Psi$ (note capitals).