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In the article on Wigner's theorem, unitary transformations ($U$) can be clearly seen as symmetric from: $T: \Psi =\{e^{ia} \Psi|a \in R \} \mapsto \Psi^{'} =\{e^{ib}U \Psi|b \in R \} $

and hence, the inner product property is preserved: $(T\Psi,T\Phi)=(\Psi,\Phi)$

But with the anti-unitary operator $A$,

$T: \Psi =\{e^{ia} \Psi|a \in R \} \mapsto \Psi^{'} =\{e^{ib}A \Psi|b \in R \} $

Since $A\Psi=\Psi^*$

Transformation $T$ associated with $A$ renders the inner product as: $(T\Psi,T\Phi)=(\Phi,\Psi)$

Clearly the inner product property is not preserved as the inner product is flipped, not same. What exactly is symmetric about anti-unitary transformations?

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    $\begingroup$ In that article, a symmetry transformation is defined in terms of preserving a ray product which takes the complex magnitude of the inner product. I think the basic idea is to preserve probability rather than probability amplitude. $\endgroup$ – G. Smith Aug 7 '19 at 2:12
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Time reversal must be unitary or antiunitary. But the standard definition further demands that it is antiunitary in particular. An antiunitary operator is a bijection $T : H \mapsto H$ that satisfies,

  1. (adjoint inverse)$\quad \, T^{*}T = T T^{*} = I$, and
  2. (antilinearity) $ \quad \,T(a\psi + b\phi) = a^{*}T \psi + b^{*}T \phi$.

It is sometimes useful to note that these conditions are together equivalent to,

  1. $\langle T \psi, T \phi \rangle = \langle \psi, \phi \rangle^{∗}$

Properties $(2)$ and $(3)$ underlie claims that time reversal ‘involves conjugation’. Theyare also slippery properties that often throw beginners (and many experts) for a loop, since they require many of the familiar properties of linear operators to be subtly adjusted.

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