# How are anti-unitary transformations symmetric?

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?

• 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. – G. Smith Aug 7 '19 at 2:12

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) $$\quad \,T(a\psi + b\phi) = a^{*}T \psi + b^{*}T \phi$$.
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.