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As the generator of a Unitary operator is a Hermitian operator, is the generator of an Anti-Unitary operator Anti-Hermitian?

Qmechanic
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Chetan Waghela
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1 Answers1

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I think you mean the following. Consider a (strongly continuous) one-parameter group of unitary operators $\mathbb R \ni t \mapsto U_t$. Then Stone's theorem implies that $$U_t= e^{itA}$$ for some self-adjoint operator $A$. Similarly, let $\mathbb R \ni t \mapsto U_t$ be a (strongly continuous) one-parameter group of anti-unitary operators. Is there a corresponding version of Stone's theorem where $$U_t= e^{itA}$$ for some antiself-adjoint operator $A$?

The answer is negative simply because it does not exist anything like a one-parameter group of anti-unitary operators. Since $U_t = U_{t/2}U_{t/2}$, every $U_t$ must be linear even if $U_{t/2}$ is antilinear (the product of two antilinear operators is linear).

This is the reason why antiunitary operators only describe discrete symmetries.

Valter Moretti
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  • I need a clarification. In C-Algebra product of two operators is rather AA and not AA so should'nt the Ut be Ut/2*Ut/2? – Chetan Waghela Oct 06 '16 at 08:42
  • I do not understand, here $C^$-algebras are quite irrelevant. However the product in a $C^$ algbera is $AA$ not $A^*A$...An one-parameter group of operators (the only notion which matters here) is a map $\mathbb R \ni s \mapsto A_s \in B(\cal H)$ such that $A_0=I$ and $A_sA_{s'}= A_{s+s'}$... – Valter Moretti Oct 06 '16 at 08:55