# Problem with time reversal operator

I am trying to understand the time reversal operator. As I read it is defined as $$\mathcal{T} = \mathcal{U}\mathcal{K},$$ where $$\mathcal{U}$$ is a unitary operator and $$\mathcal{K}$$ is a complex conjugation.

In some scripts, I read that time reversal acts on the operator as $$\mathcal{T} \mathcal{O} \mathcal{T}^{-1} = \mathcal{U} \mathcal{K} \mathcal{O} \mathcal{K} \mathcal{U}^\dagger = \mathcal{U}\mathcal{O}^* \mathcal{U}^\dagger.$$

What I understand $$\mathcal{T}^{-1} = \mathcal{K}^{-1}U^{-1} = \mathcal{K}^{-1}\mathcal{U}^\dagger$$

But, is it true that $$\mathcal{K}=\mathcal{K}^{-1}$$?

Also, as I understand the statement of operator acting on something it should be $$\mathcal{T}\mathcal{O},$$ so why there is $$\mathcal{T}^{-1}$$ on the right side of the operator $$\mathcal{O}$$?

• The second question is just a matter of language. For the first question, go back to the definition of $K$ and check if $K^2=I$... Nov 13, 2023 at 16:00
• What is a “complex conjugation”? Nov 13, 2023 at 16:01
• @ValterMoretti, I just read that $\mathcal{K} \psi = \psi^*$ Nov 13, 2023 at 16:09
• ...This is ill-defined in the general case. Do you work with the Hilbert space $L^2(\mathbb R)$ here, for example? Nov 13, 2023 at 16:10
• Same comment as Tobias’ one… Nov 13, 2023 at 16:13

The first question: yes, $$K=K^{-1}$$ (check the comment by @Tobias Fuenke). As for the second question, the meaning of "operator acting on something" in fact depends on the nature of something. Specifically in QM, if something a state vector $$| \psi \rangle$$, then indeed the operator $$\hat{O}$$ acts as just a product $$| \phi \rangle = \hat{O} | \psi \rangle$$. If the object in question is an operator itself $$\hat{V}$$, then it is acted upon by being sandwiched between the transformation operators as $$\hat{V}_O=\hat{O}^{-1} \hat{V} \hat{O}$$. Think about operators in matrix representation, and vector as (column) vectors to get an intuition.