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 $\mathcal{O}$?

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

1 Answer 1


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.


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