During a course of mine time-reversal symmetry was introduced as an anti-linear operator. One property the lecturer pointed out is that \begin{equation} \langle\phi| A\psi\rangle \neq \langle A^\dagger \phi |\psi \rangle \end{equation} but rather we have \begin{equation} \langle\phi| A\psi\rangle = \langle \psi | A^\dagger \phi \rangle = \langle A^\dagger \phi | \psi \rangle^*. \end{equation} I understand that time-reversal involves a complex conjugation, but I don't understand what makes the above equalities true.

The question is: what property of time-reversal itself makes the above hold. Is it only complex conjugation or is there anything else needed for the above to be true?


2 Answers 2


The equation you listed is the definition of the Hermitian conjugate of an antilinear map. It is true because it is defined that way.

However, I am guessing you are curious about why there must be this extra complex conjugation as compared to the case with linear operators. Suppose we tried to define Hermitian conjugation for antilinear $A$ as we do for linear operators.

$$ \langle\phi| A \psi\rangle = \langle A \phi|\psi\rangle $$

We will find it is not possible to do this in general. To see this, let $\lambda \in \mathbb{C}$ and consider the following manipulations

\begin{align} \begin{aligned} \langle\phi|A\lambda\psi\rangle &= \lambda^* \langle\phi|A\psi\rangle \\ &= \langle\lambda\phi|A\psi\rangle \\ &= \langle A^\dagger \lambda \phi | \psi\rangle \\ &= \langle \lambda^* A^\dagger\phi|\psi\rangle \\ &= \lambda \langle A^\dagger\phi|\psi\rangle \\ &= \lambda \langle \phi|A\psi\rangle \end{aligned} \end{align}

Comparing the first and last line on the right hand side, we see that if $\lambda \neq \lambda^*$, $\langle\phi|A\psi\rangle = 0$. Since this holds for any two states $\phi, \psi$, only $A = 0$ fits the bill! We clearly want other operators besides zero to have Hermitian conjugates. The true definition for antilinear operators that you showed above succeeds in that.

  • $\begingroup$ But here you assume that $A^\dagger$ is anti-linear too, no? Is this obvious? $\endgroup$ Jun 5, 2022 at 7:19
  • $\begingroup$ @JasonFunderberker Yes, I am assuming, or rather defining, $A^\dagger$ to be antilinear. Good catch! I'm not sure whether it's obvious but I think it makes sense. Mathematically we like operations to be closed, and physically we shouldn't expect time-reversal to become linear when acting on bras instead of kets. $\endgroup$
    – Jacob
    Jun 5, 2022 at 16:08
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    $\begingroup$ From that post, it looks like you can derive that $A^\dagger$ is antilinear from the correct definition. You could view my math as a proof that $A^\dagger$ is linear under the "wrong" definition. That leads into other issues such as $(A^\dagger)^\dagger \neq A$. $\endgroup$
    – Jacob
    Jun 5, 2022 at 16:26
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    $\begingroup$ I wanted to extend the comment, that's why I deleted it. Indeed, I think that you necessarily obtain that $A^\dagger$ is anti-linear. But, as far as I can see, to even show its existence via the Riesz representation theorem, you need to complex conjugate the inner product. So what is wrong with your argument, I think, is that the existence of $A^\dagger$ is not clear from the very beginning. In other words: $\langle \phi|A \psi\rangle$ is not a linear functional of $\psi$, as your first line shows. $\endgroup$ Jun 5, 2022 at 16:33
  • $\begingroup$ Yes, I think what you've shown is that if $A^\dagger$ exists and is defined as you've defined it, then it cannot be anti-linear. If it is linear instead, then its adjoint will be linear (due to the definition of the adjoint) and cannot be equal to an anti-linear operator, hence it wouldn't be an involution. Is that correct? If so, it might be worth to point this out in the answer itself. Because right now, at least for me, this seems a bit mixed up, since you did not really specify what you assume and hence the conclusion from your reductio ad absurdum is not clear, IMHO. $\endgroup$ Jun 5, 2022 at 17:48

This is a pedantic point, but in general, I would be very careful about using the language $A^\dagger$ when antihermitian operators are involved. The reason I say this is that antihermitian $A$ is not even a linear operator on the Hilbert space, so it's complete nonsense to talk about it as though it were. For the same reason, bra-ket notation is somewhat broken when taking the dual - it is never clear whether the complex conjugation inherent in an antiunitary acts on the bra, the ket or neither.

It is much easier to simply say $A$ with the understanding that if an operator is antiunitary, then

$$\langle A \psi| A \phi\rangle = \langle \psi| \phi \rangle^*$$

and derive all other results you need from the fact that $A^2$ is unitary. See here for a more in-depth discussion.

  • $\begingroup$ Good point. I checked out the blog post and it was really helpful. Thanks. $\endgroup$
    – zltn.guba
    Nov 13, 2022 at 23:22

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