What is an antiunitary operator? In field theory one can define a time reversal operator $T$ such that $T^{-1} \phi (x) T = \phi (\mathcal T x)$. It is then proved that $T$ must be antiunitary: $T^{-1} i T = -i$.

How is this equation to be understood? If $i$ is just the unit complex number, why don't we have $T^{-1} i T = i T^{-1} T$ which is just the identity times $i$?

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    $\begingroup$ $T$ is an antilinear (or conjugate-linear) map, cf. en.wikipedia.org/wiki/Antilinear_map $\endgroup$
    – Qmechanic
    Oct 25, 2011 at 21:28
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    $\begingroup$ One can think of $i$ as a multiplication operator that commutes with complex linear operators but anticommutes with complex anti-linear operators. Hence, $Ti+iT=0$. One can think of $i$ as a 2-by-2 matrix, if one wants to be very concrete. $\endgroup$ Oct 25, 2011 at 22:05
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    $\begingroup$ Hey, Unitarians are awesome! At least, all the other operators think so, which is why they don't talk to $T$ anymore ;-) I fixed the question to read "antiunitary". $\endgroup$
    – David Z
    Oct 26, 2011 at 1:27

3 Answers 3


As Qmechanic noted, $T$ is antilinear (this is part of the definition of being antiunitary). Of course, $T^{-1}$ must be antilinear as well because $T$ is. Thus, for any vector in this Hilbert space $v$, $T^{-1}(iv)=-iT^{-1}(v)$. The $i$ pops out as a $-i$. Applying this to your equation, we easily have that $$ T^{-1}iT=-iT^{-1}T=-i. $$

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    $\begingroup$ It is unfortunately the case that many sources refer to antiunitary operators as just "operators" (e.g., the "CPT operator") even though in almost all other contexts "operator" implies linearity rather than antilinearity. $\endgroup$ Feb 11 at 22:25

If I correctly understood your misunderstanding, the answer is: operator is not always a matrix. Technically, action of time inversion operator contains complex conjugation. E.g., in spin up/spin down basis it is written as $-i\sigma_y\mathcal{K}$, where $\mathcal{K}$ is complex conjugation.

  • $\begingroup$ I think that has been the source of my misunderstanding. I have considered T to be a matrix commutating with $i$. But it seems it is a more generic kind of operator. $\endgroup$
    – Whelp
    Oct 26, 2011 at 8:59

I think that just wrong (the wikipedia thing)

  1. $i \in \mathbb C$ so how can an operator act on a scalar?
  2. Even if it was right, one knows that $T = T^\dagger = T^{-1} = -T^*$. So take what you read

$$TiT^{-1} = -i$$

and now treat $i$ as an scalar (what is! not a pseudo-!) and it turns out

$$Ti(-T)^* = -T^2 i = -i \Rightarrow T^2 = 1$$

that is what we know about $T$ so it should be wrong that thingy of Wikipedia... ;).

The only real explanation could be a phase, not that $i$ changes it sign when you tossed it with $T$.

  • $\begingroup$ What's wrong with my argument/calculation? $\endgroup$ Oct 26, 2011 at 9:09
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    $\begingroup$ The argument is on a deeper level: $T$ isn't a linear operator, so most of what you've written above doesn't work. You need to go back to Wigner's theorem: a symmetry can either act on your Hilbert space as some unitary operator (the normal case), or as an antiunitary operator $U$ such that $(U \Phi, U \Psi) = (\Phi, \Psi)^*$ (i.e. which flips the complex inner product on your space). If this is the case, $U$ is necessarily antilinear, i.e. $U(a \Psi + b \Phi) = a^* U \Psi + b^* U \Phi$. $\endgroup$
    – Gerben
    Oct 26, 2011 at 16:42

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