# What is an antiunitary operator?

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 $i$?

• $T$ is an antilinear (or conjugate-linear) map, cf. en.wikipedia.org/wiki/Antilinear_map Oct 25, 2011 at 21:28
• 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. Oct 25, 2011 at 22:05
• 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". Oct 26, 2011 at 1:27

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 $$T^{-1}iT=-iT^{-1}T=-i.$$

• 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. 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.

• 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. 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 $T$.

• What's wrong with my argument/calculation? Oct 26, 2011 at 9:09
• 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. $U(a \Psi + b \Phi) = a^* U \Psi + b^* U \Phi$. Oct 26, 2011 at 16:42