Why is time-reversal represented by an antilinear and antiunitary operator? Operators related to physical transformations in quantum mechanics are usually unitary and linear except time-reversal which is both antiunitary and antilinear. What is the explanation for this difference?
 A: The statement is true if the Hamiltonian operator has spectrum bounded below ( due to thermodynamical reasons) but not above  as it happens in many relevant physical systems. Here is the proof.
The time reversal operator $T$ satisfies, by definition,
$$T e^{iHt} = e^{-iHt} T\tag{1}$$
for every real $t$
and must be either unitary or anti unitary due to Wigner's theorem. The identity above implies, according to the nature of $T$,
$$ e^{\pm iTH T^{-1}t} = e^{-iHt}$$
for all reals $t$. Stone theorem yields
$$\mp THT^{-1} = H.\tag{2}$$
If the operator were unitary, we would have
$$THT^{-1} = - H.$$
Since unitary transformations preserve the spectrum, the found identity would also imply
$$\sigma(H) = -\sigma(H)$$
that is not permitted since the spectrum we are considering is not symmetric under change of sign by hypothesis.
In general, unitary time reversal is possible for Hamiltonian operators with symmetric spectrum. However, since they must have spectrum bounded below, it is possible for Hamiltonian operators with symmetric and bounded spectrum.
ADDENDUM. As remarked by Elio Fabri, since (pure) states are unit vectors $\psi$ up to phases, (1) is a too restrictive condition and it has to be relaxed into
$$T e^{iHt}\psi = e^{ic_\psi(t)}e^{-iHt} T\psi \tag{1'}\:.$$
It is not difficult to prove, taking advantage of Stone theorem,  that $c_\psi(t)$ does not depend on $\psi$ and that  $c(t) = ct$ is the only possibility for the phase $c(t)$. Therefore, taking the derivative of both sides we have
$$\mp THT^{-1}= -H + cI\:.$$
If $T$ is unitary and $\sigma(H)$ is bounded below but not above, we have a contradiction $\sigma(H) = -\sigma(H) +c$. The only possibility is that $T$ is antiunitary so that
$$\sigma(H) = \sigma(H) - c$$
Since both sides must have finite $\inf$, we conclude that $c=0$ and
$$THT^{-1} = H$$ and the time reversal operation satisfies however
$$T e^{iHt} = e^{-iHt} T\:.$$
If $T$ is unitary,  $\sigma(T)$ must satisfy
$$\sigma(H) = -\sigma(H) +c$$
for some $c\in \mathbb R$. So, if $T$ is unitary,  $\sigma(H)$ is bounded below if and only if it is bounded above and it is symmetric with respect to some point which is not necessarily $0$.
