Why goes $i\rightarrow-i$ under $\mathcal{PT}$-transformation? Question in the title.
What I understand is that under $\mathcal{PT}$ reversal $\hat{p}\rightarrow\hat{p}$ and $\hat{x}\rightarrow-\hat{x}$ and then since the commutation $[\hat{x},\hat{p}]=i\hbar$ "should still hold after  $\mathcal{PT}$ reversal" it requires $i\rightarrow-i$.
I cannot follow that argument. I do not see how a "constant" should change when we invert space and reverse time. Can someone enlighten me?
 A: Applying PT everywhere
$$
[ PT \hat{p} , PT \hat{x} ] = PT ( i \hbar)\\
[\hat{p} , - \hat{x} ] = - [ \hat{p} , \hat{x} ] = - i \hbar\\
= PT (i \hbar)
$$
is that argument. Think what happens to the operator $e^{- i t H / \hbar}$ upon time reversal to $e^{i t H / \hbar}$. This is absorbed into changing $i \to -i$
Aside: If you have seen some Galois theory this makes even more sense. The constants that you start with are just the reals. You extend to the complex numbers, but as far as the real numbers go, you can't tell the difference between $i$ and $-i$ in any formula. They satisfy all the same equations in the real world.
A: Symmetries in a quantum theory are represented either by linear unitary operators or by anti-linear anti-unitary operators (Wigner's theorem). 
Linear operators satisfy
$$
{\cal O}(a |\psi\rangle) = a {\cal O}|\psi\rangle 
$$
whereas anti-linear operators satisfy
$$
{\cal T}(a |\psi\rangle) = a^* {\cal T}|\psi\rangle 
$$
For every symmetry, the first step is to determine whether the corresponding operator is linear or anti-linear. If a symmetry is continuously connected to the identity, then it must be linear since the identity operator is linear. Thus, only disconnected symmetries could possibly be anti-linear. 
Now, let us work with the example at hand. We have two discrete symmetries $P$ and $T$. We denote their corresponding operators by ${\cal P}$ and ${\cal T}$. How do we decide if these operators are linear or anti-linear? They way to do this is to start with the group property. Let $U(a)$ be the unitary linear operator that generates a translation of the system. The group multiplication law implies that we must have
$$
{\cal P} U(a) {\cal P}^{-1} = U(P a )  , \qquad {\cal T} U(a) {\cal T}^{-1} = U( T a ) .
$$
Now, consider we recall that $U(a) = \exp ( i a^\mu P_\mu )$. We now take $a^\mu = (\epsilon,0,0,0)$ and expand to linear order in $\epsilon$. Then, since $P^0 = H$, the Hamiltonian, we must have
$$
{\cal P} (i H) {\cal P}^{-1} = i H , \qquad {\cal T} (i H) {\cal T}^{-1} = - i H \qquad \qquad (1)
$$
where we have used the fact that $P(\epsilon,0,0,0)=(\epsilon,0,0,0)$ and $T(\epsilon,0,0,0) = (-\epsilon,0,0,0)$.
We are now ready to make our decision. Let us start with ${\cal P}$. If ${\cal P}$ is anti-linear, then
$$
{\cal P}  H  {\cal P}^{-1} =- H
$$
Now, if $|E\rangle$ is a state with energy $E$ (i.e. $H | E \rangle = E | E \rangle$) then ${\cal P}^{-1} | E \rangle$ has energy $-E$. But this breaks a fundamental tenet of quantum mechanics which states that the spectrum of the Hamiltonian must be bounded from below! Thus, ${\cal P}$ must be chosen to be a linear operator. Following the same argument, you can show that ${\cal T}$ must be an anti-linear operator. This means that 
$$
{\cal T} ( i |\psi\rangle ) = - i {\cal T} |\psi\rangle,
$$
or in words, if you attempt to move the operator ${\cal T}$ across a complex number, the number transforms to its complex conjugate.
Thus, ${\cal P}$ is linear and ${\cal T}$ is anti-linear. Thus, ${\cal P} {\cal T}$ is anti-linear. 
