$CP$-transformation for spinor field. $C$ and $P$ do not commute? I am bothered by an exercise about CP transformations where I get the result that CP acting on a Dirac spinor field is not the same as the PC transformation. The exercise states the following transformations of C and P:$$ C \psi(x) C^{-1}=-i\gamma^2 \psi ^* (x),\quad P \psi (x) P^{-1}=\gamma^0 \psi (\tilde{x}) $$ where $\tilde{x}=(x^0,-\vec{x})^T$. If I compute the CP conjugate I get: $$ P^{-1}C^{-1} \psi (x)CP= P^{-1}(-i\gamma^2\psi^* (x))P=-i\gamma^0\gamma^2\psi^* (\tilde{x})$$ But if I compute the PC conjugate, I should get:
$$C^{-1}P^{-1} \psi (x)PC= C^{-1}\gamma^0\psi (\tilde{x})C=-i\gamma^2\gamma^{0*}\psi^* (\tilde{x}) $$ Since $\gamma^{0*}=\gamma^{0T}$ and furthermore $C^{-1} \gamma^0 C=-\gamma^{0T}$ , this should yield the following:
$$C^{-1}P^{-1} \psi (x)PC=i\gamma^2C^{-1} \gamma^0 C\psi^* (\tilde{x}) $$ Since $C=-i\gamma^2=C^\dagger$ is hermitian and unitary:
$$C^{-1}P^{-1} \psi (x)PC=-i\gamma^2\gamma^2 \gamma^0 \gamma^2\psi^* (\tilde{x}) =i \gamma^0 \gamma^2\psi^* (\tilde{x})$$ Where the last step uses the Clifford-algebra of the gammas. Therefore there remains a sign difference in CP and PC transformation. I am pretty sure that I have made an error somewhere, but I don't know where. Another thought of mine was if the $-$ sign is just an unobservable phase for the Dirac field.
I hope someone can tell me exactly where the error occurs in the calculation.
 A: They do commute. The error is in the second calculation, which is the third line of equations in the OP. Here's a copy of that calculation, with an equation number for reference:
\begin{align}
 C^{-1} P^{-1}\psi(x)PC 
 &= C^{-1}\gamma^0\psi(\tilde x) C
\\
 &{\color{red}\neq} -i \gamma^2(\gamma^0)^*\psi^*(\tilde x).
\tag{1}
\end{align}
I change the last equal-sign to an unequal-sign, because they're not equal. The rest of this answer explains why.
The key is to remember that $C$ is a linear transformation of the  operator algebra. In symbols, for any operator $A$ on the Hilbert space and for any complex number $z$, we have
$$
 C^{-1} (z A) C = z C^{-1} A C.
\tag{2}
$$
For reference, the definition of $C$ shown in the OP is essentially
$$
 C^{-1} \psi(x) C = -i\gamma^2\psi^*(x),
\tag{3}
$$
except that I switched the $C$/$C^{-1}$ convention to be consistent with the convention used in the calculations. In this definition, we're implicitly choosing a set of linearly independent generators for the operator algebra, namely the field operators $\psi_a(x)$, which are the components of $\psi(x)$ in a particular basis. Equation (3) says that $C$ replaces each of those operators $\psi_a(x)$ by a special linear combination of the adjoint operators $\psi_a^*(x)$. But $C$ is still a linear operator, so we have
$$
 C^{-1} z\psi(x) C = z C^{-1} \psi(x) C 
\tag{4}
$$
for all complex numbers $z$. In components,
$$
 C^{-1} z\psi_a(x) C 
 =  zC^{-1}\psi_a(x) C 
 = -iz\sum_b(\gamma^2)_{ab} \psi_b^*(x).
\tag{5}
$$
This implies
$$
  C^{-1}\gamma^0\psi(x) C
 = \gamma^0 C^{-1}\psi(x) C
 = -i\gamma^0\gamma^2\psi^*(x).
\tag{6}
$$
Altogether, the corrected version of equation (1) is
\begin{align}
 C^{-1} P^{-1}\psi(x)PC 
 &= C^{-1}\gamma^0\psi(\tilde x) C
\\
 &= \gamma^0 C^{-1}\psi(\tilde x)C
\\
 &= -i \gamma^0\gamma^2\psi^*(\tilde x),
\tag{7}
\end{align}
which agrees with the first (already correct) calculation in the OP.
The important message is that $C$ is a linear transformation of operators on the Hilbert space. A Dirac matrix is not an operator on the Hilbert space. It's just an array of coefficients.
