# $CP$-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.

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.