Why for the complex scalar field
$$
\hat\phi = \int \frac{d^3p}{(2\pi)^{3/2}(2E_{\vec{p}})^{1/2}}\left(\hat{a}_{\vec{p}}e^{-p \cdot x} + \hat{b}_{\vec{p}}^\dagger e^{p \cdot x}\right),
$$
the commutation relation $[\hat\phi(x),\hat{\phi}^\dagger(y)]=0$, but using the non-relativistic limit for the fields $\phi(\vec{x},t)\rightarrow\Psi(\vec{x},t)e^{-imc^2t/\hbar}$, $[\hat\Psi(\vec{x}), \hat\Psi(\vec{x})^\dagger] = \delta(\vec{x} – \vec{y})$, this commutator is different of zero?
(All commutators are taken at equal times).
I had the idea that from the relativistic commutation relation you could derive the non-relativistic one, but it doesn't seem to be the case, unless one has to be careful when taking the limit.
Best Answer
Instead of showing how to derive the nonrelativistic limit by rote calculation, I'll list some conceptual points that should help dissolve the apparent paradox. To reduce clutter, I'll omit the hats on the operators.
In a relativistic model, local observables cannot annihilate the vacuum state (Reeh-Schlieder theorem). In a nonrelativistic model, they can. Therefore, we should expect the relationship between $\phi$ and $\psi$ to be nonlocal.
As written in the OP, $\phi$ is the sum of two parts: one involving $ a$, and one involving $ b$. This can be expressed as $\phi(x)=\phi_a(x)+\phi_b(x)$, where the $a$-part and the $b$-part are distinguished from each other by the sign of the frequency.
For the relativistic scalar field shown in the OP, the particle is distinct from its antiparticle. That means we need two field operators $\psi_a$ and $\psi_b$ (and their adjoints) in the nonrelativistic approximation. These will correspond to the (nonrelativistic versions of) the operators $\phi_a(x)$ and $ \phi_b(x)$.
The operator $\phi(x)$ is local by definition, so the operators $\phi_a(x)$ and $\phi_b(x)$ cannot be local. This follows the square-root in the relationship $\omega=\pm(m^2+\vec p^2)^{1/2}$ that is involved in defining the positive and negative frequency parts. Expanding this relationship in powers of $\vec p^2/m^2$ is the key to deriving the non-relativistic approximation explicitly.
The operator $\phi_a$ does not commute with its adjoint $\phi_a^\dagger$, not even at spacelike separation. (These operators are not local!) Similarly, $\phi_b$ does not commute with its adjoint $\phi_b^\dagger$. This is possible even though $\phi$ does commute with its adjoint, because the non-zero commutator bewteen $\phi_a$ and $\phi_a^\dagger$ cancels the non-zero commutator between $\phi_b$ and $\phi_b^\dagger$.
In the nonrelativistic approximation $\phi_a\to\psi_a$, the operators $\psi_a$ and $\psi_a^\dagger$ commute with each other at equal time with nonzero spatial separation (this is a consequence of expanding $\omega$ in powers of $\vec p^2/m^2$ and truncating the expansion), but they still don't commute with each other at zero separation. The noncommutativity of $\psi_a$ and $\psi_a^\dagger$ at zero separation is a remnant of the noncommutativity of $\phi_a$ and $\phi_a^\dagger$ at all separations. The same comments apply to the antiparticle operators $\phi_b$ and $\psi_b$.