Q:

Each elementary particle has its own field. What about antiparticles?

I understand how particles correspond to their respective fields. What about antiparticles though? Do they have separate fields as well?

A:

Both particles and antiparticles arise from the same quantum field.
Particles (and antiparticles) are obtained from the Fourier mode expansion of the free quantum field - for a scalar, it is
$$ \phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$
and in the process of quantization, $a,a^\dagger$ become the annihilation and creation operators for the particle, while $b,b^\dagger$ become the annihilation and creation operators for the antiparticle associated with the quantum field. (For real---not complex---fields, these coincide, meaning the particle is "its own antiparticle")

A:

Particle and anti-particle are described by the same field.
Let's look at the Dirac field:
\begin{equation}
\psi\left(x\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{p}}}\sum_{s}\left(a_{\vec{p}}^{s}u^{s}\left(p\right)e^{-ip\cdot x}+b_{\vec{p}}^{s\dagger}v^{s}\left(p\right)e^{ip\cdot x}\right)
\end{equation}
where $a_{\vec{p}}^{s}$ annihilates a fermion, while $b_{\vec{p}}^{s\dagger}$
creates an anti-fermion. So the effect of the $\psi\left(x\right)$
is to annihilates a fermion or creates an anti-fermion. While if you
look at the Hermition conjugate of the above field, $\psi^{\dagger}\left(x\right)$
you'll find that it can creates a fermion or annihilates an anti-fermion.

