The field and the wavefunction look similar, but they don't really have much to do with each other. The main point of the field is to group the creation and annihilation operators in a convenient way, which we can use to construct observables. As usual I will start with the free theory.

If we want to find a connection to non-relativistic QM, the field equation is not the way to go. Rather, we should look at the states and the Hamiltonian, which are the basic ingredients of the Schrödinger equation. Let's look at the Hamiltonian first. The usual procedure is to start with the Lagrangian for the free scalar field, pass to the Hamiltonian, write the field in terms of $a$ and $a^\dagger$, and plug that into $H$. I will assume you know all this (it's done in every chapter on second quantization in every QFT book), and just use the result:

$$H = \int \frac{d^3 p}{(2\pi)^3}\, \omega_p\, a^\dagger_p a_p$$

where $\omega_p = \sqrt{p^2+m^2}$. There's also a momentum operator $P_i$, which turns out to be

$$P_i = \int \frac{d^3 p}{(2\pi)^3}\, p_i\, a^\dagger_p a_p$$

Using the commutation relations it is straightforward to calculate the square of the momentum, which we will need later:

$$P^2 = P_i P_i = \int \frac{d^3 p}{(2\pi)^3}\, p^2\, a^\dagger_p a_p + \text{something}$$

where $\text{something}$ gives zero when applied to one particle states, because it has two annihilation operators next to each other.

Now let's see how to take the non-relativistic limit. We will assume that we are dealing only with one-particle states. (I don't know how much loss of generality this is; the free theory doesn't change particle number so it shouldn't a big deal, and also we usually assume a fixed number of particles in regular QM.) Let's say that in the Schrödinger picture we have a state that at some point is written as $|\psi\rangle = \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle$, where $|k\rangle$ is a state with three-momentum $\mathbf{k}$. $f(k)$ should be nonzero only for $k \ll m$. Now look what happens if we apply the Hamiltonian. Since we only have low momentum, over the range of integration we can approximate $\omega_p$ as $m+p^2/2m$ and ignore the constant rest energy $m$.

$$H|\psi\rangle = \int \frac{d^3p}{(2\pi)^3} \frac{p^2}{2m} a^\dagger_p a_p \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle \\
= \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) a^\dagger_p a_p |k\rangle \\
= \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) (2\pi)^3 \delta(p-k) |k\rangle \\
= \int \frac{d^3 k}{(2\pi)^3} \frac{k^2}{2m} f(k) |k\rangle = \frac{P^2}{2m} |\psi\rangle
$$

So if $|\psi\rangle$ is any one-particle state (which it is because the states of definite momentum form a basis), we have that $H|\psi\rangle = P^2/2m |\psi\rangle$. In other words, on the space of one-particle states, $H = P^2/2m$. The Schrödinger equation is still valid in QFT, so we can immediately write

$$\frac{P^2}{2m} |\psi\rangle = i \frac{d}{dt} |\psi\rangle$$

This is the Schrödinger equation for a free, non-relativistic particle. You will notice that I kept some concepts from QFT, particularly the creation and annihilation operators. You can do this no problem, but working with $a$ and $a^\dagger$ in QM is not particularly useful because they create and destroy particles, and we have assumed the energy is not high enough to do that.

Handling interactions is more complicated, and I fully admit I'm not sure how to include them here in a natural way. I think part of the issue is that interactions in QFT are quite limited in their form. We would have to start with the full QED Lagrangian, remove the $F_{\mu\nu}F^{\mu\nu}$ term since we aren't interested in the dynamics of the EM field itself, maybe set $A_i = 0$ if we don't care about magnetic fields, and see what happens to the Hamiltonian. Right now I'm not up to the task.

I hope I can convinced you that this newfangled formalism reduces to QM in a meaningful way. A noteworthy message is that the fields themselves don't carry a lot of physical meaning; they're just convenient tools to set up the states we want and calculate correlation functions. I learned this from reading Weinberg; if you're interested in these kinds of questions, I recommend you do so too after you've become more comfortable with QFT.

It has to do with Feynman-Stuckelberg interpretation of negative energy solutions as outgoing antiparticles.

A plane wave mode with negative energy is given by
$$a(\vec p)e^{-i(Et-\vec p\cdot\vec x)},\quad E<0,$$
which can be written as
$$a(\vec p)e^{i(Et+\vec p\cdot\vec x)},\quad E>0.$$
Since the plane wave expansion is a sum over all 3-momenta, we can write it as
$$\int d^3pa(\vec p)e^{i(Et+\vec p\cdot\vec x)}=\int d^3pa(-\vec p)e^{i(Et-\vec p\cdot\vec x)}=\int d^3pa(\vec p)d^3pe^{ipx},\quad E>0.$$

However, all particles in the theory must have positive energy so we need to given an interpretation to negative energy/frequency modes. Changing sign of time is equivalent to changing sign of charge and reflecting 3-momentum of the particle. Hence, if we second quantize the theory and the coefficients $a,a^\dagger$ are associated to particles, then the coefficients corresponding to negative frequencies are associated to antiparticles, $b,b^\dagger$. Moreover, these modes correspond to a reflected 3-momentum (negative time) and we interpret this as outgoing antiparticles as long as we interpret the positive frequency modes as incoming particles. By outgoing we mean a particle that is created in the system and then leaves it, therefore must be associated to a creation operator. An incoming particle enters the system and is absorbed by it, thus being associated to a annihilation operator.

## Best Answer

Although the wave function $\phi_0$ in the old formalism and the field operator $\phi$ in QFT both satisfy the K-G equation, their consequences are very different. As a wave function, the expansion of $\phi_0$ in energy eigenstates has the form $\phi_0(x)=\Sigma c_n(x) e^{-iE_n t}$, So a term like $a^*(p)e^{ip\cdot x}$ means the existence of negative energy. But in QFT, the operator (where $a^\dagger_c$ corresponds to the antiparticle) $$\phi(x)\propto\int \frac{d^3 p}{\sqrt{2E}}\left[a(p) e^{-ip\cdot x}+a_c^\dagger(p) e^{ip\cdot x})\right]$$ has no such interpretation. In free theory, if we act $\phi^\dagger$ on the vacuum state: $$\phi^\dagger(x)|0\rangle\propto\int \frac{d^3 p}{\sqrt{2E}} e^{ip\cdot x}|\vec{p}\rangle=\int \frac{d^3 p}{\sqrt{2E}} e^{-i\vec{p}\cdot \vec{x}}\left(e^{iEt}|\vec{p}\rangle\right)$$ we get the superposition of momentum eigenstates in the non-relativistic limit, with weighting factor $e^{-i\vec{p}\cdot\vec{x}}$. So $\phi^\dagger(x)|0\rangle$ is just the position eigenstate of a particle in the Heisenberg picture. Similarly, $\phi(x)|0\rangle$ is the position eigenstate of an antiparticle. Therefore the wave function for a particle should be $$\langle x|\psi\rangle_{particle}=\langle0|\phi(x)|\psi\rangle_{particle}$$ which satisfies the K-G equaiton but only $a(p)$ and positive energy are involved here. And the wave function for an antiparticle is $$\langle x|\psi\rangle_{antiparticle}=\langle0|\phi^\dagger(x)|\psi\rangle_{antiparticle}$$ where only $a_c(p)$ and positive energy are involved.

I'm not sure what the exact meaning of your second question is , maybe Section 5.2 in

The Quantum Theory of Fieldsby Weinberg about the construction of the scalar field is what you're seeking.