I have read the book on quantum field theory for some time, but I still do not get the physics underline those tedious calculations. The thing confused me most is how quantum mechanics relates to quantum field theory as an approximation in low energy limit. Take a free scalar field $\psi$ for example. It describes free spin 0 boson. And it satisfies Klein-Gordon equation $(\partial_{\mu}\partial^{\mu}+m^2)\psi=0$. In quantum mechanics, the wave function $\psi^{\prime}$ of a free particle without spin should also obey the Klein-Gordon equation or its classical limit Schrodinger equation. So the field operator $\psi$ and the wave function $\psi^{\prime}$ must have some relations. If we take canonical quantization into account, we may assume $\psi(x,t)=\frac{1}{\sqrt{2}}[a(x,t)+a^{\dagger}(x,t)]$ and interpret $a^{\dagger}$ as a creation operator. But I do not know the next step to deduce the Klein-Gordon equation or Schrodinger equation for particle's wave function $\psi^{\prime}$ in quantum mechanics if we start from quantum field theory and take low energy limit.
Quantum Mechanics – How to Derive Quantum Mechanics from Quantum Field Theory
klein-gordon-equationquantum mechanicsquantum-field-theoryschroedinger equation
Related Solutions
Spin is a property of the representation of the rotation group $SO(3)$ that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation.
The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also accounts for discrete symmetries).
The components of every free field satistfy the Klein-Gordon equation, irrespective of their spin. In particular, every component of the Dirac equations solves the Klein-Gordon equation. Indeed, the Klein-Gordon equation only expresses the mass shell constraint and nothing else. Spin comes in when one looks at what happens to the components.
A rotation (and more generally a Lorentz transformation) mixes the components of the Dirac field (or any other field not composed of spin 0 fields only), while on a $k$-component spin 0 field, it will transform each component separately.
In general, a Lorentz transformation given as a $4\times 4$ matrix $\Lambda$ changes a $k$-component field $F(x)$ into $F_\Lambda(\Lambda x)$, where $F_\Lambda=D(\Lambda)F$ with a $k\times k$ matrix $D(\Lambda)$ that depends on the representation. The components are spin 0 fields if and only if $D(\Lambda)$ is always the identity.
Let's understand this statement in Hamiltonian formalism, where KG equation is equivalent to having the free scalar field hamiltonian and the Heisenberg equations of motion for the free fields.
Then $\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( a_p e^{ipx} + a^\dagger_p e^{-ipx}\right)$ and the canonical conjugate $\pi(\vec{x},t) = \dot{\phi}(\vec{x},t)$, are the most general solution.
Now let's consider an interacting Hamiltonian $H = H_0 + \lambda V$, and DEFINE $$\Phi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \phi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$ $$\Pi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \pi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$ Then it is straight forward to show that $\Phi$ and $\Pi$ satisfy the canonical commutation relations, as well as the new interacting Heisenberg equations (notice that in the definition we use $H(\phi,\pi)$ and in the Heisenberg equations $H(\Phi,\Pi)$, we are allowed to do so because both are equal!). (Hint: to see that the new fields satisfy the full heisenberg equations notice that $\Phi(\vec{x},t) = e^{iHt}\phi(\vec{x},0)e^{-iHt}$) So in this sense they are the interacting fields, written in terms of the free fields.
Then we conclude that $$\Phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( \mathbb{a}(t)_p e^{ipx} + \mathbb{a}(t)^\dagger_p e^{-ipx}\right)$$ Where $$\mathbb{a}(t)_p\equiv e^{iHt}e^{-iH_0 t} a_p e^{iH_0 t}e^{-iHt}$$ and $\mathbb{a}_p(t)$ satisfies the required commutation relations as a consequence of its parents $\Phi$ and $\Pi$ doing so, or as can be verified directly using those of $a_p$.
Notice that this description is particularly useful for weakly coupled theories, since then $\mathbb{a}_p = a_p + \mathcal{O}(\lambda,a_p^2)$, then all our particle spectrum can be inferred from that of the free theory, unlike when this expansion is no longer valid, and the new creation operator can create states completely different in nature from what's contained in the free theory.
Best Answer
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.