I just started learning QFT and I was wondering if one is able to quantize the Schrödinger field similar to the way one is able to quantize electromagnetic or elastic mechanical wave modes. E.g. phonons are acquired by solving the classical Newton's equation for the crystal lattice in normal coordinates, and then each normal mode is considered as a harmonic oscillator which is able to gain or lose energy in quanta of $\hbar \omega$. Is there a similar way for the Schrödinger field? E.g. by solving the Schrödinger equation for the required potential and then considering each wave mode as a harmonic oscillator. By direct calculation one is able to see that this does not give the correct energies (e.g. this way one doesn't get the $\approx0.5~MeV$ energy of the electron), but maybe with some small correction, it could. Any ideas?
[Physics] Naive quantization of Schrödinger field
quantum-field-theoryschroedinger equationsecond-quantization
Related Solutions
Is this not simply the closure relation $\sum_{i}\psi _{i}(\mathbf{r},\xi )\psi _{i}^{\ast }(\mathbf{r}\prime ,\xi \prime )=<\mathbf{r,\xi }|{\{}\sum_{i}|\psi _{i}><\psi _{i}|{\}}|% \mathbf{r}\prime ,\xi \prime >=<\mathbf{r,\xi }|\mathbf{r% }\prime ,\xi \prime >$?
$\renewcommand{ket}[1]{|#1\rangle}$ Item #4 in your list is best thought of as the definition of the word "particle".
Consider a classical vibrating string. Suppose it has a set of normal modes denoted $\{A, B, C, \ldots\}$. To specify the state of the string, you write it as a Fourier series
$$f(x) = \sum_{\text{mode } n=\in \{A,B,C,\ldots \}} c_n [\text{shape of mode }n](x) \, .$$
In the typical case, $[\text{shape of mode }n](x)$ is something like $\sin(n\pi x / L)$ where $L$ is the length of the string. Anyway, the point is that you describe the string by enumerating its possible modes and specifying the amount by which each mode is excited by giving the $c_n$ values.
Suppose mode $A$ has one unit of energy, mode $C$ has two units of energy, and all the other modes have zero units of energy. There are two ways you could describe this situation.
Enumerate the modes (good)
The first option is like the Fourier series: you enumerate the modes and give each one's excitation level: $$|1\rangle_A, |2\rangle_C \, .$$ This is like second quantization; we describe the system by saying how many units of excitation are in each mode. In quantum mechanics, we use the word "particle" instead of the phrase "unit of excitation". This is mostly because historically we first understood "units of excitations" as things we could detect with a cloud chamber or Geiger counter. To be honest, I think "particle" is a pretty awful word given how we now understand things.
Label the units of excitation (bad)
The second way is to give each unit of excitation a label, and then say which mode each excitation is in. Let's call the excitations $x$, $y$, and $z$. Then in this notation the state of the system would be $$\ket{A}_x, \ket{C}_y, \ket{C}_z \, .$$ This is like first quantization. We've now labelled the "particles" and described the system by saying which state each particle is in. This is a terrible notation though, because the state we wrote is equivalent to this one $$\ket{A}_y, \ket{C}_x, \ket{C}_z \, .$$ In fact, any permutation of $x,y,z$ gives the same state of the string. This is why first quantization is terrible: particles are units of excitation so it is completely meaningless to give them labels.
Traditionally, this terribleness of notation was fixed by symmetrizing or anti-symmetrizing the first-quantized wave functions. This has the effect of removing the information we injected by labeling the particles, but you're way better off just not labeling them at all and using second quantization.
Meaning of 2$^{\text{nd}}$ quantization
Going back to the second quantization notation, our string was written $$\ket{1}_A, \ket{2}_C$$ meaning one excitation (particle) in $A$ and two excitations (particles) in $C$. Another way to write this could be to write a single ket and just list all the excitation numbers for each mode: $$\ket{\underbrace{1}_A \underbrace{0}_B \underbrace{2}_C \ldots}$$ which is how second quantization is actually written (without the underbraces). Then you can realize that $$\ket{000\ldots \underbrace{N}_{\text{mode }n} \ldots000} = \frac{(a_n^\dagger)^N}{\sqrt{N!}} \ket{0}$$ and just write all states as strings of creation operators acting on the vacuum state.
Anyway, the interpretation of second quantization is just that it's telling you how many excitation units ("quanta" or "particles") are in each mode in exactly the same way you would do it in classical physics.
Comments on #4 from OP
In introductory quantum we learn about systems with a single particle, say, in a 1D box. That particle can be excited to a variety of different energy levels denoted $\ket{0}, \ket{1},\ldots$. We refer to this system as having "a single particle" regardless of which state the system is in. This may seem to run contrary to the statements made above in this answer in which we said that the various levels of excitation are referred to as zero, one, two particles. However, it's actually perfectly consistent as we now discuss.
Let's write the equivalent first and second quantized notations for the the single particle being in each state: $$\begin{array}{lllll} \text{second quantization:} & \ket{1,0,0,\ldots}, & \ket{0,1,0,\ldots}, & \ket{0,0,1,\ldots} & \ldots \\ \text{first quantization:} &\ket{0}, &\ket{1}, &\ket{2}, & \ldots \end{array} $$ Although it's not at all obvious in the first quantized notation, the second quantized notation makes clear that the various first quantized states involve the particle occupying different modes of the system. This is actually pretty obvious if we think about the wave functions associated to the various states, e.g. using first quantized notation for a box of length $L$ \begin{align} \langle x | 0 \rangle & \propto \sin(\pi x / L) \\ \langle x | 1 \rangle & \propto \sin(2\pi x / L) \, . \end{align} These are just like the various modes of the vibrating string. Anyway, calling the first quantized states $\ket{0}$, $\ket{1}$ etc. "single particle states" is consistent with the idea that a particle is a unit of excitation of a mode because each of these states has one total excitation when you sum over all the modes. This is really obvious in second quantized notation.
Best Answer
Indeed, the ordinary Schrödinger equation can be second-quantized, yielding non-relativistic QFT.
To do so, one first rewrites the Schrödinger equation as a "classical" field theory:
the phase space of this field theory is the Hilbert space $\mathcal{K}$ of the first-quantized theory
with (complex) coordinates $a_k(\psi) = \left\langle e_k \middle| \psi \right\rangle$ for some ONB $\left(e_k\right)_k$ of $\mathcal{K}$
obeying the Poisson brackets $\left\{ a_k, a^*_l \right\} = -i \delta_kl$
and with Hamiltonian function(al) $\mathcal{H}(\psi) = \left\langle \psi \middle| H \middle| \psi \right\rangle$, where $H$ denotes the first-quantized Hamiltonian operator (that this Hamiltonian indeed encodes the first-quantized Schrödinger equation as can be checked from $\frac{d}{dt} a_k(\psi) = \{ a_k, \mathcal{H} \}$).
Remarkably, the canonical (second-)quantization of this field theory describes precisely an arbitrary number of undistinguishable (bosonic) particles, which each obey the first-quantized theory:
the second-quantized Hilbert space is the Fock space $\mathcal{F} = \bigoplus_{N \geq 0} \mathcal{K}^{\otimes N, \text{sym}}$, where $\mathcal{K}^{\otimes N, \text{sym}}$ is the symmetric subspace of $\mathcal{K}^{(1)} \otimes \dots \otimes \mathcal{K}^{(N)}$, which describes $N$ undistinguishable particles
the quantization of $a_k$, resp. $a^*_l$, is the annihilation operator $\hat{a}_k$ for a particle of "type $e_k$", resp. the creation operation $\hat{a}^+_l$, whose commutator is indeed $\left[ \hat{a}_k, \hat{a}^+_l \right] = \delta_kl$
the second-quantized Hamiltonian, obtained by quantizing $\mathcal{H}$, is $\hat{H} = \bigoplus_{N \geq 0} \sum_{n=1}^N 1^{(1)} \otimes \dots \otimes H^{(n)} \otimes \dots \otimes 1^{(N)}$ (this is easier to show if the ONB $\left(e_k\right)_k$ is chosen as the eigenbasis of the first-quantized Hamiltonian $H$, in which case $\hat{H} = \sum_k E_k \hat{a}^+_k \hat{a}_k$), meaning that each particle evolves independently, according to the first-quantized Hamiltonian H.
Of course, the Schrödinger equation is non-relativistic, so if $e_k$ is an energy eigenstate for, say, a single electron, the corresponding $E_k$ will correspond to the NR limit $E = \sqrt{m^2 c^4 + p^2 c^2} \approx m c^2 + \frac{p^2}{2m}$. Although the constant term $mc^2$ is often dropped from the 1st quantized theory, one should keep it here if we want the creation/annihilation operator to change the total energy by the correct order of magnitude.
I like this example as it cleanly demonstrates how the 2 aspects of second-quantization tie together, namely:
describing an arbitrary number of particles obeying this first-quantized theory;
taking a (first-quantized) wave-equation, and thinking of it as a "classical" field theory to be quantized again, hence the name "second-quantization": although the physical interpretation is a priori different, the mathematics of going from the 1st to the 2nd-quantized theory happen to be exactly the same as the ones to go from the classical theory to the 1st-quantized one!
In a typical introduction to QFT, which goes directly to the relativistic case and deals with the second-quantization of the Dirac/Klein-Gordon wave-equations, this remarkable match is somewhat obscured by the lack of a well-defined first-quantized theory (due to the pathologies of these relativistic wave-equations).
In addition, the above holds for any first-quantized theory, so one can see the magic happens already by working out the second-quantization of a simple spin $1/2$: there, the phase space to be (second-)quantized is just 4 dimensional (thus bypassing field theoretic subtleties arising from an infinite-dimensional phase space), with two independent complex variables, each of which can be identified with the $z = x + ip$ variable of an harmonic oscillator. For reference purposes, the more difficult case of 2nd-quantizing the Schrödinger equation of a non-relativistic particle is treated in details eg. in Schiff L.I. "Quantum mechanics" book (sections 45 & 46 of chapter XIII).