OK, I'll sketch the trailmap for you, but it's all math, of the most abstract kind, since it asks you (and all students) to suspend/abrogate their suppositious intuition from QM when transitioning to QFT.
Instead of transitioning by thinking about wavefunctions, I will, perhaps perversely, beg you to completely ignore any and all wavefunctions, which I have often found confuse students (and chemists) more than illuminate them (with illusory intuition).
I'll just use the standard Heisenberg/Dirac matrix mechanics picture that provides all answers for the oscillator in matrix mechanics, based on operators and their matrix elements. Non-dimensionalize ℏ=1 throughout, for simplicity, and work in one space dimension.
$$ [a,a^\dagger ]=1, \qquad H=\omega ( a^\dagger a +1/2), \leadsto \\
[H, a^\dagger]=\omega a^\dagger, \qquad a|0\rangle, \qquad (a^\dagger)^n|0\rangle = |n\rangle ,
$$
etc... The spectrum, the time evolution , and all matrix elements, etc, can be derived without any mention of wavefunctions, whose subordinate features are given in the above link; but you never really need to know about Hermite functions and stuff to do anything. (In fact, these (eigen)wavefunctions, $\psi_n(x)=\langle x| n\rangle$, are but a mere formal basis in building up your states; but, as you must appreciate, Schrödinger's wavepacket states that truly remind you of a classical oscillator, the so-called coherent states, are something much simpler than superpositions of confounded Hermite functions!)
- QFT (in 1D for simplicity, but easy to generalize to 3D):
Starting from classical field theory (classical strings that you simulate as an infinity of coupled oscillators, whose uncoupled normal modes you find by Fourier transformation to momentum space), you are impelled to consider the quantum linear combination of an infinity of decoupled oscillator generators, $a_p$ and $a^\dagger_p$,
$$
[a_p,a_p'^\dagger]=2\pi \delta(p-p'), \qquad \int \!\! dp~~{\omega_p \over 2\pi} (a_p^\dagger a_p + [a_p, a_p^\dagger]/2 ),\\
[H,a_p^\dagger]=\omega_p a_p^\dagger, \qquad ...
$$
where $\omega^2_p\equiv m^2 +p^2$, and the commutator term in the hamiltonian is an infinite constant $\propto \delta (0)$, ignored, like all such constants in such systems, where we are only interested in energy differences.
- This is but an infinity of oscillators with idiosyncratic energies.
The actual quantized (real, scalar) field operator is
$$
\phi(x)=\int\!\!{ dp\over 2\pi \omega_p}(a_p e^{ipx}+ a_p^\dagger e^{-ipx})\\
= \int\!\!{ dp\over 2\pi \omega_p}(a_p + a_{-p}^\dagger )e^{ipx},
$$
and so on; it satisfies interesting commutation relations of its own, and produces all the answers acting on $|0\rangle$, thus generating the Fock space; but this is not the place to understand QFT and its computational techniques. Time dependence trivially follows QΜ: $\phi(x,t)=e^{iHt}\phi(x)e^{-iHt}$. This is a counterintuitive operator whose functions' matrix elements in vacuum should not be thought of as wavefunctions, even when localized in a peculiar way—see linked question in the comments and below.
The point of the non-answer to your misbegotten question is to illustrate that you really do not need the treacherous intuition of wavefunctions to appreciate the essence of the QM $\mapsto$ QFT transition.
There are lots of murky wavefunctions' illustrations in the market, and you could concoct your own, given the links above; but they almost always misdirect you rather than help you.
- QFT runs on decoupled oscillator operators in momentum space. Each state $a^\dagger_p|0\rangle$ represents a particle of momentum p. The state $(a^\dagger_p)^2/\sqrt{2}|0\rangle$ two particles of momentum p, etc... No need for wavefunctions.
