A Ensembles in quantum field theory

[vanhees71]

I still don't get, what is so difficult.

The standard description in the Heisenberg picture of the Potel paper is that the field operators move according to the Heisenberg operator equations of motion. Since it's a linear system in this simplified treatment, the equations are solved as if the operators where simple real-valued functions, i.e., it looks like the solution of the classical equations.

The state in this description is time-independent and a single-particle state describing a silver atom emitted from the oven through a hole.

Of course, the result is the same as in the first-quantization formalism described in the paper by Potel et al.

 

[A. Neumaier]

I still don't get, what is so difficult.

I don't think that anything here is difficult. I just want to get your agreement on what seems to be the findings of the discussion so far, or an explanation of where you think my description in post #60 needs to be changed.

The standard description in the Heisenberg picture

In post #60 I had already switched to the interaction picture since in this picture no separate discussion of boundary conditions is needed. Please confirm agreement or let me know where you don't agree.

of the Potel paper [...]

The state in this description is time-independent and a single-particle state describing a silver atom emitted from the oven through a hole.

But this thread is about a quantum field description, not a 1-particle description, and I just want to know whether you agree that #60 is an adequate QFT formulation.

 

[vanhees71]

I don't understand, why you think there are equations of motion for some potential. I can't say, whether I agree or agree not with something, I don't understand.

In the interaction picture the state evolves according to
$$\hat{\rho}(t)=\hat{C}(t) \hat{\rho}(0) \hat{C}^{\dagger}(t),$$
where
$$\dot{\hat{C}}(t)=-\mathrm{i} \hat{H}_I(t) \hat{C}(t) \; \Rightarrow \; \hat{C}(t) =\mathcal{T}_c \exp[-\mathrm{i} \int_{0}^{t} \mathrm{d} t' \hat{H}_I(t')]$$
and the operators representing observables
$$\hat{O}(t) = \hat{A}(t) \hat{O}(0) \hat{A}^{\dagger}(t),$$
where
$$\dot{\hat{A}}(T) = \mathrm{i} \hat{H}_0(t) \hat{A}(t) \; \Rightarrow \; \hat{A}(t) = \mathcal{T}_c \exp[+\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_0(t')].$$
You have
$$\hat{H}=\hat{H}_0+\hat{H}_I=\frac{\vec{p}^2}{2m} + \mu_{\text{B}} g_s \hat{\vec{s}} \cdot \vec{B}(\hat{\vec{x}}).$$
A nice model field is
$$\vec{B}=(B_0+\beta z) \vec{e}_3 -\beta y \vec{e}_2.$$
It's realized by the usual magnet depicted in the Wikipedia article

https://commons.wikimedia.org/wiki/...#/media/File:Stern-Gerlach_experiment_svg.svg

for particles close to the axis marked by the beam in that picture.

A good decomposition here is not the naive "interaction-picture" one but
$$\hat{H}_0=\frac{\hat{\vec{p}}^2}{2m} + \mu_{\text{B}} g_s \hat{s}_z (B_0+\beta z)$$
and
$$\hat{H}_I=-mu_{\text{B}} g_s \hat{s}_y \beta y.$$
You can solve the equation of motion for $\hat{H}_0$ analytically and treat the "interaction" $\hat{H}_I$ as perturbation, because for a large $B_0$ the $y$-component of the spin is rapidly precessing and thus the influence on the much less rapid momentum change is negligible, since it averages out to 0 approximately, and that's the interesting case to describe the SGE.

 

[A. Neumaier]

I don't understand, why you think there are equations of motion for some potential. I can't say, whether I agree or agree not with something, I don't understand.

OK; then I'll explain again. I take the interaction picture with respect to the QFT Hamiltonian associated with the 1-particle Hamiltonian
$$\frac{\vec{p}^2}{2m}-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator. Then

$$\hat{O}(t) = \hat{A}(t) \hat{O}(0) \hat{A}^{\dagger}(t),$$
where
$$\dot{\hat{A}}(T) = \mathrm{i} \hat{H}_0(t) \hat{A}(t) \; \Rightarrow \; \hat{A}(t) = \mathcal{T}_c \exp[+\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_0(t')].$$

can be evaluated with (2) from #52 and gives (3) from #54, where $V$ the potential $V$ satisfies an equation of motion because it is transformed in this way by your general recipe. This follows from the general properties of how linear combinations of creation and annihilation operators transform under conjugation with $\hat{A}(t)$. Doing the calculations shows that $V$ is as claimed in #60.

 

[A. Neumaier]

I don't understand, why you think there are equations of motion for some potential.

A more intuitive way to understand this is to first consider in (3) a general potential. It is surely completely determined by the way the beam is prepared.

Since a local source can prepare only some boundary condition at the source itself (at some fixed $z<0$), so how the potential changes with $z$ must be determined by the forces acting on the beam. These forces therefore determine the potential at all other $z$, and the only sensible way to do so is by means of some differential equation. Working out the details when the source is an electromagnetic field leads necessarily to what I wrote in #60.

 

[vanhees71]

OK; then I'll explain again. I take the interaction picture with respect to the QFT Hamiltonian associated with the 1-particle Hamiltonian
$$\frac{\vec{p}^2}{2m}-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator. Then

can be evaluated with (2) from #52 and gives (3) from #54, where $V$ the potential $V$ satisfies an equation of motion because it is transformed in this way by your general recipe. This follows from the general properties of how linear combinations of creation and annihilation operators transform under conjugation with $\hat{A}(t)$. Doing the calculations shows that $V$ is as claimed in #60.

But the potential is given by plugging in the operators with the time-dependence solved exactly with $\hat{H}_0$ as the generator for the time-evolution of the "observable operators".

I've unfortunately not the time right now to do this additional calculation for the time-evolution of the states. There it should be sufficient to use the first-order perturbation theory because of the rapid precession of the $\sigma_y$ component (using the simplified magnetic field given above), but this should be straight forward.

 

[A. Neumaier]

But the potential is given by plugging in the operators with the time-dependence solved exactly with $\hat{H}_0$ as the generator for the time-evolution of the "observable operators".

Yes. That's what I did, with the results I claimed. No approximation is involved.

I've unfortunately not the time right now to do this additional calculation for the time-evolution of the states.

I'll give explicit details for deriving #60, later today or tomorrow.

 

[vanhees71]

Then it should be sufficient to use the 1st-order approximation for $\hat{C}$:
$$\hat{C}(t)=1-\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_I(t').$$
The 1st-order correction term should stay hopefully "small".

 

[A. Neumaier]

I'll give explicit details for deriving #60, later today or tomorrow.

See the attached pdf. I gave a complete derivation of all formulas used, but I guess that one can find them also in books such as

• A.L. Fetter and J.D. Walecka, Quantum theory of many-particle systems. Courier Corporation (2012).

While writing up the details I noticed that I had mixed up bosons and fermions in part of the earlier arguments. As a result, the operators (2) and (3) vanish since squares of odd operators are zero. To model a nontrivial source, one therefore needs to use in the interaction picture the more general form
$$V_I(t)=\int dxdx' w(t,x,x')a(x)a(x')+ h.c.$$
for the interaction, where $w(t,x,x')$ satisfies the 2-particle Schrödinger equation for two independent particles in an external magnetic field.

Then it should be sufficient to use the 1st-order approximation for $\hat{C}$:
$$\hat{C}(t)=1-\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_I(t').$$
The 1st-order correction term should stay hopefully "small".

Since the results are nice and exact, I derived everything without using perturbation theory. Of course you may check the validity by comparing with the results of first order perturbation theory.

 

[vanhees71]

What should be given at the end of the calculation is the distribution function of the silver atoms at the plate. I don't see this in your pdf.

I understand sect. 1, and it looks correct. I have no idea what's done in sect. 2.

 

[A. Neumaier]

What should be given at the end of the calculation is the distribution function of the silver atoms at the plate. I don't see this in your pdf.

I had only promised the derivation of #60, where there is no second plate at $x_3>0$ where the silver would be absorbed.

Before modeling the measurement at a second plate I first want to be sure of your agreement to this part. For if we cannot agree on the physics without the measurement we'll never agree on stuff that depends on this.

I understand sect. 1, and it looks correct. I have no idea what's done in sect. 2.

Section 2 is a special case of Section 1, where I apply it to the situation discussed in posts #1-#60. All formulas are standard and have a complete derivation, with only very elementary steps omitted. Thus you'd be easily able to verify its correctness.

 

[vanhees71]

As I said, I don't understand Sect. 2. I can't judge, whether it's right or wrong.

 

[A. Neumaier]

As I said, I don't understand Sect. 2. I can't judge, whether it's right or wrong.

Where is the first formula which you don't understand? Then I can explain....

 

[vanhees71]

I don't understand anything of Sect. 2. Do you want to work in the 1st or 2nd-quantization formalism?

In the 2nd-quantization formalism the single-particle Hamiltonian looks like (in the Heisenberg picture)
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \sum_{\sigma} \hat{\psi}_{\sigma}^{\dagger}(t,\vec{x}) \left [-\frac{\hbar^2}{2m} \Delta + \mu_{\text{B}} g_s \vec{S} \cdot \vec{B}(\vec{x}) \right] \hat{\psi}(t,\vec{x}).$$
The non-zero equal-time (anti-)commutation relations are
$$[\hat{\psi}_{\sigma}(t,\vec{x}),\hat{\psi}_{\sigma'}^{\dagger}(t,\vec{x}')]_{\pm}=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma \sigma'}.$$

 

[A. Neumaier]

I don't understand anything of Sect. 2. Do you want to work in the 1st or 2nd-quantization formalism?

I work in second quantization, just with a simplified notation compared to yours.

In the 2nd-quantization formalism the single-particle Hamiltonian looks like (in the Heisenberg picture)
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \sum_{\sigma} \hat{\psi}_{\sigma}^{\dagger}(t,\vec{x}) \left [-\frac{\hbar^2}{2m} \Delta + \mu_{\text{B}} g_s \vec{S} \cdot \vec{B}(\vec{x}) \right] \hat{\psi}(t,\vec{x}).$$

This is exactly equation (3) from Section 2,
$$H_0:=\int dx a(x)^*H_1(t,x,\widehat p)a(x),$$
just written with far fewer (in the latex 43 rather than your 214) characters, and I used the Schrödinger picture. (Mathematicians are lazy, and simplify their typography to the essentials.) The integration domain, the dimension, and the spin indices are suppressed (to avoid index errors as in your formula), so are your hats and vecs. The annihilation operators are denoted by $a$ rather than $\hat\psi$, which is as in https://en.wikipedia.org/wiki/Creation_and_annihilation_operators . The daggers are replaced by stars,
and the detailed form of the Hamiltonian is not described (but $\widehat p$ is the momentum operator) . Does (3) now make sense?

The non-zero equal-time (anti-)commutation relations are
$$[\hat{\psi}_{\sigma}(t,\vec{x}),\hat{\psi}_{\sigma'}^{\dagger}(t,\vec{x}')]_{\pm}=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma \sigma'}.$$

Since silver is Fermionic, this is exactly my third formula in Section 2,
$$a_j(x)a_k(y)^*+a_k(y)^*a_j(x)=\delta_{jk}\delta(x-y)$$
but with $j,k$ in place of your Greek spinor indices.

With this dictionary, the remainder is now perhaps understandable, or where do you need further explanations?

 

[vanhees71]

Ok, the only difference is that in the Schrödinger picture the $\hat{\psi}_{\sigma}(\vec{x})$ have no time argument. Otherwise you can condense your notation such that it gets unrecognizable. I know that mathematicians tend to do this, but when it comes to do really calculations saving this little effort in notation often leads to a huge efford to disentangle the so introduced confusion ;-)).

The $\hat{a}$'s usually denote annihilation operators for single-particle energy eigenstates.

 

[A. Neumaier]

Ok, the only difference is that in the Schrödinger picture the $\hat{\psi}_{\sigma}(\vec{x})$ have no time argument. Otherwise you can condense your notation such that it gets unrecognizable.

So, do you now understand the remainder of Section 2, including (6) and the explanations afterwards? Or what else needs explanation?

The $\hat{a}$'s usually denote annihilation operators for single-particle energy eigenstates.

Weinberg uses $a$'s for annihilation operators in QFT.

 

[vanhees71]

I don't understand the remainder of Sect. 2. I've no clue, what your aim is. If the notation is so different from the standard notation, it needs more explaining text.

Weinberg uses standard notation. The $\hat{a}$'s are annihilation operators for energy eigenstates and occur in the corresponding mode decomposition of the free-particle operators in the interaction picture.

 

[A. Neumaier]

I don't understand the remainder of Sect. 2. I've no clue, what your aim is.

May aim is to prove the final sentence of Section 2, which is the statement given in #69 about the form of the Hamiltonian in the interaction picture.

If the notation is so different from the standard notation, it needs more explaining text.

I can add explaining text only if I know what more to explain.

In the sentence containing (3) I say what what $H_1$ is, what $a(x)$ is, and I list the commutation rules in the next sentence. Then I give a definition of what I mean by $a(f)$ - which is the same as in the Wikipedia link given in #75. Similarly I explain everything else.

Thus in my view I explained every bit of notation. It is standard in mathematical physics; e.g., Derezinski and Gerard.

So please let me know where more explanation is needed, and I'll add it to make it more understandable.

 

[vanhees71]

I don't have the time right now to learn a completely new notation, and I don't know what you want to achieve. The Hamiltonian for particles moving in an external magnetic field is given in #74. It looks the same in all pictures of time evolution. I don't know, which kind of interaction (two-body?) Hamiltonian you want to add. A simple two-body interaction potential contribution reads
$$\hat{H}_{\text{int}}=\frac{1}{2} \sum_{\sigma_1,\sigma_2} \int_{\mathbb{R}^3} \mathrm{d}^3 x_1 \int_{\mathbb{R}^3} \mathrm{d}^3 x_2 \hat{\psi}_{\sigma_1}^{\dagger}(t,\vec{x}_1) \hat{\psi}_{\sigma_2}^{\dagger}(t,\vec{x}_2) V_{\sigma_1 \sigma_2}(\vec{x}_1-\vec{x}_2) \hat{\psi}_{\sigma_2}(\vec{x}_2) \hat{\psi}_{\sigma_1}(\vec{x}_1).$$

 

[A. Neumaier]

The Hamiltonian for particles moving in an external magnetic field is given in #74. It looks the same in all pictures of time evolution. I don't know, which kind of interaction (two-body) Hamiltonian you want to add.

An interaction representing the presence of a source. We had already discussed it in posts #41 and #52; it is needed since we model only the halfspace $x_3\ge 0$; the remainder of the lab (in $x_3<0$) is too complex to be modeled.

 

[vanhees71]

But this isn't described by a two-body interaction. Maybe it's something like $J(x) \hat{\psi}(x)+\text{h.c.}$ with a c-number source $J$.

 

[A. Neumaier]

But this isn't described by a two-body interaction. Maybe it's something like $J(x) \hat{\psi}(x)+\text{h.c.}$ with a c-number source $J$.

But a Hamiltonian cannot have terms linear in an odd operator; we discussed this before. Thus a 2-body term is the simplest possibility!

 

[A. Neumaier]

But this isn't described by a two-body interaction. Maybe it's something like $J(x) \hat{\psi}(x)+\text{h.c.}$ with a c-number source $J$.

But a Hamiltonian cannot have terms linear in an odd operator; we discussed this before. Thus a 2-body term is the simplest possibility!

I found the proper way to do it. Thus the description in #60 should be updated as follows:

A silver source outside the beam, switched on at time $t=0$, produces silver particles, approximately one particle per second that quickly (in much less than a second) proceed along a beam initially along the $z$ axis. The beam is subject to a weak external magnetic field $B(t,x)$ that may be switched on gradually. No other forces are assumed to be present; in particular no air or obstacle. Since silver is heavy, we consider a nonrelativistic quantum field description.

For this particular preparation procedure, the potential is given in the interaction picture by
$$V_I(t):=\int dx v(t,x)a(x)+h.c.,$$
where the source
$$v(t,x)=\sum_h v_h(t,x)b_h,$$
is a linear combination of hole annihilators $b_h$. It is proved in Section 2 of the attached text that each $v_h(t,x)$ satisfies the single-particle Pauli equation
$$i\hbar \partial_t v_h(t,x)=V_I(t,x) v_h(t,x)$$
with the single-particle interaction
$$V_I(t,x):=-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator.

The preparation considered also implies that, before the magnetic field is switched on, we may assume that each $v_h(t,x)$ is a narrow complex Gaussians peaked at $x_3=0$ at second $h$ that moves along the $x_3$ axis and flattens with time (as described in Section 1.2 of your lecture notes).

Do you now agree?

 

[Peter Morgan]

Disclaimer: I've only scanned through the many comments here. I'm responding to the initial post in a way that I didn't notice reflected anywhere else. My apologies if it is.

My theoretical approach to how I see that question has been to take any quantum model to be constructed as an algebra of operators that are used as models for statistical trials. The algebra is generated by a finite set of operators $\hat\phi_1, \hat\phi_2, ..., \hat\phi_n$, then we introduce a state over those (measurement) operators, which allows us to GNS-construct a Hilbert space that contains other states that we can use to describe different experiments. I take it we can only know what we can measure, so the measurement operators we have in hand (or "in algebra") is the limit of what we can say about the state.

For QFT, I see this modified only by taking the algebra to be generated by a finite set of test functions, which give an indication of how each experiment is placed in space and time relative to others. The index set is a finite set of test functions instead of a finite set of integers, so we have $\hat\phi_{f_1}, \hat\phi_{f_2}, ..., \hat\phi_{f_n}$. The state that we introduce is, by convention and axiom, maximally symmetric, which for a Wightman field requires the "vacuum" state to be manifestly Poincaré invariant. For free fields for which the vacuum state is Gaussian, it's enough to require the 2-operator function $\langle 0|\hat\phi_f^\dagger\hat\phi_g|0\rangle$ to be a positive semi-definite manifestly Poincaré invariant bilinear functional of the test functions (which completely fixes the Gaussian state). [ If we want to construct the usual idealized quantized Klein-Gordon field, with the Fock space as its Hilbert space, we can take the "direct" or "inductive" limit, in which, in very broad brush terms, the set of test functions is increased without limit until it is the Schwartz space of test functions that are defined on Minkowski space and that are smooth and have a smooth Fourier transform. ]
We have to ask how we know what test functions we should use for a given experiment. I think the answer to that is not easy: insofar as it exists, the rules are given in thick textbooks on quantum optics and AMO and other physics. The ultimate test will always be consistency, which will evolve over decades.
To me, it seems clear that if the mathematics we use as models for "measurements" (which I think I prefer to call "statistical trials") is that of operators and operator algebras, as an algebraic presentation of a specific class of Generalized Probability Theories, then we will always be able to write down some finite index set and the operators will be $\hat\phi_1, \hat\phi_2, ..., \hat\phi_n$. If we do more experiments, we will introduce more operators and the Hilbert space we construct will be bigger. For the idealized QFT case, we take the index set to be the Schwartz space on Minkowski space, which introduces much more extensive and detailed operational questions.
2¢.

 

[vanhees71]

The only caveat is that in QFT Poincare invariance in the present form is realized by the microcausality constraint, i.e., the field operators themselves, whose equations of motion are defined, in the Heisenberg picture, by local unitary representations of the proper orthochronous Poincare group, usually are not representing observables, but the observables are defined by operator products of these field operators fulfilling the microcausality constraint.

For gauge theories in addition these local "observable operators" also have to be gauge-covariant. That's why, e.g., there is no consistent definition of a "photon-number-density operator", and that's why there is no naive particle interpretation for the photon (in addition to the fact that there are no bona fide position observables for them and thus the localizability of single photons is even less possible than for massive "particles") and that's why the true "intensity measure" of the electromagnetic field is the (gauge-invariante Belinfante) energy density rather than "photon-number density".

This is of course also to be taken with a grain of salt since a mathematically rigorous formulation of interacting quantum fields in (1+3) dimensions has still not been achieved. The above thus has to taken as realized in the usual "perturbative way" with all the regularization/renormalization procedures which circumvent all the fundamental quibbles related to, e.g., Haag's theorem (non-existence of the interaction picture).

 

[Peter Morgan]

This is of course also to be taken with a grain of salt since a mathematically rigorous formulation of interacting quantum fields in (1+3) dimensions has still not been achieved. The above thus has to taken as realized in the usual "perturbative way" with all the regularization/renormalization procedures which circumvent all the fundamental quibbles related to, e.g., Haag's theorem (non-existence of the interaction picture).

Going a little off-topic in response to your third paragraph, I venture to say that I've given a mathematically rigorous formulation of a class of interacting quantum fields in (1+n)-dimensions in my "A source fragmentation approach to interacting quantum field theory", https://arxiv.org/abs/2109.04412. I can't show that it's empirically supported, so there's that (though you didn't insist on it), and I choose to weaken one of the Wightman axioms so that $\hat\phi_f$ is a nonlinear function of the test function $f(x)$, so there's also that, but I'm pretty sure it's rigorous and interacting, et cetera. The models I exhibit are all toy models, so I'll forgive you if you don't even have a look (though it's only a short paper, so go on).
Concerning Haag's theorem, did you see the preprint that dropped yesterday, "How Haag-tied is QFT, really?", https://arxiv.org/abs/2212.06977?

 

[vanhees71]

Yes, that's a nice article about Haag's theorem. I'm more at the side of the "phenomenologists", i.e., take the usual perturbative approach with renormalization is the "solution" of the problem with Haag's theorem. It's of course a disappointment that there's not a rigorous formulation along the goals of the axiomatic QFT program yet. On the other hand, all our physical theories are so far "effective theories", and we don't have a theory of everything yet, and it's not guaranteed that we'll ever find one!

I'll have a look at your paper, but I'm not so familiar with the axiomatic approach.

 

[Peter Morgan]

Yes, that's a nice article about Haag's theorem. I'm more at the side of the "phenomenologists", i.e., take the usual perturbative approach with renormalization is the "solution" of the problem with Haag's theorem. It's of course a disappointment that there's not a rigorous formulation along the goals of the axiomatic QFT program yet. On the other hand, all our physical theories are so far "effective theories", and we don't have a theory of everything yet, and it's not guaranteed that we'll ever find one!

I'll have a look at your paper, but I'm not so familiar with the axiomatic approach.

Renormalization works. Yes. I think we can make it more rigorous, however, by paying much closer attention to how we use the test functions that describe an experiment. As we have had it, that description of an experiment has been taken to control the regularization and renormalization scales implicitly and with hand-waving, only saying, very crudely, that a given experiment probes the atomic scale or weak scale (or Planck scale), say, but I think there's good reason for mathematical physicists to insist on explicitness and detail. Showing that this is mathematically workable needs the Reeh-Schlieder theorem, I think, but the idea doesn't really need an axiomatic approach.
I'm trying to write a new version, which has been slow going for the last year but I've started to feel up to improving it, so that I have a new figure that I hope might be helpful:

 

[vanhees71]

Is this somehow related to the Epstein-Glaser approach (as used in Scharff's book "finite quantum electrodynamics"? That's of course also very understandable for a "phenomenologist": Measurements are always made with real-world devices, which are always "coarse graining". You never measure the position of a "particle" at a precise point in space and time but you register a particle at a finite (small) region in space (e.g., a "pixel" at a silicon chip) and time. The same holds for momenta. That's all encoded also in the sloppy formalism used by "phenomenologists": Even on this sloppy level of the mathematics a plane-wave (momentum eigenstate) is not a true state but a distribution, and real states have to be square integrable and thus you don't have, e.g., a "photon at a precise momentum and energy" but a finite-width spectral function (although this width can be made very narrow of course).

Another thing is that this hierarchy of resolution is a gift for physicists, given the historical development of science, i.e., they could deal with pretty classical physics first, before they discovered that underlying this "effective model of the world" there's need for a more abstract and less familiar "quantum model of the world", which however leads to the conclusion that the "classical description" is valid only as an "effective theory" for coarse-grained macroscopic observables.

The next level then was "atomic physics". At least as long as you only deal with the lightest atoms, a non-relativistic quantum theoretical description of electrons and a classical treatment of the electromagnetic (for the most simple cases just static Coulomb fields) is sufficient and pretty accurate. Then came relativistic corrections, which were nicely treatable within perturbation theory, needed to explain, e.g., the "fine structure" of atomic spectra.

Field quantization, ironically, was rejected by the physics community as "overdoing it". Indeed, in the 1st paper by Born and Jordan, where they put Heisenberg's famous "Helgoland idea" to a more solid ground ("matrix mechanics"), there's a chapter about the quantization of the electromagnetic field is contained, but lead to the above mentioned reaction of many quantum physicists. That's why Dirac had to come to the same conclusion a few years later, when he again quantized the e.m. field (within his operator formalism, of course) to explain "spontaneous emission" needed to get the black-body Planck spectrum and to understand the "natural line width" of atomic spectral lines, i.e., the instability of the atomic excited states due to coupling to the quantized em. field.

Of course, the divergence of higher-order perturbation theory was immediately apparent, and it took some years, to discover the idea of renormalization. This was again triggered by the necessity to explain the "Lamb shift" for the hydrogen atom. The first calculation was by Bethe treating the atom non-relativistically and then it was also figured out for the full relativistic theory (Feynman, Schwinger) and also for "scalar QED" (Weisskopf, Pauli).

In this sense the fact that Nature doesn't need to be explained by a "theory of everything" but in steps of "ever finer resolution" (or observing at "ever higher energies") is a gift for model building, and it's pretty sure that also all our currently used QFTs (including the Standard Model on the yet most "fundamental" level but also the effective theories needed to describe hadrons like (unitarized) chiral perturbation theory) are indeed "effective theories" as well.

If we are unlucky there's realy the "great dessert", i.e., the next energy scale of new physics may indeed be the Planck scale, and then it will be very difficult to get glimpses of "physics beyond the Standard Model" leading theory into the right direction.