# What is the basic scheme of quantum field theories?

• I
Demystifier
Gold Member
WHy is that each one of you gives me a different version?
Because there are many equivalent ways to represent QFT. The one I mentioned is the closest to the Schrodinger picture of ordinary QM. But it is not very convenient in practice, so practical physicists usually ignore it.

jonjacson
Because there are many equivalent ways to represent QFT. The one I mentioned is the closest to the Schrodinger picture of ordinary QM. But it is not very convenient in practice, so practical physicists usually ignore it.
I understand, thanks.

https://www.amazon.com/dp/0201360799/?tag=pfamazon01-20&tag=pfamazon01-20 talks about the Schroedinger picture wave functional formulation (mentioned by @Demystifier), and its equivalence to the more usual Heisenberg picture formulation of QFT.
Well, the book looks ambitious, it says they present string theory assuming you don't know quantum field theory so they explain you in the first part of the book. That has to be very interesting.

DarMM
Gold Member
WHy is that each one of you gives me a different version?

I will search info about wave functionals.

I appreciate any comment really, thank you so much.
As mentioned above there are many ways of formulating Quantum Field Theory.

The Wavefunctional approach is the form most closely related to the way people normally do non-relativistic QM. However solving for the Wave Functional or Wave Functional eigenvalue problems are virtually impossible outside of free field theories, so people usually use the method of calculating the correlation functions I mentioned. Non-relativistic QM can also be expressed via correlation functions just to be clear.

One thing I didn't mention in my post is renormalization. When we write down QFT we often make use of expressions like ##\phi^{3}(x)##. However because the object ##\phi(x)## is so singular, it turns out that ##\phi^{3}(x)## isn't a well defined mathematical expression. So one has to attempt to define a quantity similar to ##\phi^{3}(x)## that is well defined: ##[\phi^{3}(x)]_R##.

I'll give an example using plain old functions. Say you have the Heaviside function ##\Theta(x)## and the function ##\frac{1}{x}##. We have their multiple ##\frac{\Theta(x)}{x}##. However consider the following integral:
$$\int_{\mathbb{R}}{\frac{\Theta(x)}{x}g(x)dx} = \int^{\infty}_{0}{\frac{g(x)}{x}dx}$$
with ##g## some function. The ##\Theta## function has been used to restrict the integral in the second expression.
In general this will diverge if ##g## is non-zero about the point ##x = 0##.
However the following modified version of the integral is well-defined:
$$\int^{a}_{0}{\frac{g(x) - g(0)}{x}dx} + \int^{\infty}_{a}{\frac{g(x)}{x}dx}$$
So we isolate the problematic region and add a term that removes the divergence.
This can be rewritten as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \int^{a}_{0}{\frac{g(0)}{x}dx}$$
And since ##g(0)## can be obtained from a delta function, we can further rewrite it as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\int{\delta(x)g(x)dx}$$

Finally if we put back in the Heaviside function, this can be written as:
$$\int_{\mathbb{R}}{\left[\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)\right]g(x)dx}$$

So we see that:
$$\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
is a modified version of:
$$\frac{\Theta(x)}{x}$$
that produces well defined integrals.

The physicist way of doing this is to call
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
a counterterm and to consider
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)$$
to define an "infinite constant" ##C_a##. So they would write:

$$\left[\frac{\Theta(x)}{x}\right]_R = \frac{\Theta(x)}{x} - C_a\delta(x)$$

In physics language you obtain a version of the product of the Heaviside step function and ##\frac{1}{x}## that behaves well under integrals by subtracting off a counterterm which has an infinite constant. This "well behaved" version is called "renormalized".

For some Quantum Field Theories the terms you have to add are no problem. For instance ##\phi^{4}## theory in three dimensions requires:
$$\left[\phi^{4}\right]_R = \phi^{4} - \delta m^{2}\phi^{2}$$

That is making the ##\phi^{4}## interaction well-defined requires adding a counterterm with the ##\phi^{2}## term. However since the Lagrangian already has a ##\phi^{2}## this is no problem, it doesn't change the character of the theory.

However some theories require the addition of new terms to their Lagrangian to be well defined. If this process doesn't stop, i.e. the new terms then need further new terms, the theory is called non-renormalizable.

lowlize, dextercioby and jonjacson
As mentioned above there are many ways of formulating Quantum Field Theory.

The Wavefunctional approach is the form most closely related to the way people normally do non-relativistic QM. However solving for the Wave Functional or Wave Functional eigenvalue problems are virtually impossible outside of free field theories, so people usually use the method of calculating the correlation functions I mentioned. Non-relativistic QM can also be expressed via correlation functions just to be clear.

One thing I didn't mention in my post is renormalization. When we write down QFT we often make use of expressions like ##\phi^{3}(x)##. However because the object ##\phi(x)## is so singular, it turns out that ##\phi^{3}(x)## isn't a well defined mathematical expression. So one has to attempt to define a quantity similar to ##\phi^{3}(x)## that is well defined: ##[\phi^{3}(x)]_R##.

I'll give an example using plain old functions. Say you have the Heaviside function ##\Theta(x)## and the function ##\frac{1}{x}##. We have their multiple ##\frac{\Theta(x)}{x}##. However consider the following integral:
$$\int_{\mathbb{R}}{\frac{\Theta(x)}{x}g(x)dx} = \int^{\infty}_{0}{\frac{g(x)}{x}dx}$$
with ##g## some function. The ##\Theta## function has been used to restrict the integral in the second expression.
In general this will diverge if ##g## is non-zero about the point ##x = 0##.
However the following modified version of the integral is well-defined:
$$\int^{a}_{0}{\frac{g(x) - g(0)}{x}dx} + \int^{\infty}_{a}{\frac{g(x)}{x}dx}$$
So we isolate the problematic region and add a term that removes the divergence.
This can be rewritten as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \int^{a}_{0}{\frac{g(0)}{x}dx}$$
And since ##g(0)## can be obtained from a delta function, we can further rewrite it as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\int{\delta(x)g(x)dx}$$

Finally if we put back in the Heaviside function, this can be written as:
$$\int_{\mathbb{R}}{\left[\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)\right]g(x)dx}$$

So we see that:
$$\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
is a modified version of:
$$\frac{\Theta(x)}{x}$$
that produces well defined integrals.

The physicist way of doing this is to call
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
a counterterm and to consider
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)$$
to define an "infinite constant" ##C_a##. So they would write:

$$\left[\frac{\Theta(x)}{x}\right]_R = \frac{\Theta(x)}{x} - C_a\delta(x)$$

In physics language you obtain a version of the product of the Heaviside step function and ##\frac{1}{x}## that behaves well under integrals by subtracting off a counterterm which has an infinite constant. This "well behaved" version is called "renormalized".

For some Quantum Field Theories the terms you have to add are no problem. For instance ##\phi^{4}## theory in three dimensions requires:
$$\left[\phi^{4}\right]_R = \phi^{4} - \delta m^{2}\phi^{2}$$

That is making the ##\phi^{4}## interaction well-defined requires adding a counterterm with the ##\phi^{2}## term. However since the Lagrangian already has a ##\phi^{2}## this is no problem, it doesn't change the character of the theory.

However some theories require the addition of new terms to their Lagrangian to be well defined. If this process doesn't stop, i.e. the new terms then need further new terms, the theory is called non-renormalizable.
And do we know the physical reason behind a theory being or not being renormalizable?

DarMM
Gold Member
And do we know the physical reason behind a theory being or not being renormalizable?
I've presented renormalization from a very mathematical point of view, there are other ways of viewing it.

So people approach nonrenormalizibility differently. Some say it simply means the theory isn't well defined, i.e. there's no physical reason as such, it's just that some classic Lagrangians/Hamiltonians that you can write down don't have a well defined quantum version.

Others view all Quantum Field Theories as coming with an intrinsic cut-off, i.e. a maximum length scale at which they can no longer be used due to new physics. In that case non-renormalizible theories are those field theories where the details of the new physics cannot be ignored, you end up with an infinite amount of terms with constants that encapsulate effects from the higher theory, rather than a simple short Lagrangian to describe low energy physics with.

The two views aren't opposed either. You might say a theory doesn't exist because it can only be defined with a cutoff as some approximation of a higher theory, where as if a theory fundamentally exists due to being renormalizable it's because the higher effect don't matter too much to it and it can define a complete (although ultimately incorrect) theory on its own.

bhobba and atyy
A. Neumaier
2019 Award
Please explain why this is a response to my comment. Srednicki discusses no lattice approach to QED - only to strongly coupled gauge theories with confinement.

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dextercioby
bhobba
Mentor
And do we know the physical reason behind a theory being or not being renormalizable?
Yes - in the modern effective field theory approach. Things have moved on since the time Feynman etc worked out renormalization. We now know you can start with just about any QFT, and at low energies the theory will look renormalizable. Such a theory is called an effective field theory. This of course is quite interesting and useful. But if you go to energies that are high enough, renormalizability may break down, and you may not have a quantum field theory at all. So the reason for theories being renormalizable is, at least at the moment, we do not have the technology to reach the energy where it is not.

Wilson sorted it out:
https://quantumfrontiers.com/2013/06/18/we-are-all-wilsonians-now/

There is quite a bit of (advanced) material on EFT to be found on the internet eg:
http://pages.physics.cornell.edu/~ajd268/Notes/EffectiveFieldTheories.pdf

Its actually an interesting area with unexpected phenomena such as triviality occurring. I wish I knew more than I do, and it is an area I am investigating myself.

Thanks
Bill

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Demystifier
bhobba
Mentor
So people approach nonrenormalizibility differently. Some say it simply means the theory isn't well defined, i.e. there's no physical reason as such, it's just that some classic Lagrangians/Hamiltonians that you can write down don't have a well defined quantum version.
That's true, its the old view of renormalisation. But you have to ask yourself - why is renormalization so fundamental that only such theories are valid? Wilsons view that its simply the low energy limit of a possibly non-renormalizeable theory makes more sense. Of course nature making sense to me may not be how nature works.

Thanks
Bill

Peter Morgan
A. Neumaier
2019 Award
people approach nonrenormalizibility differently. Some say it simply means the theory isn't well defined, i.e. there's no physical reason as such, it's just that some classic Lagrangians/Hamiltonians that you can write down don't have a well defined quantum version.
Well, they don't have a unique quantum version. But they have a parameterized family of quantum versions with an infinite number of parameters, most of them being irrelevant except at super high energies. We had discussed this in another thread; see https://www.physicsforums.com/threa...um-field-theory-comments.925220/#post-5844345 and the discussion following that post.

king vitamin
Gold Member
Can anybody tell me a similar basic scheme for quantum field theory? For example, in QED, Is there a basic differential equation? What about QCD?
I'll add yet another perspective, since you asked about formulating QFT in terms of differential equations. As mentioned by @DarMM in multiple posts here, the main observables of a QFT are the correlation functions, which are related to experimental quantities. It turns out that the correlation functions of a quantum field theory satisfy a (generally infinite) set of differential equations called the Schwinger-Dyson equations. In principle, one could consider the solution to these equations as giving you the same information as any other method of calculating correlation functions. Whether this is actually the easiest approach depends on the problem, but sometimes numerically solving a truncated subset of them or solving them in some approximation scheme can be useful.

jonjacson
DarMM
Gold Member
Well, they don't have a unique quantum version. But they have a parameterized family of quantum versions with an infinite number of parameters, most of them being irrelevant except at super high energies. We had discussed this in another thread; see https://www.physicsforums.com/threa...um-field-theory-comments.925220/#post-5844345 and the discussion following that post.
I know this and it is true, but I think this just depends on what you call a "quantum version", i.e. for ##\phi^{6}## in 4D there simply isn't a self-adjoint Hamiltonian whose highest power of the scalar field is the sixth power and thus the quantum version/analogue of that classical theory doesn't exist.

That one can order by order enlarge the space of terms to include ##\phi^{8}, \phi^{10}, \cdots## while keeping the theory finite perturbatively is true, but it isn't what this sort of nonexistence refers to. It refers to there simply being no quantum theory with a Hamiltonian of that form.

dextercioby
A. Neumaier
2019 Award
It refers to there simply being no quantum theory with a Hamiltonian of that form.
Well, renormalization destroys this form anyway. Your example of Gross-Neveu shows that ''form'' is not something sensible to discuss.

But you have to ask yourself - why is renormalization so fundamental that only such theories are valid?
Renormalizable theories are simply distinguished by having a natural parameterization with few parameters.

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A. Neumaier
2019 Award
For example, in QED, Is there a basic differential equation?
What remains of these after renormalization runs under the name of operator product expansions.

dextercioby and jonjacson
A. Neumaier
2019 Award
From the calculated S-matrix you can extract scattering cross-sections and energies/lifetimes of bound states. That's all you can do with the textbook QFT.

"The more one thinks about this situation, the more one is led to the conclusion that one should not insist on a detailed description of the system in time. From the physical point of view, this is not so surprising, because in contrast to non-relativistic quantum mechanics, the time behavior of a relativistic system with creation and annihilation of particles is unobservable. Essentially only scattering experiments are possible, therefore we retreat to scattering theory. One learns modesty in field theory." G. Scharf, Finite quantum electrodynamics. The causal approach, 1995.
You may think this is the case.

But on the level of rigor customary in theoretical physics, quantum field dynamics at finite time is actually well-defined in terms of the so-called closed time path (CTP) approach; see, e.g.,

E. Calzetta and B.L. Hu, Nonequilibrium quantum field theory, Cambridge Univ. Press, New York 2008.

dextercioby
DarMM
Gold Member
Well, renormalization destroys this form anyway.
Does it? For example in Glimm's construction of ##\phi^{4}_{3}## the Hamiltonian on the interacting Hilbert space still has the same form. What is super-renormalization on Fock space turns out to be simple Wick ordering on the interacting Hilbert space.

But on the level of rigor customary in theoretical physics, quantum field dynamics at finite time is actually well-defined in terms of the so-called closed time path (CTP) approach; see, e.g.,

E. Calzetta and B.L. Hu, Nonequilibrium quantum field theory, Cambridge Univ. Press, New York 2008.
I briefly looked into this book and didn't find what I was looking for. I am interested in a QFT Hamiltonian that would generate the time evolution of simple interacting systems, e.g., two charges. So that using this Hamiltonian one can describe time-dependent wave functions of colliding particles in the interaction region in addition to the usual S-matrix. As far as I understand, standard QFT does not permit such a description. (Such a description is possible in the dressed particle approach to QFT; see vol. 3 of my book.)

It seems that the book you recommended is not interested in such simple few-particle systems. The authors are focused on applying QFT to nonequilibrium statistical physics and such complex phenomena as heavy ion collisions, early universe cosmology, Bose-Einstein condensation, dissipation, entropy, decoherence. All this is interesting and exciting, but I would like to stick to the old-fashioned approach: learn simple systems (mechanics of few particles) first before turning to complicated ones (nonequilibrium statistical physics and cosmology).

I couldn't understand how CPT approach can replace the good old Hamiltonian as a method for describing time evolution. Perhaps you can explain that? In my understanding, if we want a unitary and relativistically-invariant time evolution, then there is nothing better than a Hamiltonian satisfying the usual commutation relations with other Poincare generators. If you describe the time evolution in any other way, then you lose either unitarity or relativistic invariance or both.

Eugene.

DarMM
Gold Member
That's true, its the old view of renormalisation. But you have to ask yourself - why is renormalization so fundamental that only such theories are valid? Wilsons view that its simply the low energy limit of a possibly non-renormalizeable theory makes more sense. Of course nature making sense to me may not be how nature works.
It's actually quite simple in this view. Renormalization is simply a procedure of correctly constructing the Hamiltonian or Path Integral measure. Only those theories for which this procedure works (i.e. are renormalizable) exist.

For example in ##d = 3## let's say the equation:
$$(\partial^{2} - m^{2})\phi = \frac{\lambda}{3!}\left[\phi^{3}\right]_R$$
has a solution, but
$$(\partial^{2} - m^{2})\phi = \frac{\lambda}{7!}\left[\phi^{7}\right]_R$$
doesn't. Perturbatively this nonexistence will show up as non-renormalizability, but "fundamentally" it's simply that the equations don't have solutions. Quantum Fields are so singular that the space of PDEs with solutions is smaller than for Classical Fields. And that's all there is to it in this view.

dextercioby and bhobba
You may think this is the case.

But on the level of rigor customary in theoretical physics, quantum field dynamics at finite time is actually well-defined in terms of the so-called closed time path (CTP) approach; see, e.g.,

E. Calzetta and B.L. Hu, Nonequilibrium quantum field theory, Cambridge Univ. Press, New York 2008.
Trying to describe the time evolution of a relativistic quantum system contradicts the most fundamental principles of quantum mechanics + relativity:

"The momentum can figure in a consistent theory only for free particles; for these it is conserved, and can therefore be measured with any desired accuracy. This indicates that the theory will not consider the time dependence of particle interaction processes. It will show that in these processes there are no characteristics precisely definable (even within the usual limitations of quantum mechanics); the description of such a process as occurring in the course of time is therefore just as unreal as the classical paths are in non-relativistic quantum mechanics. The only observable quantities are the properties (momenta, polarisations) of free particles: the initial particles which come into interaction, and the final particles which result from the process"
Berestetskii Lifshitz Quantum Electrodynamics Section 1.
Section 1 is devoted to justifying this.

QFT can only calculate the S-matrix, which is a mapping of states from the infinite past to the infinite future. From the calculated S-matrix you can extract scattering cross-sections and energies/lifetimes of bound states. That's all you can do with the textbook QFT.

"The more one thinks about this situation, the more one is led to the conclusion that one should not insist on a detailed description of the system in time. From the physical point of view, this is not so surprising, because in contrast to non-relativistic quantum mechanics, the time behavior of a relativistic system with creation and annihilation of particles is unobservable. Essentially only scattering experiments are possible, therefore we retreat to scattering theory. One learns modesty in field theory." G. Scharf, Finite quantum electrodynamics. The causal approach, 1995.

Eugene.

As far as I understand, standard QFT does not permit such a description.
.
The book above justifies this in section 1 as referenced.

In classical mechanics I would say:
a particle can have any initial position and velocities, newton laws give you the evolution of the particle:
F=ma , is the basic equation, if you know the forces acting on the particle by solving this equation you get the future values for velocity and position

In quantum mechanics:
The "particle" is characterized by a wavefunction that can be used to calculate the probability of finding the particle at certain position or momentum.
If you know the wavefunction you plug it into the Schrodinger equation and you get the evolution on time of this wavefunction.

Can anybody tell me a similar basic scheme for quantum field theory? For example, in QED, Is there a basic differential equation? What about QCD?

The basic scheme is still just the Schrodinger equation, for relativistic Hamiltonians, however relativity forces one to restrict the probability distribution interpretation of wave functions to momentum space for free particles and to work with wave functions for variable numbers of particles (section 1 cited above), for which second quantization is most convenient - and the Heisenberg picture for second quantization operators even more convenient, the reference above being a good start.

Peter Morgan
Gold Member
Quite a morass of constructions above. To go back to basics of "What is the basic scheme of quantum field theories?", the basics of a free field bosonic QFT can be presented by a generating function for time-ordered free quantum field as $$\langle 0|T[\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}]|0\rangle=\mathrm{e}^{-\lambda^2(f^*,f)_F/2},$$ where ##(f,g)_F## is the Feynman propagator smeared antilinearly by ##f## and linearly by ##g##, so that the n-th derivative w.r.t. ##\lambda## of this at ##\lambda=0## will give ##\langle 0|T[\hat\phi_f^{\,n}]|0\rangle##. I'll mention that ##(f,g)_F## is also proportional to Planck's constant.
An aside: the subscript ##f## in ##\hat\phi_f## is usually called a test function (take ##f## to be a smooth function that also has a smooth fourier transform), which is an index or coordinate for a measurement. We can think loosely of the usual quantum field object ##\hat\phi(x)## as what we would get if we took ##f## to be a Dirac delta function at ##x##, but it's better to avoid doing that because a Dirac delta function is not a smooth function. We can think of ##f## as a modulation that is applied to a vacuum state or as like the window functions of signal analysis, depending on how we use it.​
The time-ordered form, however, erases information about the algebraic structure, which we can take a first step towards by presenting a generating function form without time-ordering, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}|0\rangle=\mathrm{e}^{-\lambda^2(f^*,f)/2}.$$ Everything about the free quantum field algebra can be fixed by a single equation that generalizes the single operator case, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\phi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\phi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\phi_{f_n}}|0\rangle=\exp\left[-\sum_{i,j}\lambda_i\lambda_j(f_i^*,f_j)/2-\sum_{i<j}\left[(f^*_i,f_j)-(f^*_j,f_i)\right]/2\right],$$ where in the exponent on the r.h.s. the first term is what would be classically called noise and the second term is the measurement incompatibility that characterizes quantum theory.
For an interacting quantum field ##\hat\xi_f##, in principle all the Feynman diagrams, regularization, and renormalization are trying to do is to deform this expression to give us a new expression, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\xi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\xi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\xi_{f_n}}|0\rangle=\,\cdots,$$ a function of all the ##\lambda_i##'s and ##f_i##'s, that we can use to generate any VEV (vacuum expectation Value) and which fixes the algebraic structure for ##\hat\xi_f##, just as we saw above for ##\hat\phi_f##. That gets complicated because for at least the last 70 years the Feynman integral formalism has insisted on using powers of the operator-valued distribution ##\hat\phi(x)##, which is frankly a mathematically stupid thing to do and requires all sorts of desperate first-aid to fix the resulting problems, instead of figuring out ways to construct deformations that use only the well-defined objects, the test functions ##f_i##. In principle the VEVs that are generated by $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\xi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\xi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\xi_{f_n}}|0\rangle$$ are commonly supposed to be measurable by experiment (unless someone insists that only the S-matrix can be measured, which describes how a state changes between times ##t=\pm\infty##, so go figure why anyone would think that), although they are not exactly close to the recorded measurements of a signal voltage on a wire that is attached to some kind of fancy material coupled to a region of space-time deep within an experimental apparatus.
I'm not sure whether that will be helpful for you or for anyone else, but I find it helpful for me. I particularly find it helpful to think of QFT, free or interacting, as a signal analysis formalism, because all our experimental raw data comes into a computer as voltages on signal lines, which are converted to a binary representation and stored.
All the calculations above are essentially easy to check by using a Baker-Campbell-Haussdorf identity, setting ##\hat\phi_f=a_{f^*}+a_f^\dagger## in terms of creation and annihilation operators and using the commutation relation ##[a_f,a_g^\dagger]=(f,g)##.

jonjacson
king vitamin
Gold Member
Trying to describe the time evolution of a relativistic quantum system contradicts the most fundamental principles of quantum mechanics + relativity:

Section 1 is devoted to justifying this.

The book above justifies this in section 1 as referenced.

The basic scheme is still just the Schrodinger equation, for relativistic Hamiltonians, however relativity forces one to restrict the probability distribution interpretation of wave functions to momentum space for free particles and to work with wave functions for variable numbers of particles (section 1 cited above), for which second quantization is most convenient - and the Heisenberg picture for second quantization operators even more convenient, the reference above being a good start.
It seems to me that these considerations only make sense if one either (1) defines "particles" in a strict sense which is not useful in an interacting QFT (which generically does not have particles in the sense often meant), or (2) demands that the theory is "relativistic" in the sense that one does not have a UV cutoff, in which case I do not know of any realistic interacting relativistic QFT in (3+1) dimensions.

If I define a quantum field theory with the appropriate regulators, I can describe time-dependent phenomena (equilibrium or non-equilibrium) just fine. You may say that I have broken Lorentz invariance, which is true (after all Lorentz invariant theories have not been proven to exist), but for suitably large cutoffs all low-energy phenomena exhibit emergent Lorentz invariance anyways. And these calculations agree with experiment (see the many thousands of papers using time-dependent methods in QFT published every year).

(I also don't trust a QFT textbook/lecture which does not cover the renormalization group. This field has evolved in the past 50 years!)

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And these calculations agree with experiment (see the many thousands of papers using time-dependent methods in QFT published every year).
Perhaps you can mention a couple of papers where a full renormalized relativistic QFT (i.e., whose Hamiltonian has cutoff-dependent divergent counterterms) was used to calculate time-dependent processes, and the results agree with experiment?

Thanks.
Eugene.

I'll add yet another perspective, since you asked about formulating QFT in terms of differential equations. As mentioned by @DarMM in multiple posts here, the main observables of a QFT are the correlation functions, which are related to experimental quantities. It turns out that the correlation functions of a quantum field theory satisfy a (generally infinite) set of differential equations called the Schwinger-Dyson equations. In principle, one could consider the solution to these equations as giving you the same information as any other method of calculating correlation functions. Whether this is actually the easiest approach depends on the problem, but sometimes numerically solving a truncated subset of them or solving them in some approximation scheme can be useful.
So interesting, thanks. An infinite set of differential equations looks like something not easy to solve definitely :(

What remains of these after renormalization runs under the name of operator product expansions.
Thanks.

Quite a morass of constructions above. To go back to basics of "What is the basic scheme of quantum field theories?", the basics of a free field bosonic QFT can be presented by a generating function for time-ordered free quantum field as $$\langle 0|T[\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}]|0\rangle=\mathrm{e}^{-\lambda^2(f^*,f)_F/2},$$ where ##(f,g)_F## is the Feynman propagator smeared antilinearly by ##f## and linearly by ##g##, so that the n-th derivative w.r.t. ##\lambda## of this at ##\lambda=0## will give ##\langle 0|T[\hat\phi_f^{\,n}]|0\rangle##. I'll mention that ##(f,g)_F## is also proportional to Planck's constant.
An aside: the subscript ##f## in ##\hat\phi_f## is usually called a test function (take ##f## to be a smooth function that also has a smooth fourier transform), which is an index or coordinate for a measurement. We can think loosely of the usual quantum field object ##\hat\phi(x)## as what we would get if we took ##f## to be a Dirac delta function at ##x##, but it's better to avoid doing that because a Dirac delta function is not a smooth function. We can think of ##f## as a modulation that is applied to a vacuum state or as like the window functions of signal analysis, depending on how we use it.​
The time-ordered form, however, erases information about the algebraic structure, which we can take a first step towards by presenting a generating function form without time-ordering, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}|0\rangle=\mathrm{e}^{-\lambda^2(f^*,f)/2}.$$ Everything about the free quantum field algebra can be fixed by a single equation that generalizes the single operator case, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\phi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\phi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\phi_{f_n}}|0\rangle=\exp\left[-\sum_{i,j}\lambda_i\lambda_j(f_i^*,f_j)/2-\sum_{i<j}\left[(f^*_i,f_j)-(f^*_j,f_i)\right]/2\right],$$ where in the exponent on the r.h.s. the first term is what would be classically called noise and the second term is the measurement incompatibility that characterizes quantum theory.
For an interacting quantum field ##\hat\xi_f##, in principle all the Feynman diagrams, regularization, and renormalization are trying to do is to deform this expression to give us a new expression, $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\xi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\xi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\xi_{f_n}}|0\rangle=\,\cdots,$$ a function of all the ##\lambda_i##'s and ##f_i##'s, that we can use to generate any VEV (vacuum expectation Value) and which fixes the algebraic structure for ##\hat\xi_f##, just as we saw above for ##\hat\phi_f##. That gets complicated because for at least the last 70 years the Feynman integral formalism has insisted on using powers of the operator-valued distribution ##\hat\phi(x)##, which is frankly a mathematically stupid thing to do and requires all sorts of desperate first-aid to fix the resulting problems, instead of figuring out ways to construct deformations that use only the well-defined objects, the test functions ##f_i##. In principle the VEVs that are generated by $$\langle 0|\mathrm{e}^{\mathrm{i}\lambda_1\hat\xi_{f_1}}\mathrm{e}^{\mathrm{i}\lambda_2\hat\xi_{f_2}}\cdots\mathrm{e}^{\mathrm{i}\lambda_n\hat\xi_{f_n}}|0\rangle$$ are commonly supposed to be measurable by experiment (unless someone insists that only the S-matrix can be measured, which describes how a state changes between times ##t=\pm\infty##, so go figure why anyone would think that), although they are not exactly close to the recorded measurements of a signal voltage on a wire that is attached to some kind of fancy material coupled to a region of space-time deep within an experimental apparatus.
I'm not sure whether that will be helpful for you or for anyone else, but I find it helpful for me. I particularly find it helpful to think of QFT, free or interacting, as a signal analysis formalism, because all our experimental raw data comes into a computer as voltages on signal lines, which are converted to a binary representation and stored.
All the calculations above are essentially easy to check by using a Baker-Campbell-Haussdorf identity, setting ##\hat\phi_f=a_{f^*}+a_f^\dagger## in terms of creation and annihilation operators and using the commutation relation ##[a_f,a_g^\dagger]=(f,g)##.
Thanks.

Why do you think it hasn't changed in 70 years?

DarMM
Gold Member
Why do you think it hasn't changed in 70 years?
I would say because it clearly works. We know rigorously that defining products of operator valued distributions like ##\phi^{4}## is tricky, but it is well-defined and the standard methods of doing things have been proven to be valid in any rigorous construction we've achieved.

So we know the "textbook" way of doing things is quite good.

Doing things rigorously/correctly is extremely difficult and abstract and hard to get numbers out of and even when you do it they're pretty much the same numbers. The well-defined objects that you should be using, local operator algebras, are very abstract and not easy to work with.

king vitamin
Gold Member
Perhaps you can mention a couple of papers where a full renormalized relativistic QFT (i.e., whose Hamiltonian has cutoff-dependent divergent counterterms) was used to calculate time-dependent processes, and the results agree with experiment?

Thanks.
Eugene.
What issues do you have with the Calzetta & Hu book cited by Arnold Neumaier? It has about 1000 references in it, and the book contains many applications to relativistic heavy ions collisions, cosmology, chiral condensates, and condensed matter (the latter often has a relativistic low-energy description). I just searched through my copy and found many examples where agreement with experiments was cited.

I might be exaggerating with "thousands," but calculating dynamics in a QFT seems very standard to me. It might come down to what you consider a time-dependent calculation. Does linear response count? I suppose you wouldn't count calculations of relaxation rates or decay times? Would the measurement of a spectral function (basically the Fourier transform of a two-point function) count? I'd consider all of these to be dynamical info about time-evolution of the theory, but perhaps it is simply not what you mean with your statement.

Also, although you asked about theories with "divergent counterterms" (implying perturbation theory), there are exactly solvable interacting relativistic QFTs where we can specify the dynamics exactly. The only experimental applications I know of are condensed matter systems (probed at long wavelengths), but my point is that the question is well-posed.

dextercioby and protonsarecool