# The construction of particles in QFT

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## Summary:

The question discusses the wave packet representation of particles in QFT.

## Main Question or Discussion Point

In all books about QFT I have seen I can not find anything about what a localized particle concept is. Suddenly I found this note in Zee's 'QFT in nutshell' page 4:
"As usual, we can form wave packets by superposing eigenmodes. When we quantize the theory, these wave packets behave like particles, in the same way that electromagnetic wave packets when quantized behave like particles called photons."
So he states that a particle in QFT is a localized wavepacket as far as I interpret the citation.
1. in QM is long well known that superposing of wavefunctions of different k can not form a stable packet because of the different velocities in vacuum. Is it something different here and why?
2. If the packet is stable and well localized how is the double slit results accounted for?
3. As is always represented particles of a non-interacting field are connected with (ascribed) single value of k. How is one supposed to construct one particle of many particles from the same type?

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PeterDonis
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in QM is long well known that superposing of wavefunctions of different k can not form a stable packet because of the different velocities in vacuum. Is it something different here and why?
Yes, it's something different. The wave packets in QFT are not superpositions of wave functions, they're superpositions of quantum field states.

If the packet is stable and well localized how is the double slit results accounted for?
By not using a single localized wave packet to model that experiment. If you want to use a "particle" approximation for the double slit, you have to use two wave packets, one for each slit. If you object that "a particle can't be in two places at once", the answer is that you are trying to use a classical concept of "particle" and that concept is not what we're talking about here.

As is always represented particles of a non-interacting field are connected with (ascribed) single value of k.
Where is this "always represented"? (Hint: In the very texbook you reference, is there any place where Zee says this?)

Yes, it's something different. The wave packets in QFT are not superpositions of wave functions, they're superpositions of quantum field states.
But the wave function is also a field state. So this is just a word game. The QM wavefunction (WF) is a function in real space. And so is any field state. (at least nowhere is stated the opposite - e.g. quantum fields are fields in real space and hence they could not form stable packets as many WF can not.

By not using a single localized wave packet to model that experiment. If you want to use a "particle" approximation for the double slit, you have to use two wave packets, one for each slit. If you object that "a particle can't be in two places at once", the answer is that you are trying to use a classical concept of "particle" and that concept is not what we're talking about here.
So it turns out that when you need to account for DSE you throw the whole construction away and take an opposite construction just for that case practically.
Where is this "always represented"? (Hint: In the very texbook you reference, is there any place where Zee says this?)
I dont know about Zee. But you always start with the Born Jordan procedure. Free Field = plane waves of k = oscillators in Fourier space = quantization in box = particles. So the modes are interpreted like particles in every textbook (b.e. Ryder).

Also Zee immediately abandons the notion of this localized packet after adopting Feynman path integral formulation. It is sure the packet can not move as a whole and also cover every path.

vanhees71
Gold Member
2019 Award
Peskin&Schroeder have a nice discussion about the derivation of the cross-section formula in terms of the S-matrix elements using wave packets in the in-state and then taking the plane-wave limit. That's a bit more complicated than the usual "box-regularization argument", but I found it very helpful for the physical understanding.

PeterDonis
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2019 Award
the wave function is also a field state
No, it isn't. They are two different mathematical objects.

The QM wavefunction (WF) is a function in real space.
No, it isn't. It's a function on the configuration space of the quantum system. The only case where that is the same as "real space" is a single non-interacting particle with zero spin.

so is any field state
No, a quantum field state is an operator at a particular point in spacetime, not space.

You seem to have some serious misunderstandings about the math underlying QM and QFT.

PeterDonis
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2019 Award
it turns out that when you need to account for DSE you throw the whole construction away and take an opposite construction
That is not at all what I described. I don't know where you are getting this from.

Also note that nothing requires you to use a "localized particle concept" in QFT at all, for explaining the double slit experiment or anything else. That concept appears nowhere in the fundamental construction of the theory. It is simply a name we give to certain particular kinds of solutions that are useful in certain kinds of problems (scattering problems are the most common csae), which are approximations anyway. You should not expect those approximations to always work.

Demystifier
1. in QM is long well known that superposing of wavefunctions of different k can not form a stable packet because of the different velocities in vacuum. Is it something different here and why?
It's not much different.

2. If the packet is stable and well localized how is the double slit results accounted for?
It's usually not stable, unless there are some interactions that stabilize it.

3. As is always represented particles of a non-interacting field are connected with (ascribed) single value of k. How is one supposed to construct one particle of many particles from the same type?
As in 1, that is, by superposition. A superposition of one-particle states with different $k$ is a one-particle state with uncertain $k$.

vanhees71
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2019 Award
No, a quantum field state is an operator at a particular point in spacetime, not space.
This is also not entirely correct. You have to distinguish between the "first-quantization formalism" (which only works for non-relativistic QT for the cases where no "particles" can be destroyed and produced) and the "second-quantization formalism" (or the quantum-field theoretical formulation) which works for both relativistic and non-relativistic theories.

Let's discuss the latter, because here seems to be the most confusion.

The mathematical building blocks for a QFT as for any quantum theory are as follows:

You have a fundamental set of operators, from which all the operators having some physical meaning are built. For a QFT these are some fundamental field operators (e.g., a Dirac field), which need not necessarily represent any observables or states.

Then the field operators are used to express the operators representing observables. E.g., for the Dirac field the electric charge-current densities for a free field are given by (in Heaviside-Lorentz units with $\hbar=c=1$)
$$\hat{j}^{\mu}(x)=q \hat{\bar{\psi}}(x) \gamma^{\mu} \psi(x).$$

The state is represented by the statistical operator $\hat{\rho}$ of the system, which can also be expressed in terms of the field operators. Expectation values of observables are then given by
$$\langle O \rangle=\mathrm{Tr}(\hat{O} \hat{\rho}).$$
Usually the statistical operator is not a local operator (i.e., it's not dependent on space-time arguments).

An important example for a statistical operator is the canonical equilibrium ensemble operator,
$$\hat{\rho}=\frac{1}{Z} \exp(-\hat{H}/T), \quad Z=\mathrm{Tr} \exp(-\hat{H}/T),$$
where the Hamiltonian is given as the integral over entire space over the energy density (which is a Lorentz scalar if and only if $\partial_{\mu} \Theta^{\mu \nu}=0$.

No, it isn't. They are two different mathematical objects.
In coordinate representation WF is function of (x,t) and if we define it as a two component
vector (Re, Im) then we have two function Re= f(x,t) and Im=g(x,t). (There is a prove from Schroedinger
that a wave packet of Re very soon decays.) Now what is the difference with QFT starting point (before quantization and operators - you also have some functions of x,t (which Zee said build a localized packet).
The differences between WF and quantum fields come later after introducing operators. What I meant is
this first stage namely.

No, it isn't. It's a function on the configuration space of the quantum system. The only case where that is the same as "real space" is a single non-interacting particle with zero spin.
Do you mean a multiparticle case? My example is just for single particle and not the general case.

No, a quantum field state is an operator at a particular point in spacetime, not space.
I can not see why not operator in space changing in time? In Zee is also a lettice of x and the function fi(x) depend on parameter t. But yes more corect is spacetime.

You seem to have some serious misunderstandings about the math underlying QM and QFT.
I dont think I have math problems whatsoever, but indeed many problems with the physical
meaning of these schemes. I dont deny that they work on some cases to some extent but obviously
math is taking over physics.

It's not much different.
I also can't see any reason to be different. So here is a conclusion - the QFT wave packets decay rapidly (my memory from this is some thousands of a second).

It's usually not stable, unless there are some interactions that stabilize it.
Ok but there is nothing in vaccuum to stabilize it.
Now one can account for DSE but how is then justified to apply wavepackets in say LHC where the particle travels km before one will apply the machinery of the Scattering matrix.
As in 1, that is, by superposition. A superposition of one-particle states with different $k$ is a one-particle state with uncertain $k$.
Mathematically not a problem. But where do they live. In measurement we observe just one.

Demystifier
Ok but there is nothing in vaccuum to stabilize it.
Now one can account for DSE but how is then justified to apply wavepackets in say LHC where the particle travels km before one will apply the machinery of the Scattering matrix.
In LHC there are magnets that focus the beam.

Mathematically not a problem. But where do they live. In measurement we observe just one.
When the momentum is measured, then wave function "collapses" to a state with a single definite momentum. How exactly that happens is the central question of the so called measurement problem in quantum mechanics. https://en.wikipedia.org/wiki/Measurement_problem

EPR
Gold Member
When you accelerate a proton to 100 GeV in the LHC, does it make sense to talk of wave packets given the relativistic mass of the object?

In LHC there are magnets that focus the beam.
But they stabilized them in say z, y directions. They don't stabilize them along the path x so the packet would spread in x and can give very different times to the target.

When the momentum is measured, then wave function "collapses" to a state with a single definite momentum. How exactly that happens is the central question of the so called measurement problem in quantum mechanics. https://en.wikipedia.org/wiki/Measurement_as problem
Please note that this question is not about QM but QFT. As far as know the field is real and collapse is not considered in QFT. (or it is a silent convention? I never encountered anything about collapse in QFT. I think that if QFT is regarded as almost ultimate physical model of the world (and not just calculation apparatus) this should be a central point in any book of QFT but there is not even a mention anywhere I know)

PeterDonis
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what is the difference with QFT starting point (before quantization and operators
Before quantization and operators you don' t have QFT. So the fact that before quantization and operators, you have a function, is irrelevant to this discussion since we are talking about QFT.

My example is just for single particle
A single noninteracting particle with zero spin doesn't tell you anything useful, and the properties of this special case that you are trying to focus on (such as the wave function being a function on "real space") do not generalize to any other case. So focusing on the details of this case is a waste of time if you are trying to actually understand how QFT works and what makes it different from non-relativistic QM.

why not operator in space changing in time?
Because this requires a particular splitting of spacetime into space and time. In a relativistically invariant theory, nothing in the actual physics can depend on how you split up spacetime into space and time (i.e., on your choice of coordinates). Only invariant quantities have physical meaning.

obviously
math is taking over physics
Physics has always been done in math. If you don't like the particular way the math is presented in one textbook, you can always find another that presents it a different way. Zee's textbook is pretty relaxed about rigor; he is not trying to prove mathematical theorems but to give you working tools that are good enough for working physicists. If you want a more rigorous viewpoint you might be better off with a different textbook.

PeterDonis
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2019 Award
collapse is not considered in QFT.
It is, but it's less obvious how. In QFT you always have to define some kind of boundary condition or initial and final condition: for example, in a scattering problem you specify that there are one or more sources emitting quantum objects, and one or more sinks detecting them. Those boundary/initial/final conditions are the QFT counterpart of collapse in non-relativistic QM.

It is, but it's less obvious how. In QFT you always have to define some kind of boundary condition or initial and final condition: for example, in a scattering problem you specify that there are one or more sources emitting quantum objects, and one or more sinks detecting them. Those boundary/initial/final conditions are the QFT counterpart of collapse in non-relativistic QM.
Math in QFT is bringing much artificial renormalizations in the end and enormous calculations. Surely the nature does it not that way.
I have a question apart from mathematical formulation. Is the QFT field a field of probabilities? Operator valued field is a field in math but what I want to know is what is it in reality? The collapse is possible only for probabilities.

Demystifier
As far as know the field is real and collapse is not considered in QFT.
That's not true. Fermionic fields are certainly not "real". Bosonic fields can be considered "real" in a certain sense, but it doesn't mean that there is no "collapse" of the quantum state in QFT. For analogy, in nonrelativistic QM the particle position can be considered "real" in a certain sense, but it doesn't mean that there is no "collapse" of the quantum state in QM.

Demystifier
Is the QFT field a field of probabilities?
No. A field for QFT is what a particle position is for QM.

PeterDonis
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2019 Award
Math in QFT is bringing much artificial renormalizations in the end and enormous calculations. Surely the nature does it not that way.
You are welcome to your personal speculation about how "nature does it", but we don't discuss that here. This thread is about QFT.

Math in QFT is bringing much artificial renormalizations in the end and enormous calculations. Surely the nature does it not that way.
That's not really the way to think about it in my opinion. Essentially what's going on (as far as I can tell) is that we start by writing down a "first draft" of the Lagrangian based on symmetries and what-not, after which we apply the regularization and renormalization procedures to find the correct parameterization of the Lagrangian that yields sensible perturbation results. Nature isn't "doing" any of this stuff, it's just a way to find a good (perturbative) quantum theory with the desired symmetries and degrees of fredom.

As far as know the field is real and collapse is not considered in QFT.
It is, but it's less obvious how. In QFT you always have to define some kind of boundary condition or initial and final condition: for example, in a scattering problem you specify that there are one or more sources emitting quantum objects, and one or more sinks detecting them. Those boundary/initial/final conditions are the QFT counterpart of collapse in non-relativistic QM.
Do you mean that the excitation from some source is spreading out in every direction but we neglect (are not interested in) the part not going into a sink, but that part really exists. As soon as the sink detects the part of the excitation reaching it, the other part of the excitation vanishes in the whole field immediately?

Is the quantum field a probability field?
No. A field for QFT is what a particle position is for QM.
I don't understand this. I think QM is about waves (ascribed to objects regarded poinlike). In Copenhagen interpretation particles dont have positions. So...
What I thought is: Since after quantization in QFT one has operators in every point, one has its eigenvalues in that point, which will appear with certain probabilities when measured.

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A. Neumaier
2019 Award
In QFT you always have to define some kind of boundary condition or initial and final condition: for example, in a scattering problem you specify that there are one or more sources emitting quantum objects, and one or more sinks detecting them. Those boundary/initial/final conditions are the QFT counterpart of collapse in non-relativistic QM.
Not really since the boundary conditions are at times plus/minus infinity. Thus no collapse at finite times.

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A. Neumaier