- #1

ftr

- 624

- 47

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter ftr
- Start date

In summary: The correct explanation is you need re-normalization which basically says well this is obviously wrong so let's cutoff the energy scale - but what...That's a bit beyond the scope of this thread. More information on re-normalization can be found in other threads on the forum.

- #1

ftr

- 624

- 47

Physics news on Phys.org

- #2

strangerep

Science Advisor

- 3,765

- 2,212

The latter. If one redefines the fields in QED such that the electron is dressed with a particular configuration of EM field (such that it gives the correct Coulomb field), then the usual IR divergences in QED scattering are banished to all orders in perturbation theory, iiuc.ftr said:So it seams the electron does not have the intrinsic property of charge, either that or the EM is integral to the electron. Any idea?

The dressed electron+EM composite field also has the nice property of being gauge-invariant.

- #3

ftr

- 624

- 47

- #4

strangerep

Science Advisor

- 3,765

- 2,212

Yes, it's just a problem of terminology (and possibly interpretation). I know that Arnold is totally familiar with the field dressing I mentioned. The dressing is not additive, but rather involves products of field operators.ftr said:

Depending on how deeply you want to look into this, I could probably dig out some references again.

- #5

ftr

- 624

- 47

yes, I appreciate it but if it is only minor trouble to you. It is not an urgent problem for me, it just looked an interesting awkward(however successful) setup.strangerep said:Depending on how deeply you want to look into this, I could probably dig out some references again.

- #6

strangerep

Science Advisor

- 3,765

- 2,212

https://www.physicsforums.com/threa...-virtual-particles.485597/page-3#post-3222521

https://www.physicsforums.com/threads/cancellation-of-infrared-divergence.562554/#post-3686015

The first one references a seminal paper by Dirac which shows the basic idea of how dressing the electron field can yield a Lagrangian that involves only explicitly gauge-invariant fields. It also shows how the Coulomb field associated with the dressed electron field emerges. But it's a very difficult paper to follow (or so I found).

The other references in those 2 threads cover lots of the stuff about banishing IR divergences in QED.

- #7

ftr

- 624

- 47

- #8

strangerep

Science Advisor

- 3,765

- 2,212

It's the ordinary electric field of an electron.ftr said:Is the coulomb field the same as the gauge field,

- #9

ftr

- 624

- 47

so is the electron field the same as ITS electric field.

- #10

PeterDonis

Mentor

- 47,415

- 23,697

ftr said:is the electron field the same as ITS electric field.

No. We are getting into material that is really too much for an "I" level thread; you need a graduate level background in quantum field theory, and specifically quantum electrodynamics (QED), which is the quantum field theory of electrons/positrons and photons (the EM field). But briefly: in QED there is an electron field, and there is a photon (or electromagnetic) field, and these two fields interact. These are quantum fields; they aren't what you are used to thinking of as "fields". And the interaction between them is a fundamental part of the equations; as far as QED is concerned, there is no such thing as a "free" electron that doesn't interact with photons (which means the "free" Dirac equation you referred to, without the EM interaction term, does not describe anything that actually exists). The coupling constant of the interaction between electrons and photons in QED is the electron's electric charge. So the answer to the title question of this thread is "yes".

- #11

bhobba

Mentor

- 10,818

- 3,688

strangerep said:The latter. If one redefines the fields in QED such that the electron is dressed with a particular configuration of EM field (such that it gives the correct Coulomb field), then the usual IR divergences in QED scattering are banished to all orders in perturbation theory. The dressed electron+EM composite field also has the nice property of being gauge-invariant.

What was it Meatloaf said - you took the words right out of my mouth.

Just to elaborate a bit more, heuristically (meaning virtual particles do not really exist - its just an aid to intuition) the sea of virtual electrons and positrons around the electron screens the charge so the closer you get the larger the charge is. In fact Landau showed it was infinite if you get too close:

https://en.wikipedia.org/wiki/Landau_pole

That is obviously a BIG problem, but don't worry long before the Landau pole is reached the electro-weak theory takes over.

The correct explanation is you need re-normalization which basically says well this is obviously wrong so let's cutoff the energy scale - but what cut off do we use. The interesting thing about QED (and other re-normalizeable theories) is it doesn't really matter what cutoff you use - you can even take the limit to infinity if you want - you get the same answer for anything you can actually measure. The trick is to realize the charge is cutoff dependent - it called the bare charge. What you do is you actually measure it at some energy level - such is called the re-normalized charge. You then write your equations in terms of the re-normalized charge - vola - no problems

I give a bit more technical detail here:

https://www.physicsforums.com/insights/renormalisation-made-easy/

Thanks

Bill.

Last edited:

- #12

- 24,488

- 15,028

Well, of course the physical electron comes with its Coulomb field. In the case of QED the true asymptotic states are also not the naive "plane wave states" usually used in perturbation theory but the electron is accompanied by a "cloud of soft photons" (to be more strict it's a kind of coherent state of the em. field). Using these correct asymptotic states solves the problem with IR divergences as mentioned by @strangerep in #2.ftr said:

P. Kulish and L. Faddeev, Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys., 4 (1970), p. 745.

http://dx.doi.org/10.1007/BF01066485

In the standard approach to perturbative QFT that's achieved by the appropriate soft-photon resummations a la Bloch and Nordsieck (or for the non-Abelian case accoring to Kinoshita, Lee, and Nauenberg). For a good treatment of these issues in the conventional approach, see

S. Weinberg, Quantum Theory of Fields vol. I, Cambridge University Press.

- #13

- 13,362

- 3,500

ftr said:

Excellent point and question raised. Indeed, the Dirac equation is spelled out in terms of only one species of fields - the specially relativistic generalization of Schroedinger's wave functions describing electrons in the presence of 1/2 spin which is obtained as a consequence of linearization in momentum of the Hamiltonian - and there's no trace of e (or generically q) anywhere. We now know that Dirac's model of single type of wave function type is incorrect - it won't describe experimentally proven antiparticles. We need quantized fields. 2 species of fields. The need for the 2nd species (which is usually denoted by psi bar) comes from the need to account for interactions of the "electron" (can be a taon or muon, too) field with other fields, such as a gauge field (a single one = QED, a collection = QCD). The electron's intrinsic properties are mass and spin. The electric charge is extrinsic, comes into play only if the electron interacts electromagnetically with other electrons. The electric charge comes from the gauge symmetry imposed to the Dirac field, the spin and mass come from the global restricted Poincare invariance. These two symmetries mix only trivially in the SM, as per the Coleman-Mandula no-go theorem.

- #14

PeterDonis

Mentor

- 47,415

- 23,697

dextercioby said:The electric charge is extrinsic, comes into play only if the electron interacts electromagnetically with other electrons.

This is not correct. There is no such thing as an electron that does not interact electromagnetically.

It is possible to use the Dirac equation to describe a spin-1/2 particle that does not have electric charge (for example, a neutrino), but you can't do that with an electron.

- #15

- 13,362

- 3,500

Let me remind you that at the level of 1928, there were only two particles known, the electron and the proton. Dirac's equation for a free electron made perfect sense (ironically, it wasn't realized then that it could describe a free proton, too).

- #16

radium

Science Advisor

Education Advisor

- 765

- 243

- #17

PeterDonis

Mentor

- 47,415

- 23,697

dextercioby said:The free Dirac Hamiltonian could very well describe a free electron

But there is no such thing as a free electron in this sense. Maybe that wasn't fully understood in 1928 (though I'm not sure even about that), but it is now, and our best current understanding is what we are trying to explain to the OP.

- #18

- 24,488

- 15,028

A particle description is only possible in the sense of asymptotic free states, and thus also it's only the S matrix (in vacuum QFT) that encodes the observable properties of these "particles". Indeed, it leads to decay widths of unstable "particles" (strictly speaking they are no particles but resonances), cross sections of various scattering processes, etc.

In the case of QED (or any QFT with non-confined long-range forces) the naive association of asymptotic free particle states with plane-wave Fock states is problematic and indeed leads to problems with infrared and collinear divergences. The most simple example treated in almost all textbooks about QED is bremsstrahlung in scattering a light particle (electron) at a very heavy particle (nucleus) that can be approximated as an external electrostatic field. Here it becomes clear that you cannot distinguish a naive asymptotic free electron from a state, where you have such a naive electron and an ultrasoft photon, which you don't detect since any real detector has only finite energy resolution. So you have to rearrange the Feynman diagrams with appropriately "soft-photon resummed" diagrams.

An alternative way to understand this is to use correct asymptotic free states, which can be interpreted (in a somewhat handwavy way) as a naive asymptotic free electron surrounded by a coherent em. field. For details see

P. Kulish and L. Faddeev, Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys.,

- #19

PeterDonis

Mentor

- 47,415

- 23,697

vanhees71 said:I think, it's correct that the free Dirac field has a conserved charge (from invariance under global phase changes, i.e., a global U(1) symmetry)

The "free" Dirac field in the sense the OP is using that term means a field ##\psi## whose equation of motion is

$$

\left( \gamma^\mu \partial_\mu - m \right) \psi = 0

$$

This field does not have a global U(1) symmetry, does it?

- #20

strangerep

Science Advisor

- 3,765

- 2,212

Er,... why not? (I'm assuming "global" means that the phase factor is not spacetime-dependent.) I.e., ##\psi \to \psi' = e^{i\theta}\psi##.PeterDonis said:The "free" Dirac field in the sense the OP is using that term means a field ##\psi## whose equation of motion is

$$\left( \gamma^\mu \partial_\mu - m \right) \psi = 0$$This field does not have a global U(1) symmetry, does it?

Or am I missing something?

- #21

PeterDonis

Mentor

- 47,415

- 23,697

strangerep said:I.e., ##\psi \to \psi' = e^{i\theta}\psi##.

This is true of any quantum theory--multiplying everything by a constant phase changes nothing. So if that's what's meant by global U(1) symmetry, I don't see how the concept is of any use. I also don't see how you get a conserved charge from it.

- #22

bhobba

Mentor

- 10,818

- 3,688

PeterDonis said:This is true of any quantum theory--multiplying everything by a constant phase changes nothing. So if that's what's meant by global U(1) symmetry, I don't see how the concept is of any use. I also don't see how you get a conserved charge from it.

I am scratching my head about this particular direction of discussing global symmetries. For EM is local U(1) symmetry that's important. You try to write a Lagrangian for an interaction and its local symmetries that constrain it and more or less determine its form.

Thanks

Bill

Last edited:

- #23

strangerep

Science Advisor

- 3,765

- 2,212

Well, umm,... the way I've always understood it is like this...PeterDonis said:This is true of any quantum theory--multiplying everything by a constant phase changes nothing. So if that's what's meant by global U(1) symmetry, I don't see how the concept is of any use. I also don't see how you get a conserved charge from it.

Following the conventions of Peskin & Schroeder pp70-71, the discrete transformation known as "

So we have 2 representations of U(1) symmetry: for a given abstract element of U(1), we can implement it through multiplying by ##e^{i\theta}##, and another where we multiply by ##e^{-i\theta}##. There's no admissible analytic transformation between a function space and its conjugate space, in general, so we have 2 distinct reps -- which can only be mapped between each other using the discrete charge-conjugation operator ##C##.

But ##[C,H]=0##, (where ##H## is the Hamiltonian), so in this sense we have conserved charge: time evolution won't spontaneously flip the charge.

[And now I wait for @samalkhaiat to stop by and tell me I've got it wrong. ]

Last edited:

- #24

radium

Science Advisor

Education Advisor

- 765

- 243

In order to make this a local symmetry you must add a gauge field to form the covariant derivative so derivatives of the fields transform the same way as the fields under a gauge transformation. The action under this transformation will no longer be invariant and you will get the integral over the arbitrary gauge transformation function of spacetime time and the 4-divergence of the current. However, when the equations of motion are satisfied, this boundary term is zero and you can see the current is conserved since the variation is an arbitrary function of space time.

For non-Abelian symmetries this is more complicated since you have a

- #25

samalkhaiat

Science Advisor

- 1,802

- 1,200

No, forPeterDonis said:This is true of any quantum theory--multiplying everything by a constant phase changes nothing.

The fact thatSo if that's what's meant by global U(1) symmetry, I don't see how the concept is of any use. I also don't see how you get a conserved charge from it.

- #26

samalkhaiat

Science Advisor

- 1,802

- 1,200

strangerep said:Well, umm,... the way I've always understood it is like this...

Following the conventions of Peskin & Schroeder pp70-71, the discrete transformation known as "charge conjugation" acts as $$C \psi C = -i \gamma^2 \psi^* ~.$$ I.e., ##C## essentially takes ##\psi## to its complex-conjugate (and fiddles around with the 4-spinor components, but that's irrelevant here, iiuc).

So we have 2 representations of U(1) symmetry: for a given abstract element of U(1), we can implement it through multiplying by ##e^{i\theta}##, and another where we multiply by ##e^{-i\theta}##. There's no admissible analytic transformation between a function space and its conjugate space, in general, so we have 2 distinct reps -- which can only be mapped between each other using the discrete charge-conjugation operator ##C##.

But ##[C,H]=0##, (where ##H## is the Hamiltonian), so in this sense we have conserved charge: time evolution won't spontaneously flip the charge.

[And now I wait for @samalkhaiat to stop by and tell me I've got it wrong. ]

Charge-conjugation is not the correct operation to consider here, because

(1) the unitary charge-conjugation operator [itex]\mathscr{C}[/itex] is always defined up to phase [itex]e^{i\theta}[/itex],

(2) even for the real photon field [itex]A_{\mu}[/itex], you can construct [tex]\mathscr{C} = \exp \left( i\pi \int \frac{d^{3}k}{(2\pi)^{3}2\omega} \sum_{\sigma = 1}^{2} a^{\dagger}(k , \sigma) a(k , \sigma) \right) ,[/tex] such that [tex]\mathscr{C} A^{\mu}(x) \mathscr{C}^{-1} = - A^{\mu}(x).[/tex]

- #27

samalkhaiat

Science Advisor

- 1,802

- 1,200

[itex]\frac{dQ}{dt} = 0[/itex] even when the symmetry isradium said:This U(1) symmetry is an internal symmetry and the Lagrangian is invariant under it (since the phase is just a constant) as long as the charge is conserved (if the field has a nonzero vacuum expectation value the symmetry is spontaneously broken)

- #28

A. Neumaier

Science Advisor

- 8,633

- 4,679

A single physical electron is the special case of QED in a state of charge ##e##, in which we just have a single physical particle (excited state state of the electron field) and the electromagnetic field belonging to it, in the form of a coherent state. (Only this combination is gauge invariant and an eigenstate of the momentum 4-vector.) This e/m field is just ''the'' e/m field, not something that could be separated from other e/m fields, since there is only one e/m field.ftr said:the other day Arnold told me that he has never heard of "associated" em field and the two fields are different and cannot be add.

Once more than one electron is present, this single electron description only holds asymptotically for ##t\to\pm\infty## in directions where no other particles are. For the purposes of asymptotic analysis (scattering in the Bloch-Nordsieck formalism) one can consider these asymptotic electrons as being dressed with ''their'' asymptotic e/m fields - but the latter is part of the overall electromagnetic field, and cannot be separated by any experimental means - they are just contributions to a theoretically convenient decomposition.

Last edited:

- #29

PeterDonis

Mentor

- 47,415

- 23,697

samalkhaiat said:forsingle realscalar field ##\phi(x)##, the transformation ##\phi(x) \to e^{i\theta} \phi(x)## does not make any sense.

Hm, ok, I was thinking of quantum states as vectors in a Hilbert space over the complex numbers, but I see what you mean, quantum fields are different kinds of objects.

samalkhaiat said:For the free Dirac Lagrangian, this gives you ##J^{\mu} = \bar{\psi} \gamma^{\mu}\psi##.

Ah, got it.

- #30

samalkhaiat

Science Advisor

- 1,802

- 1,200

Woops, I should have said some thing about this. You don't have 2 representations. Recall the group axiom: For any [itex]g \in G[/itex], there exists [itex]g^{-1} \in G[/itex]. So, by parametrizing the group elements by [itex]\theta \in \mathbb{R}[/itex], you get [itex]g^{-1}(\theta) = g(-\theta) \in G[/itex].strangerep said:So we have 2 representations of U(1) symmetry: for a given abstract element of U(1), we can implement it through multiplying by ##e^{i\theta}##, and another where we multiply by ##e^{-i\theta}##. There's no admissible analytic transformation between a function space and its conjugate space, in general, so we have 2 distinct reps

- #31

radium

Science Advisor

Education Advisor

- 765

- 243

- #32

ftr

- 624

- 47

samalkhaiat said:(Dirac) Lagrangian is invariant under the continuous group U(1)

As an internal symmetry the U(1) still indicates an interaction, Otherwise U(1) is global and it only indicates conserved probability, correct?

- #33

bhobba

Mentor

- 10,818

- 3,688

ftr said:As an internal symmetry the U(1) still indicates an interaction, Otherwise U(1) is global and it only indicates conserved probability, correct?

I wouldn't put it that way - but local U(1) gauge symmetry is the basis of EM.

The details are not suitable for a thread - you need to study it. I like (see chapter 7):

https://www.amazon.com/dp/3319192000/?tag=pfamazon01-20

Basically symmetry determines the Lagrangian's for spin 0, spin 1/2 and spin one particles (the fields are complex numbers). However you notice, for say, spin 0 particles, it has a global U(1) symmetry. But when you check local U(1) symmetry a term appears added to the Lagrangian - you need something to cancel the term if you want it locally U(1) invariant. Then you look at spin 1 particles and you notice it has a local symmetry by adding the derivative of an arbitrary function to it. You compare the two and notice easily how to get this term you add to it to cancel the term that appears in the spin 0 Lagrangian so you write down a combined Lagrangian that is locally U(1) invariant. This is the EM Lagrangian from which Maxwell's equations and the Lorentz force law easily follows. The way its done in the above text can be simplified quite a bit - its a an interesting exercise doing it.

Now of relevance here is there is this number q that appears multiplying the interaction term in the Lagrangian and gives the strength of the coupling of the two fields - it called the coupling constant. This pretty obviously is related to charge in EM. But it turns out, and this is the twist in the answer to your question, its value depends on the energy scale you are probing:

https://en.wikipedia.org/wiki/Coupling_constant

Not taking that things like mass and charge in fact depend on that, which mathematically means you are introducing a cutoff in your theory, is what lead to the infinities that plagues QED. Once you introduce a cutoff, then get rid of the cutoff terms by replacing them with actually measured values (called the re-normalized values) by the trick in my paper on simple re-normalization you can get finite answers in your calculations.

Thanks

Bill

Last edited by a moderator:

- #34

strangerep

Science Advisor

- 3,765

- 2,212

Thanks Sam. I had a definite suspicion I was on the wrong track.samalkhaiat said:Charge-conjugation is not the correct operation to consider here, because [...]

It's always beneficial to receive another,... er,... suppository of wisdom.

Hopefully all these sorts of things will be explained in your (eventual) book?

Last edited:

- #35

bhobba

Mentor

- 10,818

- 3,688

strangerep said:Hopefully all these sorts of things will be explained in your (eventual) book?

I think I will want a copy to.

I just wish Samalkhaiat would post more - his posts are always GOOD.

Thanks

Bill

- Replies
- 13

- Views
- 2K

- Replies
- 4

- Views
- 2K

- Replies
- 4

- Views
- 839

- Replies
- 1

- Views
- 1K

- Replies
- 19

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 0

- Views
- 643

- Replies
- 36

- Views
- 3K

- Replies
- 2

- Views
- 2K

- Replies
- 21

- Views
- 2K

Share: