# I Does the electron have charge?

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1. Jan 15, 2017

### ftr

I mean for the free electron Dirac equation the charge (e or q) does not appear in the equation. It only appears after its interaction with EM field. 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?

2. Jan 15, 2017

### strangerep

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.

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

3. Jan 15, 2017

### ftr

Ok, Thanks. That is a bit confusing, because 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. I guess it just a matter of words like most of the time.

4. Jan 15, 2017

### strangerep

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.

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

5. Jan 15, 2017

### ftr

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.

6. Jan 15, 2017

### strangerep

Try looking at these old threads first, wherein I posted more extensive explanations and references.

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

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. Jan 15, 2017

### ftr

Thanks for the links. Is the coulomb field the same as the gauge field, and shouldn't field be related to real particles.

8. Jan 15, 2017

### strangerep

It's the ordinary electric field of an electron.

9. Jan 15, 2017

### ftr

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

10. Jan 15, 2017

### Staff: Mentor

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. Jan 16, 2017

### Staff: Mentor

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 lets 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:

Thanks
Bill.

Last edited: Jan 16, 2017
12. Jan 16, 2017

### vanhees71

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.

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. Jan 16, 2017

### dextercioby

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. Jan 16, 2017

### Staff: Mentor

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. Jan 16, 2017

### dextercioby

Well. for the sake of arguing, you are not properly addressing my statement which you quoted. The free Dirac Hamiltonian could very well describe a free electron - which is thus free, it will not interact. It is the analogue of the Laplacian of the Schroedinger theory in the absence of any potential.

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 realised then that it could describe a free proton, too).

16. Jan 16, 2017

I think the proper way to say it is that the electron is charged under the U(1) gauge field, and if you use the path integral, you can derive the Ward identity which provides a relationship between correlators that tells you that the charge renormalization is completely determined by the EM field renormalization and only comes from the vacuum polarization diagram. This shows you that charge is conserved, relative charges remain the same. The proof is completely non-perturbative.

17. Jan 16, 2017

### Staff: Mentor

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. Jan 17, 2017

### vanhees71

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), but that this is the electric charge of an electron (or any other charged Dirac particle) is only defined at the moment you make the theory interacting by coupling it to the photon field. One successful recipe is to make the global symmetry local, which introduces an in this case Abelian gauge field, which can be identified with the electromagnetic field.

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., 4 (1970), p. 745.

19. Jan 17, 2017

### Staff: Mentor

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. Jan 17, 2017

### strangerep

Er,... why not? (I'm assuming "global" means that the phase factor is not spacetime-dependent.) I.e., $\psi \to \psi' = e^{i\theta}\psi$.

Or am I missing something?

21. Jan 17, 2017

### Staff: Mentor

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. Jan 17, 2017

### Staff: Mentor

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: Jan 18, 2017
23. Jan 17, 2017

### strangerep

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. ]

Last edited: Jan 17, 2017
24. Jan 17, 2017

For complex fields (scalar, fermions), under an infinitesimal constant U(1) global phase rotation you can treat the two fields independently and vary the action with respect to both to get a conserved current. 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)

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 covariantly conserved current

25. Jan 18, 2017

### samalkhaiat

No, for single real scalar field $\phi(x)$, the transformation $\phi(x) \to e^{i\theta} \phi(x)$ does not make any sense. Also, such $U(1)$ transformation does not leave the corresponding (scalar field) Lagrangian invariant. Thus, a one-component real field cannot carry a $U(1)$ charge.
The fact that both $\psi(x)$ and $e^{i\theta}\psi(x)$ solve the same equation of motion (here the Dirac equation) mean that the (Dirac) Lagrangian is invariant under the continuous group $U(1)$. This, in turn, leads (via Noether theorem) to conserved symmetry current: Since $U(1)$ is a connected Lie group, we can consider infinitesimal transformations $\delta \psi = i \epsilon \psi$ and $\delta \bar{\psi} = -i \epsilon \bar{\psi}$, then $$\delta \mathcal{L} = 0 \ \Rightarrow \ \partial_{\mu} J^{\mu}(x) = 0 ,$$ where $$J^{\mu} (x) \equiv \delta\bar{\psi} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\bar{\psi})} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)} \delta \psi .$$ For the free Dirac Lagrangian, this gives you $J^{\mu} = \bar{\psi} \gamma^{\mu}\psi$. From this current, you obtain the following time-independent and Lorentz scalar charge $$Q = e \int d^{3}x J^{0}(x) = e \int d^{3}x \psi^{\dagger}(x) \psi(x) .$$ Now, if you use the equal-time anti-commutation relations, you can easily show that $$[Q , \psi (y)] = -e \int d^{3}x \{ \psi^{\dagger}(x) , \psi (y) \} \psi(x) = - e \psi (y) .$$ Thus, if $|q \rangle$ is an eigen-state of $Q$ with eigen-value $q$, then $\psi(x)|q\rangle$ is also an eigen-state of $Q$ with eigen-value $(q - e)$, and correspondingly $\psi^{\dagger}(x)|q\rangle$ belongs to the eigen-value $(q + e)$. This is consistent with the interpretation of the field operators: $\psi(x)$ creates particles of charge $(-e)$ or absorbs particles of charge $(+e)$, while $\psi^{\dagger}(x)$ creates particles of charge $(+e)$ or absorbs particles of charge $(-e)$.