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Discussion from "Does a real electron possess a certain spin"

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Discussion moved from answer to Does a real electron possess a certain spin?.

I have a question about a "boosted Coulomb field". In the rest frame of the real electron and in absence of real photons, the state must have a certain mass $m_0$, I guess. Then, if we look at such an electron from a moving RF, we will see a variable electromagnetic field, which can be expanded in harmonics, etc. But this state is still without real photons, so the mass remains $m_0$ and the energy/momentum are certain too.

Uncertainty in energy/momentum appears during and after interaction which gives birth to an uncertain number of soft photons. It means the function $\rho(m)$ is a process-dependent or so. Is it right?

asked Nov 9, 2014 in Chat by Vladimir Kalitvianski (22 points) [ revision history ]
edited Mar 27, 2015 by dimension10

5 Answers

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"the function ρ(m) is a process-dependent or so. Is it right?''

No. ρ(m) is characteristic of the infraparticle field itself, it figures in the field commutation rules, and hence appears in every state, including states of rest. m0 is the lower limit (branch point) of the mass spectrum, and is the substitute for the rest mass.

The infraparticle field gets its electromagnetic structure from nontrivial commutation rules with the e/m field. Because of gauge invariance a physical asymptotic infraparticle state automatically contains a coherent electromagnetic field - the e/m field created by the electron. In terms of photons, the coherent electromagnetic field consists (like any coherent state) of an indefinite number of soft virtual photons. There are no real photons involved as long as the asymptotic conditions prevail; real photons have their own asymptotic states.

For the electron branch cut and its relations to anomalous exponents see:

T. Appelquist and J. Carazzone, Infrared singularities and massive fields, Phys. Rev. D11 (1975), 2856--2861.

answered Nov 9, 2014 by Arnold Neumaier (12,425 points) [ revision history ]

Thank you, Arnold. I will read it. But I am not sure about the coherent states for a still electron and with no real photons. It seems to me there is some ambiguity here.

I augmented my post to address this.

Probably, I have to understand this in the following way: the total EM field created with electron motion is a coherent state (such is the exact solution for the field in case the electron motion is known). But a uniformly moving electron does not create real photons, so the coherent state describes just a boosted Coulomb field, a "near" field.

Yes, this is also my understanding.

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 ''Asymptotic electron is already constructed, according to you. What I want to see is it explicit expression''

I had told you already: In the absence of other particles and fields (i.e., where the field theoretic description reduces to a single-particle description) it is given by the tensor product of an infraparticle spinor state with a generalized free dynamics, and a coherent state of the electromagnetic field. 

But once the electron makes a non-free motion because of interaction with a neutral particle, the description given is no longer adequate: The electron then no longer behaves like an asymptotic QED state, and the back reaction of the accelerated electron to the quantum e/m field (and the corresponding radiation generated) is not accounted for. This is why in QED, one cannot treat electrons as particles except when they behave freely - i.e., before or after any scattering, not only the scattering off an e/m field. To get a correct account of what happens you need to incorporate all interactions into terms defining the action.

Your electronium-like models are very far from a field theory.

answered Nov 11, 2014 by Arnold Neumaier (12,425 points) [ no revision ]

Thank you, Arnold.

It is rather strange that the matrix element of scattering when the initial and final states of the target are known, and interaction potential too, cannot be evaluated. It is also strange that "the back reaction of the accelerated electron to the quantum e/m field (and the corresponding radiation generated) is not accounted for."

In my sorry theory the situation is the opposite.

The reference you gave is not free of charge (T. Appelquist and J. Carazzone, Infrared singularities and massive fields, Phys. Rev. D11 (1975), 2856--2861.)

I see too many "impossibilities" in your argumentation, so I am loosing the interest to it, sorry.

"the back reaction of the accelerated electron to the quantum e/m field (and the corresponding radiation generated) is not accounted for." This is because the asymptotic electron is asymptotic only when essentially free, i.e., noninteracting.

Interacting electrons can also be treated but need a full field theoretic context. This is why I asked you to provide an action that conveys in field theoretic terms which interaction you want to study. This interaction can the be treated in a 1-loop approximation, or if it only consists of classical external fields in a ladder approximation, to give the results you are interested in. 

If you want to stay on the particle level, you need effective interactions with effective electrons - this is then related to your electronium models. But it is far from a fundamental treatment, unless one can derive the effective interactions from a field theory or writes down the most general possibility in terms of invariance principles. This is what I did in http://www.physicsoverflow.org/23566/form-factors-self-energy-and-dynamics-single-spin-particle.

But you also lost interest there. You seem to be only interested in your electronium, not in the state of the art of how the world thinks about the basic structure of matter.

Arnold, you overcomplicate things. For example, radiation of the electron due to instantly pushing it is calculated in CED and QED. In the latter case, there is a Glauber approach giving the coherent states. What is missing there is the energy-momentum conservation because in those coherent states all frequencies are present. In my electronium construction the spectrum of exited states is finite just because the energy conservation law is implemented in the electronium construction..

The fact that the QED electron radiates whenever it is pushed shows the permanent coupling to the EMF degrees of freedom, so the coupling is on everywhere, even in asymptotic regions. If somebody uses decoupled things in the asymptotic region and takes everything into account perturbatively during calculations, it is not the state of the art. The radiation of a classical current and pair creation in an external field are (extreme) cases where the coupling is on permanently. These cases are not truly exact, but approximate; however the possibility of taking the interaction into account non perturbatively is a step forward with respect to the initially decoupled things.

Thank you for sending me the article. I will read it.

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Thank you, Arnold, but what about a neutral incident projectile like a neutron, neutrino, etc., which is not involved in $\langle A(x)\rangle$?

Besides, I do not understand why you include the incident photon into the target asymptotic state. They are separated in asymptotic regions. Moreover, I thought the coupling with the "proper" e/m field was already included in the notion of the real electron.

answered Nov 10, 2014 by Vladimir Kalitvianski (22 points) [ no revision ]

This cannot be defined in QED, hence is outside the scope of my setting given in the answer to your original question. You are changing the subject too frequently; it would be better to ask new questions whenever you change the focus.

Within the standard model, the asymptotic electron state is a product of an infraparticle state, a coherent state of the e/m field, a coherent state of the gluon field, and (assuming neutrinos are massless) a coherent state of the neutrino field. 

Ellectron-neutrino scattering can be understood in terms of an external neutrino field similar to the electron-photon interaction, while electron-neutron interactions are true scattering processes that need the full quantum field machinery.

''why you include the incident photon into the target asymptotic state. They are separated in asymptotic regions.'' 

The physical (asymptotic) electron state contains all soft photon contributions in the form of a coherent state. An asymptotic photon can still be in a coherent state encoding an arbitrary transversal classical field, and a spherical such field is part of the outgoing electron state. Without this coherent e/m state, the electron state would not be stable. The physical (asymptotic) photon states only comprise hard photons (except for higher order corrections). Only these are asymptotically separated. 

The reason for all this is that the state space of QED is far from a Fock space. Therefore the usual particle picture (which doesn't have a concept of an associated field) is of only limited applicability. In the literature, this is discussed partially as the superselection structure of QED, partially as scattering in terms of modified Moeller operators.

@ArnoldNeumaier Would it be better if Vladimir's new questions were asked in a separate answer on this chat discussion?

It depends. In general, if he can formulate the question such that it makes sense independent of the present discussion; a normal Q&A question is appropriate; if reference needs to be made to the chat then a separate answer would be right. For the present question, it makes sense if you move his question as a new answer here, and my answer as a comment on his answer.

You are changing the subject too frequently; it would be better to ask new questions whenever you change the focus.

Within the standard model, the asymptotic electron state is a product of an infraparticle state, a coherent state of the e/m field, a coherent state of the gluon field, and (assuming neutrinos are massless) a coherent state of the neutrino field. 

No, I do not change the subject, at least as I understand it. Our subject is the real electron state as a target for some projectile. As well, I am speaking mostly of coupling the constituent electron and its EMF. We can add electron-positron pairs to that in order to deal with full QED, but it is not necessary to involve the Standard Model with gluons, etc. QED as a model is self-sufficient for our goal. Finally, let us limit ourselves with the first Born approximation for the scattering processes. We need to know the initial target state. What is it - that's the question. Whatever projectile is, it serves just to transfer a momentum, to push the electron and change the target state.

For some unknown to me reason, you only admit a hard incident photon (a neutral projectile) as separated from the target in asymptotic scattering regions. I do not understand the difference between hard and soft incident photons. The incident neutral particle is not a part of the target in any way, by definition. And it should not radiate in order not to intervene into the resulting target radiation, that is why I choose it neutral.

You say the target state is a product of this and that. OK. In my electronium model the target is also a product and in the first Born approximation the EMF oscillators get excited. Their final state represents nearly coherent states. Nearly, because there is the energy-momentum conservation and "too hard" modes of the target oscillators cannot get excited. The nearly coherent states do not contain excited modes of all energies, but only those  "accessible" energetically. Radiated photons carry not only energy-momentum, but also an angular momentum. The angular momentum of the target in the final state changes.

I see that the coupling to its EMF influences the electron motion in my electronium in a specific way - it smears quantum mechanically the electron charge or position, whatever. My electronium is spinless for the time being. How should I introduce the spin? As a total angular momentum of the target or as a spin of the constituent electron? In the latter case the coupling must modify the spin state too. This is a conceptual question/problem - what the spin of a real electron is and how one should build the theory.

Could you be more specific and give me the target state explicitly, as a product you mentioned above, please? I would like to see what "belongs" to the constituent electron, what belongs to its field, and where the spin is.

The spin is in the infraparticle part of the electron; the self-field in the coherent part. The infraparticle part of a single electron is a standard spin 1/2 particle but with all momenta with $p^2\ge m_e^2$ present, and an inner product that integrates the standard inner product (over a mass shell) in addition over momenta, with a weight determined by the anomalous exponent, as described in one of the references I gave.

The form of a physical electron state - being defined as an asymptotic bound state of the theory  - depends on how it is embedded in a quantum field theory. In this sense, QFT respects Bohr's requirement that everything relevant to the experiment under discussion must be embedded in the model. 

QED contains only photons, electrons and positrons; thus neutrons and neutrinos are outside the scope of QED, hence a different subject from what I treated in my answer. (Note that I didn't discuss your electronium model - this is not the topic off this thread - but an electron according to current QFT.)

So if you want me to discuss an interaction not present in vanilla QED, please give me the action that modifies the bare QED action, and I'll be able to tell you its impact on the renormalized asymptotic electron.

So if you want me to discuss an interaction not present in vanilla QED, please give me the action that modifies the bare QED action, and I'll be able to tell you its impact on the renormalized asymptotic electron.

Arnold, let us simplify things. There is no need in bare stuff here. Asymptotic electron is already constructed, according to you. What I want to see is it explicit expression. Then I can use it in the first Born approximation to see what effect is produced while scattering a neurtal projectile acting on the electron. There is no need to go farther than the first Born approximation. There is no need to incorporate a projectile in the initial and final target states. What I want is a result like formulas (12), (13) from http://arxiv.org/abs/0806.2635 where I have the initial and final atomic states (also products) and impact of a short-range potential describing qualitatively "pushing" the dressed nucleus.

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My worry is about the phenomenon described in the phrase:

''According to QED an electron continuously emits and absorbs virtual photons (see the leading order diagram in Fig. 8) and as a result its electric charge is spread over a finite volume instead of being pointlike.''

This "emitting and absorbing" makes the Dirac equation non free, so it might influence the spin state. Before switching the interaction on, the spin state could be certain, and with this permanent interaction the spin state is unclear to me. I would even say it becomes a mixture.

answered Nov 9, 2014 by Vladimir Kalitvianski (22 points) [ no revision ]

As you know very well, talking about the exchange of virtual photons is purely figurative speech, used to illustrate a phenomenon, not to talk precisely about it. The spread of the charge is quantitatively covered by the form factors, not by anything depending on virtual photons.

Yes, it is a figurative speech, but I do not mean the size in this thread. I worry about the spin state.

Well, the charge spread was part of what you quoted, so I had thought you referred to that.

The spin state is unrelated to the charge distribution. One can have exactly the same charge distribution with spin up electrons or with completely unpolarized electrons. Nevertheless, the spin (as a  characterization of the representation space of the electron) is always 1/2. The spin state vector or density matrix is of course arbitrary and may be either certain (with respect to a given measurement setup) or uncertain.

Nothing guarantees that anything involving the spin is conserved. But it is always a 2-diemsnional state, which is what is expressed by saying that the electron has spin 1/2. This has nothing to do with conservation or not.

OK, apparently I am unable to explain my worry. Then, let us consider a simple two level spin system, a spin in a constant external magnetic field. There are two distinct energy levels for spin 1/2. There are the corresponding experiments.

Now, let us add an additional magnetic field, which varies with time in an arbitrary way. It will perturb the external field and the corresponding levels. Now experiments will not show those distinct energy levels. It will be difficult to judge about spin then.

The permanent interaction of electron with its own electromagnetic field may be modelled as that arbitrarily varying additional magnetic field. That is my worry.

So I think the spin is not a property of a "bare" or "free" electron, but of the real electron, i.e., a property of a compound (interacting) system. Attempts to construct a real electron from bare particles are therefore somewhat inconsistent.

Your spin is a property of the state of the spin 1/2 electron, independent of whether the electron is bare or physical, and independent of whether there are interactions. Both can be in arbitrary spin 1/2 states, which form a 2-dimensional state space.

Only how the state changes in time, and which state is stationary depends on the interaction.

A real electron alone in the universe will not notice its electromagnetic field, since it sees everything as if being at rest.

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I disagree with your last statement. Because if you push the electron, it will radiate. This is possible only if the electron is permanently coupled to the radiated field degrees of freedom, so it sees them. In the reference you gave recently, it is (awkwardly) described as continuously emitting and absorbing virtual photons, but the meaning is right - the electron sees its own field. That is why the true (meaningful) external lines are always "hairy" rather than "free".

answered Nov 9, 2014 by Vladimir Kalitvianski (22 points) [ no revision ]

But pushing is an additional interaction. The asymptotic picture I gave of the electron is for the asymptotic 1-particle state of QED, where there is no additional interaction, hence no pushing.

If you include an additional interaction by an external field, the asymptotic changes, and the coherent state acccompanying the infraparticle electron will contain radiation, and not just a pure Coulomb field.

I do not understand why the asymptotic changes if you include a scattering. The initial target state (that of the real electron) does not change until the scattering happens, for example, scattering a photon from the electron or a neutron or something neutral with short-range interaction potential. The projectile itself does not radiate. The real electron radiates because it is coupled to the EMF degrees of freedom. The coupling does not get "activated" only during scattering, it is always on. I tried to explain this mechanism on a toy model, if you do not feel the coupling physics as I feel it.

There are also questions like: "What is the spin of the target after scattering"? What is its average spin? What is its square average value? "Discrete"? "Continuous"? "Uncertain?"

Coupling to the e/m degrees of freedom is also an interaction, and it changes the asymptotics. The coherent state associated with the same infraparticle state changes depending on the e/m field incident on the infraparticle. The reason is that the electron state (i.e.,  coherent state times  infraparticle state) is determined by the quadratic part of the (renormalized) action, and this part depends on $\langle A(x)\rangle$. The precise asymptotic state therefore depends on $\langle A(x)\rangle$, too (or rather on the superselection sector determined by it). This dependence accounts for radiation in all context where the electron radiates. 

Classical thinking does not suffice to understand this since the above is the effect of subtracting a classical field (namely $\langle A(x)\rangle$) from the quantum field and treating the classical part as external.

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