Shape of elementary particles in QFT, etc?

[A. Neumaier]

Maybe in classical electrodynamics. But in QED it is certainly common to say that $A^{\mu}$ is a vector field with spin-1.

But this is loose, formal but physically meaningless talk, since it is a gauge dependent quantity and gauge fixing is an arbitrary, unphysical procedure. Only gauge invariant quantities have a physical meaning.

 

[secur]

the borderline where one switsches from one language to the other is fluent.

That borderline may be fluid (or 'flexible', 'fuzzy', 'uncertain', 'indeterminate', 'ill-defined', etc), but I doubt it's fluent - in language or anything else. Unless you have a new theory of "conscious borderlines"? Given your belief that consciousness is a meaningless superstition, that would be quite a switsch!

 

[vanhees71]

Maybe in classical electrodynamics. But in QED it is certainly common to say that $A^{\mu}$ is a vector field with spin-1.

It is very important to say that it is a massless vector field, which together with locality implies that it must be a U(1) gauge field. See, e.g., Weinberg, Quantum Theory of Fields, vol. 1.

 

[vanhees71]

I just asserted that this is the meaning with which the term is used by those who use it, by extrapolation form the interpretation that is valid for larger objects.

In the macroscopic and nonrelativistic limit it becomes the mass density which defines very obviously the shape, and since it can be defined for all objects no matter how large it is a meaningful physical concent, even though one can measure it only for sufficently robust objects such as single atoms on a surface or in a cavity, or mesoscopic or even macroscopic objects. (There is lots of nonmeasurable stuff that has a clear physical meaning, for example the detailed mass distribution in the interior of the sun or in the Andromeda galaxy.)

True, but these quantities are measurable in principle. It's only not within our technical possibilities to do so.

That shape is frame-dependent is already well-known form classical relativity (length contraction). It has to be so.

This is a misconception you find astonishingly often. Of course, for a system with non-zero mass there's a preferred frame of reference, the center-momentum frame, and this frame is used to define intrinsic quantities of this system. If you want you can call the energy-density distribution in this frame the "shape" of the object, although I've never seen this anywhere in the literature. You can define this quantity in a frame-independent way as a scalar density. For a fluid it's given by $$\epsilon=u_{\mu} u_{\nu} \Theta^{\mu \nu},$$ where $u^{\mu}$ is the four-velocity field of the fluid (normalized such that $u_{\mu} u^{\mu}=1$) and $\Theta^{\mu \nu}$ is the energy-momentum tensor of the fluid.

 

[A. Neumaier]

If you want you can call the energy-density distribution in this frame the "shape" of the object,

This doesn't work for a photon, whereas the energy-density distribution in the lab frame always exists. This is the one of interest to the experimenter. http://arxiv.org/abs/1605.00023, http://arxiv.org/abs/1601.07142, http://arxiv.org/abs/1512.08213, http://link.aps.org/abs/10.1103/PhysRevAccelBeams.19.021304 are some recent references to photon shape.

 

[vanhees71]

Yes, that's because a single photon is a massless state, and there is no restframe for it. That's why, I don't like to call that quantity "shape". It's just proportional to the probability distribution for detecting a photon. From reading the abstracts of the cited papers, I think that's also the common understanding among quantum opticians.

 

[Arjun Wasan]

In string theory classic point like particles are in 0 dimensions.