0
$\begingroup$

I was reading a paper¹, and it states:

" Therefore, the gauge fields themselves cannot be entities of the physical reality, as any observations should be independent of the chosen gauge"

I'm trying to understand what he wanted to say. Because, as far as I know, W bosons and Z bosons ARE PART of the Physical reality as much as photons and electrons.... Someone could help me figure it out what the author was trying to say?

¹ http://arxiv.org/abs/1106.3942

$\endgroup$
3
  • 3
    $\begingroup$ This is referring to the concept of gauge redundancy and this has been answered many times before. See physics.stackexchange.com/q/18640 and physics.stackexchange.com/q/13870. Essentially, a gauge symmetry tells you about the invariance of a Lagrangian under that gauge condition and performing calculations under one gauge may not necessarily agree under another gauge and this is not very good practice. So, purely physical quantities are independent on how you constrain your theory. $\endgroup$
    – user106422
    Commented May 24, 2016 at 0:02
  • $\begingroup$ This statement, and especially the following statement "Consequently, any particle-like excitations associated with the gauge fields, also cannot represent physical observable particles." seems too strong. It doesn't apply to photons, because transverse photon states are gauge-invariant. It doesn't apply to W and Z bosons because there the gauge symmetry is broken by the Higgs mechanism, I think. $\endgroup$
    – knzhou
    Commented May 24, 2016 at 0:07
  • $\begingroup$ The more restrictive statement I think he wants to make is, e.g. you can't point at a "blue" gluon, because color is not gauge invariant. But maybe he means something stronger? $\endgroup$
    – knzhou
    Commented May 24, 2016 at 0:08

1 Answer 1

2
$\begingroup$

This is just very sloppy language on the paper's part. As you say, gauge bosons are very real and their existence has physically measurable consequences (otherwise, why would we ever waste time talking about them?). (By the way "photons and electrons" are not good examples of non-gauge particles, because photons are also gauge bosons :) .)

The paper just means that the only physically measurable properties of the gauge fields are the gauge-independent ones. For example, a quantity like $\langle A_\mu^a \rangle$ is not physically measurable because its value depends on your choice of gauge. On the other hand, the quantity $\langle \partial_\mu A_\nu - \partial_\nu A_\mu + i g [A_\mu, A_\nu] \rangle$ is gauge invariant and can be directly measured.

It's like (a slightly more complex version of) the fact that the potential energy $U({\bf x})$ is not physically measurable because it is only defined up to an overall additive constant. Only its derivative ${\bf F(x)} = -{\bf \nabla} U({\bf x})$ is physically measurable, but that doesn't mean that the potential energy function has no physical consequence at all. The gauge fields contain some gauge-dependencies which make them not directly physical, but the appropriate derivatives are directly physical.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.