How can gauge fields be associated with real particles?

[Pradyuman]

TL;DR Summary

Gauge fields, bosons

When we make our lagrangian invariant by U(1) symmetry we employ the fact that nature doesn't care how I describe it, but, how come that I can associate the real physical particles with the coordinates I use to describe? Even though gauge symmetry is not a physical Symmetry,

 

[vanhees71]

It's not clear to me, what you are asking. The most simple paradigmatic example of a gauge field is the electromagnetic field. The gauge group is U(1), and indeed you are right, gauge invariance is not a symmetry of physics, because it's rather a redundancy in describing a physical situation.

In electromagnetism that's pretty intuitive. In classical electromagnetism what's physically observable is the electromagneticmagnetic field, $(\vec{E},\vec{B})$. It can, in principle, be determined by observing the motion of test charges in this field.

On the other hand the homogeneous Maxwell equations, which mathematically are constraint equations on $(\vec{E},\vec{B})$ can be fulfilled identically by introducing a scalar and a vector potential,
$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
but for a given physical situation, i.e., for a given physical field $(\vec{E},\vec{B})$ the potentials are not uniquely determined, because you can always introduce new potentials by a "gauge transformation",
$$\vec{A}'=\vec{A} -\vec{\nabla} \chi, \quad \Phi'=\Phi+\partial_t \chi,$$
because then
$$-\partial_t \vec{A}'-\vec{\nabla} \Phi'=-\partial_t \vec{A}'+\partial_t \vec{\nabla} \chi - \vec{\nabla} \Phi - \vec{\nabla} \partial_t \chi=-\partial_t \vec{A}-\vec{\nabla} \Phi=\vec{E}$$
and
$$\vec{\nabla} \times \vec{A}'=\vec{\nabla} \times \vec{A} - \vec{\nabla} \times \vec{\nabla} \Phi=\vec{\nabla} \times \vec{A}=\vec{B},$$
i.e., the physical situation is not uniquely described by the potentials $(\Phi,\vec{A})$ but by "$(\Phi,\vec{A})$ modulo an arbitrary gauge transformation".

When quantizing a gauge-field theory you run into characteristic problems due to this "gauge freedom", and indeed for the electromagnetic field a naive particle picture is even less useful than when quantizing massive fields. To understand this, it's useful to remember the origin of the fields to describe representations of the Poincare group, and it turns out that for massless fields the entire business changes compared to the case of massive fields. Indeed, the notion of massless and massive fields of course also originates from the representation theory of the Poincare group since mass from this point of view is a Casimir operator of the corresponding Lie algebra, given by $P_{\mu} P^{\mu}=m^2 c^2$, where $P_{\mu}$ (the "four-momentum operators") are the generators of space-time translations.

It turns out that in the case of massless fields you don't have spin in the usual sense but only "helicity", i.e., the projection of the total angular momentum to the direction of momentum, and for massless particles of "spin" $s$ you don't have $(2s+1)$ values for a spin component (corresponding to $(2s+1)$ field degrees of freedom) but only two helicity values $\pm s$. For photons, i.e., the quanta of the em. field, you have only the two helicities $\pm 1$, corresponding to left- and right-circular polarized field modes instead of three spin-degrees of freedom as you get for massive spin-1 fields.

This also indicates that you cannot describe a massless spin-1 field in a naive way by "wave functions". If you try to realize them with local field equations of motion, as needed to implement the microcausality constraint on the quantized theory, you are let necessarily to the description by a gauge theory as in classical electrodynamics.

For the details on the representation theory of the Poincare group, see

R. U. Sexl and H. K. Urbantke, Relativity, Groups, Particles, Springer, Wien (2001).

 

[Demystifier]

Instead of giving an answer that can be found in any textbook (see e.g. the post above by @vanhees71 ), I will give a simple analogy. Consider one free classical non-relativistic particle moving in one dimension. It's action can be written as
$$A=\int dt\, \frac{mv^2(t)}{2}$$
where $v(t)=\dot{x}(t)$ is the velocity of the particle and $x(t)$ is its position. The action is invariant under the transformation
$$x(t) \to x'(t)=x(t)+a$$
where $a$ is an arbitrary constant. Physically, this is the translation invariance reflecting the fact that we can choose the origin of the position coordinate as we want, without changing physics. With a little abuse of language, let me call this invariance the "gauge invariance". In this language, we can say that $v(t)$ is gauge invariant, i.e.
$$v(t) \to v'(t)=v(t)$$
The equation of motion resulting from the action above is
$$\dot{v}(t)=0$$
which is a gauge invariant equation because $v(t)$ is a gauge invariant quantity.

The "gauge invariant" theory of a particle above is somewhat analogous to the gauge invariant theory of free electromagnetic field. The $x$ is analogous to the electromagnetic potential $A_{\mu}$, the $v$ is analogous to the field strength $F_{\mu\nu}$, the transformation $x'=x+a$ is analogous to the transformation $A'_{\mu}=A_{\mu}+\partial_\mu\lambda$, and the equation of motion $\dot{v}=0$ is analogous to $\partial_{\mu}F^{\mu\nu}=0$.

Now consider the following. Even though we can choose $a$ at will, i.e. we can take the origin of coordinates wherever we want, the particle is real, it's somewhere. To specify where it is, we must choose some coordinates with an origin at a definite place. But the particle is somewhere even if we don't choose the origin, it's just that we don't have a way to say where it is.

Likewise, the electromagnetic field is real even if we don't choose the gauge, but we must choose a gauge in order to speak of the electromagnetic field in terms of $A_{\mu}$. The physics does not depend on this choice, but some choice must be made, otherwise we cannot describe it.

 

[vanhees71]

But that example is NOT an example for a gauge invariance, because it's not a "redundancy in describing a physical situation" but a true symmetry. Indeed the free-particle Langrangian is describing a theory which has a true symmetry, namely symmetry under spatial translations (among others, which we simply ignore here). That it's a true symmetry is also reflected in the fact that it has an interpretation in the sense of an "active transformation", i.e., it predicts that when going with an entire real-world experiment to another place, you'll get the same results.

This is in contradistinction to the gauge invariance of electrodynamics. There a gauge transformation of the potentials doesn't have an interpretation in terms of an "active transformation", i.e., you cannot realize the gauge transformation as any manipulation on the setup of an experiment (or any physical situation). It's just the non-uniqueness of the potentials given any physical situation (concerning electromagnetic phenomena).

 

[Demystifier]

But that example is NOT an example for a gauge invariance, because it's not a "redundancy in describing a physical situation" but a true symmetry.

I agree. But I believe it can still help to answer the OP's question, i.e. to understand conceptually how can something be real despite "the fact that nature doesn't care how I describe it".

 

[vanhees71]

I think, it's very important to distinguish between symmetries and gauge invariance. The understanding is hindered by the fact that even in the best textbooks (including my favorites by Weinberg) that's not done properly!

A very good clarification on hand of the example of BCS theory of superconductivity is given here:

https://arxiv.org/abs/cond-mat/0503400
https://doi.org/10.1016/j.aop.2005.03.008

 

[Demystifier]

I think, it's very important to distinguish between symmetries and gauge invariance.

Yes, it's important, but I think that the OP struggles here with more elementary concepts.

 

[vanhees71]

I think students struggle with these elementary concepts, because these subtleties are only sloppily discussed even in the best textbooks.

Examples are:

-a particle interpretation in relativistic QFT is only possible for asymptotic free Fock states, and the meaning is not that of some "miniature billiard ball". In other words: already, massive "particles" cannot be arbitrarily precisely localized in relativistic QFTs although they still admit the definition of a position observable; massless particles with spin $\geq 1$ don't even allow the definition of a position observable.

-local gauge invariance is not describing a physical symmetry

-gauge dependent quantities cannot be observed and don't have a direct physical meaning

-consequently local "gauge symmetries" cannot be "spontaneously broken" (but rather only be "Higgsed") and that's why there are no Goldstone bosons in the physical spectrum of a "Higgsed gauge theory". It's better to call it "hidden gauge invariance" rather than "spontaneously broken gauge symmetry"

 

[Pradyuman]

It's not clear to me, what you are asking. The most simple paradigmatic example of a gauge field is the electromagnetic field. The gauge group is U(1), and indeed you are right, gauge invariance is not a symmetry of physics, because it's rather a redundancy in describing a physical situation.

In electromagnetism that's pretty intuitive. In classical electromagnetism what's physically observable is the electromagneticmagnetic field, $(\vec{E},\vec{B})$. It can, in principle, be determined by observing the motion of test charges in this field.

On the other hand the homogeneous Maxwell equations, which mathematically are constraint equations on $(\vec{E},\vec{B})$ can be fulfilled identically by introducing a scalar and a vector potential,
$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
but for a given physical situation, i.e., for a given physical field $(\vec{E},\vec{B})$ the potentials are not uniquely determined, because you can always introduce new potentials by a "gauge transformation",
$$\vec{A}'=\vec{A} -\vec{\nabla} \chi, \quad \Phi'=\Phi+\partial_t \chi,$$
because then
$$-\partial_t \vec{A}'-\vec{\nabla} \Phi'=-\partial_t \vec{A}'+\partial_t \vec{\nabla} \chi - \vec{\nabla} \Phi - \vec{\nabla} \partial_t \chi=-\partial_t \vec{A}-\vec{\nabla} \Phi=\vec{E}$$
and
$$\vec{\nabla} \times \vec{A}'=\vec{\nabla} \times \vec{A} - \vec{\nabla} \times \vec{\nabla} \Phi=\vec{\nabla} \times \vec{A}=\vec{B},$$
i.e., the physical situation is not uniquely described by the potentials $(\Phi,\vec{A})$ but by "$(\Phi,\vec{A})$ modulo an arbitrary gauge transformation".

When quantizing a gauge-field theory you run into characteristic problems due to this "gauge freedom", and indeed for the electromagnetic field a naive particle picture is even less useful than when quantizing massive fields. To understand this, it's useful to remember the origin of the fields to describe representations of the Poincare group, and it turns out that for massless fields the entire business changes compared to the case of massive fields. Indeed, the notion of massless and massive fields of course also originates from the representation theory of the Poincare group since mass from this point of view is a Casimir operator of the corresponding Lie algebra, given by $P_{\mu} P^{\mu}=m^2 c^2$, where $P_{\mu}$ (the "four-momentum operators") are the generators of space-time translations.

It turns out that in the case of massless fields you don't have spin in the usual sense but only "helicity", i.e., the projection of the total angular momentum to the direction of momentum, and for massless particles of "spin" $s$ you don't have $(2s+1)$ values for a spin component (corresponding to $(2s+1)$ field degrees of freedom) but only two helicity values $\pm s$. For photons, i.e., the quanta of the em. field, you have only the two helicities $\pm 1$, corresponding to left- and right-circular polarized field modes instead of three spin-degrees of freedom as you get for massive spin-1 fields.

This also indicates that you cannot describe a massless spin-1 field in a naive way by "wave functions". If you try to realize them with local field equations of motion, as needed to implement the microcausality constraint on the quantized theory, you are let necessarily to the description by a gauge theory as in classical electrodynamics.

For the details on the representation theory of the Poincare group, see

R. U. Sexl and H. K. Urbantke, Relativity, Groups, Particles, Springer, Wien (2001).

Suppose I finished making my lagrangian invariant and i found ohh I have to add this term to my lagrangian, to make it invariant, now how come I can say oh that's the term associated with EM fields,

 

[vanhees71]

Which Lagrangian are you talking about? It's really difficult to guess what you are talking about. A clear formulation of the problem is often more than half the way to full understanding!

 

[Pradyuman]

I am currently studying qft and in that I considered the spinor lagrangian, i made it local gauge invariant I had to replace derivative with covariant derivative, and added #F_{\mu\nu}# term but how do I say that this gauge field correspond to the EM field

 

[Demystifier]

I am currently studying qft and in that I considered the spinor lagrangian, i made it local gauge invariant I had to replace derivative with covariant derivative, and added #F_{\mu\nu}# term but how do I say that this gauge field correspond to the EM field

You derive the equation of motion for the gauge field and find that it satisfies the Maxwell equations. Since you already know that the EM field satisfies the Maxwell equations, and since you don't know any other field in Nature that satisfies the Maxwell equations, you make a hypothesis that the gauge field is in fact the EM field. Then you make experiments that test this hypothesis (and by "you" I mean the whole community of physicists), see e.g. https://edu.itp.phys.ethz.ch/hs10/ppp1/PPP1_6.pdf , and when experiments turn out to confirm the hypothesis you conclude that the hypothesis is true.

 

[WernerQH]

how come that I can associate the real physical particles with the coordinates I use to describe?

Many think of electrons and photons as real (even fundamental) particles. But in my opinion QED does not treat them as separate "objects". Rather, it only describes interactions between them. More precisely, the correlations of localized, short-lived microscopic currents (events). For me, QED is more of a phenomenological ("effective") theory than something fundamental. Lagrangians are chosen to "fit the facts", or "save the phenomena", much like what is done to describe the numerous fields excitations that crop up in condensed matter physics.

 

[vanhees71]

Physical theories are always aiming at describing what's observed, and Lagrangians are found by using the concepts of symmetry to describe empirical facts, and it is indeed amazing, from how little input you can describe all known particles and their interactions with the Standard Model of elementary-particle physics.