# Why is gauge symmetry not a true symmetry?

samalkhaiat
In the classical theory what's measurable are only the field strengths, i.e., ##F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}##, and you can express everything in terms of this gauge-invariant field-strength tensor.
It would be so nice if this is true. Unfortunately, the interaction Lagrangian $A_{\mu}J^{\mu}$ cannot be expressed in terms of the $F_{\mu\nu}$.

ShayanJ
Gold Member
What is more “physical” than a principle that gives birth to all known force carriers (the vector bosons)? With no extra input other than local gauge invariance, we are able to determine the form of all known interactions, i.e., demanding local gauge invariance leads to $A_{\mu}^{a}J^{\mu}_{a}$ as the only possible renormalizable and Lorentz invariant (matter-vector fields) interaction Lagrangians. Furthermore, the explicit form of the matter fields physical currents $J^{\mu}_{a}$ is determined by the very same local gauge invariance through the second Noether theorem. If this is not a “physical symmetry” (whatever that means) then we should be able to replace it. The question then is: Can we do physics with out it?
My advice to you (if you like) is to ignore this “redundancy” thing and think of gauge transformations as the symmetries of constraint systems.
This seems better.
Is this paper a good explanation of what you say?

samalkhaiat
This seems better.
Is this paper a good explanation of what you say?
I don't realy know the paper. But, my answer is yes, provided that they give proper answer to the questions they posed:

How many gauge transformations (with independent gauge parameters) are there for a given action?
What is the structure of the gauge generators (how many time derivatives they contain) for a
given action? What is the structure of an arbitrary symmetry of the action of a singular theory?
Is there a constructive procedure to find all the gauge transformations for a given action?
How can one relate the constraint structure in the Hamiltonian formulation with the symmetry
structure of the Lagrangian action?
I must say though, that understanding the second Noether theorem is the key point in here.

vanhees71
Gold Member
2019 Award
It would be so nice if this is true. Unfortunately, the interaction Lagrangian $A_{\mu}J^{\mu}$ cannot be expressed in terms of the $F_{\mu\nu}$.
That's true, but in the classical theory all you need are the equations of motion, and these depend only on the field strengths ##F_{\mu \nu}## or ##(\vec{E},\vec{B})## in the 1+3-formalism.

stevendaryl
Staff Emeritus
That's true, but in the classical theory all you need are the equations of motion, and these depend only on the field strengths ##F_{\mu \nu}## or ##(\vec{E},\vec{B})## in the 1+3-formalism.
The equations of motion don't involve potentials, but the Lagrangian does. I guess there is a sense in which the Lagrangian is less physical than the equations of motion, because the lagrangian isn't unique.

vanhees71
Gold Member
2019 Award
That's true, but in quantum theory the field strength tensor is also not enough (at least not within local QFT). An example is the Aharonov-Bohm effect, where a non-integrable phase ##\propto \oint_{\partial F} \mathrm{d} \vec{r} \cdot \vec{A} = \int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{B}## leads to observable effects. Of course, although anything observable is gauge invariant, and that's the case here too, because the phase is gauge invariant when written in the second form as the magnetic flux, but in this form it looks non-local, because it integrates over a surface, where the particles don't feel any force (in a classical picture), while the first way shows that the interaction is local in the sense of QFT, because it's along an arbitrary trajectory around the "classically forbidden region".

Last but not least the analysis of the representation theory of the Poincare group a la Wigner (1939) leads to the conclusion that massless vector fields must be represented as gauge fields, and the photon is to very high accuracy known to be massless. In this sense it is quite natural to describe electromagnetism as a gauge theory.

ShayanJ
Gold Member
Last but not least the analysis of the representation theory of the Poincare group a la Wigner (1939) leads to the conclusion that massless vector fields must be represented as gauge fields, and the photon is to very high accuracy known to be massless. In this sense it is quite natural to describe electromagnetism as a gauge theory.
As I understand it, we know that a massless spin 1 particle has two degrees of freedom but a massless vector field has three, so we need gauge invariance to eliminate one degree of freedom. It seems to me its a natural thing to suggest that we should look for a mathematical structure that by itself accommodates only two degrees of freedom and also behaves like a vector field in the ways we want, but now that we don't have such a thing, we use vector fields. Maybe gauge invariance is needed because of this "lack of technology". But if there is no such structure in mathematics and it doesn't make sense to look for such a thing, then maybe you should think that this "lack of technology" and the need for gauge invariance have some physical meaning, as Sam suggested.

vanhees71
Gold Member
2019 Award
No, it's really the math of the Poincare group. In short the argument goes as follows: You start with the representation of space-time translations, which leads to eigenvectors of energy and momentum. Now you want irreducible representations (defining what's an elementary particle). This leads you to look for the Casimir operators of the Poincare group. With energy and momentum you can build ##p_{\mu} p^{\mu}=m^2##, where ##m## is the (invariant) mass of the particle.

Now since the representation is irreducible, you can use the energy-momentum states at fixed mass, and further you can define a standard momentum ##K## with ##K^2=m^2## and all other momentum eigenstates posssible must be reachable by application of the unitary operations on the eigenvectors of energy-momementum eigenvalues ##K##. All the basis vectors at fixed standard momentum ##K## must be reachable with POincare transformations either, because we have an irreducible representation. In other words these vectors build a representation of the subgroup which leave the standard momentum unchanged.

Now take the two physically relevant cases ##m^2>0## and ##m=0## (##m^2<0##, socalled tachyons have trouble, and as far as I know there are no sensible models with interacting tachyons known, let alone that no tachyon has been ever seen in experiment). For ##m>0##, you can choose the standard momentum ##K=(m,0,0,0)##. Then the subgroup of the Lorentz group which leaves this momentum invariant is the SO(3) acting on the spatial part of the momentum, and since ##\vec{K}=0##, these rotations all leave ##K## unchanged. Now you only need to find the irreps. of SO(3) (or its covering group SU(2)), which leads to particles with spin quantum number ##s \in \{0,1/2,1,3/2,\ldots \}##. Then the spin-##z## component (i.e., the angular momentum of a particle with a standard momentum, which here means it's the particle at rest) takes the ##2s+1## values ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##. E.g., a massive vector particle belongs to the representation with ##s=1## (by definition) and thus has three spin-degrees of freedom. Particle states with non-zero momentum can be defined by applying rotation free Lorentz boosts on the states with standard momentum (defining the socalled Wigner basis and with it the full unitary irrep of the Poincare group).

What's different for ##m=0## is that the standard momentum must be light-like. The usual choice is to use a particle running in ##z## direction, i.e., the standard momentum is taken as ##K=(k,0,0,k)##. Now the subgroup of the Lorentz group leaving this standard vector unchanged (the socalled "null rotations") turns out to be non-compact and equivalent to the group ISO(2) of the Euclidean plane. This group consists of translations (two degrees of freedom) and a rotation in the plane O(2). A generall irrep of this little group to the standard momentum ##K## will define an infinite dimensional Hilbert space (as in quantum theory of a particle in a plane the generators of the translations have the entire ##\mathbb{R}^2## as a spectrum), but here this implies that you'd have something like a particle with an infinite number of intrinsic quantum states like the spin of massive particles. Such a thing is also not known to exist in nature, and thus we restrict ourselves to representations, where the translations are represented trivially. Then you are only left with the O(2) rotations, i.e., in the physical space the rotations around the 3 axis. Since in the end you want to have representations of the entire Poincare group, including all rotations and boosts of the Lorentz group, the rotations around the 3-axis must also be represented by ##\exp(\pm \mathrm{i} \lambda \varphi)## with ##\lambda \in \{0,\pm1/2,\pm 1,\ldots \}##. Thus all massless particles have only two polarization degrees of freedom (or one if you have a scalar particle) (here we used the helicity ##\lambda##, i.e., the projection of the particles angular momentum to the direction of its momenum).

Now we want local quantum field theories, i.e., we like to work with field operators that transform like classical fields, e.g., for a vector field we want to have $$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x),$$
where ##\Lambda## is an arbitrary proper orthochronous Lorentz transformation. This shows that for the massless case, in order to have the null rotations represented trivially, the ##A^{\mu}## must be defined modulo to gauge transformations, i.e., with ##A_{\mu}## also ##A_{\mu}+\partial_{\mu} \chi## with an arbitrary scalar field ##\chi## represents the very same field.

For details, see Appendix B of my lecture notes on QFT:

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

• ShayanJ
samalkhaiat
No, it's really the math of the Poincare group. In short the argument goes as follows: You start with the representation of space-time translations, which leads to eigenvectors of energy and momentum. Now you want irreducible representations (defining what's an elementary particle). This leads you to look for the Casimir operators of the Poincare group. With energy and momentum you can build ##p_{\mu} p^{\mu}=m^2##, where ##m## is the (invariant) mass of the particle.

Now since the representation is irreducible, you can use the energy-momentum states at fixed mass, and further you can define a standard momentum ##K## with ##K^2=m^2## and all other momentum eigenstates posssible must be reachable by application of the unitary operations on the eigenvectors of energy-momementum eigenvalues ##K##. All the basis vectors at fixed standard momentum ##K## must be reachable with POincare transformations either, because we have an irreducible representation. In other words these vectors build a representation of the subgroup which leave the standard momentum unchanged.

Now take the two physically relevant cases ##m^2>0## and ##m=0## (##m^2<0##, socalled tachyons have trouble, and as far as I know there are no sensible models with interacting tachyons known, let alone that no tachyon has been ever seen in experiment). For ##m>0##, you can choose the standard momentum ##K=(m,0,0,0)##. Then the subgroup of the Lorentz group which leaves this momentum invariant is the SO(3) acting on the spatial part of the momentum, and since ##\vec{K}=0##, these rotations all leave ##K## unchanged. Now you only need to find the irreps. of SO(3) (or its covering group SU(2)), which leads to particles with spin quantum number ##s \in \{0,1/2,1,3/2,\ldots \}##. Then the spin-##z## component (i.e., the angular momentum of a particle with a standard momentum, which here means it's the particle at rest) takes the ##2s+1## values ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##. E.g., a massive vector particle belongs to the representation with ##s=1## (by definition) and thus has three spin-degrees of freedom. Particle states with non-zero momentum can be defined by applying rotation free Lorentz boosts on the states with standard momentum (defining the socalled Wigner basis and with it the full unitary irrep of the Poincare group).

What's different for ##m=0## is that the standard momentum must be light-like. The usual choice is to use a particle running in ##z## direction, i.e., the standard momentum is taken as ##K=(k,0,0,k)##. Now the subgroup of the Lorentz group leaving this standard vector unchanged (the socalled "null rotations") turns out to be non-compact and equivalent to the group ISO(2) of the Euclidean plane. This group consists of translations (two degrees of freedom) and a rotation in the plane O(2). A generall irrep of this little group to the standard momentum ##K## will define an infinite dimensional Hilbert space (as in quantum theory of a particle in a plane the generators of the translations have the entire ##\mathbb{R}^2## as a spectrum), but here this implies that you'd have something like a particle with an infinite number of intrinsic quantum states like the spin of massive particles. Such a thing is also not known to exist in nature, and thus we restrict ourselves to representations, where the translations are represented trivially. Then you are only left with the O(2) rotations, i.e., in the physical space the rotations around the 3 axis. Since in the end you want to have representations of the entire Poincare group, including all rotations and boosts of the Lorentz group, the rotations around the 3-axis must also be represented by ##\exp(\pm \mathrm{i} \lambda \varphi)## with ##\lambda \in \{0,\pm1/2,\pm 1,\ldots \}##. Thus all massless particles have only two polarization degrees of freedom (or one if you have a scalar particle) (here we used the helicity ##\lambda##, i.e., the projection of the particles angular momentum to the direction of its momenum).

Now we want local quantum field theories, i.e., we like to work with field operators that transform like classical fields, e.g., for a vector field we want to have $$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x),$$
where ##\Lambda## is an arbitrary proper orthochronous Lorentz transformation. This shows that for the massless case, in order to have the null rotations represented trivially, the ##A^{\mu}## must be defined modulo to gauge transformations, i.e., with ##A_{\mu}## also ##A_{\mu}+\partial_{\mu} \chi## with an arbitrary scalar field ##\chi## represents the very same field.

For details, see Appendix B of my lecture notes on QFT:

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf
Okay, the thread below translates Vanhees71's words into math. Have a look at it.

• vanhees71
king vitamin
Gold Member
Symmetries are mathematical transformations, on the independent and/or dependent variables of the system, which leave the action integral unchanged.
If you use this as your definition of symmetry, then gauge transformations count, but so do anomalous symmetries. So it's not a good definition for quantum systems. E.g. pure Yang Mills is classically scale invariant but it isn't a symmetry for the quantized theory.

I believe saying that gauge transformations aren't symmetries or are "redundancies" makes sense because they do not transform distinct states into each other. If you take an arbitrary state and apply a gauge transformation to it, you end up with the same physical state. Whereas global symmetries do transform between states, for example it transforms between states in an irreducible multiplet. Or as another example, maybe every energy eigenstate is in a 1D irrep of a symmetry, but a linear combination of two states in different irreps will transform to a different linear combination.

vanhees71
Gold Member
2019 Award
If you use this as your definition of symmetry, then gauge transformations count, but so do anomalous symmetries. So it's not a good definition for quantum systems. E.g. pure Yang Mills is classically scale invariant but it isn't a symmetry for the quantized theory.

I believe saying that gauge transformations aren't symmetries or are "redundancies" makes sense because they do not transform distinct states into each other. If you take an arbitrary state and apply a gauge transformation to it, you end up with the same physical state. Whereas global symmetries do transform between states, for example it transforms between states in an irreducible multiplet. Or as another example, maybe every energy eigenstate is in a 1D irrep of a symmetry, but a linear combination of two states in different irreps will transform to a different linear combination.
Ok, to have anomalies included, just substitute "effective action" instead of action, and everything is fine. In the classical theory it's by the way sufficient that the variation of the action functional obeys the symmetry. I'm not so sure about the quantum case. For the validity of the Ward-Takahashi identities of proper vertex functions, excluding vacuum diagrams, it should be sufficient, but I'm not sure concerning the effective action itself, which gets a physical observable in the equilibrium case, where it is a thermodynamical potential.

DrDu
However, you could imagine an alternate history of science in which Maxwell first discovered the laws of electromagnetism in terms of $\vec{A}$ and $\phi$, and then later discovered that his laws were invariant under the above transformations. What reason would physicists have for concluding that this was a gauge symmetry, and not a true physical symmetry?
A and phi would still not be observables as we can't measure them unambiguously.

I asked here not a long time ago how to recognize a gauge theory if one only looks at the observables. I won't claim I understood the answers, but A. Neumaier cited some interesting papers who showed how to eliminate non-observables from some simple model-gauge theories and quantize the system:

https://www.physicsforums.com/threa...-gauge-theory-in-terms-of-observables.839208/

DrDu
Anyway, my original comment is still applicable. As gauge invariance is not a physical symmetry and is only a redundancy in our description, this seems really strange that a non-physical thing is determining the physics. For me, either gauge invariance is somehow physical or there is something really physical that can be used instead of gauge invariance.
I agree 100% with you. However, we face this almost everywhere in physics. For example, enumerating the positions of identical particles is also such a kind of redundancy. It is more physical, to use only a set of relative position vectors pointing from e.g. one position to the other and restricting the position vector to lie in one half space, identifying points on the boundary. If you do this, you get a multiply connected space and it turns out that the usual classification of particles in terms of permutation symmetry is related to the topological different paths in that space. I would love to see something similar for e.g. the gauge group U(1) of electromagnetism.

DrDu
It would be so nice if this is true. Unfortunately, the interaction Lagrangian $A_{\mu}J^{\mu}$ cannot be expressed in terms of the $F_{\mu\nu}$.
I don't take this to be god given: E.g. you could use for the interaction ##F_{\mu\nu}P^{\mu\nu}## with the polarisation P obeying ##\partial_\mu P^{\mu\nu}=J^\nu##. The two expressions can be shown to be equivalent up to a total derivative.

RUTA
You can also view gauge invariance as the result of relationalism. Rovelli writes, "Gauge is ubiquitous. It is not unphysical redundancy of our mathematics. It reveals the relational structure of our world." Rovelli, C.: Why Gauge? (2013) http://arxiv.org/pdf/1308.5599v1.pdf, p 7. Mathematically speaking in terms of a path integral for lattice gauge theory, that the difference matrix K (derivatives are differences in discrete form) for the action on the spacetime lattice has a non-trivial null space (at least one eigenvalue = 0) means its rows are not linearly independent. Fadeev-Popov gauge fixing is just the restriction of the path integral to the row space of K (subspace with non-zero eigenvalues). When you restrict the source vector J to the row space of K you get conserved J. All that follows from simple relationalism, e.g., stating that I'm behind you in a line at the grocery store entails that you're in front of me (you can see the "redundancy" here). In the same way, constructing the rows for the difference matrix K entails a redundancy that leads to the non-trivial null space.

samalkhaiat
I don't take this to be god given
No, it is forced on you by local gauge symmetry, with very accurate experimental predictions.
E.g. you could use for the interaction ##F_{\mu\nu}P^{\mu\nu}## with the polarisation P obeying ##\partial_\mu P^{\mu\nu}=J^\nu##.
Who orderd this P for you? Plus, if both tensors P and F satisfy the same field equation $\partial P = \partial F = J$, then $F \propto P$ So, your "interaction term" is proportional to the free electromagnetic Lagrangian $F^{2}$.

DrDu
This form of the Lagrangian involving the polarisation is quite common in solid state and molecular physics.

DrDu
Thinking about it, I forgot an epsilon tensor in the definition of P.

I think the most natural way to think of symmetries in a quantum system is to think about the transformation of the state. Symmetries and gauge invariance can have profound effects in many quantum mechanical systems which are not immediately obvious in the classical theory, so you cannot look at just the action since you have things like dimensional transmutation from loop effects.

Going back to the subject of gauge invariance, the way to think about it is in terms of the Hilbert space which relates to gauge orbits etc. This is actually pretty clear in a Z2 lattice gauge theory. If you formulate the theory in terms of a superposition a of loops, you will see just by counting arguments that your Hilbert space is overcomplete. This means that many states you have are actually equivalent by some gauge transformation and must be projected out using some gauge fixing condition to get the physical state of the systems. In this case you actually have a richer phenomena of projective symmetry. This means that a combination of symmetry operations equal to the identity in the symmetry group are actually only the same up to some gauge transformation.

I think the best way to phrase all this is that while the choice of gauge is not observable, gauge invariance most definitely has physical consequences.

Some more examples: boundary conditions become incredibly important.
Some examples are Chern-Simons theory and 2+1D QED. In the first case, you have a term which is a total derivative. If you add a boundary to this system, the bulk theory is no longer gauge invariant, you must consider the whole system together to preserve gauge invariance. This can explain the edge states in quantum hall systems/topological insulators. In compact 2+1D QED, you have additional "large" gauge transformations allowed which are topologically nontrivial. This causes the charge to be quantized. By contrast, charge is not quantized for non compact 2+1D QED. Overall, gauge invariance is very important for systems with topological/symmetry protected topological order.

• king vitamin and vanhees71
king vitamin
Gold Member
Ok, to have anomalies included, just substitute "effective action" instead of action, and everything is fine. In the classical theory it's by the way sufficient that the variation of the action functional obeys the symmetry. I'm not so sure about the quantum case. For the validity of the Ward-Takahashi identities of proper vertex functions, excluding vacuum diagrams, it should be sufficient, but I'm not sure concerning the effective action itself, which gets a physical observable in the equilibrium case, where it is a thermodynamical potential.
It's been known for a long time that the effective action is gauge-dependent (I believe Jackiw is the original reference: http://journals.aps.org/prd/abstract/10.1103/PhysRevD.9.1686 ). Perhaps only demanding that the extrema are gauge invariant would work? I think this is still a major topic of research, see e.g. http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.113.241801 from last year (which already has several references).

vanhees71